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Chapter 9

Matrices

Class - 10 RS Aggarwal Mathematics Solutions

Exercise 9A

Question 1

Write the order of each of the following matrices :

(i) [3−57−694]\begin{bmatrix} 3 & -5 & 7 \\ -6 & 9 & 4 \end{bmatrix}

(ii) [25−361−8]\begin{bmatrix} 2 & 5 \\ -3 & 6 \\ 1 & -8 \end{bmatrix}

(iii) [64−2]\begin{bmatrix} 6 \\ 4 \\ -2 \end{bmatrix}

(iv) [8−3]\begin{bmatrix} 8 & -3 \end{bmatrix}

(v) [11−836]\begin{bmatrix} 11 & -8 \\ 3 & 6 \end{bmatrix}

(vi) [10]\begin{bmatrix} 10 \end{bmatrix}

Answer

(i) The matrix has 2 rows and 3 columns.

Hence, the order of matrix is (2 × 3).

(ii) The matrix has 3 rows and 2 columns.

Hence, the order of matrix is (3 × 2).

(iii) The matrix has 3 rows and 1 column.

Hence, the order of matrix is (3 × 1).

(iv) The matrix has 1 row and 2 columns.

Hence, the order of matrix is (1 × 2).

(v) The matrix has 2 rows and 2 columns.

Hence, the order of matrix is (2 × 2).

(vi) The matrix has 1 row and 1 column.

Hence, the order of matrix is (1 × 1).

Question 2

Classify the following matrices :

(i) [93−4]\begin{bmatrix} 9 \\ 3 \\ -4 \end{bmatrix}

(ii) [7−894]\begin{bmatrix} 7 & -8 \\ 9 & 4 \end{bmatrix}

(iii) [8−72]\begin{bmatrix} 8 & -7 & 2 \end{bmatrix}

(iv) [1001]\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}

(v) [60003000−2]\begin{bmatrix} 6 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & -2 \end{bmatrix}

(vi) [000000]\begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}

Answer

(i) Since, there is only once column and three rows in the matrix.

Hence, it is column matrix of order 3 x 1.

(ii) Since, no. of rows = no. of columns in the matrix.

Hence, it is a square matrix of order 2.

(iii) Since, there is only one row and 3 columns in the matrix.

Hence, it is a row matrix of order 1 x 3.

(iv) Since, there are 2 rows and 2 columns in the matrix.

Also, only the main diagonal element is 1 and the rest elements are zero.

Hence, it is a unit matrix of order 2.

(v) Since, there are 3 rows and 3 columns in the matrix.

Also, all the elements apart from main diagonal is zero.

Hence , it is a diagonal matrix of order 3.

(vi) Since, there are 2 rows and 3 columns in the matrix and all elements are zero.

Hence , it is a zero matrix of order 2 x 3.

Question 3

Construct a (2 × 3) matrix, whose elements aij are given by aij = (3i − j).

Answer

Given,

aij = (3i − j).

∴ a11 = [3(1) - 1] = 2, a12 = [3(1) - 2] = 1, a13 = [3(1) - 3] = 0.

a21 = [3(2) - 1] = 5, a22 = [3(2) - 2]= 4, a23 = [3(2) - 3] = 3.

Hence, required matrix = [210543]\begin{bmatrix} 2 & 1 & 0 \\ 5 & 4 & 3 \end{bmatrix}.

Question 4

Construct a (3 × 2) matrix [aij]3×2 for which aij = (i × j).

Answer

Given,

aij = (i × j)

⇒ a11 = 1 × 1 = 1, a12 = 1 × 2 = 2.

⇒ a21 = 2 × 1 = 2, a22 = 2 × 2 = 4.

⇒ a31 = 3 × 1 = 3, a32 = 3 × 2 = 6.

Hence, required matrix = [122436]\begin{bmatrix} 1 & 2 \\ 2 & 4 \\ 3 & 6 \end{bmatrix}.

Question 5

If a matrix has 4 elements, what are possible orders it can have?

Answer

If a matrix has 4 elements the possible orders are, 1 x 4, 2 x 2, 4 x 1.

Hence, possible orders are 1 x 4, 2 x 2, 4 x 1.

Question 6

If a matrix has 6 elements, what are the possible orders it can have ?

Answer

If a matrix has 6 elements the possible orders are, 1 × 6, 6 × 1, 2 × 3, 3 × 2.

Hence, possible orders are 1 × 6, 6 × 1, 2 × 3, 3 × 2.

Question 7

Find the values of a, b, c and d when [a+34−2b−6]=[−1c−3d+12]\begin{bmatrix} a + 3 & 4 \\ -2 & b - 6 \end{bmatrix}= \begin{bmatrix} -1 & c - 3 \\ d + 1 & 2 \end{bmatrix}

Answer

Given,

[a+34−2b−6]=[−1c−3d+12]\begin{bmatrix} a + 3 & 4 \\ -2 & b - 6 \end{bmatrix} =\begin{bmatrix} -1 & c - 3 \\ d + 1 & 2 \end{bmatrix}

∴ a + 3 = -1

⇒ a = -1 - 3 = -4.

∴ b - 6 = 2

⇒ b = 6 + 2 = 8

∴ c - 3 = 4

⇒ c = 4 + 3 = 7

∴ d + 1 = -2

⇒ d = -2 - 1 = -3

Hence , a = -4, b = 8, c = 7, d = -3.

Question 8

Find the values of x and y when [5x+3y2x−y]=[127]\begin{bmatrix} 5x + 3y \\ 2x - y \end{bmatrix} =\begin{bmatrix} 12 \\ 7 \end{bmatrix}

Answer

Given,

[5x+3y2x−y]=[127]\begin{bmatrix} 5x + 3y \\ 2x - y \end{bmatrix} =\begin{bmatrix} 12 \\ 7 \end{bmatrix}

∴ 5x + 3y = 12....(1)

∴ 2x - y = 7

y = 2x - 7....(2)

Substituting value of y from equation (2) in equation (1), we get :

⇒ 5x + 3(2x - 7) = 12

⇒ 5x + 6x - 21 = 12

⇒ 11x = 12 + 21

⇒ 11x = 33

⇒ x = 3311\dfrac{33}{11} = 3.

Substitute value of x in equation (2):

⇒ y = 2x - 7

⇒ y = 2(3) - 7

⇒ y = 6 - 7

⇒ y = -1.

Hence , x = 3 and y = -1.

Question 9

Find the values of x, y, a and b when [x+ya−ba+b2x−3y]=[53−1−5]\begin{bmatrix} x + y & a - b \\ a + b & 2x - 3y \end{bmatrix} =\begin{bmatrix} 5 & 3 \\ -1 & -5 \end{bmatrix}

Answer

Given,

[x+ya−ba+b2x−3y]=[53−1−5]\begin{bmatrix} x + y & a - b \\ a + b & 2x - 3y \end{bmatrix} =\begin{bmatrix} 5 & 3 \\ -1 & -5 \end{bmatrix}

∴ x + y = 5

⇒ y = 5 - x ........(1)

∴ 2x - 3y = -5 ........(2)

Substituting value of y from equation(1) in (2), we get :

⇒ 2x - 3(5 - x) = -5

⇒ 2x - 15 + 3x = -5

⇒ 5x - 15 = -5

⇒ 5x = -5 + 15

⇒ 5x = 10

⇒ x = 105\dfrac{10}{5} = 2.

Substitute value of x in equation 1 :

⇒ y = 5 - x

⇒ y = 5 - 2

⇒ y = 3.

∴ a - b = 3

⇒ a = 3 + b .........(3)

∴ a + b = -1 .........(4)

Substituting value of a from equation (3) in (4), we get :

⇒ 3 + b + b = -1

⇒ 2b = -1 - 3

⇒ 2b = -4

⇒ b = −42\dfrac{-4}{2} = -2.

Substitute value of b in equation (3), we get :

⇒ a = -2 + 3

⇒ a = 1.

Hence, x = 2, y = 3, a = 1, b = -2.

Question 10

Find the transpose of each of the matrices given below:

(i) A=[235−4]A = \begin{bmatrix} 2 & 3 \\ 5 & -4 \end{bmatrix}

(ii) B=[57−3]B = \begin{bmatrix} 5 & 7 & -3 \end{bmatrix}

(iii) C=[−26]C = \begin{bmatrix} -2 \\ 6 \end{bmatrix}

Answer

The matrix otained by interchanging the rows and column of a matrix is called the transpose of the matrix.

(i) Given,

⇒A=[235−4]⇒AT=[253−4].\Rightarrow A = \begin{bmatrix} 2 & 3 \\ 5 & -4 \end{bmatrix} \\[1em] \Rightarrow A^T = \begin{bmatrix} 2 & 5 \\ 3 & -4 \end{bmatrix}.

Hence, the transpose of matrix A =[253−4]= \begin{bmatrix} 2 & 5 \\ 3 & -4 \end{bmatrix}.

(ii) Given,

⇒B=[57−3]⇒BT=[57−3].\Rightarrow B = \begin{bmatrix} 5 & 7 & -3 \end{bmatrix} \\[1em] \Rightarrow B^T = \begin{bmatrix} 5 \\ 7 \\ -3 \end{bmatrix}.

Hence, the transpose of matrix B = [57−3]\begin{bmatrix} 5 \\ 7 \\ -3 \end{bmatrix}

(iii) Given,

⇒C=[−26]⇒CT=[−26]\Rightarrow C = \begin{bmatrix} -2 \\ 6 \end{bmatrix} \\[1em] \Rightarrow C^T = \begin{bmatrix} -2 & 6 \\ \end{bmatrix}

Hence, the transpose of matrix C =[−26]= \begin{bmatrix} -2 & 6 \\ \end{bmatrix}.

Exercise 9B

Question 1

If, A=[2−34]A = \begin{bmatrix} 2 & -3 & 4 \end{bmatrix}, find :

(i) 5A

(ii) (−4)A

(iii) −A

Answer

(i) 5A

⇒5×[2−34]⇒[10−1520]\Rightarrow 5 \times \begin{bmatrix} 2 & -3 & 4 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 10 & -15 & 20 \end{bmatrix}

Hence, 5A = [10−1520]\begin{bmatrix} 10 & -15 & 20 \end{bmatrix}.

(ii) (-4)A

⇒−4×[2−34]⇒[−812−16]\Rightarrow -4 \times \begin{bmatrix} 2 & -3 & 4 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} -8 & 12 & -16 \end{bmatrix}

Hence, (-4)A = [−812−16]\begin{bmatrix} -8 & 12 & -16 \end{bmatrix}.

(iii) −A

⇒−1×[2−34]⇒[−23−4]\Rightarrow -1 \times \begin{bmatrix} 2 & -3 & 4 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} -2 & 3 & -4 \end{bmatrix}

Hence, −A = [−23−4]\begin{bmatrix} -2 & 3 & -4 \end{bmatrix}.

Question 2

If M=[6−4]M = \begin{bmatrix} 6 & -4 \end{bmatrix}, find:

(i) 3M

(ii) 12M\dfrac{1}{2}M

(iii) −2M

Answer

(i) 3M

⇒3×[6−4]⇒[18−12]\Rightarrow 3 \times \begin{bmatrix} 6 & -4 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 18 & -12 \end{bmatrix}

Hence, 3M = [18−12]\begin{bmatrix} 18 & -12 \end{bmatrix}.

(ii) 12M\dfrac{1}{2}M

⇒12×[6−4]⇒[3−2]\Rightarrow \dfrac{1}{2} \times \begin{bmatrix} 6 & -4 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 3 & -2 \end{bmatrix}

Hence, 12M=[3−2]\dfrac{1}{2}M = \begin{bmatrix} 3 & -2 \end{bmatrix}.

(iii) −2M

⇒−2×[6−4]⇒[−128]\Rightarrow -2 \times \begin{bmatrix} 6 & -4 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} -12 & 8 \end{bmatrix}

Hence, −2M = [−128]\begin{bmatrix} -12 & 8 \end{bmatrix}.

Question 3

If C=[3−609]C = \begin{bmatrix} 3 & -6 \\ 0 & 9 \end{bmatrix}, find :

(i) 2C

(ii) 13C\dfrac{1}{3}C

(iii) −C

Answer

(i) 2C

⇒2×[3−609]⇒[6−12018]\Rightarrow 2 \times\begin{bmatrix} 3 & -6 \\ 0 & 9 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 6 & -12 \\ 0 & 18 \end{bmatrix}

Hence, 2C = [6−12018]\begin{bmatrix} 6 & -12 \\ 0 & 18 \end{bmatrix}.

(ii) 13C\dfrac{1}{3}C

⇒13×[3−609]⇒[1−203]\Rightarrow \dfrac{1}{3} \times\begin{bmatrix} 3 & -6 \\ 0 & 9 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 1 & -2 \\ 0 & 3 \end{bmatrix}

Hence, 13C=[1−203]\dfrac{1}{3}C = \begin{bmatrix} 1 & -2 \\ 0 & 3 \end{bmatrix}.

(iii) −C

⇒−1×[3−609]⇒[−360−9].\Rightarrow -1 \times\begin{bmatrix} 3 & -6 \\ 0 & 9 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} -3 & 6 \\ 0 & -9 \end{bmatrix}.

Hence, -C = [−360−9]\begin{bmatrix} -3 & 6 \\ 0 & -9 \end{bmatrix}.

Question 4

If, 6M=[06−1224]6M = \begin{bmatrix} 0 & 6 \\ -12 & 24 \end{bmatrix}, find M.

Answer

Given,

⇒6M=[06−1224]⇒M=16×[06−1224]⇒M=[01−24].\Rightarrow 6M = \begin{bmatrix} 0 & 6 \\ -12 & 24 \end{bmatrix} \\[1em] \Rightarrow M = \dfrac{1}{6} \times \begin{bmatrix} 0 & 6 \\ -12 & 24 \end{bmatrix} \\[1em] \Rightarrow M = \begin{bmatrix} 0 & 1 \\ -2 & 4 \end{bmatrix}.

Hence, M = [01−24]\begin{bmatrix} 0 & 1 \\ -2 & 4 \end{bmatrix}.

Question 5

If A=[25−37]A = \begin{bmatrix} 2 & 5 \\ -3 & 7 \end{bmatrix} and B=[1−325]B = \begin{bmatrix} 1 & -3 \\ 2 & 5 \end{bmatrix}, find :

(i) A + B

(ii) A − B

(iii) B − A

Answer

(i) Given,

A=[25−37]A = \begin{bmatrix} 2 & 5 \\ -3 & 7 \end{bmatrix}

B=[1−325]B = \begin{bmatrix} 1 & -3 \\ 2 & 5 \end{bmatrix}

Solving,

⇒A+B=[25−37]+[1−325]⇒A+B=[2+15+(−3)−3+27+5]⇒A+B=[32−112].\Rightarrow A + B = \begin{bmatrix} 2 & 5 \\ -3 & 7 \end{bmatrix} + \begin{bmatrix} 1 & -3 \\ 2 & 5 \end{bmatrix} \\[1em] \Rightarrow A + B = \begin{bmatrix} 2 + 1 & 5 + (-3) \\ -3 + 2 & 7 + 5 \end{bmatrix}\\[1em] \Rightarrow A + B = \begin{bmatrix} 3 & 2 \\ -1 & 12 \end{bmatrix}.

Hence, A + B = [32−112]\begin{bmatrix} 3 & 2 \\ -1 & 12 \end{bmatrix}.

(ii) Given,

A=[25−37]A = \begin{bmatrix} 2 & 5 \\ -3 & 7 \end{bmatrix}

B=[1−325]B = \begin{bmatrix} 1 & -3 \\ 2 & 5 \end{bmatrix}

Solving,

⇒A−B=[25−37]−[1−325]⇒[2−15−(−3)−3−27−5]⇒[18−52].\Rightarrow A - B = \begin{bmatrix} 2 & 5 \\ -3 & 7 \end{bmatrix} - \begin{bmatrix} 1 & -3 \\ 2 & 5 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 2 - 1 & 5 - (-3) \\ -3 - 2 & 7 - 5 \end{bmatrix}\\[1em] \Rightarrow \begin{bmatrix} 1 & 8 \\ -5 & 2 \end{bmatrix}.

Hence, A - B = [18−52]\begin{bmatrix} 1 & 8 \\ -5 & 2 \end{bmatrix}.

(iii) Given,

A=[25−37]A = \begin{bmatrix} 2 & 5 \\ -3 & 7 \end{bmatrix}

B=[1−325]B = \begin{bmatrix} 1 & -3 \\ 2 & 5 \end{bmatrix}

Solving,

⇒B−A=[1−325]−[25−37]⇒[1−2−3−52−(−3)5−7]⇒[−1−85−2].\Rightarrow B - A = \begin{bmatrix} 1 & -3 \\ 2 & 5 \end{bmatrix} - \begin{bmatrix} 2 & 5 \\ -3 & 7 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 1 - 2 & -3 - 5 \\ 2 - (-3) & 5 - 7 \end{bmatrix}\\[1em] \Rightarrow \begin{bmatrix} -1 & -8 \\ 5 & -2 \end{bmatrix}.

Hence, B - A = [−1−85−2]\begin{bmatrix} -1 & -8 \\ 5 & -2 \end{bmatrix}.

Question 6

If M=[2−345]M = \begin{bmatrix} 2 & -3 \\ 4 & 5 \end{bmatrix} and N=[−1632]N = \begin{bmatrix} -1 & 6 \\ 3 & 2 \end{bmatrix}, find :

(i) 2M + 5N

(ii) 4N − 3M

Answer

(i) 2M + 5N

Given,

M=[2−345]M = \begin{bmatrix} 2 & -3 \\ 4 & 5 \end{bmatrix}

N=[−1632]N = \begin{bmatrix} -1 & 6 \\ 3 & 2 \end{bmatrix}

Solving,

⇒2M+5N=2×[2−345]+5×[−1632]⇒[4−6810]+[−5301510]⇒[4−5−6+308+1510+10]⇒[−1242320].\Rightarrow 2M + 5N = 2 \times \begin{bmatrix} 2 & -3 \\ 4 & 5 \end{bmatrix} + 5 \times \begin{bmatrix} -1 & 6 \\ 3 & 2 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 4 & -6 \\ 8 & 10 \end{bmatrix} + \begin{bmatrix} -5 & 30 \\ 15 & 10 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 4 - 5 & -6 + 30 \\ 8 + 15 & 10 + 10 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} -1 & 24 \\ 23 & 20 \end{bmatrix}.

Hence, 2M + 5N = [−1242320]\begin{bmatrix} -1 & 24 \\ 23 & 20 \end{bmatrix}.

(ii) 4N − 3M

Given,

M=[2−345]M = \begin{bmatrix} 2 & -3 \\ 4 & 5 \end{bmatrix}

N=[−1632]N = \begin{bmatrix} -1 & 6 \\ 3 & 2 \end{bmatrix}

Solving,

⇒4N−3M=4×[−1632]−3×[2−345]⇒[−424128]−[6−91215]⇒[−4−624−(−9)12−128−15]⇒[−10330−7].\Rightarrow 4N − 3M = 4 \times \begin{bmatrix} -1 & 6 \\ 3 & 2 \end{bmatrix} - 3 \times \begin{bmatrix} 2 & -3 \\ 4 & 5 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} -4 & 24 \\ 12 & 8 \end{bmatrix} - \begin{bmatrix} 6 & -9 \\ 12 & 15 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} - 4 - 6 & 24 - (-9) \\ 12 - 12 & 8 - 15 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} -10 & 33 \\ 0 & -7 \end{bmatrix}.

Hence, 4N − 3M = [−10330−7]\begin{bmatrix} -10 & 33 \\ 0 & -7 \end{bmatrix}

Question 7

Given, that

A=[2432],B=[13−25],C=[−2534]A = \begin{bmatrix} 2 & 4 \\ 3 & 2 \end{bmatrix}, B = \begin{bmatrix} 1 & 3 \\ -2 & 5 \end{bmatrix}, C = \begin{bmatrix} -2 & 5 \\ 3 & 4 \end{bmatrix}, find :

(i) 2A + 3B

(ii) 3B − 2C

(iii) 3A − 2B + 4C

Answer

(i) 2A + 3B

Given,

A=[2432]A = \begin{bmatrix} 2 & 4 \\ 3 & 2 \end{bmatrix}

B=[13−25]B = \begin{bmatrix} 1 & 3 \\ -2 & 5 \end{bmatrix}

⇒2A+3B=2×[2432]+3×[13−25]=[4864]+[39−615]=[4+38+96−64+15]=[717019].\Rightarrow 2A + 3B = 2 \times \begin{bmatrix} 2 & 4 \\ 3 & 2 \end{bmatrix} + 3 \times \begin{bmatrix} 1 & 3 \\ -2 & 5 \end{bmatrix} \\[1em] = \begin{bmatrix} 4 & 8 \\ 6 & 4 \end{bmatrix} + \begin{bmatrix} 3 & 9 \\ -6 & 15 \end{bmatrix} \\[1em] = \begin{bmatrix} 4 + 3 & 8 + 9 \\ 6 - 6 & 4 + 15 \end{bmatrix} \\[1em] = \begin{bmatrix} 7 & 17 \\ 0 & 19 \end{bmatrix}.

Hence, 2A + 3B = [717019]\begin{bmatrix} 7 & 17 \\ 0 & 19 \end{bmatrix}.

(ii) 3B − 2C

Given,

B=[13−25]B = \begin{bmatrix} 1 & 3 \\ -2 & 5 \end{bmatrix}

C=[−2534]C = \begin{bmatrix} -2 & 5 \\ 3 & 4 \end{bmatrix}

Solving,

⇒3B−2C=3×[13−25]−2×[−2534]⇒[39−615]−[−41068]⇒[3−(−4)9−10−6−615−8]⇒[7−1−127]\Rightarrow 3B − 2C = 3 \times \begin{bmatrix} 1 & 3 \\ -2 & 5 \end{bmatrix} - 2 \times \begin{bmatrix} -2 & 5 \\ 3 & 4 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 3 & 9 \\ -6 & 15 \end{bmatrix} - \begin{bmatrix} -4 & 10 \\ 6 & 8 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 3 - (-4) & 9 - 10 \\ -6 - 6 & 15 - 8 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 7 & -1 \\ -12 & 7 \end{bmatrix}

Hence, 3B − 2C = [7−1−127]\begin{bmatrix} 7 & -1 \\ -12 & 7 \end{bmatrix}.

(iii) 3A − 2B + 4C

Given,

A=[2432]A = \begin{bmatrix} 2 & 4 \\ 3 & 2 \end{bmatrix}

B=[13−25]B = \begin{bmatrix} 1 & 3 \\ -2 & 5 \end{bmatrix}

C=[−2534]C = \begin{bmatrix} -2 & 5 \\ 3 & 4 \end{bmatrix}

Solving,

⇒3A−2B+4C=3×[2432]−2×[13−25]+4×[−2534]⇒[61296]−[26−410]+[−8201216]⇒[6−212−69−(−4)6−10]+[−8201216]⇒[4613−4]+[−8201216]⇒[4−86+2013+12−4+16]⇒[−4262512].\Rightarrow 3A − 2B + 4C = 3 \times \begin{bmatrix} 2 & 4 \\ 3 & 2 \end{bmatrix} - 2 \times \begin{bmatrix} 1 & 3 \\ -2 & 5 \end{bmatrix} + 4 \times \begin{bmatrix} -2 & 5 \\ 3 & 4 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 6 & 12 \\ 9 & 6 \end{bmatrix} - \begin{bmatrix} 2 & 6 \\ -4 & 10 \end{bmatrix} + \begin{bmatrix} -8 & 20 \\ 12 & 16 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 6 - 2 & 12 - 6 \\ 9 - (-4) & 6 - 10 \end{bmatrix} + \begin{bmatrix} -8 & 20 \\ 12 & 16 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 4 & 6 \\ 13 & -4 \end{bmatrix} + \begin{bmatrix} -8 & 20 \\ 12 & 16 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 4 - 8 & 6 + 20 \\ 13 + 12 & -4 + 16 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} -4 & 26 \\ 25 & 12 \end{bmatrix}.

Hence, 3A − 2B + 4C = [−4262512]\begin{bmatrix} -4 & 26 \\ 25 & 12 \end{bmatrix}.

Question 8

Let A=[015−1],B=[1−30−2],C=[2−540].A = \begin{bmatrix} 0 & 1 \\ 5 & -1 \end{bmatrix}, B = \begin{bmatrix} 1 & -3 \\ 0 & -2 \end{bmatrix}, C = \begin{bmatrix} 2 & -5 \\ 4 & 0 \end{bmatrix}. Find (3A + 4B − 5C).

Answer

Given,

A=[015−1]A = \begin{bmatrix} 0 & 1 \\ 5 & -1 \end{bmatrix}

B=[1−30−2]B = \begin{bmatrix} 1 & -3 \\ 0 & -2 \end{bmatrix}

C=[2−540]C = \begin{bmatrix} 2 & -5 \\ 4 & 0 \end{bmatrix}

⇒3A+4B−5C=3×[015−1]+4×[1−30−2]−5×[2−540]⇒[0315−3]+[4−120−8]−[10−25200]⇒[0+43−1215+0−3−8]−[10−25200]⇒[4−915−11]−[10−25200]⇒[4−10−9−(−25)15−20−11−0]⇒[−616−5−11].\Rightarrow 3A + 4B − 5C = 3 \times \begin{bmatrix} 0 & 1 \\ 5 & -1 \end{bmatrix} + 4 \times \begin{bmatrix} 1 & -3 \\ 0 & -2 \end{bmatrix} - 5 \times \begin{bmatrix} 2 & -5 \\ 4 & 0 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 0 & 3 \\ 15 & -3 \end{bmatrix} + \begin{bmatrix} 4 & -12 \\ 0 & -8 \end{bmatrix} - \begin{bmatrix} 10 & -25 \\ 20 & 0 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 0 + 4 & 3 - 12 \\ 15 + 0 & -3 - 8 \end{bmatrix} - \begin{bmatrix} 10 & -25 \\ 20 & 0 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 4 & -9 \\ 15 & -11 \end{bmatrix} - \begin{bmatrix} 10 & -25 \\ 20 & 0 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 4 - 10 & -9 - (-25) \\ 15 - 20 & -11 - 0 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} -6 & 16 \\ -5 & -11 \end{bmatrix}.

Hence, 3A + 4B − 5C = [−616−5−11]\begin{bmatrix} -6 & 16 \\ -5 & -11 \end{bmatrix}.

Question 9

Find a matrix X such that X+[46−37]=[31−5−2].X + \begin{bmatrix} 4 & 6 \\ -3 & 7 \end{bmatrix} = \begin{bmatrix} 3 & 1 \\ -5 & -2 \end{bmatrix}.

Answer

Given,

⇒X+[46−37]=[31−5−2]⇒X=[31−5−2]−[46−37]=[3−41−6−5−(−3)−2−7]=[−1−5−2−9].\Rightarrow X + \begin{bmatrix} 4 & 6 \\ -3 & 7 \end{bmatrix} = \begin{bmatrix} 3 & 1 \\ -5 & -2 \end{bmatrix} \\[1em] \Rightarrow X = \begin{bmatrix} 3 & 1 \\ -5 & -2 \end{bmatrix} - \begin{bmatrix} 4 & 6 \\ -3 & 7 \end{bmatrix} \\[1em] = \begin{bmatrix} 3 - 4& 1 - 6 \\ -5 - (-3) & -2 - 7 \end{bmatrix} \\[1em] = \begin{bmatrix} -1 & -5 \\ -2 & -9 \end{bmatrix}.

Hence, X = [−1−5−2−9]\begin{bmatrix} -1 & -5 \\ -2 & -9 \end{bmatrix}.

Question 10

If A=[−2345]A = \begin{bmatrix} -2 & 3 \\ 4 & 5 \end{bmatrix} and B=[52−73]B = \begin{bmatrix} 5 & 2 \\ -7 & 3 \end{bmatrix}, find a matrix C such that A + B − C = 0.

Answer

Given,

A=[−2345]A = \begin{bmatrix} -2 & 3 \\ 4 & 5 \end{bmatrix}

B=[52−73]B = \begin{bmatrix} 5 & 2 \\ -7 & 3 \end{bmatrix}

Since,

⇒ A + B - C = 0

⇒ C = A + B

⇒C=[−2345]+[52−73]=[−2+53+24−75+3]=[35−38]\Rightarrow C = \begin{bmatrix} -2 & 3 \\ 4 & 5 \end{bmatrix} + \begin{bmatrix} 5 & 2 \\ -7 & 3 \end{bmatrix} \\[1em] = \begin{bmatrix} -2 + 5 & 3 + 2 \\ 4 - 7 & 5 + 3 \end{bmatrix} \\[1em] = \begin{bmatrix} 3 & 5 \\ -3 & 8 \end{bmatrix}

Hence, C = [35−38]\begin{bmatrix} 3 & 5 \\ -3 & 8 \end{bmatrix}

Question 11(i)

Given A=[2−120],B=[−3240],C=[1002],A = \begin{bmatrix} 2 & -1 \\ 2 & 0 \end{bmatrix}, B = \begin{bmatrix} -3 & 2 \\ 4 & 0 \end{bmatrix}, C = \begin{bmatrix} 1 & 0 \\ 0 & 2 \end{bmatrix}, find the matrix X such that A + X = 2B + C.

Answer

Given,

A=[2−120],B=[−3240],C=[1002]A = \begin{bmatrix} 2 & -1 \\ 2 & 0 \end{bmatrix}, B = \begin{bmatrix} -3 & 2 \\ 4 & 0 \end{bmatrix}, C = \begin{bmatrix} 1 & 0 \\ 0 & 2 \end{bmatrix}

⇒ A + X = 2B + C

⇒ X = 2B + C - A

⇒X=2×[−3240]+[1002]−[2−120]=[−6480]+[1002]−[2−120]=[−6+14+08+00+2]−[2−120]=[−5482]−[2−120]=[−5−24−(−1)(8−2)2−0]=[−7562].\Rightarrow X = 2 \times \begin{bmatrix} -3 & 2 \\ 4 & 0 \end{bmatrix} + \begin{bmatrix} 1 & 0 \\ 0 & 2 \end{bmatrix} - \begin{bmatrix} 2 & -1 \\ 2 & 0 \end{bmatrix} \\[1em] = \begin{bmatrix} -6 & 4 \\ 8 & 0 \end{bmatrix} + \begin{bmatrix} 1 & 0 \\ 0 & 2 \end{bmatrix} - \begin{bmatrix} 2 & -1 \\ 2 & 0 \end{bmatrix} \\[1em] = \begin{bmatrix} -6 + 1 & 4 + 0 \\ 8 + 0 & 0 + 2 \end{bmatrix} - \begin{bmatrix} 2 & -1 \\ 2 & 0 \end{bmatrix} \\[1em] = \begin{bmatrix} -5 & 4 \\ 8 & 2 \end{bmatrix} - \begin{bmatrix} 2 & -1 \\ 2 & 0 \end{bmatrix} \\[1em] = \begin{bmatrix} -5 - 2 & 4 - (-1) \\ (8 - 2) & 2 - 0 \end{bmatrix} \\[1em] = \begin{bmatrix} -7 & 5 \\ 6 & 2 \end{bmatrix}.

Hence, X = [−7562].\begin{bmatrix} -7 & 5 \\ 6 & 2 \end{bmatrix}.

Question 11(ii)

If [14−23]+2M=3[320−3],\begin{bmatrix} 1 & 4 \\ -2 & 3 \end{bmatrix} + 2M = 3 \begin{bmatrix} 3 & 2 \\ 0 & -3 \end{bmatrix}, find the matrix M.

Answer

Solving,

⇒[14−23]+2M=3[320−3]⇒[14−23]+2M=[960−9]⇒2M=[960−9]−[14−23]⇒M=12[9−16−40−(−2)−9−3]⇒M=12[822−12]⇒M=[411−6].\Rightarrow \begin{bmatrix} 1 & 4 \\ -2 & 3 \end{bmatrix} + 2M = 3\begin{bmatrix} 3 & 2 \\ 0 & -3 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 1 & 4 \\ -2 & 3 \end{bmatrix} + 2M = \begin{bmatrix} 9 & 6 \\ 0 & -9 \end{bmatrix} \\[1em] \Rightarrow 2M = \begin{bmatrix} 9 & 6 \\ 0 & -9 \end{bmatrix} - \begin{bmatrix} 1 & 4 \\ -2 & 3 \end{bmatrix} \\[1em] \Rightarrow M = \dfrac{1}{2} \begin{bmatrix} 9 - 1 & 6 - 4 \\ 0 - (-2) & -9 - 3 \end{bmatrix} \\[1em] \Rightarrow M = \dfrac{1}{2} \begin{bmatrix} 8 & 2 \\ 2 & -12 \end{bmatrix} \\[1em] \Rightarrow M = \begin{bmatrix} 4 & 1 \\ 1 & -6 \end{bmatrix}.

Hence, M = [411−6]\begin{bmatrix} 4 & 1 \\ 1 & -6 \end{bmatrix}.

Question 12

If A=[625−4] and B=[12−51],A = \begin{bmatrix} 6 & 2 \\ 5 & -4 \end{bmatrix} \text{ and } B = \begin{bmatrix} 1 & 2 \\ -5 & 1 \end{bmatrix}, find a matrix X such that 2A + 3B − 5X = 0.

Answer

Given,

A=[625−4] and B=[12−51]A = \begin{bmatrix} 6 & 2 \\ 5 & -4 \end{bmatrix} \text{ and } B = \begin{bmatrix} 1 & 2 \\ -5 & 1 \end{bmatrix}.

⇒ 2A + 3B − 5X = 0

⇒ 2A + 3B = 5X

⇒ X = 15\dfrac{1}{5} (2A + 3B)

Substituting values of A and B, we get :

⇒X=15(2×[625−4]+3×[12−51])=15([12410−8]+[36−153])=15([12+34+610+(−15)−8+3])=15[1510−5−5]=[32−1−1].\Rightarrow X = \dfrac{1}{5} \Big(2 \times \begin{bmatrix} 6 & 2 \\ 5 & -4 \end{bmatrix} + 3 \times \begin{bmatrix} 1 & 2 \\ -5 & 1 \end{bmatrix}\Big) \\[1em] = \dfrac{1}{5} \Big(\begin{bmatrix} 12 & 4 \\ 10 & -8 \end{bmatrix} + \begin{bmatrix} 3 & 6 \\ -15 & 3 \end{bmatrix}\Big) \\[1em] = \dfrac{1}{5} \Big(\begin{bmatrix} 12 + 3 & 4 + 6 \\ 10 + (-15) & -8 + 3 \end{bmatrix}\Big) \\[1em] = \dfrac{1}{5}\begin{bmatrix} 15 & 10 \\ -5 & -5 \end{bmatrix} \\[1em] = \begin{bmatrix} 3 & 2 \\ -1 & -1 \end{bmatrix}.

Hence, X = [32−1−1]\begin{bmatrix} 3 & 2 \\ -1 & -1 \end{bmatrix}.

Question 13

If Y=[12−15]Y = \begin{bmatrix} 1 & 2 \\ -1 & 5 \end{bmatrix},

Find a matrix X such that 2X + Y = [50−33].\begin{bmatrix} 5 & 0 \\ -3 & 3 \end{bmatrix}.

Answer

Given,

Y=[12−15]⇒2X+Y=[50−33]⇒2X=[50−33]−Y⇒X=12([50−33]−Y)Y = \begin{bmatrix} 1 & 2 \\ -1 & 5 \end{bmatrix} \\[1em] \Rightarrow 2X + Y = \begin{bmatrix} 5 & 0 \\ -3 & 3 \end{bmatrix} \\[1em] \Rightarrow 2X = \begin{bmatrix} 5 & 0 \\ -3 & 3 \end{bmatrix} - Y \\[1em] \Rightarrow X = \dfrac{1}{2} \Big(\begin{bmatrix} 5 & 0 \\ -3 & 3 \end{bmatrix} - Y\Big)

Substituting value of Y, we get :

⇒X=12([50−33]−[12−15])=12[5−10−2−3−(−1)3−5]=12[4−2−2−2]=[2−1−1−1].\Rightarrow X = \dfrac{1}{2} \Big(\begin{bmatrix} 5 & 0 \\ -3 & 3 \end{bmatrix} - \begin{bmatrix} 1 & 2 \\ -1 & 5 \end{bmatrix}\Big) \\[1em] = \dfrac{1}{2}\begin{bmatrix} 5 - 1 & 0 - 2 \\ -3 - (-1) & 3 - 5 \end{bmatrix} \\[1em] = \dfrac{1}{2}\begin{bmatrix} 4 & -2 \\ -2 & -2 \end{bmatrix} \\[1em] = \begin{bmatrix} 2 & -1 \\ -1 & -1 \end{bmatrix}.

Hence, X = [2−1−1−1].\begin{bmatrix} 2 & -1 \\ -1 & -1 \end{bmatrix}.

Question 14

Find matrices A and B such that

A+B=[5473] and A−B=[112−17].A + B = \begin{bmatrix} 5 & 4 \\ 7 & 3 \end{bmatrix} \text{ and } A - B = \begin{bmatrix} 11 & 2 \\ -1 & 7 \end{bmatrix}.

Answer

Given,

A+B=[5473]....(1)A + B = \begin{bmatrix} 5 & 4 \\ 7 & 3 \end{bmatrix}....(1)

A−B=[112−17]....(2)A - B = \begin{bmatrix} 11 & 2 \\ -1 & 7 \end{bmatrix}....(2)

Adding equations (1) and (2), we get :

⇒(A+B)+(A−B)=[5473]+[112−17]⇒A+B+A−B=[5+114+27+(−1)3+7]⇒2A=[166610]⇒A=12[166610]⇒A=[8335].\Rightarrow (A + B) + (A - B) = \begin{bmatrix} 5 & 4 \\ 7 & 3 \end{bmatrix} + \begin{bmatrix} 11 & 2 \\ -1 & 7 \end{bmatrix} \\[1em] \Rightarrow A + B + A - B = \begin{bmatrix} 5 + 11 & 4 + 2 \\ 7 + (-1) & 3 + 7 \end{bmatrix} \\[1em] \Rightarrow 2A = \begin{bmatrix} 16 & 6 \\ 6 & 10 \end{bmatrix} \\[1em] \Rightarrow A = \dfrac{1}{2} \begin{bmatrix} 16 & 6 \\ 6 & 10 \end{bmatrix} \\[1em] \Rightarrow A = \begin{bmatrix} 8 & 3 \\ 3 & 5 \end{bmatrix}.

Substituting value of A in equation (1), we get :

⇒[8335]+B=[5473]⇒B=[5473]−[8335]⇒B=[5−84−37−33−5]⇒B=[−314−2].\Rightarrow \begin{bmatrix} 8 & 3 \\ 3 & 5 \end{bmatrix} + B = \begin{bmatrix} 5 & 4 \\ 7 & 3 \end{bmatrix} \\[1em] \Rightarrow B = \begin{bmatrix} 5 & 4 \\ 7 & 3 \end{bmatrix} - \begin{bmatrix} 8 & 3 \\ 3 & 5 \end{bmatrix} \\[1em] \Rightarrow B = \begin{bmatrix} 5 - 8 & 4 - 3 \\ 7 - 3 & 3 - 5 \end{bmatrix} \\[1em] \Rightarrow B = \begin{bmatrix} -3 & 1 \\ 4 & -2 \end{bmatrix}.

Hence, A = [8335]\begin{bmatrix} 8 & 3 \\ 3 & 5 \end{bmatrix} and B = [−314−2]\begin{bmatrix} -3 & 1 \\ 4 & -2 \end{bmatrix}.

Question 15

Compute: 6[4−532]−3[2−3−14]6 \begin{bmatrix} 4 & -5 \\ 3 & 2 \end{bmatrix} - 3\begin{bmatrix} 2 & -3 \\ -1 & 4 \end{bmatrix}

Answer

Solving,

⇒6[4−532]−3[2−3−14]=[24−301812]−[6−9−312]=[24−6−30−(−9)18−(−3)12−12]=[18−21210].\Rightarrow 6 \begin{bmatrix} 4 & -5 \\ 3 & 2 \end{bmatrix} - 3\begin{bmatrix} 2 & -3 \\ -1 & 4 \end{bmatrix} \\[1em] = \begin{bmatrix} 24 & -30 \\ 18 & 12 \end{bmatrix} - \begin{bmatrix} 6 & -9 \\ -3 & 12 \end{bmatrix} \\[1em] = \begin{bmatrix} 24 - 6 & -30 - (-9) \\ 18 - (-3) & 12 - 12 \end{bmatrix} \\[1em] = \begin{bmatrix} 18 & -21 \\ 21 & 0 \end{bmatrix}.

Hence, resultant matrix = [18−21210].\begin{bmatrix} 18 & -21 \\ 21 & 0 \end{bmatrix}.

Question 16

Simplify : sin⁡A[sin⁡A−cos⁡Acos⁡Asin⁡A]+cos⁡A[cos⁡Asin⁡A−sin⁡Acos⁡A]\sin A \begin{bmatrix} \sin A & -\cos A \\ \cos A & \sin A \end{bmatrix} + \cos A \begin{bmatrix} \cos A & \sin A \\ -\sin A & \cos A \end{bmatrix}.

Answer

Solving,

⇒sin⁡A[sin⁡A−cos⁡Acos⁡Asin⁡A]+cos⁡A[cos⁡Asin⁡A−sin⁡Acos⁡A]⇒[sin⁡2A−cos⁡Asin⁡Asin⁡Acos⁡Asin⁡2A]+[cos⁡2Asin⁡Acos⁡A−sin⁡Acos⁡Acos⁡2A]⇒[sin⁡2A+cos⁡2A−cos⁡Asin⁡A+sin⁡Acos⁡Asin⁡Acos⁡A−sin⁡Acos⁡Asin⁡2A+cos⁡2A]⇒[1001].\Rightarrow \sin A \begin{bmatrix} \sin A & -\cos A \\ \cos A & \sin A \end{bmatrix} + \cos A \begin{bmatrix} \cos A & \sin A \\ -\sin A & \cos A \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} \sin^2 A & -\cos A \sin A \\ \sin A \cos A & \sin^2 A \end{bmatrix} + \begin{bmatrix} \cos^2 A & \sin A \cos A \\ -\sin A \cos A & \cos^2 A \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} \sin^2 A + \cos^2 A & -\cos A \sin A + \sin A \cos A \\ \sin A \cos A - \sin A \cos A & \sin^2 A + \cos^2 A \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}.

Hence, resultant matrix = [1001]\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}.

Question 17

Show that :

cos⁡θ[cos⁡θsin⁡θ−sin⁡θcos⁡θ]+sin⁡θ[sin⁡θ−cos⁡θcos⁡θsin⁡θ]=[1001]\cos \theta \begin{bmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix} + \sin \theta \begin{bmatrix} \sin \theta & -\cos \theta \\ \cos \theta & \sin \theta \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}

Answer

Solving L.H.S.,

⇒cos⁡θ[cos⁡θsin⁡θ−sin⁡θcos⁡θ]+sin⁡θ[sin⁡θ−cos⁡θcos⁡θsin⁡θ]⇒[cos⁡2θsin⁡θcos⁡θ−sin⁡θcos⁡θcos⁡2θ]+[sin⁡2θ−cos⁡θsin⁡θsin⁡θcos⁡θsin⁡2θ]⇒[cos⁡2θ+sin⁡2θsin⁡θcos⁡θ−cos⁡θsin⁡θ−sin⁡θcos⁡θ+sin⁡θcos⁡θcos⁡2θ+sin⁡2θ]⇒[1001].\Rightarrow \cos \theta \begin{bmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix} + \sin \theta \begin{bmatrix} \sin \theta & -\cos \theta \\ \cos \theta & \sin \theta \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} \cos^2 \theta & \sin \theta \cos \theta \\ -\sin \theta \cos \theta & \cos^2 \theta \end{bmatrix} + \begin{bmatrix} \sin^2 \theta & -\cos \theta \sin \theta \\ \sin \theta \cos \theta & \sin^2 \theta \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} \cos^2 \theta + \sin^2 \theta & \sin \theta \cos \theta - \cos \theta \sin \theta \\ -\sin \theta \cos \theta + \sin \theta \cos \theta & \cos^2 \theta + \sin^2 \theta \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}.

Hence, resultant matrix = [1001].\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}.

Question 18

If, 3[2x10]+2[43y2]=[z−3154]3 \begin{bmatrix} 2 & x \\ 1 & 0 \end{bmatrix} + 2 \begin{bmatrix} 4 & 3 \\ y & 2 \end{bmatrix} =\begin{bmatrix} z & -3 \\ 15 & 4 \end{bmatrix}, find the values of x, y, and z.

Answer

Solving,

⇒3[2x10]+2[43y2]=[z−3154]⇒[63x30]+[862y4]=[z−3154]⇒[6+83x+63+2y0+4]=[z−3154]⇒[143x+63+2y4]=[z−3154].\Rightarrow 3 \begin{bmatrix} 2 & x \\ 1 & 0 \end{bmatrix} + 2 \begin{bmatrix} 4 & 3 \\ y & 2 \end{bmatrix} =\begin{bmatrix} z & -3 \\ 15 & 4 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 6 & 3x \\ 3 & 0 \end{bmatrix} + \begin{bmatrix} 8 & 6 \\ 2y & 4 \end{bmatrix} =\begin{bmatrix} z & -3 \\ 15 & 4 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 6 + 8 & 3x + 6 \\ 3 + 2y & 0 + 4 \end{bmatrix} =\begin{bmatrix} z & -3 \\ 15 & 4 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 14 & 3x + 6 \\ 3 + 2y & 4 \end{bmatrix} =\begin{bmatrix} z & -3 \\ 15 & 4 \end{bmatrix}.

∴ z = 14

∴ 3x + 6 = -3

⇒ 3x = -3 - 6

⇒ 3x = -9

⇒ x = −93\dfrac{-9}{3}

⇒ x = -3

∴ 3 + 2y = 15

⇒ 2y = 15 - 3

⇒ 2y = 12

⇒ y = 122\dfrac{12}{2}

⇒ y = 6.

Hence, x = -3, y = 6 and z = 14.

Question 19(i)

If 3[562x]−[8y08]=[7867]3\begin{bmatrix} 5 & 6 \\ 2 & x \end{bmatrix} - \begin{bmatrix} 8 & y \\ 0 & 8 \end{bmatrix} = \begin{bmatrix} 7 & 8 \\ 6 & 7 \end{bmatrix} find the values of x and y.

Answer

Given,

3[562x]−[8y08]=[7867]3\begin{bmatrix} 5 & 6 \\ 2 & x \end{bmatrix} - \begin{bmatrix} 8 & y \\ 0 & 8 \end{bmatrix} = \begin{bmatrix} 7 & 8 \\ 6 & 7 \end{bmatrix}

Solving,

⇒3[562x]−[8y08]=[7867]⇒[151863x]−[8y08]=[7867]⇒[15−818−y6−03x−8]=[7867]⇒[718−y63x−8]=[7867].\Rightarrow 3\begin{bmatrix} 5 & 6 \\ 2 & x \end{bmatrix} - \begin{bmatrix} 8 & y \\ 0 & 8 \end{bmatrix} = \begin{bmatrix} 7 & 8 \\ 6 & 7 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 15 & 18 \\ 6 & 3x \end{bmatrix} - \begin{bmatrix} 8 & y \\ 0 & 8 \end{bmatrix} = \begin{bmatrix} 7 & 8 \\ 6 & 7 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 15 - 8 & 18 - y \\ 6 - 0 & 3x - 8 \end{bmatrix} = \begin{bmatrix} 7 & 8 \\ 6 & 7 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 7 & 18 - y \\ 6 & 3x - 8 \end{bmatrix} = \begin{bmatrix} 7 & 8 \\ 6 & 7 \end{bmatrix}.

∴ 18 - y = 8

⇒ y = 18 - 8

⇒ y = 10.

∴ 3x - 8 = 7

⇒ 3x = 7 + 8

⇒ 3x = 15

⇒ x = 153\dfrac{15}{3}

⇒ x = 5.

Hence, x = 5, y = 10.

Question 19(ii)

If 2[345x]+[1y01]=[70105]2\begin{bmatrix} 3 & 4 \\ 5 & x \end{bmatrix} + \begin{bmatrix} 1 & y \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 7 & 0 \\ 10 & 5 \end{bmatrix} find the values of x and y.

Answer

Given,

2[345x]+[1y01]=[70105]2\begin{bmatrix} 3 & 4 \\ 5 & x \end{bmatrix} + \begin{bmatrix} 1 & y \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 7 & 0 \\ 10 & 5 \end{bmatrix}

Solving,

⇒[68102x]+[1y01]=[70105]⇒[6+18+y10+02x+1]=[70105]⇒[78+y102x+1]=[70105].\Rightarrow \begin{bmatrix} 6 & 8 \\ 10 & 2x \end{bmatrix} + \begin{bmatrix} 1 & y \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 7 & 0 \\ 10 & 5 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 6 + 1 & 8 + y \\ 10 + 0 & 2x + 1 \end{bmatrix} = \begin{bmatrix} 7 & 0 \\ 10 & 5 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 7 & 8 + y \\ 10 & 2x + 1 \end{bmatrix} = \begin{bmatrix} 7 & 0 \\ 10 & 5 \end{bmatrix}.

∴ 8 + y = 0

⇒ y = -8

∴ 2x + 1 = 5

⇒ 2x = 5 - 1

⇒ 2x = 4

⇒ x = 42\dfrac{4}{2}

⇒ x = 2.

Hence, x = 2, y = -8.

Question 19(iii)

Find the value of x and y if 2[x79y−5]+[6−745]=[1072215].2 \begin{bmatrix} x & 7 \\ 9 & y - 5 \end{bmatrix} + \begin{bmatrix} 6 & -7 \\ 4 & 5 \end{bmatrix} = \begin{bmatrix} 10 & 7 \\ 22 & 15 \end{bmatrix}.

Answer

Given,

2[x79y−5]+[6−745]=[1072215].2 \begin{bmatrix} x & 7 \\ 9 & y - 5 \end{bmatrix} + \begin{bmatrix} 6 & -7 \\ 4 & 5 \end{bmatrix} = \begin{bmatrix} 10 & 7 \\ 22 & 15 \end{bmatrix}.

Solving,

⇒[2x14182y−10]+[6−745]=[1072215]⇒[2x+614−718+42y−10+5]=[1072215]⇒[2x+67222y−5]=[1072215].\Rightarrow \begin{bmatrix} 2x & 14 \\ 18 & 2y - 10 \end{bmatrix} + \begin{bmatrix} 6 & -7 \\ 4 & 5 \end{bmatrix} = \begin{bmatrix} 10 & 7 \\ 22 & 15 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 2x + 6 & 14 - 7 \\ 18 + 4 & 2y - 10 + 5 \end{bmatrix} = \begin{bmatrix} 10 & 7 \\ 22 & 15 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 2x + 6 & 7 \\ 22 & 2y - 5 \end{bmatrix} = \begin{bmatrix} 10 & 7 \\ 22 & 15 \end{bmatrix}.

∴ 2x + 6 = 10

⇒ 2x = 10 - 6

⇒ 2x = 4

⇒ x = 42\dfrac{4}{2}

⇒ x = 2.

∴ 2y - 5 = 15

⇒ 2y = 15 + 5

⇒ 2y = 20

⇒ y = 202\dfrac{20}{2}

⇒ y = 10.

Hence, x = 2, y = 10.

Exercise 9C

Question 1

Given A=[1−21213]A = \begin{bmatrix} 1 & -2 & 1 \\ 2 & 1 & 3 \end{bmatrix} and B=[213211]B = \begin{bmatrix} 2 & 1 \\ 3 & 2 \\ 1 & 1 \end{bmatrix},

(i) Write down the product matrix AB.

(ii) Would it be possible to form the product matrix BA? If so, compute BA; if not, give reasons why it is not possible.

Answer

(i) Given,

A=[1−21213]A = \begin{bmatrix} 1 & -2 & 1 \\ 2 & 1 & 3 \end{bmatrix} and B=[213211]B = \begin{bmatrix} 2 & 1 \\ 3 & 2 \\ 1 & 1 \end{bmatrix}

Solving,

AB=[1−21213]×[213211]=[(1)(2)+(−2)(3)+(1)(1)(1)(1)+(−2)(2)+(1)(1)(2)(2)+(1)(3)+(3)(1)(2)(1)+(1)(2)+(3)(1)]=[2−6+11−4+14+3+32+2+3]=[−3−2107].AB = \begin{bmatrix} 1 & -2 & 1 \\ 2 & 1 & 3 \end{bmatrix} \times \begin{bmatrix} 2 & 1 \\ 3 & 2 \\ 1 & 1 \end{bmatrix} \\[1em] = \begin{bmatrix} (1)(2) + (-2)(3) + (1)(1) & (1)(1) + (-2)(2) + (1)(1) \\ (2)(2) + (1)(3) + (3)(1) & (2)(1) + (1)(2) + (3)(1) \end{bmatrix} \\[1em] = \begin{bmatrix} 2 - 6 + 1 & 1 - 4 + 1 \\ 4 + 3 + 3 & 2 + 2 + 3 \end{bmatrix} \\[1em] = \begin{bmatrix} -3 & -2 \\ 10 & 7 \end{bmatrix}.

Hence, AB = [−3−2107]\begin{bmatrix} -3 & -2 \\ 10 & 7 \end{bmatrix}.

(ii) Yes, it is possible.

The number of columns in B equals the number of rows in A . The resulting matrix BA will be a 3×3 matrix.

BA=[213211]×[1−21213]=[(2)(1)+(1)(2)(2)(−2)+(1)(1)(2)(1)+(1)(3)(3)(1)+(2)(2)(3)(−2)+(2)(1)(3)(1)+(2)(3)(1)(1)+(1)(2)(1)(−2)+(1)(1)(1)(1)+(1)(3)]=[2+2−4+12+33+4−6+23+61+2−2+11+3]=[4−357−493−14].BA = \begin{bmatrix} 2 & 1 \\ 3 & 2 \\ 1 & 1 \end{bmatrix} \times \begin{bmatrix} 1 & -2 & 1 \\ 2 & 1 & 3 \end{bmatrix} \\[1em] = \begin{bmatrix} (2)(1) + (1)(2) & (2)(-2) + (1)(1) & (2)(1) + (1)(3) \\ (3)(1) + (2)(2) & (3)(-2) + (2)(1) & (3)(1) + (2)(3) \\ (1)(1) + (1)(2) & (1)(-2) + (1)(1) & (1)(1) + (1)(3) \end{bmatrix} \\[1em] = \begin{bmatrix} 2 + 2 & -4 + 1 & 2 + 3 \\ 3 + 4 & -6 + 2 & 3 + 6 \\ 1 + 2 & -2 + 1 & 1 + 3 \end{bmatrix} \\[1em] = \begin{bmatrix} 4 & -3 & 5 \\ 7 & -4 & 9 \\ 3 & -1 & 4 \end{bmatrix}.

Hence, BA = [4−357−493−14]\begin{bmatrix} 4 & -3 & 5 \\ 7 & -4 & 9 \\ 3 & -1 & 4 \end{bmatrix}.

Question 2

Let A = [132−1]\begin{bmatrix} 1 & 3 \\ 2 & -1 \end{bmatrix} and B = [2−3]\begin{bmatrix} 2 \\ -3 \end{bmatrix}

(i) Show that AB exists and write its order.

(ii) Find AB.

Answer

(i) Given,

A = [132−1]\begin{bmatrix} 1 & 3 \\ 2 & -1 \end{bmatrix} (order 2 × 2)

B = [2−3]\begin{bmatrix} 2 \\ -3 \end{bmatrix} (order 2 × 1)

For matrix multiplication :

The number of columns in the first matrix (A) must equal the number of rows in the second matrix(B).

Since the number of columns in A (2) equals the number of rows in B (2), the product AB exists.

Resultant matrix order = No. of rows in A × No. of columns in B.

Hence, order of matrix AB 2 × 1.

(ii) AB=[132−1]×[2−3]=[(1)(2)+(3)(−3)(2)(2)+(−1)(−3)]=[2−94+3]=[−77].AB = \begin{bmatrix} 1 & 3 \\ 2 & -1 \end{bmatrix} \times \begin{bmatrix} 2 \\ -3 \end{bmatrix} \\[1em] = \begin{bmatrix} (1)(2) + (3)(-3) \\ (2)(2) + (-1)(-3) \end{bmatrix} \\[1em] = \begin{bmatrix} 2 - 9 \\ 4 + 3 \end{bmatrix} \\[1em] = \begin{bmatrix} -7 \\ 7 \end{bmatrix}.

Hence, AB = [−77].\begin{bmatrix} -7 \\ 7 \end{bmatrix}.

Question 3

Let M = [1−2]\begin{bmatrix} 1 & -2 \end{bmatrix} and N = [21−12]\begin{bmatrix} 2 & 1 \\ -1 & 2 \end{bmatrix}

(i) Show that MN exists and write its order.

(ii) Find MN.

(iii) Does NM exist? Give reasons.

Answer

(i) Given,

For matrix multiplication :

Number of columns in the first matrix must be equal to number of rows in the second matrix.

We know that,

The number of rows in resultant matrix equals to number of rows in first matrix and number of columns in resultant matrix equals to number of columns in second matrix.

M1×2×N2×2=MN1×2M_{1 \times 2} \times N_{2 \times 2} = MN_{1 \times 2}

Hence, MN exists and order of MN = 1 × 2.

(ii) MN=[1−2]×[21−12]=[(1)(2)+(−2)(−1)(1)(1)+(−2)(2)]=[2+21−4]=[4−3].MN = \begin{bmatrix} 1 & -2 \end{bmatrix} \times \begin{bmatrix} 2 & 1 \\ -1 & 2 \end{bmatrix} \\[1em] = \begin{bmatrix} (1)(2) + (-2)(-1) & (1)(1) + (-2)(2) \end{bmatrix} \\[1em] = \begin{bmatrix} 2 + 2 & 1 - 4 \end{bmatrix} \\[1em] = \begin{bmatrix} 4 & -3 \end{bmatrix}.

Hence, MN = [4−3].\begin{bmatrix} 4 & -3 \end{bmatrix}.

(iii)

For matrix multiplication :

Number of columns in the second matrix must be equal to number of rows in the first matrix.

N2×2×M1×2N_{2 \times 2} \times M_{1 \times 2}

Not possible because Number of columns in the second matrix ≠Number of rows in the first matrix.

Hence, NM does not exists.

Question 4

Let A = [23]\begin{bmatrix} 2 & 3 \end{bmatrix} and B = [−21]\begin{bmatrix} -2 \\ 1 \end{bmatrix}

Show that AB and BA both exist. Write the order of each.

(i) Find AB.

(ii) Find BA.

Answer

Given,

A = [23]\begin{bmatrix} 2 & 3 \end{bmatrix} B = [−21]\begin{bmatrix} -2 \\ 1 \end{bmatrix}

For matrix multiplication :

Number of columns in the first matrix (A) = Number of rows in the second matrix(B).

Since the number of columns in A (2) equals the number of rows in B (2), the product AB exists.

Resultant matrix order = No. of rows in A × No. of columns in B.

A1×2×B2×1=AB1×1A_{1 \times 2} \times B_{2 \times 1} = AB_{1 \times 1}

∴ Order of matrix AB will be 1 × 1.

Since the number of columns in B (1) equals the number of rows in A (1), the product AB exists.

Resultant matrix order = No. of rows in B × No. of columns in A.

∴ m = 2 and n = 2.

Hence, order of matrix AB 1 × 1 and BA 2 × 2.

(i) AB=[23]×[−21]=[(2)(−2)+(3)(1)]=[−4+3]=[−1].AB = \begin{bmatrix} 2 & 3 \end{bmatrix} \times \begin{bmatrix} -2 \\ 1 \end{bmatrix} \\[1em] = \begin{bmatrix} (2)(-2) + (3)(1) \end{bmatrix} \\[1em] = \begin{bmatrix} -4 + 3 \end{bmatrix} \\[1em] = \begin{bmatrix} -1 \end{bmatrix}.

Hence, AB = [−1]\begin{bmatrix} -1 \end{bmatrix}.

(ii) For matrix multiplication :

Number of columns in the first matrix (B) = Number of rows in the second matrix(A).

Since the number of columns in B (1) equals the number of rows in A (1), the product BA exists.

AB2×1×B1×2=BA2×2AB_{2 \times 1} \times B_{1 \times 2} = BA_{2 \times 2}

BA=[−21]×[23]=[(−2)(2)(−2)(3)(1)(2)(1)(3)]=[−4−623]BA = \begin{bmatrix} -2 \\ 1 \end{bmatrix} \times \begin{bmatrix} 2 & 3 \end{bmatrix} \\[1em] = \begin{bmatrix} (-2)(2) & (-2)(3) \\ (1)(2) & (1)(3) \end{bmatrix} \\[1em] = \begin{bmatrix} -4 & -6 \\ 2 & 3 \end{bmatrix} \\[1em]

Hence, BA = [−4−623]\begin{bmatrix} -4 & -6 \\ 2 & 3 \end{bmatrix}.

Question 5

Let A = [132−1]\begin{bmatrix} 1 & 3 \\ 2 & -1 \end{bmatrix} and B = [2103]\begin{bmatrix} 2 & 1 \\ 0 & 3 \end{bmatrix}

(i) Find AB.

(ii) Find BA.

(iii) Is AB = BA?

Answer

(i) Solving AB,

⇒[132−1]×[2103]⇒[1×2+3×01×1+3×32×2+(−1)×02×1+(−1)×3]⇒[2+01+94+02+(−3)]⇒[2104−1].\Rightarrow \begin{bmatrix} 1 & 3 \\ 2 & -1 \end{bmatrix} \times \begin{bmatrix} 2 & 1 \\ 0 & 3 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 1 \times 2 + 3 \times 0 & 1 \times 1 + 3 \times 3 \\ 2 \times 2 + (-1) \times 0 & 2 \times 1 + (-1) \times 3 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 2 + 0 & 1 + 9 \\ 4 + 0 & 2 + (-3) \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 2 & 10 \\ 4 & -1 \end{bmatrix}.

Hence, AB = [2104−1]\begin{bmatrix} 2 & 10 \\ 4 & -1 \end{bmatrix}.

(ii) Solving BA,

⇒[2103]×[132−1]⇒[2×1+1×22×3+1×(−1)0×1+3×20×3+3×(−1)]⇒[2+26+(−1)0+60+(−3)]⇒[456−3].\Rightarrow \begin{bmatrix} 2 & 1 \\ 0 & 3 \end{bmatrix} \times \begin{bmatrix} 1 & 3 \\ 2 & -1 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 2 \times 1 + 1 \times 2 & 2 \times 3 + 1 \times (-1) \\ 0 \times 1 + 3 \times 2 & 0 \times 3 + 3 \times (-1) \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 2 + 2 & 6 + (-1) \\ 0 + 6 & 0 + (-3) \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 4 & 5 \\ 6 & -3 \end{bmatrix}.

Hence, BA = [456−3]\begin{bmatrix} 4 & 5 \\ 6 & -3 \end{bmatrix}.

(iii) As calculated above,

AB ≠ BA

Hence, AB ≠ BA.

Question 6

Let A = [5−162]\begin{bmatrix} 5 & -1 \\ 6 & 2 \end{bmatrix} and B = [21−34]\begin{bmatrix} 2 & 1 \\ -3 & 4 \end{bmatrix}, show that AB ≠ BA.

Answer

Solving AB,

⇒[5−162]×[21−34]⇒[5×2+(−1)×(−3)5×1+(−1)×46×2+2×(−3)6×1+2×4]⇒[10+35+(−4)12+(−6)6+8]⇒[131614].\Rightarrow \begin{bmatrix} 5 & -1 \\ 6 & 2 \end{bmatrix} \times \begin{bmatrix} 2 & 1 \\ -3 & 4 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 5 \times 2 + (-1) \times (-3) & 5 \times 1 + (-1) \times 4 \\ 6 \times 2 + 2 \times (-3) & 6 \times 1 + 2 \times 4 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 10 + 3 & 5 + (-4) \\ 12 + (-6) & 6 + 8 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 13 & 1 \\ 6 & 14 \end{bmatrix}.

AB = [131614]\begin{bmatrix} 13 & 1 \\ 6 & 14 \end{bmatrix}

Solving BA,

⇒[21−34]×[5−162]⇒[2×5+1×62×(−1)+1×2−3×5+4×6−3×(−1)+4×2]⇒[10+6−2+2−15+243+8]⇒[160911].\Rightarrow \begin{bmatrix} 2 & 1 \\ -3 & 4 \end{bmatrix} \times \begin{bmatrix} 5 & -1 \\ 6 & 2 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 2 \times 5 + 1 \times 6 & 2 \times (-1) + 1 \times 2 \\ -3 \times 5 + 4 \times 6 & -3 \times (-1) + 4 \times 2 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 10 + 6 & -2 + 2 \\ -15 + 24 & 3 + 8 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 16 & 0 \\ 9 & 11 \end{bmatrix}.

BA = [160911]\begin{bmatrix} 16 & 0 \\ 9 & 11 \end{bmatrix}

AB ≠ BA.

Hence, proved that AB ≠ BA.

Question 7

Let A = [−1234]\begin{bmatrix} -1 & 2 \\ 3 & 4 \end{bmatrix} and B = [3−2−45]\begin{bmatrix} 3 & -2 \\ -4 & 5 \end{bmatrix}. Find the matrix (AB + BA).

Answer

Solving AB,

⇒[−1234]×[3−2−45]⇒[−1×3+2×(−4)−1×(−2)+2×53×3+4×(−4)3×(−2)+4×5]⇒[−3+(−8)2+109+(−16)(−6)+20]⇒[−1112−714].\Rightarrow \begin{bmatrix} -1 & 2 \\ 3 & 4 \end{bmatrix} \times \begin{bmatrix} 3 & -2 \\ -4 & 5 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} -1 \times 3 + 2 \times (-4) & -1 \times (-2) + 2 \times 5 \\ 3 \times 3 + 4 \times (-4) & 3 \times (-2) + 4 \times 5 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} -3 + (-8) & 2 + 10 \\ 9 + (-16) & (-6) + 20 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} -11 & 12 \\ -7 & 14 \end{bmatrix}.

AB = [−1112−714]\begin{bmatrix} -11 & 12 \\ -7 & 14 \end{bmatrix}

Solving BA,

⇒[3−2−45]×[−1234]⇒[3×(−1)+(−2)×33×2+(−2)×4−4×(−1)+5×3−4×2+5×4]⇒[−3+(−6)6+(−8)4+15−8+20]⇒[−9−21912].\Rightarrow \begin{bmatrix} 3 & -2 \\ -4 & 5 \end{bmatrix} \times \begin{bmatrix} -1 & 2 \\ 3 & 4 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 3 \times (-1) + (-2) \times 3 & 3 \times 2 + (-2) \times 4 \\ -4 \times (-1) + 5 \times 3 & -4 \times 2 + 5 \times 4 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} -3 + (-6) & 6 + (-8) \\ 4 + 15 & -8 + 20 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} -9 & -2 \\ 19 & 12 \end{bmatrix}.

BA = [−9−21912]\begin{bmatrix} -9 & -2 \\ 19 & 12 \end{bmatrix}

Now, AB + BA

⇒[−1112−714]+[−9−21912]⇒[−11+(−9)12+(−2)−7+1914+12]⇒[−20101226].\Rightarrow \begin{bmatrix} -11 & 12 \\ -7 & 14 \end{bmatrix} + \begin{bmatrix} -9 & -2 \\ 19 & 12 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} -11 + (-9) & 12 + (-2) \\ -7 + 19 & 14 + 12 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} -20 & 10 \\ 12 & 26 \end{bmatrix}.

Hence, (AB + BA) = [−20101226]\begin{bmatrix} -20 & 10 \\ 12 & 26 \end{bmatrix}.

Question 8

If A = [1224]\begin{bmatrix} 1 & 2 \\ 2 & 4 \end{bmatrix}, B = [2132]\begin{bmatrix} 2 & 1 \\ 3 & 2 \end{bmatrix}, and C = [−275−1]\begin{bmatrix} -2 & 7 \\ 5 & -1 \end{bmatrix}, show that AB = AC.

Answer

Let's find AB,

⇒[1224]×[2132]⇒[1×2+2×31×1+2×22×2+4×32×1+4×2]⇒[2+61+44+122+8]⇒[851610].\Rightarrow \begin{bmatrix} 1 & 2 \\ 2 & 4 \end{bmatrix} \times \begin{bmatrix} 2 & 1 \\ 3 & 2 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 1 \times 2 + 2 \times 3 & 1 \times 1 + 2 \times 2 \\ 2 \times 2 + 4 \times 3 & 2 \times 1 + 4 \times 2 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 2 + 6 & 1 + 4 \\ 4 + 12 & 2 + 8 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 8 & 5 \\ 16 & 10 \end{bmatrix}.

Let's find AC,

⇒[1224]×[−275−1]⇒[1×(−2)+2×51×7+2×(−1)2×(−2)+4×52×7+4×(−1)]⇒[−2+107+(−2)−4+2014+(−4)]⇒[851610].\Rightarrow \begin{bmatrix} 1 & 2 \\ 2 & 4 \end{bmatrix} \times \begin{bmatrix} -2 & 7 \\ 5 & -1 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 1 \times (-2) + 2 \times 5 & 1 \times 7 + 2 \times (-1) \\ 2 \times (-2) + 4 \times 5 & 2 \times 7 + 4 \times (-1) \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} -2 + 10 & 7 + (-2) \\ -4 + 20 & 14 + (-4) \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 8 & 5 \\ 16 & 10 \end{bmatrix}.

Hence, proved that AB = AC.

Question 9

Let A = [4123]\begin{bmatrix} 4 & 1 \\ 2 & 3 \end{bmatrix}, B = [21−30]\begin{bmatrix} 2 & 1 \\ -3 & 0 \end{bmatrix}, and C = [−3−122]\begin{bmatrix} -3 & -1 \\ 2 & 2 \end{bmatrix}.

Verify that: (AB)C = A(BC).

Answer

Solving (AB)C,

⇒([4123]×[21−30])×[−3−122]⇒[4×2+1×(−3)4×1+1×02×2+3×(−3)2×1+3×0]×[−3−122]⇒[8+(−3)4+04+(−9)2+0]×[−3−122]⇒[54−52]×[−3−122]⇒[5×(−3)+4×25×(−1)+4×2−5×(−3)+2×2−5×(−1)+2×2]⇒[−15+8−5+815+45+4]⇒[−73199].∴(AB)C=[−73199]\Rightarrow \Big(\begin{bmatrix} 4 & 1 \\ 2 & 3 \end{bmatrix} \times \begin{bmatrix} 2 & 1 \\ -3 & 0 \end{bmatrix}\Big) \times \begin{bmatrix} -3 & -1 \\ 2 & 2 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 4 \times 2 + 1 \times (-3) & 4 \times 1 + 1 \times 0 \\ 2 \times 2 + 3 \times (-3) & 2 \times 1 + 3 \times 0 \end{bmatrix} \times \begin{bmatrix} -3 & -1 \\ 2 & 2 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 8 + (-3) & 4 + 0 \\ 4 + (-9) & 2 + 0 \end{bmatrix} \times \begin{bmatrix} -3 & -1 \\ 2 & 2 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 5 & 4 \\ -5 & 2 \end{bmatrix} \times \begin{bmatrix} -3 & -1 \\ 2 & 2 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 5 \times (-3) + 4 \times 2 & 5 \times (-1) + 4 \times 2 \\ -5 \times (-3) + 2 \times 2 & -5 \times (-1) + 2 \times 2 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} -15 + 8 & -5 + 8 \\ 15 + 4 & 5 + 4 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} -7 & 3 \\ 19 & 9 \end{bmatrix}. \\[1em] \therefore (AB)C = \begin{bmatrix} -7 & 3 \\ 19 & 9 \end{bmatrix}

Solving A(BC),

⇒[4123]×([21−30]×[−3−122])\[1em]⇒[4123]×[2×(−3)+1×22×(−1)+1×2−3×(−3)+0×2−3×(−1)+0×2]\[1em]⇒[4123]×[−6+2−2+29+03+0]\[1em]⇒[4123]×[−4093]⇒[4×(−4)+1×94×0+1×32×(−4)+3×92×0+3×3]⇒[−16+90+3−8+270+9]⇒[−73199].∴A(BC)=[−73199]\Rightarrow \begin{bmatrix} 4 & 1 \\ 2 & 3 \end{bmatrix} \times \Big(\begin{bmatrix} 2 & 1 \\ -3 & 0 \end{bmatrix} \times \begin{bmatrix} -3 & -1 \\ 2 & 2 \end{bmatrix}\Big) \[1em] \Rightarrow \begin{bmatrix} 4 & 1 \\ 2 & 3 \end{bmatrix} \times \begin{bmatrix} 2 \times (-3) + 1 \times 2 & 2 \times (-1) + 1 \times 2 \\ -3 \times (-3) + 0 \times 2 & -3 \times (-1) + 0 \times 2 \end{bmatrix} \[1em] \Rightarrow \begin{bmatrix} 4 & 1 \\ 2 & 3 \end{bmatrix} \times \begin{bmatrix} -6 + 2 & -2 + 2 \\ 9 + 0 & 3 + 0 \end{bmatrix} \[1em] \Rightarrow \begin{bmatrix} 4 & 1 \\ 2 & 3 \end{bmatrix} \times \begin{bmatrix} -4 & 0 \\ 9 & 3 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 4 \times (-4) + 1 \times 9 & 4 \times 0 + 1 \times 3 \\ 2 \times (-4) + 3 \times 9 & 2 \times 0 + 3 \times 3 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} -16 + 9 & 0 + 3 \\ -8 + 27 & 0 + 9 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} -7 & 3 \\ 19 & 9 \end{bmatrix}. \\[1em] \therefore A(BC) = \begin{bmatrix} -7 & 3 \\ 19 & 9 \end{bmatrix}

Hence, proved that (AB)C = A(BC).

Question 10

Let A = [34−12]\begin{bmatrix} 3 & 4 \\ -1 & 2 \end{bmatrix}, B = [502−3]\begin{bmatrix} 5 & 0 \\ 2 & -3 \end{bmatrix}, and C = [2−140]\begin{bmatrix} 2 & -1 \\ 4 & 0 \end{bmatrix}

Compute:

(i) A(B + C)

(ii) (AB + AC)

(iii) Is A(B + C) = (AB + AC)?

Answer

(i) A(B + C)

⇒[34−12]×([502−3]+[2−140])⇒[34−12]×[5+20−12+4−3+0]⇒[34−12]×[7−16−3]⇒[3×7+4×63×(−1)+4×(−3)−1×7+2×6−1×(−1)+2×(−3)]⇒[21+24−3+(−12)−7+121+(−6)]=[45−155−5].\Rightarrow \begin{bmatrix} 3 & 4 \\ -1 & 2 \end{bmatrix} \times \Big(\begin{bmatrix} 5 & 0 \\ 2 & -3 \end{bmatrix} + \begin{bmatrix} 2 & -1 \\ 4 & 0 \end{bmatrix}\Big) \\[1em] \Rightarrow \begin{bmatrix} 3 & 4 \\ -1 & 2 \end{bmatrix} \times \begin{bmatrix} 5 + 2 & 0 - 1 \\ 2 + 4 & -3 + 0 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 3 & 4 \\ -1 & 2 \end{bmatrix} \times \begin{bmatrix} 7 & -1 \\ 6 & -3 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 3 \times 7 + 4 \times 6 & 3 \times (-1) + 4 \times (-3) \\ -1 \times 7 + 2 \times 6 & -1 \times (-1) + 2 \times (-3) \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 21 + 24 & -3 + (-12) \\ -7 + 12 & 1 + (-6) \end{bmatrix} \\[1em] = \begin{bmatrix} 45 & -15 \\ 5 & -5 \end{bmatrix}.

Hence, A(B + C) = [45−155−5]\begin{bmatrix} 45 & -15 \\ 5 & -5 \end{bmatrix}

(ii) (AB + AC)

⇒AB=[34−12]×[502−3]=[3×5+4×23×0+4×(−3)−1×5+2×2−1×0+2×(−3)]=[15+80+(−12)−5+40+(−6)]=[23−12−1−6]⇒AC=[34−12]×[2−140]=[3×2+4×43×(−1)+4×0−1×2+2×4−1×(−1)+2×0]=[6+16−3+0−2+81+0]=[22−361]⇒AB+AC=[23−12−1−6]+[22−361]=[23+22−12−3−1+6−6+1]=[45−155−5].\Rightarrow AB = \begin{bmatrix} 3 & 4 \\ -1 & 2 \end{bmatrix} \times \begin{bmatrix} 5 & 0 \\ 2 & -3 \end{bmatrix} \\[1em] = \begin{bmatrix} 3 \times 5 + 4 \times 2 & 3 \times 0 + 4 \times (-3) \\ -1 \times 5 + 2 \times 2 & -1 \times 0 + 2\times(-3) \end{bmatrix} \\[1em] = \begin{bmatrix} 15 + 8 & 0 + (-12) \\ -5 + 4 & 0 + (-6) \end{bmatrix}\\[1em] = \begin{bmatrix} 23 & -12 \\ -1 & -6 \end{bmatrix} \\[1em] \Rightarrow AC = \begin{bmatrix} 3 & 4 \\ -1 & 2 \end{bmatrix} \times \begin{bmatrix} 2 & -1 \\ 4 & 0 \end{bmatrix} \\[1em] = \begin{bmatrix} 3 \times 2 + 4 \times 4 & 3\times(-1) + 4 \times 0 \\ -1 \times 2 + 2 \times 4 & -1\times(-1) + 2 \times 0 \end{bmatrix} \\[1em] = \begin{bmatrix} 6 + 16 & -3 + 0 \\ -2 + 8 & 1 + 0 \end{bmatrix} \\[1em] = \begin{bmatrix} 22 & -3 \\ 6 & 1 \end{bmatrix} \\[1em] \Rightarrow AB + AC = \begin{bmatrix} 23 & -12 \\ -1 & -6 \end{bmatrix} + \begin{bmatrix} 22 & -3 \\ 6 & 1 \end{bmatrix} \\[1em] = \begin{bmatrix} 23 + 22 & -12 - 3 \\ -1 + 6 & -6 + 1 \end{bmatrix} \\[1em] = \begin{bmatrix} 45 & -15 \\ 5 & -5 \end{bmatrix}.

Hence, (AB + AC) = [45−155−5]\begin{bmatrix} 45 & -15 \\ 5 & -5 \end{bmatrix}

(iii) As calculated,

A(B+C)=(AB+AC)=[45−155−5].A(B + C) = (AB + AC)= \begin{bmatrix} 45 & -15 \\ 5 & -5 \end{bmatrix}.

Hence, A(B + C) = (AB + AC).

Question 11

Let A = [5732]\begin{bmatrix} 5 & 7 \\ 3 & 2 \end{bmatrix}, B = [3−2−40]\begin{bmatrix} 3 & -2 \\ -4 & 0 \end{bmatrix} and C = [2−314]\begin{bmatrix} 2 & -3 \\ 1 & 4 \end{bmatrix}

Compute :

(i) (A - B)C

(ii) AC - BC

(iii) Is (A - B)C = AC - BC ?

Answer

(i) (A - B)C

⇒([5732]−[3−2−40])×[2−314]⇒[5−37−(−2)3−(−4)2−0]×[2−314]⇒[2972]×[2−314]⇒[(2)(2)+(9)(1)(2)(−3)+(9)(4)(7)(2)+(2)(1)(7)(−3)+(2)(4)]⇒[4+9−6+3614+2−21+8]⇒[133016−13].\Rightarrow \Big(\begin{bmatrix} 5 & 7 \\ 3 & 2 \end{bmatrix} - \begin{bmatrix} 3 & -2 \\ -4 & 0 \end{bmatrix}\Big) \times \begin{bmatrix} 2 & -3 \\ 1 & 4 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 5 - 3 & 7 - (-2) \\ 3 - (-4) & 2 - 0 \end{bmatrix} \times \begin{bmatrix} 2 & -3 \\ 1 & 4 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 2 & 9 \\ 7 & 2 \end{bmatrix} \times \begin{bmatrix} 2 & -3 \\ 1 & 4 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} (2)(2) + (9)(1) & (2)(-3) + (9)(4) \\ (7)(2) + (2)(1) & (7)(-3) + (2)(4) \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 4 + 9 & -6 + 36 \\ 14 + 2 & -21 + 8 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 13 & 30 \\ 16 & -13 \end{bmatrix}.

Hence, (A - B)C =[133016−13].\begin{bmatrix} 13 & 30 \\ 16 & -13 \end{bmatrix}.

(ii) AC - BC

⇒AC=[5732]×[2−314]=[(5)(2)+(7)(1)(5)(−3)+(7)(4)(3)(2)+(2)(1)(3)(−3)+(2)(4)]=[10+7−15+286+2−9+8]=[17138−1]⇒BC=[3−2−40]×[2−314]=[(3)(2)+(−2)(1)(3)(−3)+(−2)(4)(−4)(2)+(0)(1)(−4)(−3)+(0)(4)]=[6−2−9−8−8+012+0]=[4−17−812]⇒AC−BC=[17138−1]−[4−17−812]=[17−413−(−17)8−(−8)−1−12]=[133016−13].\Rightarrow AC = \begin{bmatrix} 5 & 7 \\ 3 & 2 \end{bmatrix} \times \begin{bmatrix} 2 & -3 \\ 1 & 4 \end{bmatrix} \\[1em] = \begin{bmatrix} (5)(2) + (7)(1) & (5)(-3) + (7)(4) \\ (3)(2) + (2)(1) & (3)(-3) + (2)(4) \end{bmatrix} \\[1em] = \begin{bmatrix} 10 + 7 & -15 + 28 \\ 6 + 2 & -9 + 8 \end{bmatrix} \\[1em] = \begin{bmatrix} 17 & 13 \\ 8 & -1 \end{bmatrix} \\[1em] \Rightarrow BC = \begin{bmatrix} 3 & -2 \\ -4 & 0 \end{bmatrix} \times \begin{bmatrix} 2 & -3 \\ 1 & 4 \end{bmatrix} \\[1em] = \begin{bmatrix} (3)(2) + (-2)(1) & (3)(-3) + (-2)(4) \\ (-4)(2) + (0)(1) & (-4)(-3) + (0)(4) \end{bmatrix} \\[1em] = \begin{bmatrix} 6 - 2 & -9 - 8 \\ -8 + 0 & 12 + 0 \end{bmatrix} \\[1em] = \begin{bmatrix} 4 & -17 \\ -8 & 12 \end{bmatrix} \\[1em] \Rightarrow AC - BC = \begin{bmatrix} 17 & 13 \\ 8 & -1 \end{bmatrix} - \begin{bmatrix} 4 & -17 \\ -8 & 12 \end{bmatrix} \\[1em] = \begin{bmatrix} 17 - 4 & 13 - (-17) \\ 8 - (-8) & -1 - 12 \end{bmatrix} \\[1em] = \begin{bmatrix} 13 & 30 \\ 16 & -13 \end{bmatrix}.

Hence, AC - BC =[133016−13].\begin{bmatrix} 13 & 30 \\ 16 & -13 \end{bmatrix}.

(ii) Yes, (A - B)C = AC - BC = [133016−13].\begin{bmatrix} 13 & 30 \\ 16 & -13 \end{bmatrix}.

Hence, (A - B)C = AC - BC .

Question 12

Given A = [3−2−14]\begin{bmatrix} 3 & -2 \\ -1 & 4 \end{bmatrix}, B = [61]\begin{bmatrix} 6 \\ 1 \end{bmatrix}, C = [−45]\begin{bmatrix} -4 \\ 5 \end{bmatrix}, and D = [22]\begin{bmatrix} 2 \\ 2 \end{bmatrix}. Find AB + 2C – 4D.

Answer

Solving for AB,

⇒AB=[3−2−14]×[61]=[(3)(6)+(−2)(1)(−1)(6)+(4)(1)]=[18−2−6+4]=[16−2].\Rightarrow AB = \begin{bmatrix} 3 & -2 \\ -1 & 4 \end{bmatrix} \times \begin{bmatrix} 6 \\ 1 \end{bmatrix} \\[1em] = \begin{bmatrix} (3)(6) + (-2)(1) \\ (-1)(6) + (4)(1) \end{bmatrix} \\[1em] = \begin{bmatrix} 18 - 2 \\ -6 + 4 \end{bmatrix} \\[1em] = \begin{bmatrix} 16 \\ -2 \end{bmatrix}.

Substituting values in AB + 2C – 4D,we get :

⇒[16−2]+2×[−45]−4×[22]=[16−2]+[−810]−[88]=[16+(−8)−2+10]−[88]=[88]−[88]=[00].\Rightarrow \begin{bmatrix} 16 \\ -2 \end{bmatrix} + 2 \times \begin{bmatrix} -4 \\ 5 \end{bmatrix} - 4 \times \begin{bmatrix} 2 \\ 2 \end{bmatrix} \\[1em] = \begin{bmatrix} 16 \\ -2 \end{bmatrix} + \begin{bmatrix} -8 \\ 10 \end{bmatrix} - \begin{bmatrix} 8 \\ 8 \end{bmatrix} \\[1em] = \begin{bmatrix} 16 + (-8) \\ -2 + 10 \end{bmatrix} - \begin{bmatrix} 8 \\ 8 \end{bmatrix} \\[1em] = \begin{bmatrix} 8 \\ 8 \end{bmatrix} - \begin{bmatrix} 8 \\ 8 \end{bmatrix} \\[1em] = \begin{bmatrix} 0 \\ 0 \end{bmatrix}.

Hence, AB + 2C – 4D = [00].\begin{bmatrix} 0 \\ 0 \end{bmatrix}.

Question 13

Given A = [1021]\begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix} and B = [23−10]\begin{bmatrix} 2 & 3 \\ -1 & 0 \end{bmatrix} .Find (A2 + AB + B2).

Answer

Solving A2,

⇒A2=[1021]×[1021]=[(1)(1)+(0)(2)(1)(0)+(0)(1)(2)(1)+(1)(2)(2)(0)+(1)(1)]=[1+00+02+20+1]=[1041].\Rightarrow A^2 = \begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix} \times \begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix} \\[1em] = \begin{bmatrix} (1)(1) + (0)(2) & (1)(0) + (0)(1) \\ (2)(1) + (1)(2) & (2)(0) + (1)(1) \end{bmatrix} \\[1em] = \begin{bmatrix} 1 + 0 & 0 + 0 \\ 2 + 2 & 0 + 1 \end{bmatrix} \\[1em] = \begin{bmatrix} 1 & 0 \\ 4 & 1 \end{bmatrix}.

Solving AB,

⇒AB=[1021]×[23−10]=[(1)(2)+(0)(−1)(1)(3)+(0)(0)(2)(2)+(1)(−1)(2)(3)+(1)(0)]=[2+03+04−16+0]=[2336].\Rightarrow AB = \begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix} \times \begin{bmatrix} 2 & 3 \\ -1 & 0 \end{bmatrix} \\[1em] = \begin{bmatrix} (1)(2) + (0)(-1) & (1)(3) + (0)(0) \\ (2)(2) + (1)(-1) & (2)(3) + (1)(0) \end{bmatrix} \\[1em] = \begin{bmatrix} 2 + 0 & 3 + 0 \\ 4 - 1 & 6 + 0 \end{bmatrix} \\[1em] = \begin{bmatrix} 2 & 3 \\ 3 & 6 \end{bmatrix}.

Solving B2,

⇒B2=[23−10]×[23−10]=[(2)(2)+(3)(−1)(2)(3)+(3)(0)(−1)(2)+(0)(−1)(−1)(3)+(0)(0)]=[4−36+0−2+0−3+0]=[16−2−3].\Rightarrow B^2 = \begin{bmatrix} 2 & 3 \\ -1 & 0 \end{bmatrix} \times \begin{bmatrix} 2 & 3 \\ -1 & 0 \end{bmatrix} \\[1em] = \begin{bmatrix} (2)(2) + (3)(-1) & (2)(3) + (3)(0) \\ (-1)(2) + (0)(-1) & (-1)(3) + (0)(0) \end{bmatrix} \\[1em] = \begin{bmatrix} 4 - 3 & 6 + 0 \\ -2 + 0 & -3 + 0 \end{bmatrix} \\[1em] = \begin{bmatrix} 1 & 6 \\ -2 & -3 \end{bmatrix}.

A2 + AB + B2

⇒[1041]+[2336]+[16−2−3]⇒[1+2+10+3+64+3+(−2)1+6+(−3)]⇒[4954].\Rightarrow \begin{bmatrix} 1 & 0 \\ 4 & 1 \end{bmatrix} + \begin{bmatrix} 2 & 3 \\ 3 & 6 \end{bmatrix} + \begin{bmatrix} 1 & 6 \\ -2 & -3 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 1 + 2 + 1 & 0 + 3 + 6 \\ 4 + 3 + (-2) & 1 + 6 + (-3) \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 4 & 9 \\ 5 & 4 \end{bmatrix}.

Hence, A2 + AB + B2 = [4954].\begin{bmatrix} 4 & 9 \\ 5 & 4 \end{bmatrix}.

Question 14(i)

Let A = [4−26−3]\begin{bmatrix} 4 & -2 \\ 6 & -3 \end{bmatrix}, B = [021−1]\begin{bmatrix} 0 & 2 \\ 1 & -1 \end{bmatrix}, and C = [−231−1]\begin{bmatrix} -2 & 3 \\ 1 & -1 \end{bmatrix}

Find (A2 – A + BC).

Answer

(i) A2 – A + BC

Solving A2,

⇒A2=[4−26−3]×[4−26−3]⇒[4×4+(−2)×64×(−2)+(−2)×(−3)6×4+(−3)×66×(−2)+(−3)×(−3)]⇒[16−12−8+624−18−12+9]⇒[4−26−3].\Rightarrow A^2 = \begin{bmatrix} 4 & -2 \\ 6 & -3 \end{bmatrix} \times \begin{bmatrix} 4 & -2 \\ 6 & -3 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 4\times4 + (-2)\times6 & 4\times(-2) + (-2)\times(-3) \\ 6\times4 + (-3)\times6 & 6\times(-2) + (-3)\times(-3) \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 16 - 12 & -8 + 6 \\ 24 - 18 & -12 + 9 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 4 & -2 \\ 6 & -3 \end{bmatrix}.

Solving BC,

⇒BC=[021−1]×[−231−1]⇒[0×(−2)+2×10×3+2×(−1)1×(−2)+(−1)×11×3+(−1)×(−1)]⇒[2−2−34].\Rightarrow BC = \begin{bmatrix} 0 & 2 \\ 1 & -1 \end{bmatrix} \times \begin{bmatrix} -2 & 3 \\ 1 & -1 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 0\times(-2) + 2 \times 1 & 0 \times 3 + 2\times(-1) \\ 1\times(-2) + (-1) \times 1 & 1 \times 3 + (-1)\times(-1) \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 2 & -2 \\ -3 & 4 \end{bmatrix}.

Solving A2 – A + BC

⇒[4−26−3]−[4−26−3]+[2−2−34]⇒[0000]+[2−2−34]⇒[2−2−34].\Rightarrow \begin{bmatrix} 4 & -2 \\ 6 & -3 \end{bmatrix} - \begin{bmatrix} 4 & -2 \\ 6 & -3 \end{bmatrix} + \begin{bmatrix} 2 & -2 \\ -3 & 4 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} + \begin{bmatrix} 2 & -2 \\ -3 & 4 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 2 & -2 \\ -3 & 4 \end{bmatrix}.

Hence, (A2 – A + BC) = [2−2−34]\begin{bmatrix} 2 & -2 \\ -3 & 4 \end{bmatrix}

Question 14(ii)

If A = [2357]\begin{bmatrix} 2 & 3 \\ 5 & 7 \end{bmatrix}, B = [04−17]\begin{bmatrix} 0 & 4 \\ -1 & 7 \end{bmatrix}, and C = [10−14]\begin{bmatrix} 1 & 0 \\ -1 & 4 \end{bmatrix},

Find AC + B2 – 10C.

Answer

Solving AC,

⇒AC=[2357]×[10−14]⇒[2×1+3×(−1)2×0+3×45×1+7×(−1)5×0+7×4]⇒[2−30+125−70+28]⇒[−112−228].\Rightarrow AC = \begin{bmatrix} 2 & 3 \\ 5 & 7 \end{bmatrix} \times \begin{bmatrix} 1 & 0 \\ -1 & 4 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 2 \times 1 + 3 \times (-1) & 2 \times 0 + 3 \times 4 \\ 5 \times 1 + 7 \times (-1) & 5 \times 0 + 7 \times 4 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 2 - 3 & 0 + 12 \\ 5 - 7 & 0 + 28 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} -1 & 12 \\ -2 & 28 \end{bmatrix}.

AC = [−112−228]\begin{bmatrix} -1 & 12 \\ -2 & 28 \end{bmatrix}

Solving B2,

⇒B2=[04−17]×[04−17]⇒[0×0+4×(−1)0×4+4×7−1×0+7×(−1)−1×4+7×7]⇒[−428−7−4+49]⇒[−428−745].\Rightarrow B^2 = \begin{bmatrix} 0 & 4 \\ -1 & 7 \end{bmatrix} \times \begin{bmatrix} 0 & 4 \\ -1 & 7 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 0 \times 0 + 4 \times (-1) & 0 \times 4 + 4 \times 7 \\ -1 \times 0 + 7 \times (-1) & -1 \times 4 + 7 \times 7 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} -4 & 28 \\ -7 & -4 + 49 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} -4 & 28 \\ -7 & 45 \end{bmatrix}.

Solving 10C,

⇒10C=10×[10−14]⇒[100−1040].\Rightarrow 10C = 10 \times \begin{bmatrix} 1 & 0 \\ -1 & 4 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 10 & 0 \\ -10 & 40 \end{bmatrix}.

Now, AC + B2 – 10C

⇒[−112−228]+[−428−745]−[100−1040]⇒[−540−973]−[100−1040]⇒[−5−1040−0−9−(−10)73−40]⇒[−1540133].\Rightarrow \begin{bmatrix} -1 & 12 \\ -2 & 28 \end{bmatrix} + \begin{bmatrix} -4 & 28 \\ -7 & 45 \end{bmatrix} - \begin{bmatrix} 10 & 0 \\ -10 & 40 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} -5 & 40 \\ -9 & 73 \end{bmatrix} - \begin{bmatrix} 10 & 0 \\ -10 & 40 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} -5 - 10 & 40 - 0 \\ -9 - (-10) & 73 - 40 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} -15 & 40 \\ 1 & 33 \end{bmatrix}.

Hence, (AC + B2 – 10C) = [−1540133]\begin{bmatrix} -15 & 40 \\ 1 & 33 \end{bmatrix}

Question 15

If A = [2513]\begin{bmatrix} 2 & 5 \\ 1 & 3 \end{bmatrix}, B = [4−2−13]\begin{bmatrix} 4 & -2 \\ -1 & 3 \end{bmatrix}, and I is the identity matrix of the same order and At is the transpose of A, find (AtB + BI).

Answer

Given,

A = [2513]\begin{bmatrix} 2 & 5 \\ 1 & 3 \end{bmatrix}

AT=[2153]A^T = \begin{bmatrix} 2 & 1 \\ 5 & 3 \end{bmatrix}

Solving ATB,

⇒[2153]×[4−2−13]⇒[2×4+1×(−1)2×(−2)+1×35×4+3×(−1)5×(−2)+3×3]⇒[7−117−1].\Rightarrow \begin{bmatrix} 2 & 1 \\ 5 & 3 \end{bmatrix} \times \begin{bmatrix} 4 & -2 \\ -1 & 3 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 2 \times 4 + 1 \times (-1) & 2 \times (-2) + 1 \times 3 \\ 5 \times 4 + 3 \times (-1) & 5 \times (-2) + 3 \times 3 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 7 & -1 \\ 17 & -1 \end{bmatrix}.

Solving BI,

⇒[4−2−13]×I⇒[4−2−13]×[1001]⇒[4×1+(−2)×04×0+(−2)×1(−1)×1+3×0(−1)×0+3×1]⇒[4−2−13].\Rightarrow \begin{bmatrix} 4 & -2 \\ -1 & 3 \end{bmatrix} \times I \\[1em] \Rightarrow \begin{bmatrix} 4 & -2 \\ -1 & 3 \end{bmatrix} \times \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 4 \times 1 + (-2) \times 0 & 4 \times 0 + (-2) \times 1 \\ (-1) \times 1 + 3 \times 0 & (-1) \times 0 + 3 \times 1 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 4 & -2 \\ -1 & 3 \end{bmatrix}.

Now, ATB + BI

⇒[7−117−1]+[4−2−13]⇒[7+4−1+(−2)17+(−1)−1+3]⇒[11−3162].\Rightarrow \begin{bmatrix} 7 & -1 \\ 17 & -1 \end{bmatrix} + \begin{bmatrix} 4 & -2 \\ -1 & 3 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 7 + 4 & -1 + (-2) \\ 17 + (-1) & -1 + 3 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 11 & -3 \\ 16 & 2 \end{bmatrix}.

Hence, (ATB + BI) = [11−3162]\begin{bmatrix} 11 & -3 \\ 16 & 2 \end{bmatrix}

Question 16

If A = [31−12]\begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}, show that A2 – 5A + 7I = 0.

Answer

Solving A2,

⇒A2=[31−12]×[31−12]⇒[3×3+1×(−1)3×1+1×2−1×3+2×(−1)−1×1+2×2]⇒[9−13+2−3−2−1+4]⇒[85−53].\Rightarrow A^2 = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} \times \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 3 \times 3 + 1 \times (-1) & 3 \times 1 + 1 \times 2 \\ -1 \times 3 + 2 \times (-1) & -1 \times 1 + 2 \times 2 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 9 - 1 & 3 + 2 \\ -3 - 2 & -1 + 4 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix}.

Solving 5A,

⇒5A=5×[31−12]⇒[155−510].\Rightarrow 5A = 5 \times \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 15 & 5 \\ -5 & 10 \end{bmatrix}.

Solving 7I,

⇒7I=7×[1001]⇒[7007].\Rightarrow 7I = 7 \times \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix}.

Now, A2 – 5A + 7I

⇒[85−53]−[155−510]+[7007]⇒[8−(15)+75−(5)+0−5−5+03−(10)+7]⇒[0000].\Rightarrow \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix} - \begin{bmatrix} 15 & 5 \\ -5 & 10 \end{bmatrix} + \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 8 - (15) + 7 & 5 - (5) + 0 \\ -5 - 5 + 0 & 3 - (10) + 7 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}.

Hence, A2 – 5A + 7I = [0000]\begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}.

Question 17

If A = [41−12]\begin{bmatrix} 4 & 1 \\ -1 & 2 \end{bmatrix}, show that 6A – A2 = 9I.

Answer

Solving A2,

⇒A2=[41−12]×[41−12]⇒[4×4+1×(−1)4×1+1×2−1×4+2×(−1)−1×1+2×2]⇒[16−14+2−4−2−1+4]⇒[156−63].\Rightarrow A^2 = \begin{bmatrix} 4 & 1 \\ -1 & 2 \end{bmatrix} \times \begin{bmatrix} 4 & 1 \\ -1 & 2 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 4 \times 4 + 1 \times (-1) & 4 \times 1 + 1 \times 2 \\ -1 \times 4 + 2 \times (-1) & -1 \times 1 + 2 \times 2 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 16 - 1 & 4 + 2 \\ -4 - 2 & -1 + 4 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 15 & 6 \\ -6 & 3 \end{bmatrix}.

Solving 6A,

⇒6A=6×[41−12]⇒[246−612].\Rightarrow 6A = 6 \times \begin{bmatrix} 4 & 1 \\ -1 & 2 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 24 & 6 \\ -6 & 12 \end{bmatrix}.

Now, 6A – A2

⇒[246−612]−[156−63]⇒[24−156−6−6−(−6)12−3]⇒[9009]⇒9[1001]=9I.\Rightarrow \begin{bmatrix} 24 & 6 \\ -6 & 12 \end{bmatrix} - \begin{bmatrix} 15 & 6 \\ -6 & 3 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 24 - 15 & 6 - 6 \\ -6 - (-6) & 12 - 3 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 9 & 0 \\ 0 & 9 \end{bmatrix} \\[1em] \Rightarrow 9\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = 9I.

Hence, proved that 6A – A2 = 9I.

Question 18

If A = [1324]\begin{bmatrix} 1 & 3 \\ 2 & 4 \end{bmatrix}, B = [1224]\begin{bmatrix} 1 & 2 \\ 2 & 4 \end{bmatrix}, C = [4115]\begin{bmatrix} 4 & 1 \\ 1 & 5 \end{bmatrix} and I = [1001]\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}, find A(B + C) – 14I.

Answer

Solving A(B + C),

⇒[1324]×([1224]+[4115])⇒[1324]×[1+42+12+14+5]⇒[1324]×[5339]⇒[1×5+3×31×3+3×92×5+4×32×3+4×9]⇒[5+93+2710+126+36]⇒[14302242].\Rightarrow \begin{bmatrix} 1 & 3 \\ 2 & 4 \end{bmatrix} \times \Big(\begin{bmatrix} 1 & 2 \\ 2 & 4 \end{bmatrix} + \begin{bmatrix} 4 & 1 \\ 1 & 5 \end{bmatrix}\Big) \\[1em] \Rightarrow \begin{bmatrix} 1 & 3 \\ 2 & 4 \end{bmatrix} \times \begin{bmatrix} 1+4 & 2+1 \\ 2+1 & 4+5 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 1 & 3 \\ 2 & 4 \end{bmatrix} \times \begin{bmatrix} 5 & 3 \\ 3 & 9 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 1 \times 5 + 3 \times 3 & 1 \times 3 + 3 \times 9 \\ 2 \times 5 + 4 \times 3 & 2 \times 3 + 4 \times 9 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 5 + 9 & 3 + 27 \\ 10 + 12 & 6 + 36 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 14 & 30 \\ 22 & 42 \end{bmatrix}.

Solving 14I,

⇒14×[1001]⇒[140014].\Rightarrow 14 \times \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 14 & 0 \\ 0 & 14 \end{bmatrix}.

Now, A(B + C) – 14I

⇒[14302242]−[140014]⇒[14−1430−022−042−14]⇒[0302228].\Rightarrow \begin{bmatrix} 14 & 30 \\ 22 & 42 \end{bmatrix} - \begin{bmatrix} 14 & 0 \\ 0 & 14 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 14 - 14 & 30 - 0 \\ 22 - 0 & 42 - 14 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 0 & 30 \\ 22 & 28 \end{bmatrix}.

Hence, A(B + C) – 14I = [0302228]\begin{bmatrix} 0 & 30 \\ 22 & 28 \end{bmatrix}.

Question 19

If A = [0110]\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}, show that A2 = I.

Answer

Solving A2,

⇒A2=[0110]×[0110]⇒[0×0+1×10×1+1×01×0+0×11×1+0×0]⇒[0+10+00+01+0]⇒[1001].\Rightarrow A^2 = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} \times \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 0 \times 0 + 1 \times 1 & 0 \times 1 + 1 \times 0 \\ 1 \times 0 + 0 \times 1 & 1 \times 1 + 0 \times 0 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 0 + 1 & 0 + 0 \\ 0 + 0 & 1 + 0 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}.

Hence, proved that A2 = I.

Question 20

If A = [abb2−a2−ab]\begin{bmatrix} ab & b^2 \\ -a^2 & -ab \end{bmatrix}, show that A2 = 0.

Answer

Solving A2,

⇒A2=[abb2−a2−ab]×[abb2−a2−ab]⇒[ab×ab+b2×(−a2)ab×b2+b2×(−ab)(−a2)×ab+(−ab)×(−a2)(−a2)×b2+(−ab)×(−ab)]⇒[a2b2−a2b2ab3−ab3−a3b+a3b−a2b2+a2b2]⇒[0000].\Rightarrow A^2 = \begin{bmatrix} ab & b^2 \\ -a^2 & -ab \end{bmatrix} \times \begin{bmatrix} ab & b^2 \\ -a^2 & -ab \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} ab \times ab + b^2 \times (-a^2) & ab \times b^2 + b^2 \times (-ab) \\ (-a^2) \times ab + (-ab) \times (-a^2) & (-a^2) \times b^2 + (-ab) \times (-ab) \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} a^2 b^2 - a^2 b^2 & a b^3 - a b^3 \\ - a^3 b + a^3 b & - a^2 b^2 + a^2 b^2 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}.

Hence, proved that A2 = 0.

Question 21

If A = [2−3ab]\begin{bmatrix} 2 & -3 \\ a & b \end{bmatrix}, find a and b so that A2 = I.

Answer

Solving A2,

⇒A2=[2−3ab]×[2−3ab]⇒[2×2+(−3)×a2×(−3)+(−3)×ba×2+b×aa×(−3)+b×b]⇒[4−3a−6−3b2a+ab−3a+b2].\Rightarrow A^2 = \begin{bmatrix} 2 & -3 \\ a & b \end{bmatrix} \times \begin{bmatrix} 2 & -3 \\ a & b \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 2 \times 2 + (-3) \times a & 2 \times (-3) + (-3) \times b \\ a \times 2 + b \times a & a \times (-3) + b \times b \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 4 - 3a & -6 - 3b \\ 2a + ab & -3a + b^2 \end{bmatrix}.

Given,

A2 = I,

[4−3a−6−3b2a+ab−3a+b2]=[1001].\begin{bmatrix} 4 - 3a & -6 - 3b \\ 2a + ab & -3a + b^2 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}.

∴ 4 - 3a = 1,

⇒ -3a = -3

⇒ a = 1.

∴ -6 - 3b = 0,

⇒ -3b = 6

⇒ b = -2.

Hence, a = 1 and b = -2.

Question 22

If A = [21201]\begin{bmatrix} 2 & 12 \\ 0 & 1 \end{bmatrix} and B = [4x01]\begin{bmatrix} 4 & x \\ 0 & 1 \end{bmatrix} and A2 = B, find the value of x.

Answer

Let's find A2,

⇒A2=[21201]×[21201]⇒[2×2+12×02×12+12×10×2+1×00×12+1×1]⇒[424+1201]⇒[43601].\Rightarrow A^2 = \begin{bmatrix} 2 & 12 \\ 0 & 1 \end{bmatrix} \times \begin{bmatrix} 2 & 12 \\ 0 & 1 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 2 \times 2 + 12 \times 0 & 2 \times 12 + 12 \times 1 \\ 0 \times 2 + 1 \times 0 & 0 \times 12 + 1 \times 1 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 4 & 24 + 12 \\ 0 & 1 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 4 & 36 \\ 0 & 1 \end{bmatrix}.

Since, A2 = B,

⇒[43601]=[4x01]\Rightarrow \begin{bmatrix} 4 & 36 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 4 & x \\ 0 & 1 \end{bmatrix}

∴ x = 36

Hence, x = 36.

Question 23

Find matrix A such that A × [2345]\begin{bmatrix} 2 & 3 \\ 4 & 5 \end{bmatrix} = [0−4103]\begin{bmatrix} 0 & -4 \\ 10 & 3 \end{bmatrix}.

Answer

Let B = [2345]\begin{bmatrix} 2 & 3 \\ 4 & 5 \end{bmatrix} Then, AB = [0−4103]\begin{bmatrix} 0 & -4 \\ 10 & 3 \end{bmatrix}

Since AB exists, we have:

Number of columns of A = Number of rows in B = 2

Number of rows of A = Number of rows in AB = 2

Order of A is 2 × 2

Let A = [abcd]\begin{bmatrix} a & b \\ c & d \end{bmatrix}.

Then,

⇒[abcd]×[2345]=[0−4103]⇒[2a+4b3a+5b2c+4d3c+5d]=[0−4103].\Rightarrow \begin{bmatrix} a & b \\ c & d \end{bmatrix} \times \begin{bmatrix} 2 & 3 \\ 4 & 5 \end{bmatrix} = \begin{bmatrix} 0 & -4 \\ 10 & 3 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 2a + 4b & 3a + 5b \\ 2c + 4d & 3c + 5d \end{bmatrix} =\begin{bmatrix} 0 & -4 \\ 10 & 3 \end{bmatrix}.

Solving for a and b:

∴ 2a + 4b = 0

⇒ 2(a + 2b) = 0

⇒ a + 2b = 0

⇒ a = -2b ......(1)

∴ 3a + 5b = -4

Substituting value of a from equation(1) in 3a + 5b = -4, we get:

⇒ 3(-2b) + 5b = -4

⇒ -6b + 5b = -4

⇒ -b = -4

⇒ b = 4.

Substituting value of b in equation 1, we get,

⇒ a = -2(4)

⇒ a = -8.

Solving for c and d:

∴ 2c + 4d = 10

⇒ 2(c + 2d) = 10

⇒ c + 2d = 5

⇒ c = 5 - 2d .......(2)

∴ 3c + 5d = 3

Substituting value of c from equation (2) in 3c + 5d = 3, we get :

⇒ 3c + 5d = 3

⇒ 3(5 - 2d) + 5d = 3

⇒ 15 - 6d + 5d = 3

⇒ 15 - d = 3

⇒ d = 15 - 3

⇒ d = 12.

Substituting value of d in equation (2), we get,

⇒ c = 5 - 2(12)

⇒ c = 5 - 24

⇒ c = -19.

∴A=[abcd]=[−84−1912]\therefore A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} = \begin{bmatrix} -8 & 4 \\ -19 & 12 \end{bmatrix}.

Hence, A = [−84−1912]\begin{bmatrix} -8 & 4 \\ -19 & 12 \end{bmatrix}.

Question 24

Find matrix X such that [5−7−23]\begin{bmatrix} 5 & -7 \\ -2 & 3 \end{bmatrix} X = [−16−672]\begin{bmatrix} -16 & -6 \\ 7 & 2 \end{bmatrix}.

Answer

Let Y = [5−7−23]\begin{bmatrix} 5 & -7 \\ -2 & 3 \end{bmatrix}.

Then, YX = [−16−672]\begin{bmatrix} -16 & -6 \\ 7 & 2 \end{bmatrix}

Since YX exists, we have:

Number of rows of X = Number of columns in Y = 2

Number of columns of X = Number of columns in YX = 2

Order of X is 2 × 2.

Let X = [abcd]\begin{bmatrix} a & b \\ c & d \end{bmatrix}.

Then,

⇒[5−7−23]×[abcd]=[−16−672]⇒[5a−7c5b−7d−2a+3c−2b+3d]=[−16−672]\Rightarrow \begin{bmatrix} 5 & -7 \\ -2 & 3 \end{bmatrix} \times \begin{bmatrix} a & b \\ c & d \end{bmatrix}= \begin{bmatrix} -16 & -6 \\ 7 & 2 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 5a - 7c & 5b - 7d \\ -2a + 3c & -2b + 3d \end{bmatrix} = \begin{bmatrix} -16 & -6 \\ 7 & 2 \end{bmatrix} \\[1em]

Solving for a and c :

∴ -2a + 3c = 7

⇒ 2a = 3c - 7

⇒ a = 3c−72\dfrac{3c - 7}{2} ...(1)

∴ 5a - 7c = -16 .......(2)

Substituting value of a from equation (1) in (2), we get :

⇒5(3c−72)−7c=−16⇒(15c−352)−7c=−16⇒15c−352−7c×22=−16⇒15c−35−14c2=−16⇒c−352=−16⇒c−35=−16×2⇒c−35=−32⇒c=−32+35⇒c=3.\Rightarrow 5\Big(\dfrac{3c - 7}{2}\Big) - 7c = -16 \\[1em] \Rightarrow \Big(\dfrac{15c - 35}{2}\Big) - 7c = -16 \\[1em] \Rightarrow \dfrac{15c - 35}{2} - \dfrac{7c \times 2}{2} = -16 \\[1em] \Rightarrow \dfrac{15c - 35 - 14c}{2} = -16 \\[1em] \Rightarrow \dfrac{c - 35}{2} = -16 \\[1em] \Rightarrow c - 35 = -16 \times 2 \\[1em] \Rightarrow c - 35 = -32 \\[1em] \Rightarrow c = -32 + 35 \\[1em] \Rightarrow c = 3.

Substituting value of c in equation (1), we get:

⇒ a = 3(3)−72\dfrac{3(3) - 7}{2}

⇒ a = 9−72\dfrac{9 - 7}{2}

⇒ a = 22\dfrac{2}{2}

⇒ a = 1.

Solving for b and d:

∴ -2b + 3d = 2

⇒ 2b = 3d - 2

⇒ b = 3d−22\dfrac{3d - 2}{2} ....(3)

∴ 5b - 7d = -6 ......(4)

Substituting value of b from equation (3) in (4), we get:

⇒5(3d−22)−7d=−6⇒(15d−102)−7d=−6⇒15d−102−7d×22=−6⇒15d−10−14d2=−6⇒d−102=−6⇒d−10=−6×2⇒d−10=−12⇒d=−12+10⇒d=−2.\Rightarrow 5\Big(\dfrac{3d - 2}{2}\Big) - 7d = -6 \\[1em] \Rightarrow \Big(\dfrac{15d - 10}{2}\Big) - 7d = -6 \\[1em] \Rightarrow \dfrac{15d - 10}{2} - \dfrac{7d \times 2}{2} = -6 \\[1em] \Rightarrow \dfrac{15d - 10 - 14d}{2} = -6 \\[1em] \Rightarrow \dfrac{d - 10}{2} = -6 \\[1em] \Rightarrow d - 10 = -6 \times 2 \\[1em] \Rightarrow d - 10 = -12 \\[1em] \Rightarrow d = -12 + 10 \\[1em] \Rightarrow d = -2.

Substituting value of d in equation (3), we get:

⇒ b = 3(−2)−22\dfrac{3(-2) - 2}{2}

⇒ b = −6−22\dfrac{-6 - 2}{2}

⇒ b = −82\dfrac{-8}{2}

⇒ b = -4.

∴X=[1−43−2]\therefore X = \begin{bmatrix} 1 & -4 \\ 3 & -2 \end{bmatrix}.

Hence, X = [1−43−2]\begin{bmatrix} 1 & -4 \\ 3 & -2 \end{bmatrix}.

Question 25

Let A be a matrix such that A × [32−15]\begin{bmatrix} 3 & 2 \\ -1 & 5 \end{bmatrix} = [9−11]\begin{bmatrix} 9 & -11 \end{bmatrix}.

(i) Write the order of A.

(ii) Find A.

Answer

(i) Let B = [32−15]\begin{bmatrix} 3 & 2 \\ -1 & 5 \end{bmatrix} Then, AB = [9−11]\begin{bmatrix} 9 & -11 \end{bmatrix}

Order of B = 2 × 2

Order of AB = 1 × 2

Since AB exists, we have:

Number of columns in A = Number of rows in B = 2

Number of rows of A = Number of rows in AB = 1

Order of A is 1 × 2.

A1×2×B2×2=AB1×2A_{1 \times 2} \times B_{2 \times 2} = AB_{1 \times 2}

Hence, order of A is 1 × 2.

(ii) Let A = [ab]\begin{bmatrix} a & b \end{bmatrix}.

Then,

⇒[ab]×[32−15]=[9−11]⇒[3a−b2a+5b]=[9−11].\Rightarrow \begin{bmatrix} a & b \end{bmatrix} \times \begin{bmatrix} 3 & 2 \\ -1 & 5 \end{bmatrix} = \begin{bmatrix} 9 & -11 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 3a - b & 2a + 5b \end{bmatrix} = \begin{bmatrix} 9 & -11 \end{bmatrix}.

Solving for a and b:

∴ 3a - b = 9

⇒ b = 3a - 9 ....(1)

∴ 2a + 5b = -11 ...(2)

Substituting value of b from equation (1) in (2), we get :

⇒ 2a + 5(3a - 9) = -11

⇒ 2a + 15a - 45 = -11

⇒ 17a = -11 + 45

⇒ 17a = 34

⇒ a = 3417\dfrac{34}{17}

⇒ a = 2.

Substituting value of a in equation (1), we get :

⇒ b = 3(2) - 9

⇒ b = 6 - 9

⇒ b = -3.

∴A=[2−3]\therefore A = \begin{bmatrix} 2 & -3 \end{bmatrix}.

Hence, A = [2−3]\begin{bmatrix} 2 & -3 \end{bmatrix}.

Question 26

Given [42−11]\begin{bmatrix} 4 & 2 \\ -1 & 1 \end{bmatrix} M = 6I, where M is a matrix and I is the unit matrix of order 2 × 2

(i) State the order of matrix M.

(ii) Find the matrix M.

Answer

(i) Let N = [42−11]\begin{bmatrix} 4 & 2 \\ -1 & 1 \end{bmatrix}

Given, NM = 6I

NM = [6006]\begin{bmatrix} 6 & 0 \\ 0 & 6 \end{bmatrix}

Order of N = 2 × 2

Order of NM = 2 × 2

Since NM exists, we have:

Number of rows of M = Number of columns in N = 2

Number of columns of M = Number of columns in NM = 2

Order of M is 2 × 2.

M2×2×N2×2=MN2×2M_{2 \times 2} \times N_{2 \times 2} = MN_{2 \times 2}

Hence, order of M is 2 × 2.

(ii) Let M = [abcd]\begin{bmatrix} a & b \\ c & d \end{bmatrix}.

Then,

⇒[42−11]×[abcd]=6[1001]⇒[4a+2c4b+2d−a+c−b+d]=[6006].\Rightarrow \begin{bmatrix} 4 & 2 \\ -1 & 1 \end{bmatrix} \times \begin{bmatrix} a & b \\ c & d \end{bmatrix} = 6\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 4a + 2c & 4b + 2d \\ -a + c & -b + d \end{bmatrix} = \begin{bmatrix} 6 & 0 \\ 0 & 6 \end{bmatrix}.

Solving for a and c :

∴ -a + c = 0

⇒ c = a....(1)

∴ 4a + 2c = 6

⇒ 2(2a + c = 3)

⇒ 2a + c = 3...(2)

Substituting value of c from equation (1) in 2a + c = 3, we get:

⇒ 2a + a = 3

⇒ 3a = 3

⇒ a = 33\dfrac{3}{3}

⇒ a = 1

Hence, a = c = 1

Solving for b and d:

∴ -b + d = 6

⇒ d = b + 6...(3)

∴ 4b + 2d = 0

⇒ 2(2b + d) = 0

⇒ 2b + d = 0...(4)

Substituting value of d from equation (3) in 2b + d = 0, we get:

⇒ 2b + b + 6 = 0

⇒ 3b + 6 = 0

⇒ 3b = -6

⇒ b = −63\dfrac{-6}{3}

⇒ b = -2

Substituting value of b in equation(3), we get:

⇒ d = -2 + 6

⇒ d = 4

Hence, M = [1−214].\begin{bmatrix} 1 & -2 \\ 1 & 4 \end{bmatrix}.

Question 27

Let A be a matrix such that [5−213]\begin{bmatrix} 5 & -2 \\ 1 & 3 \end{bmatrix} × A = [411]\begin{bmatrix} 4 \\ 11 \end{bmatrix}.

(i) Write the order of A.

(ii) Find A.

Answer

(i) Let B = [5−213]\begin{bmatrix} 5 & -2 \\ 1 & 3 \end{bmatrix} Then, BA = [411]\begin{bmatrix} 4 \\ 11 \end{bmatrix}

Order of B = 2 × 2

Order of BA = 2 × 1

Since BA exists, we have:

Number of rows of A = Number of columns in B = 2

Number of columns of A = Number of columns in BA = 1

Order of A is 2 × 1.

B2×2×A2×1=BA2×1B_{2 \times 2} \times A_{2 \times 1} = BA_{2 \times 1}

Hence, order of A is 2 × 1.

(ii) Let A = [xy]\begin{bmatrix} x \\ y \end{bmatrix}.

Then,

⇒[5−213]×[xy]=[411]⇒[5x−2yx+3y]=[411].\Rightarrow \begin{bmatrix} 5 & -2 \\ 1 & 3 \end{bmatrix} \times \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 4 \\ 11 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 5x - 2y \\ x + 3y \end{bmatrix} = \begin{bmatrix} 4 \\ 11 \end{bmatrix}.

Solving for x and y:

∴ 5x - 2y = 4... (1)

∴ x + 3y = 11

⇒ x = 11 - 3y....(2)

Substituting value of d from equation (2) in 5x - 2y = 4, we get:

⇒ 5(11 - 3y) - 2y = 4

⇒ 55 - 15y - 2y = 4

⇒ 55 - 17y = 4

⇒ 17y = 55 - 4

⇒ 17y = 51

⇒ y = 5117\dfrac{51}{17}

⇒ y = 3

Substituting value of y in equation (2), we get:

⇒ x = 11 - 3(3)

⇒ x = 11 - 9

⇒ x = 2

Hence, A = [23].\begin{bmatrix} 2 \\ 3 \end{bmatrix}.

Question 28(i)

Let A = [2−1−34]\begin{bmatrix} 2 & -1 \\ -3 & 4 \end{bmatrix} and B = [8−17]\begin{bmatrix} 8 \\ -17 \end{bmatrix}. Find a matrix C such that AC = B.

Answer

A = [2−1−34]\begin{bmatrix} 2 & -1 \\ -3 & 4 \end{bmatrix} and B = [8−17]\begin{bmatrix} 8 \\ -17 \end{bmatrix}

Given,

AC = B

Order of A = 2 × 2

Order of AC = Order of B = 2 × 1

Since AC exists, we have :

Number of rows of C = Number of columns in A = 2

Number of columns of C = Number of columns in B = 1

Order of C is 2 × 1.

Let C = [xy]\begin{bmatrix} x \\ y \end{bmatrix}

AC = B

⇒[2−1−34]×[xy]=[8−17]⇒[2x−y−3x+4y]=[8−17].\Rightarrow \begin{bmatrix} 2 & -1 \\ -3 & 4 \end{bmatrix} \times \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 8 \\ -17 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 2x - y \\ -3x + 4y \end{bmatrix} = \begin{bmatrix} 8 \\ -17 \end{bmatrix}.

∴ 2x - y = 8

⇒ y = 2x - 8 .......(1)

∴ -3x + 4y = -17

Substituting value of y from equation(1) in -3x + 4y = -17, we get:

⇒ -3x + 4(2x - 8) = -17

⇒ -3x + 8x - 32 = -17

⇒ 5x - 32 = -17

⇒ 5x = -17 + 32

⇒ 5x = 15

⇒ x = 155\dfrac{15}{5}

⇒ x = 3.

Substituting value of x in equation (1), we get :

⇒ y = 2(3) - 8

⇒ y = 6 - 8

⇒ y = -2.

Hence, C = [3−2].\begin{bmatrix} 3 \\ -2 \end{bmatrix}.

Question 28(ii)

Let A = [32−11]\begin{bmatrix} 3 & 2 \\ -1 & 1 \end{bmatrix} and B = [143 24]\begin{bmatrix} 14 & 3\ 2 & 4 \end{bmatrix}. Find a matrix C such that AC = B.

Answer

A = [32−11]\begin{bmatrix} 3 & 2 \\ -1 & 1 \end{bmatrix} and B = [143 24]\begin{bmatrix} 14 & 3\ 2 & 4 \end{bmatrix}

Given,

AC = B

Order of A = 2 × 2

Order of AC = Order of B = 2 × 2

Since AC exists, we have:

Number of rows of C = Number of columns in A = 2

Number of columns of C = Number of columns in B = 2

Order of C is 2 × 2.

Let C = [ab cd]\begin{bmatrix} a & b\ c & d \end{bmatrix}

AC = B

⇒[32−11]×[ab cd]=[143 24]⇒[3a+2c3b+2d−a+c−b+d]=[143 24].\Rightarrow \begin{bmatrix} 3 & 2 \\ -1 & 1 \end{bmatrix} \times \begin{bmatrix} a & b\ c & d \end{bmatrix} = \begin{bmatrix} 14 & 3\ 2 & 4 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 3a + 2c & 3b + 2d \\ -a + c & -b + d \end{bmatrix} = \begin{bmatrix} 14 & 3\ 2 & 4 \end{bmatrix}.

Solving for a and c:

∴ -a + c = 2

⇒ c = a + 2 ...(1)

∴ 3a + 2c = 14 ......(2)

Substituting value of c from equation(1) in (2), we get :

⇒ 3a + 2(a + 2) = 14

⇒ 3a + 2a + 4 = 14

⇒ 5a = 14 - 4

⇒ 5a = 10

⇒ a = 105\dfrac{10}{5}

⇒ a = 2.

Substituting value of a in equation (1), we get :

⇒ c = 2 + 2

⇒ c = 4.

Solving for b and d :

∴ -b + d = 4

⇒ d = b + 4 .......(3)

∴ 3b + 2d = 3 .......(4)

Substituting value of d from equation (3) in (4), we get:

⇒ 3b + 2(b + 4) = 3

⇒ 3b + 2b + 8 = 3

⇒ 5b + 8 = 3

⇒ 5b = 3 - 8

⇒ 5b = -5

⇒ b = −55\dfrac{-5}{5}

⇒ b = -1.

Substituting value of b in equation (3), we get :

⇒ d = -1 + 4

⇒ d = 3.

Hence, C = [2−1 43].\begin{bmatrix} 2 & -1\ 4 & 3 \end{bmatrix}.

Question 29(i)

If A = [1334]\begin{bmatrix} 1 & 3 \\ 3 & 4 \end{bmatrix} and B = [−21−32]\begin{bmatrix} -2 & 1 \\ -3 & 2 \end{bmatrix} and A2 – 5B2 = 5C. Find matrix C, where C is a 2 × 2 matrix.

Answer

Given,

A = [1324]\begin{bmatrix} 1 & 3 \\ 2 & 4 \end{bmatrix} and B = [−21−32]\begin{bmatrix} -2 & 1 \\ -3 & 2 \end{bmatrix}

Solving for A2:

⇒A2=[1324]×[1324]=[(1)(1)+(3)(2)(1)(3)+(3)(4)(2)(1)+(4)(2)(2)(3)+(4)(4)]=[1+63+122+86+16]=[7151022].\Rightarrow A^2 = \begin{bmatrix} 1 & 3 \\ 2 & 4 \end{bmatrix} \times \begin{bmatrix} 1 & 3 \\ 2 & 4 \end{bmatrix} \\[1em] = \begin{bmatrix} (1)(1) + (3)(2) & (1)(3) + (3)(4) \\ (2)(1) + (4)(2) & (2)(3) + (4)(4) \end{bmatrix} \\[1em] = \begin{bmatrix} 1 + 6 & 3 + 12 \\ 2 + 8 & 6 + 16 \end{bmatrix} \\[1em] = \begin{bmatrix} 7 & 15 \\ 10 & 22 \end{bmatrix}.

Solving for 5B2:

⇒5B2=5([−21−32]×[−21−32])=5([(−2)(−2)+(1)(−3)(−2)(1)+(1)(2)(−3)(−2)+(2)(−3)(−3)(1)+(2)(2)])=5[4−3−2+26−6−3+4]=5[1001]=[5005].\Rightarrow 5B^2 = 5\Big(\begin{bmatrix} -2 & 1 \\ -3 & 2 \end{bmatrix} \times \begin{bmatrix} -2 & 1 \\ -3 & 2 \end{bmatrix}\Big) \\[1em] = 5\Big(\begin{bmatrix} (-2)(-2) + (1)(-3) & (-2)(1) + (1)(2) \\ (-3)(-2) + (2)(-3) & (-3)(1) + (2)(2) \end{bmatrix}\Big) \\[1em] = 5\begin{bmatrix} 4 - 3 & -2 + 2 \\ 6 - 6 & -3 + 4 \end{bmatrix} \\[1em] = 5\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \\[1em] = \begin{bmatrix} 5 & 0 \\ 0 & 5 \end{bmatrix}.

Solving for A2 – 5B2 = 5C:

⇒[7151022]−[5005]=5C⇒[7−515−010−022−5]=5C⇒15[2151517]=C⇒[2533175]=C\Rightarrow \begin{bmatrix} 7 & 15 \\ 10 & 22 \end{bmatrix} - \begin{bmatrix} 5 & 0 \\ 0 & 5 \end{bmatrix} = 5C \\[1em] \Rightarrow \begin{bmatrix} 7 - 5 & 15 - 0 \\ 10 - 0 & 22 - 5 \end{bmatrix} = 5C \\[1em] \Rightarrow \dfrac{1}{5} \begin{bmatrix} 2 & 15 \\ 15 & 17 \end{bmatrix} = C \\[1em] \Rightarrow \begin{bmatrix} \dfrac{2}{5} & 3 \\ 3 & \dfrac{17}{5} \end{bmatrix} = C

Hence, C = [2533175]\begin{bmatrix} \dfrac{2}{5} & 3 \\ 3 & \dfrac{17}{5} \end{bmatrix}

Question 29(ii)

Given matrix B = [1183]\begin{bmatrix} 1 & 1 \\ 8 & 3 \end{bmatrix}. Find the matrix X if, X = B2 - 4B. Hence, solve for a and b given X × [ab]\begin{bmatrix} a \\ b \end{bmatrix} = [550]\begin{bmatrix} 5 \\ 50 \end{bmatrix}.

Answer

Given,

B = [1183]\begin{bmatrix} 1 & 1 \\ 8 & 3 \end{bmatrix}

X = B2 - 4B

Solving for B2:

⇒B2=[1183]×[1183]=[(1)(1)+(1)(8)(1)(1)+(1)(3)(8)(1)+(3)(8)(8)(1)+(3)(3)]=[1+81+38+248+9]=[943217].\Rightarrow B^2 = \begin{bmatrix} 1 & 1 \\ 8 & 3 \end{bmatrix} \times \begin{bmatrix} 1 & 1 \\ 8 & 3 \end{bmatrix} \\[1em] = \begin{bmatrix} (1)(1) + (1)(8) & (1)(1) + (1)(3) \\ (8)(1) + (3)(8) & (8)(1) + (3)(3) \end{bmatrix} \\[1em] = \begin{bmatrix} 1 + 8 & 1 + 3 \\ 8 + 24 & 8 + 9 \end{bmatrix} \\[1em] = \begin{bmatrix} 9 & 4 \\ 32 & 17 \end{bmatrix}.

Solving for 4B:

⇒4B=4[1183]=[443212].\Rightarrow 4B = 4\begin{bmatrix} 1 & 1 \\ 8 & 3 \end{bmatrix} \\[1em] = \begin{bmatrix} 4 & 4 \\ 32 & 12 \end{bmatrix}.

Now X = B2 - 4B:

⇒X=[943217]−[443212]=[9−44−432−3217−12]=[5005].\Rightarrow X = \begin{bmatrix} 9 & 4 \\ 32 & 17 \end{bmatrix} - \begin{bmatrix} 4 & 4 \\ 32 & 12 \end{bmatrix} \\[1em] = \begin{bmatrix} 9 - 4 & 4 - 4 \\ 32 - 32 & 17 - 12 \end{bmatrix} \\[1em] = \begin{bmatrix} 5 & 0 \\ 0 & 5 \end{bmatrix}.

Hence, X =[5005].\begin{bmatrix} 5 & 0 \\ 0 & 5 \end{bmatrix}.

Now,

⇒X×[ab]=[550]⇒[5005]×[ab]=[550]⇒[5a+00a+5b]=[550].\Rightarrow X \times \begin{bmatrix} a \\ b \end{bmatrix} = \begin{bmatrix} 5 \\ 50 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 5 & 0 \\ 0 & 5 \end{bmatrix} \times \begin{bmatrix} a \\ b \end{bmatrix} = \begin{bmatrix} 5 \\ 50 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 5a + 0 \\ 0a + 5b \end{bmatrix} = \begin{bmatrix} 5 \\ 50 \end{bmatrix}.

Solving for a and b:

∴ 5a = 5

a = 55\dfrac{5}{5}

a = 1.

∴ 5b = 50

b = 505\dfrac{50}{5}

b = 10.

Hence, a = 1 and b = 10.

Question 30

Evaluate without using tables :

[2cos⁡60∘−2sin⁡30∘−tan⁡45∘cos⁡0∘]×[cot⁡45∘cosec⁡30∘sec⁡60∘sin⁡90∘]\begin{bmatrix} 2 \cos 60^\circ & -2 \sin 30^\circ \\ -\tan 45^\circ & \cos 0^\circ \end{bmatrix} \times \begin{bmatrix} \cot 45^\circ & \cosec 30^\circ \\ \sec 60^\circ & \sin 90^\circ \end{bmatrix}

Answer

Solving,

⇒[2cos⁡60∘−2sin⁡30∘−tan⁡45∘cos⁡0∘]×[cot⁡45∘cosec⁡30∘sec⁡60∘sin⁡90∘]⇒[2×12−2×12−11]×[1221]⇒[1−1−11]×[1221]⇒[(1)(1)+(−1)(2)(1)(2)+(−1)(1)(−1)(1)+(1)(2)(−1)(2)+(1)(1)]⇒[1−22−1−1+2−2+1]⇒[−111−1].\Rightarrow \begin{bmatrix} 2 \cos 60^\circ & -2 \sin 30^\circ \\ -\tan 45^\circ & \cos 0^\circ \end{bmatrix} \times \begin{bmatrix} \cot 45^\circ & \cosec 30^\circ \\ \sec 60^\circ & \sin 90^\circ \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 2 \times \dfrac{1}{2} & -2 \times \dfrac{1}{2} \\ -1 & 1 \end{bmatrix} \times \begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix} \times \begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} (1)(1) + (-1)(2) & (1)(2) + (-1)(1) \\ (-1)(1) + (1)(2) & (-1)(2) + (1)(1) \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 1 - 2 & 2 - 1 \\ -1 + 2 & -2 + 1 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} -1 & 1 \\ 1 & -1 \end{bmatrix}.

Hence, [2cos⁡60∘−2sin⁡30∘−tan⁡45∘cos⁡0∘]×[cot⁡45∘cosec⁡30∘sec⁡60∘sin⁡90∘]=[−111−1]\begin{bmatrix} 2 \cos 60^\circ & -2 \sin 30^\circ \\ -\tan 45^\circ & \cos 0^\circ \end{bmatrix} \times \begin{bmatrix} \cot 45^\circ & \cosec 30^\circ \\ \sec 60^\circ & \sin 90^\circ \end{bmatrix} = \begin{bmatrix} -1 & 1 \\ 1 & -1 \end{bmatrix}

Question 31

If A = [2639]\begin{bmatrix} 2 & 6 \\ 3 & 9 \end{bmatrix}, B = [3xy2]\begin{bmatrix} 3 & x \\ y & 2 \end{bmatrix} and AB = 0, find the values of x and y.

Answer

Given,

A = [2639]\begin{bmatrix} 2 & 6 \\ 3 & 9 \end{bmatrix}, B = [3xy2]\begin{bmatrix} 3 & x \\ y & 2 \end{bmatrix}

AB = 0

⇒[2639]×[3xy2]=[0000]⇒[(2)(3)+6y2x+(6)(2)(3)(3)+9y3x+(9)(2)]=[0000]⇒[6+6y2x+129+9y3x+18]=[0000]\Rightarrow \begin{bmatrix} 2 & 6 \\ 3 & 9 \end{bmatrix} \times \begin{bmatrix} 3 & x \\ y & 2 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} (2)(3) + 6y & 2x + (6)(2) \\ (3)(3) + 9y & 3x + (9)(2) \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 6 + 6y & 2x + 12 \\ 9 + 9y & 3x + 18 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}

Solve for x and y:

∴ 6 + 6y = 0

⇒ 6y = -6

⇒ y = −66\dfrac{-6}{6}

⇒ y = -1.

∴ 2x + 12 = 0

⇒ 2x = -12

⇒ x = −122\dfrac{-12}{2}

⇒ x = -6.

Hence, x = -6 and y = -1.

Question 32

If [−320−5]\begin{bmatrix} -3 & 2 \\ 0 & -5 \end{bmatrix} [x2]\begin{bmatrix} x \\ 2 \end{bmatrix} = [−5y]\begin{bmatrix} -5 \\ y \end{bmatrix}, find the values of x and y.

Answer

Given,

[−320−5]\begin{bmatrix} -3 & 2 \\ 0 & -5 \end{bmatrix} [x2]\begin{bmatrix} x \\ 2 \end{bmatrix} = [−5y]\begin{bmatrix} -5 \\ y \end{bmatrix}

Solving:

⇒[(−3)(x)+(2)(2)(0)(x)+(−5)(2)]=[−5y]⇒[−3x+4−10]=[−5y].\Rightarrow \begin{bmatrix} (-3)(x) + (2)(2) \\ (0)(x) + (-5)(2) \end{bmatrix} = \begin{bmatrix} -5 \\ y \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} -3x + 4 \\ -10 \end{bmatrix} = \begin{bmatrix} -5 \\ y \end{bmatrix}.

∴ y = -10

∴ -3x + 4 = -5

⇒ -3x = -5 - 4

⇒ -3x = -9

⇒ x = −9−3\dfrac{-9}{-3}

⇒ x = 3.

Hence, x = 3 and y = -10.

Question 33

If [1233][x00y]=[x090]\begin{bmatrix} 1 & 2 \\ 3 & 3 \end{bmatrix} \begin{bmatrix} x & 0 \\ 0 & y \end{bmatrix} = \begin{bmatrix} x & 0 \\ 9 & 0 \end{bmatrix}, find the values of x and y.

Answer

⇒[1233]×[x00y]=[x090]⇒[(1)(x)+(2)(0)(1)(0)+(2)(y)(3)(x)+(3)(0)(3)(0)+(3)(y)]=[x090]⇒[x+22y3x3y]=[x090]\Rightarrow \begin{bmatrix} 1 & 2 \\ 3 & 3 \end{bmatrix} \times \begin{bmatrix} x & 0 \\ 0 & y \end{bmatrix} = \begin{bmatrix} x & 0 \\ 9 & 0 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} (1)(x) + (2)(0) & (1)(0) + (2)(y) \\ (3)(x) + (3)(0) & (3)(0) + (3)(y) \end{bmatrix} = \begin{bmatrix} x & 0 \\ 9 & 0 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} x + 2 & 2y \\ 3x & 3y \end{bmatrix} = \begin{bmatrix} x & 0 \\ 9 & 0 \end{bmatrix}

Solving for x and y:

∴ 3x = 9

⇒ x = 93\dfrac{9}{3}

⇒ x = 3.

∴ 2y = 0

⇒ y = 0.

Hence, x = 3 and y = 0.

Question 34

If [2xxy3y][32]=[169]\begin{bmatrix} 2x & x \\ y & 3y \end{bmatrix} \begin{bmatrix} 3 \\ 2 \end{bmatrix} = \begin{bmatrix} 16 \\ 9 \end{bmatrix}, find the values of x and y.

Answer

Solving,

⇒[2xxy3y][32]=[169]⇒[(2x)(3)+(x)(2)(y)(3)+(3y)(2)]=[169]⇒[6x+2x3y+6y]=[169]⇒[8x9y]=[169].\Rightarrow \begin{bmatrix} 2x & x \\ y & 3y \end{bmatrix} \begin{bmatrix} 3 \\ 2 \end{bmatrix} = \begin{bmatrix} 16 \\ 9 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} (2x)(3) + (x)(2) \\ (y)(3) + (3y)(2) \end{bmatrix} = \begin{bmatrix} 16 \\ 9 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 6x + 2x \\ 3y + 6y \end{bmatrix} = \begin{bmatrix} 16 \\ 9 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 8x \\ 9y \end{bmatrix} = \begin{bmatrix} 16 \\ 9 \end{bmatrix}.

Solving for x and y:

∴ 8x = 16

⇒ x = 168\dfrac{16}{8}

⇒ x = 2.

∴ 9y = 9

⇒ y = 99\dfrac{9}{9}

⇒ y = 1.

Hence, x = 2 and y = 1.

Question 35

Given that A = [3002]\begin{bmatrix} 3 & 0 \\ 0 & 2 \end{bmatrix}, B = [ab0c]\begin{bmatrix} a & b \\ 0 & c \end{bmatrix} and AB = (A + B), find the values of a, b, and c.

Answer

Solving AB:

⇒[3002]×[ab0c]⇒[(3)(a)+(0)(0)(3)(b)+(0)(c)(0)(a)+(2)(0)(0)(b)+(2)(c)]⇒[3a3b02c].\Rightarrow \begin{bmatrix} 3 & 0 \\ 0 & 2 \end{bmatrix} \times \begin{bmatrix} a & b \\ 0 & c \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} (3)(a) + (0)(0) & (3)(b) + (0)(c) \\ (0)(a) + (2)(0) & (0)(b) + (2)(c) \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 3a & 3b \\ 0 & 2c \end{bmatrix}.

Solving A + B:

⇒[3002]+[ab0c]⇒[3+ab02+c].\Rightarrow \begin{bmatrix} 3 & 0 \\ 0 & 2 \end{bmatrix} + \begin{bmatrix} a & b \\ 0 & c \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 3 + a & b \\ 0 & 2 + c \end{bmatrix}.

Given,

AB = A + B

⇒[3a3b02c]=[3+ab02+c]\Rightarrow \begin{bmatrix} 3a & 3b \\ 0 & 2c \end{bmatrix} = \begin{bmatrix} 3 + a & b \\ 0 & 2 + c \end{bmatrix}

∴ 3a = 3 + a

⇒ 3a - a = 3

⇒ 2a = 3

⇒ a = 32\dfrac{3}{2}

∴ 3b = b

⇒ 3b - b = 0

⇒ 2b = 0

⇒ b = 0

∴ 2c = 2 + c

⇒ 2c - c = 2

⇒ c = 2.

Hence, values of a = 32\dfrac{3}{2}, b = 0 c = 2.

Question 36

Find x and y if [2xxy3y]\begin{bmatrix} 2x & x \\ y & 3y \end{bmatrix} [32]\begin{bmatrix} 3 \\ 2 \end{bmatrix} = [169]\begin{bmatrix} 16 \\ 9 \end{bmatrix}.

Answer

Solving,

⇒[2xxy3y][32]=[169]⇒[(2x)(3)+(x)(2)(y)(3)+(3y)(2)]=[169]⇒[6x+2x3y+6y]=[169]⇒[8x9y]=[169].\Rightarrow \begin{bmatrix} 2x & x \\ y & 3y \end{bmatrix} \begin{bmatrix} 3 \\ 2 \end{bmatrix} = \begin{bmatrix} 16 \\ 9 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} (2x)(3) + (x)(2) \\ (y)(3) + (3y)(2) \end{bmatrix} = \begin{bmatrix} 16 \\ 9 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 6x + 2x \\ 3y + 6y \end{bmatrix} = \begin{bmatrix} 16 \\ 9 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 8x \\ 9y \end{bmatrix} = \begin{bmatrix} 16 \\ 9 \end{bmatrix}.

∴ 8x = 16

⇒ x = 168\dfrac{16}{8}

⇒ x = 2.

∴ 9y = 9

⇒ y = 99\dfrac{9}{9}

⇒ y = 1.

Hence, x = 2 and y = 1.

Question 37

Given A = [20−17]\begin{bmatrix} 2 & 0 \\ -1 & 7 \end{bmatrix} and I = [1001]\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} and A2 = 9A + mI. Find m.

Answer

Given,

A = [20−17]\begin{bmatrix} 2 & 0 \\ -1 & 7 \end{bmatrix} and I = [1001]\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}

⇒A2=[20−17]×[20−17]=[(2)(2)+(0)(−1)(2)(0)+(0)(7)(−1)(2)+(7)(−1)(−1)(0)+(7)(7)]=[4+00+0−2−70+49]=[40−949].\Rightarrow A^2 = \begin{bmatrix} 2 & 0 \\ -1 & 7 \end{bmatrix} \times \begin{bmatrix} 2 & 0 \\ -1 & 7 \end{bmatrix} \\[1em] = \begin{bmatrix} (2)(2) + (0)(-1) & (2)(0) + (0)(7) \\ (-1)(2) + (7)(-1) & (-1)(0) + (7)(7) \end{bmatrix} \\[1em] = \begin{bmatrix} 4 + 0 & 0 + 0 \\ -2 - 7 & 0 + 49 \end{bmatrix} \\[1em] = \begin{bmatrix} 4 & 0 \\ -9 & 49 \end{bmatrix}.

Solving for A2 = 9A + mI:

⇒[40−949]=9[20−17]+m[1001]⇒[40−949]=[180−963]+m[1001]⇒[40−949]−[180−963]=[m00m]⇒[4−180−0−9−(−9)49−63]=[m00m]⇒[−1400−14]=[m00m].\Rightarrow \begin{bmatrix} 4 & 0 \\ -9 & 49 \end{bmatrix} = 9\begin{bmatrix} 2 & 0 \\ -1 & 7 \end{bmatrix} + m \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 4 & 0 \\ -9 & 49 \end{bmatrix} = \begin{bmatrix} 18 & 0 \\ -9 & 63 \end{bmatrix} + m \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 4 & 0 \\ -9 & 49 \end{bmatrix} - \begin{bmatrix} 18 & 0 \\ -9 & 63 \end{bmatrix} = \begin{bmatrix} m & 0 \\ 0 & m \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 4 - 18 & 0 - 0 \\ -9 - (-9) & 49 - 63 \end{bmatrix} = \begin{bmatrix} m & 0 \\ 0 & m \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} -14 & 0 \\ 0 & -14 \end{bmatrix} = \begin{bmatrix} m & 0 \\ 0 & m \end{bmatrix}.

∴ m = -14.

Hence, m = -14.

Question 38

If A = [3x01]\begin{bmatrix} 3 & x \\ 0 & 1 \end{bmatrix} and B = [9−160−y]\begin{bmatrix} 9 & -16 \\ 0 & -y \end{bmatrix}, find x and y when A2 = B.

Answer

Given,

A = [3x01]\begin{bmatrix} 3 & x \\ 0 & 1 \end{bmatrix} and B = [9−160−y]\begin{bmatrix} 9 & -16 \\ 0 & -y \end{bmatrix}

Solving A2:

⇒A2=[3x01]×[3x01]=[(3)(3)+(x)(0)(3)(x)+(x)(1)(0)(3)+(1)(0)(0)(x)+(1)(1)]=[9+03x+x0+00+1]=[94x01].\Rightarrow A^2 = \begin{bmatrix} 3 & x \\ 0 & 1 \end{bmatrix} \times \begin{bmatrix} 3 & x \\ 0 & 1 \end{bmatrix} \\[1em] = \begin{bmatrix} (3)(3) + (x)(0) & (3)(x) + (x)(1) \\ (0)(3) + (1)(0) & (0)(x) + (1)(1) \end{bmatrix} \\[1em] = \begin{bmatrix} 9 + 0 & 3x + x \\ 0 + 0 & 0 + 1 \end{bmatrix} \\[1em] = \begin{bmatrix} 9 & 4x \\ 0 & 1 \end{bmatrix}.

Solving for x and y:

A2 = B

⇒[94x01]=[9−160−y]\Rightarrow \begin{bmatrix} 9 & 4x \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 9 & -16 \\ 0 & -y \end{bmatrix}

∴ 4x = -16

⇒ x = −164\dfrac{-16}{4}

⇒ x = -4.

∴ -y = 1

⇒ y = -1.

Hence, x = -4 and y = -1.

Question 39

Find x and y if [−2031][−12x]+3[−21]=2[y3]\begin{bmatrix} -2 & 0 \\ 3 & 1 \end{bmatrix} \begin{bmatrix} -1 \\ 2x \end{bmatrix} + 3 \begin{bmatrix} -2 \\ 1 \end{bmatrix} = 2 \begin{bmatrix} y \\ 3 \end{bmatrix}.

Answer

⇒[−2031][−12x]+3[−21]=2[y3]⇒[(−2)(−1)+(0)(2x)(3)(−1)+(1)(2x)]+[−63]=[2y6]⇒[2+0−3+2x]+[−63]=[2y6]⇒[22x−3]+[−63]=[2y6]⇒[2+(−6)2x−3+3]=[2y6]⇒[−42x]=[2y6].\Rightarrow \begin{bmatrix} -2 & 0 \\ 3 & 1 \end{bmatrix} \begin{bmatrix} -1 \\ 2x \end{bmatrix} + 3 \begin{bmatrix} -2 \\ 1 \end{bmatrix} = 2 \begin{bmatrix} y \\ 3 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} (-2)(-1) + (0)(2x) \\ (3)(-1) + (1)(2x) \end{bmatrix} + \begin{bmatrix} -6 \\ 3 \end{bmatrix} = \begin{bmatrix} 2y \\ 6 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 2 + 0 \\ -3 + 2x \end{bmatrix} + \begin{bmatrix} -6 \\ 3 \end{bmatrix} = \begin{bmatrix} 2y \\ 6 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 2 \\ 2x - 3 \end{bmatrix} + \begin{bmatrix} -6 \\ 3 \end{bmatrix} = \begin{bmatrix} 2y \\ 6 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 2 + (-6) \\ 2x - 3 + 3 \end{bmatrix} = \begin{bmatrix} 2y \\ 6 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} -4 \\ 2x \end{bmatrix} = \begin{bmatrix} 2y \\ 6 \end{bmatrix}.

∴ 2y = -4

⇒ y = −42\dfrac{-4}{2}

⇒ y = -2.

∴ 2x = 6

x = 62\dfrac{6}{2}

⇒ x = 3.

Hence, x = 3 and y = -2.

Question 40

If A = [3051]\begin{bmatrix} 3 & 0 \\ 5 & 1 \end{bmatrix} and B = [−4210]\begin{bmatrix} -4 & 2 \\ 1 & 0 \end{bmatrix}, find A2 – 2AB + B2.

Answer

Given,

A = [3051]\begin{bmatrix} 3 & 0 \\ 5 & 1 \end{bmatrix} and B = [−4210]\begin{bmatrix} -4 & 2 \\ 1 & 0 \end{bmatrix}

Solving for A2:

⇒A2=[3051]×[3051]=[(3)(3)+(0)(5)(3)(0)+(0)(1)(5)(3)+(1)(5)(5)(0)+(1)(1)]=[9+00+015+50+1]=[90201].\Rightarrow A^2 = \begin{bmatrix} 3 & 0 \\ 5 & 1 \end{bmatrix} \times \begin{bmatrix} 3 & 0 \\ 5 & 1 \end{bmatrix} \\[1em] = \begin{bmatrix} (3)(3) + (0)(5) & (3)(0) + (0)(1) \\ (5)(3) + (1)(5) & (5)(0) + (1)(1) \end{bmatrix} \\[1em] = \begin{bmatrix} 9 + 0 & 0 + 0 \\ 15 + 5 & 0 + 1 \end{bmatrix} \\[1em] = \begin{bmatrix} 9 & 0 \\ 20 & 1 \end{bmatrix}.

Solving for B2:

⇒B2=[−4210]×[−4210]=[(−4)(−4)+(2)(1)(−4)(2)+(2)(0)(1)(−4)+(0)(1)(1)(2)+(0)(0)]=[16+2−8+0−4+02+0]=[18−8−42].\Rightarrow B^2 = \begin{bmatrix} -4 & 2 \\ 1 & 0 \end{bmatrix} \times \begin{bmatrix} -4 & 2 \\ 1 & 0 \end{bmatrix} \\[1em] = \begin{bmatrix} (-4)(-4) + (2)(1) & (-4)(2) + (2)(0) \\ (1)(-4) + (0)(1) & (1)(2) + (0)(0) \end{bmatrix} \\[1em] = \begin{bmatrix} 16 + 2 & -8 + 0 \\ -4 + 0 & 2 + 0 \end{bmatrix} \\[1em] = \begin{bmatrix} 18 & -8 \\ -4 & 2 \end{bmatrix}.

Solving for 2AB:

⇒2AB=2([3051]×[−4210])=2[(3)(−4)+(0)(1)(3)(2)+(0)(0)(5)(−4)+(1)(1)(5)(2)+(1)(0)]=2[−12+06+0−20+110+0]=2[−126−1910]=[−2412−3820].\Rightarrow 2AB = 2 \Big(\begin{bmatrix} 3 & 0 \\ 5 & 1 \end{bmatrix} \times \begin{bmatrix} -4 & 2 \\ 1 & 0 \end{bmatrix}\Big) \\[1em] = 2\begin{bmatrix} (3)(-4) + (0)(1) & (3)(2) + (0)(0) \\ (5)(-4) + (1)(1) & (5)(2) + (1)(0) \end{bmatrix} \\[1em] = 2\begin{bmatrix} -12 + 0 & 6 + 0 \\ -20 + 1 & 10 + 0 \end{bmatrix} \\[1em] = 2\begin{bmatrix} -12 & 6 \\ -19 & 10 \end{bmatrix} \\[1em] = \begin{bmatrix} -24 & 12 \\ -38 & 20 \end{bmatrix}.

A2 – 2AB + B2

⇒[90201]−[−2412−3820]+[18−8−42]⇒[9−(−24)0−1220−(−38)1−20]+[18−8−42]⇒[33−1258−19]+[18−8−42]⇒[33+18−12−858−4−19+2]⇒[51−2054−17].\Rightarrow \begin{bmatrix} 9 & 0 \\ 20 & 1 \end{bmatrix} - \begin{bmatrix} -24 & 12 \\ -38 & 20 \end{bmatrix} + \begin{bmatrix} 18 & -8 \\ -4 & 2 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 9 - (-24) & 0 - 12 \\ 20 - (-38) & 1 - 20 \end{bmatrix} + \begin{bmatrix} 18 & -8 \\ -4 & 2 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 33 & -12 \\ 58 & -19 \end{bmatrix} + \begin{bmatrix} 18 & -8 \\ -4 & 2 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 33 + 18 & -12 - 8 \\ 58 - 4 & -19 + 2 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 51 & -20 \\ 54 & -17 \end{bmatrix}.

Hence, A2 – 2AB + B2 = [51−2054−17]\begin{bmatrix} 51 & -20 \\ 54 & -17 \end{bmatrix}

Multiple Choice Questions

Question 1

The order of the matrix having x rows and y columns is denoted by:

  1. x + y

  2. x - y

  3. x × y

  4. none of these

Answer

The order of a matrix is written as "rows by columns."

If a matrix has x rows and y columns, its order is written as x × y.

Hence, option 3 is the correct option.

Question 2

The total number of elements in an m × n matrix is:

  1. m + n

  2. mn

  3. mn

  4. 2m + n

Answer

The order of an m × n matrix indicates that it has m rows and n columns. To find the total number of elements, you multiply the number of rows by the number of columns = mn.

Hence, option 3 is the correct option.

Question 3

The matrix, A = [34−820−217−9]\begin{bmatrix} 3 & 4 & -8 \\ 2 & 0 & -2 \\ 1 & 7 & -9 \end{bmatrix}, then the value of a23 is:

  1. 2

  2. -8

  3. -9

  4. -2

Answer

i = Row

j = Column

In a23, i = 2 (Second row) j = 3 (Third column)

In given matrix:

A = [34−820−217−9]\begin{bmatrix} 3 & 4 & -8 \\ 2 & 0 & -2 \\ 1 & 7 & -9 \end{bmatrix}

The element in the second row and third column is −2.

Hence, option 4 is the correct option.

Question 4

The matrix, A = [40003000−7]\begin{bmatrix} 4 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & -7 \end{bmatrix} is a:

  1. Rectangular matrix

  2. Column matrix

  3. Diagonal matrix

  4. Identity matrix

Answer

Given,

A = [40003000−7]\begin{bmatrix} 4 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & -7 \end{bmatrix}

A diagonal matrix is a square matrix in which all the elements outside the main diagonal are zero.

Thus, A is a diagonal matrix.

Hence, option 3 is the correct option.

Question 5

A matrix of order of 3 × 1 whose elements aij are given by aij = (2i + j), is :

  1. [357]\begin{bmatrix} 3 & 5 & 7 \end{bmatrix}

  2. [357]\begin{bmatrix} 3 \\ 5 \\ 7 \end{bmatrix}

  3. [246]\begin{bmatrix} 2 & 4 & 6 \end{bmatrix}

  4. [246]\begin{bmatrix} 2 \\ 4 \\ 6 \end{bmatrix}

Answer

aij = (2i + j)

⇒ a11 = 2(1) + 1 = 3

⇒ a21 = (2)2 + 1 = 5

⇒ a31 = (2)3 + 1 = 7

Matrix A = [357]\begin{bmatrix} 3 \\ 5 \\ 7 \end{bmatrix}

Hence, option 2 is the correct option.

Question 6

A matrix of order 3 × 2 whose elements aij are given by aij = (i + j), is:

  1. [234345]\begin{bmatrix} 2 & 3 & 4 \\ 3 & 4 & 5 \end{bmatrix}

  2. [234456]\begin{bmatrix} 2 & 3 & 4 \\ 4 & 5 & 6 \end{bmatrix}

  3. [233445]\begin{bmatrix} 2 & 3 \\ 3 & 4 \\ 4 & 5 \end{bmatrix}

  4. [233456]\begin{bmatrix} 2 & 3 \\ 3 & 4 \\ 5 & 6 \end{bmatrix}

Answer

aij = i + j.

∴ a11 = 1 + 1 = 2, a12 = 1 + 2 = 3.

a21 = 2 + 1 = 3, a22 = 2 + 2 = 4.

a31 = 3 + 1 = 4, a32 = 3 + 2 = 5.

Matrix A = [233445]\begin{bmatrix} 2 & 3 \\ 3 & 4 \\ 4 & 5 \end{bmatrix}

Hence, option 3 is the correct option.

Question 7

If a matrix has 12 elements, then the total number of possible orders it can have is:

  1. 4

  2. 6

  3. 8

  4. cannot be determined

Answer

If a matrix has 12 elements the possible orders are, 1 x 12, 12 x 1, 2 x 6, 6 x 2, 3 x 4, 4 x 3.

There are 6 possible orders.

Hence, option 2 is the correct option.

Question 8

A 2 × 2 matrix whose elements are given by aij = (i+2j)22\dfrac{(i + 2j)^2}{2} is:

  1. [3252  818]\begin{bmatrix} \dfrac{3}{2} & \dfrac{5}{2} \\ \space & \space \\ 8 & 18 \end{bmatrix}

  2. [5272  818]\begin{bmatrix} \dfrac{5}{2} & \dfrac{7}{2} \\ \space & \space \\ 8 & 18 \end{bmatrix}

  3. [92152  818]\begin{bmatrix} \dfrac{9}{2} & \dfrac{15}{2} \\ \space & \space \\ 8 & 18 \end{bmatrix}

  4. [92252  818]\begin{bmatrix} \dfrac{9}{2} & \dfrac{25}{2} \\ \space & \space \\ 8 & 18 \end{bmatrix}

Answer

Given,

aij = (i+2j)22\dfrac{(i + 2j)^2}{2}

a11=[1+2(1)]22=322=92,a12=[1+2(2)]22=522=252a21=[2+2]22=422=162=8,a22=[2+2(2)]22=622=362=18.a_{11} = \dfrac{[1 + 2(1)]^2}{2} = \dfrac{3^2}{2} = \dfrac{9}{2}, a_{12} = \dfrac{[1 + 2(2)]^2}{2} = \dfrac{5^2}{2} = \dfrac{25}{2} \\[1em] a_{21} = \dfrac{[2 + 2]^2}{2} = \dfrac{4^2}{2} = \dfrac{16}{2} = 8, a_{22} = \dfrac{[2 + 2(2)]^2}{2} = \dfrac{6^2}{2} = \dfrac{36}{2} = 18.

Matrix A = [92252818]\begin{bmatrix} \dfrac{9}{2} & \dfrac{25}{2} \\ 8 & 18 \end{bmatrix}

Hence, option 4 is the correct option.

Question 9

If [53x7]=[yz17]\begin{bmatrix} 5 & 3 \\ x & 7 \end{bmatrix} = \begin{bmatrix} y & z \\ 1 & 7 \end{bmatrix}, then the value of (x + y + z) is:

  1. 9

  2. 8

  3. 11

  4. 10

Answer

Given,

[53x7]=[yz17]\begin{bmatrix} 5 & 3 \\ x & 7 \end{bmatrix} = \begin{bmatrix} y & z \\ 1 & 7 \end{bmatrix}

Solving for x, y and z:

∴ x = 1

∴ y = 5

∴ z = 3

⇒ x + y + z = 9

Hence, option 1 is the correct option.

Question 10

If [x−y2x+z2x−y3z+w]=[−15013]\begin{bmatrix} x - y & 2x + z \\ 2x - y & 3z + w \end{bmatrix} = \begin{bmatrix} -1 & 5 \\ 0 & 13 \end{bmatrix}, then the value of (x + y + z + w) is:

  1. 8

  2. 9

  3. 10

  4. 12

Answer

Given,

⇒[x−y2x+z2x−y3z+w]=[−15013]\Rightarrow \begin{bmatrix} x - y & 2x + z \\ 2x - y & 3z + w \end{bmatrix} = \begin{bmatrix} -1 & 5 \\ 0 & 13 \end{bmatrix}

Solving for x and y:

∴ 2x - y = 0

⇒ y = 2x ...(1)

∴ x - y = -1 ...(2)

Substituting value of y from equation (1) in x - y = -1, we get:

⇒ x - 2x = -1

⇒ -x = -1

⇒ x = 1.

Substituting value of x in equation(1), we get:

⇒ y = 2(1)

⇒ y = 2.

Solving for w and z:

∴ 2x + z = 5

⇒ 2(1) + z = 5

⇒ 2 + z = 5

⇒ z = 5 - 2

⇒ z = 3.

∴ 3z + w = 13

⇒ 3(3) + w = 13

⇒ 9 + w = 13

⇒ w = 13 - 9

⇒ w = 4.

∴ x + y + z + w = 1 + 2 + 3 + 4 = 10.

Hence, option 3 is the correct option.

Question 11

Which of the following statements is true for the transpose of a matrix?

  1. The number of rows is same as that of the given matrix.

  2. The number of columns is same as that of the given matrix.

  3. The order is same as that of the given matrix.

  4. The number of elements is same as that of the given matrix.

Answer

If A = m × n then AT = n × m

The total number of elements is calculated by multiplying the dimensions: m × n. Since m × n = n × m

Therefore, total number of elements in A is exactly the same as the total number of elements in AT

Hence, option 4 is the correct option.

Question 12

If [x+yx−y]=[84]\begin{bmatrix} x + y \\ x - y \end{bmatrix} = \begin{bmatrix} 8 \\ 4 \end{bmatrix}, then the value of xy is:

  1. 4

  2. 8

  3. 10

  4. 12

Answer

Given,

[x+yx−y]=[84]\begin{bmatrix} x + y \\ x - y \end{bmatrix} = \begin{bmatrix} 8 \\ 4 \end{bmatrix}

∴ x + y = 8 .....(1)

∴ x - y = 4 .....(2)

Adding equations (1) and (2), we get :

⇒ x + y + x - y = 8 + 4

⇒ 2x = 12

⇒ x = 122\dfrac{12}{2}

⇒ x = 6.

Substituting value of x in equation (1) :

⇒ x + y = 8

⇒ 6 + y = 8

⇒ y = 8 - 6

⇒ y = 2.

∴ xy = 6 × 2 = 12.

Hence, option 4 is the correct option.

Question 13

If [p−q2q2q+rp+q]=[1495]\begin{bmatrix} p - q & 2q \\ 2q + r & p + q \end{bmatrix} = \begin{bmatrix} 1 & 4 \\ 9 & 5 \end{bmatrix}, then the value of (p + q + r) is:

  1. 8

  2. 10

  3. -5

  4. -10

Answer

[p−q2q2q+rp+q]=[1495]\begin{bmatrix} p - q & 2q \\ 2q + r & p + q \end{bmatrix} = \begin{bmatrix} 1 & 4 \\ 9 & 5 \end{bmatrix}

Solving for p, q and r:

∴ 2q = 4...(1)

⇒ q = 42\dfrac{4}{2}

⇒ q = 2.

∴ p - q = 1

⇒ p - 2 = 1

⇒ p = 1 + 2

⇒ p = 3.

∴ 2q + r = 9

⇒ 2(2) + r = 9

⇒ 4 + r = 9

⇒ r = 9 - 4

⇒ r = 5.

∴ p + q + r = 3 + 2 + 5 = 10.

Hence, option 2 is the correct option.

Question 14

If [x+34y−4x+y]=[5439]\begin{bmatrix} x + 3 & 4 \\ y - 4 & x + y \end{bmatrix} = \begin{bmatrix} 5 & 4 \\ 3 & 9 \end{bmatrix}, then the values of x and y respectively are:

  1. 2, 7

  2. 7, 2

  3. 3, 5

  4. 2, -7

Answer

Given,

[x+34y−4x+y]=[5439]\begin{bmatrix} x + 3 & 4 \\ y - 4 & x + y \end{bmatrix} = \begin{bmatrix} 5 & 4 \\ 3 & 9 \end{bmatrix}

Solving for x and y :

∴ x + 3 = 5

⇒ x = 5 - 3

⇒ x = 2.

∴ y - 4 = 3

⇒ y = 3 + 4

⇒ y = 7.

Hence, option 1 is the correct option.

Question 15

If [m−2n53n]\begin{bmatrix} m - 2n & 5 \\ 3 & n \end{bmatrix} = [653−2]\begin{bmatrix} 6 & 5 \\ 3 & -2 \end{bmatrix}, then the value of mn is:

  1. 2

  2. 4

  3. -4

  4. -2

Answer

Given,

[m−2n53n]\begin{bmatrix} m - 2n & 5 \\ 3 & n \end{bmatrix} = [653−2]\begin{bmatrix} 6 & 5 \\ 3 & -2 \end{bmatrix}

∴ n = -2

∴ m - 2n = 6

⇒ m - 2(-2) = 6

⇒ m + 4 = 6

⇒ m = 6 - 4

⇒ m = 2.

∴ mn = 2(-2) = -4.

Hence, option 3 is the correct option.

Question 16

If A = [−230−5]\begin{bmatrix} -2 & 3 \\ 0 & -5 \end{bmatrix} and B = [−7−234]\begin{bmatrix} -7 & -2 \\ 3 & 4 \end{bmatrix}, then (A + B) is:

  1. [−913−1]\begin{bmatrix} -9 & 1 \\ 3 & -1 \end{bmatrix}

  2. [−9−1 3−1]\begin{bmatrix} -9 & -1 \ 3 & -1 \end{bmatrix}

  3. [9131]\begin{bmatrix} 9 && 1 \\ 3 && 1 \end{bmatrix}

  4. [−913−1]\begin{bmatrix} -9 & 1 \\ 3 & -1 \end{bmatrix}

Answer

Given,

A = [−230−5]\begin{bmatrix} -2 & 3 \\ 0 & -5 \end{bmatrix} and B = [−7−234]\begin{bmatrix} -7 & -2 \\ 3 & 4 \end{bmatrix}

Solving for A + B:

⇒[−2+(−7)3+(−2)0+3−5+4]⇒[−2−73−23−1]⇒[−913−1].\Rightarrow \begin{bmatrix} -2 + (-7) & 3 + (-2) \\ 0 + 3 & -5 + 4 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} - 2 - 7 & 3 - 2 \\ 3 & - 1 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} -9 & 1 \\ 3 & -1 \end{bmatrix}.

Hence, option 4 is the correct option.

Question 17

If A = [−543−8]\begin{bmatrix} -5 & 4 \\ 3 & -8 \end{bmatrix} and B = [2−3−14]\begin{bmatrix} 2 & -3 \\ -1 & 4 \end{bmatrix}, then (A – B) is:

  1. [77−412]\begin{bmatrix} 7 & 7 \\ -4 & 12 \end{bmatrix}

  2. [−774−12]\begin{bmatrix} -7 & 7 \\ 4 & -12 \end{bmatrix}

  3. [−7−7412]\begin{bmatrix} -7 & -7 \\ 4 & 12 \end{bmatrix}

  4. [312−4]\begin{bmatrix} 3 & 1 \\ 2 & -4 \end{bmatrix}

Answer

Given,

A = [−543−8]\begin{bmatrix} -5 & 4 \\ 3 & -8 \end{bmatrix} and B = [2−3−14]\begin{bmatrix} 2 & -3 \\ -1 & 4 \end{bmatrix}

Solving for A - B:

⇒[−5−24−(−3)3−(−1)−8−4]⇒[−774−12].\Rightarrow \begin{bmatrix} -5 - 2 & 4 - (-3) \\ 3 - (-1) & -8 - 4 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} -7 & 7 \\ 4 & -12 \end{bmatrix}.

Hence, option 1 is the correct option.

Question 18

If A = [42−3−9]\begin{bmatrix} 4 & 2 \\ -3 & -9 \end{bmatrix}, then the value of (-3)A is:

  1. [−8−4618]\begin{bmatrix} -8 & -4 \\ 6 & 18 \end{bmatrix}

  2. [−12−6918]\begin{bmatrix} -12 & -6 \\ 9 & 18 \end{bmatrix}

  3. [−12−6927]\begin{bmatrix} -12 & -6 \\ 9 & 27 \end{bmatrix}

  4. [−1269−27]\begin{bmatrix} -12 & 6 \\ 9 & -27 \end{bmatrix}

Answer

Given,

[42−3−9]\begin{bmatrix} 4 & 2 \\ -3 & -9 \end{bmatrix}

Solving for -3A:

⇒−3[42−3−9]⇒[−12−6927]\Rightarrow -3\begin{bmatrix} 4 & 2 \\ -3 & -9 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} -12 & -6 \\ 9 & 27 \end{bmatrix}

Hence, option 3 is the correct option.

Question 19

[24142−1−235131236]\begin{bmatrix} 2 & 4 & 1 & 4 & 2 \\ -1 & -2 & 3 & 5 & 1 \\ 3 & 1 & 2 & 3 & 6 \end{bmatrix} is a matrix of the order:

  1. 3 × 4

  2. 5 × 3

  3. 3 × 5

  4. none of these

Answer

Given matrix has :

3rows and 5 columns

Therefore, the order of the matrix is 3 × 5.

Hence, option 3 is the correct option.

Question 20

If A = [3241]\begin{bmatrix} 3 & 2 \\ 4 & 1 \end{bmatrix} and B = [5032]\begin{bmatrix} 5 & 0 \\ 3 & 2 \end{bmatrix}, then (3A + 2B) is equal to:

  1. [196187]\begin{bmatrix} 19 & 6 \\ 18 & 7 \end{bmatrix}

  2. [8273]\begin{bmatrix} 8 & 2 \\ 7 & 3 \end{bmatrix}

  3. [19161817]\begin{bmatrix} 19 & 16 \\ 18 & 17 \end{bmatrix}

  4. [916817]\begin{bmatrix} 9 & 16 \\ 8 & 17 \end{bmatrix}

Answer

Given,

A = [3241]\begin{bmatrix} 3 & 2 \\ 4 & 1 \end{bmatrix} and B = [5032]\begin{bmatrix} 5 & 0 \\ 3 & 2 \end{bmatrix}

Solving for 3A + 2B:

⇒3[3241]+2[5032]⇒[96123]+[10064]⇒3[3241]+2[5032]⇒[9+106+012+63+4]⇒[196187]\Rightarrow 3\begin{bmatrix} 3 & 2 \\ 4 & 1 \end{bmatrix} + 2\begin{bmatrix} 5 & 0 \\ 3 & 2 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 9 & 6 \\ 12 & 3 \end{bmatrix} + \begin{bmatrix} 10 & 0 \\ 6 & 4 \end{bmatrix} \\[1em] \Rightarrow 3\begin{bmatrix} 3 & 2 \\ 4 & 1 \end{bmatrix} + 2\begin{bmatrix} 5 & 0 \\ 3 & 2 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 9 + 10 & 6 + 0 \\ 12 + 6 & 3 + 4 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 19 & 6 \\ 18 & 7 \end{bmatrix}

Hence, option 1 is the correct option.

Question 21

Find the matrix A, if A + [46−37]\begin{bmatrix} 4 & 6 \\ -3 & 7 \end{bmatrix} = [3−65−8]\begin{bmatrix} 3 & -6 \\ 5 & -8 \end{bmatrix}

  1. [−1128−15]\begin{bmatrix} -1 & 12 \\ 8 & -15 \end{bmatrix}

  2. [−1−128−15]\begin{bmatrix} -1 & -12 \\ 8 & -15 \end{bmatrix}

  3. [−1−12−8−15]\begin{bmatrix} -1 & -12 \\ -8 & -15 \end{bmatrix}

  4. [112815]\begin{bmatrix} 1 & 12 \\ 8 & 15 \end{bmatrix}

Answer

Given,

A + [46−37]\begin{bmatrix} 4 & 6 \\ -3 & 7 \end{bmatrix} = [3−65−8]\begin{bmatrix} 3 & -6 \\ 5 & -8 \end{bmatrix}

Solving for A:

⇒A=[3−65−8]−[46−37]=[3−4−6−65−(−3)−8−7]=[−1−128−15]\Rightarrow A = \begin{bmatrix} 3 & -6 \\ 5 & -8 \end{bmatrix} - \begin{bmatrix} 4 & 6 \\ -3 & 7 \end{bmatrix} \\[1em] = \begin{bmatrix} 3 - 4 & -6 - 6 \\ 5 - (-3) & -8 - 7 \end{bmatrix} \\[1em] = \begin{bmatrix} -1 & -12 \\ 8 & -15 \end{bmatrix}

Hence, option 2 is the correct option.

Question 22

cos⁡θ⋅[cos⁡θsin⁡θ−sin⁡θcos⁡θ]+sin⁡θ⋅[sin⁡θ−cos⁡θcos⁡θsin⁡θ]\cosθ \cdot \begin{bmatrix} \cosθ & \sinθ \\ -\sinθ & \cosθ \end{bmatrix} + \sinθ \cdot \begin{bmatrix} \sinθ & -\cosθ \\ \cosθ & \sinθ \end{bmatrix} is equal to:

  1. [1111]\begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}

  2. [0110]\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}

  3. [1001]\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}

  4. none of these

Answer

Given,

cos⁡θ⋅[cos⁡θsin⁡θ−sin⁡θcos⁡θ]+sin⁡θ⋅[sin⁡θ−cos⁡θcos⁡θsin⁡θ]\cosθ \cdot \begin{bmatrix} \cosθ & \sinθ \\ -\sinθ & \cosθ \end{bmatrix} + \sinθ \cdot \begin{bmatrix} \sinθ & -\cosθ \\ \cosθ & \sinθ \end{bmatrix}

Solving:

⇒[cos⁡2θcos⁡θsin⁡θ−sin⁡θcos⁡θcos⁡2θ]+⋅[sin⁡2θ−sin⁡θcos⁡θsin⁡θcos⁡θsin⁡2θ]⇒[cos⁡2θ+sin⁡2θcos⁡θsin⁡θ−sin⁡θcos⁡θ −sin⁡θcos⁡θ+sin⁡θcos⁡θcos⁡2θ+sin⁡2θ]⇒[1001],\Rightarrow \begin{bmatrix} \cos^2θ & \cosθ\sinθ \\ -\sinθ\cosθ & \cos^2θ \end{bmatrix} + \cdot \begin{bmatrix} \sin^2θ & -\sinθ\cosθ \\ \sinθ\cosθ & \sin^2θ \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} \cos^2θ + \sin^2θ & \cosθ\sinθ - \sinθ\cosθ\ -\sinθ\cosθ + \sinθ\cosθ & \cos^2θ + \sin^2θ \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix},

Hence, option 3 is the correct option.

Question 23

If 2[345x]+[1y01]=[70105]2\begin{bmatrix} 3 & 4 \\ 5 & x \end{bmatrix} + \begin{bmatrix} 1 & y \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 7 & 0 \\ 10 & 5 \end{bmatrix}, then the value of (x – y) is:

  1. 10

  2. -10

  3. 6

  4. -6

Answer

Given,

2[345x]+[1y01]=[70105]2\begin{bmatrix} 3 & 4 \\ 5 & x \end{bmatrix} + \begin{bmatrix} 1 & y \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 7 & 0 \\ 10 & 5 \end{bmatrix}

Solving:

⇒[68102x]+[1y01]=[70105]⇒[6+18+y10+02x+1]=[70105]⇒[78+y102x+1]=[70105].\Rightarrow \begin{bmatrix} 6 & 8 \\ 10 & 2x \end{bmatrix} + \begin{bmatrix} 1 & y \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 7 & 0 \\ 10 & 5 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 6 + 1 & 8 + y \\ 10 + 0 & 2x + 1 \end{bmatrix} = \begin{bmatrix} 7 & 0 \\ 10 & 5 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 7 & 8 + y \\ 10 & 2x + 1 \end{bmatrix} = \begin{bmatrix} 7 & 0 \\ 10 & 5 \end{bmatrix}.

∴ 8 + y = 0

⇒ y = -8

∴ 2x + 1 = 5

⇒ 2x = 5 - 1

⇒ 2x = 4

⇒ x = 42\dfrac{4}{2}

⇒ x = 2

∴ x - y = 2 - (-8) = 10

Hence, option 1 is the correct option.

Question 24

If 2[130x]+[y012]=[5618]2\begin{bmatrix} 1 & 3 \\ 0 & x \end{bmatrix} + \begin{bmatrix} y & 0 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 5 & 6 \\ 1 & 8 \end{bmatrix},then the value of (x + y) is:

  1. 4

  2. -4

  3. 6

  4. 8

Answer

Given,

2[130x]+[y012]=[5618]2\begin{bmatrix} 1 & 3 \\ 0 & x \end{bmatrix} + \begin{bmatrix} y & 0 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 5 & 6 \\ 1 & 8 \end{bmatrix}

Solving:

⇒[2602x]+[y012]=[5618]⇒[2+y612x+2]=[5618].\Rightarrow \begin{bmatrix} 2 & 6 \\ 0 & 2x \end{bmatrix} + \begin{bmatrix} y & 0 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 5 & 6 \\ 1 & 8 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 2 + y & 6 \\ 1 & 2x + 2 \end{bmatrix} = \begin{bmatrix} 5 & 6 \\ 1 & 8 \end{bmatrix}.

∴ 2 + y = 5

⇒ y = 5 - 2

⇒ y = 3

∴ 2x + 2 = 8

⇒ 2x = 8 - 2

⇒ 2x = 6

⇒ x = 62\dfrac{6}{2} = 3.

∴ x + y = 3 + 3 = 6.

Hence, option 3 is the correct option.

Question 25

If [a+b25ab]=[6258]\begin{bmatrix} a + b & 2 \\ 5 & ab \end{bmatrix} = \begin{bmatrix} 6 & 2 \\ 5 & 8 \end{bmatrix}, then the values of a and b are:

  1. a = 4, b = 2

  2. a = 2, b = 4

  3. both (a) and (b)

  4. none of these

Answer

Given,

[a+b25ab]=[6258]\begin{bmatrix} a + b & 2 \\ 5 & ab \end{bmatrix} = \begin{bmatrix} 6 & 2 \\ 5 & 8 \end{bmatrix}

∴ a + b = 6

⇒ b = 6 - a...(1)

∴ ab = 8...(2)

Substituting value of b from equation(1) in ab = 8, we get:

⇒ a (6 - a) = 8

⇒ 6a - a2 = 8

⇒ a2 - 6a + 8 = 0

⇒ a2 - 4a - 2a + 8 = 0

⇒ a(a - 4) -2(a - 4) = 0

⇒ (a - 2)(a - 4) = 0

(a - 2)= 0 or (a - 4) = 0 [Using zero product rule]

⇒ a = 2 or a = 4

If a = 2, then b = 6 − 2 = 4.

If a = 4, then b = 6 − 4 = 2.

Hence, option 3 is the correct option.

Question 26

If A = [0−12103−2−30]\begin{bmatrix} 0 & -1 & 2 \\ 1 & 0 & 3 \\ -2 & -3 & 0 \end{bmatrix}, then (A + A′) is equal to :

  1. 0

  2. 2A

  3. –A′

  4. A′

Answer

Given,

A = [0−12103−2−30]\begin{bmatrix} 0 & -1 & 2 \\ 1 & 0 & 3 \\ -2 & -3 & 0 \end{bmatrix}

A′ = [01−2−10−3230]\begin{bmatrix} 0 & 1 & -2 \\ -1 & 0 & -3 \\ 2 & 3 & 0 \end{bmatrix}

(A + A′)

⇒[0−12103−2−30]+[01−2−10−3230]⇒[0+0−1+12+(−2)1+(−1)0+03+(−3)−2+2−3+30+0]⇒[000000000]\Rightarrow \begin{bmatrix} 0 & -1 & 2 \\ 1 & 0 & 3 \\ -2 & -3 & 0 \end{bmatrix} + \begin{bmatrix} 0 & 1 & -2 \\ -1 & 0 & -3 \\ 2 & 3 & 0 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 0 + 0 & -1 + 1 & 2 + (-2) \\ 1 + (-1) & 0 + 0 & 3 + (-3) \\ -2 + 2 & -3 + 3 & 0 + 0 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}

Hence, option 1 is the correct option.

Question 27

If [xy+2z−3]+[y45]=[4912]\begin{bmatrix} x & y + 2 & z - 3 \end{bmatrix} + \begin{bmatrix} y & 4 & 5 \end{bmatrix} = \begin{bmatrix} 4 & 9 & 12 \end{bmatrix}, then the value of yz is:

  1. 24

  2. 20

  3. 36

  4. 30

Answer

⇒[xy+2z−3]+[y45]=[4912]⇒[x+yy+2+4z−3+5]=[4912]⇒[x+yy+6z+2]=[4912]\Rightarrow \begin{bmatrix} x & y + 2 & z - 3 \end{bmatrix} + \begin{bmatrix} y & 4 & 5 \end{bmatrix} = \begin{bmatrix} 4 & 9 & 12 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} x + y & y + 2 + 4 & z - 3 + 5 \end{bmatrix} = \begin{bmatrix} 4 & 9 & 12 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} x + y & y + 6 & z + 2 \end{bmatrix} = \begin{bmatrix} 4 & 9 & 12 \end{bmatrix} \\[1em]

Solving for y and z:

∴ y + 6 = 9

⇒ y = 9 - 6

⇒ y = 3.

∴ z + 2 = 12

⇒ z = 12 - 2

⇒ z = 10.

∴ yz = 3(10) = 30.

Hence, option 4 is the correct option.

Question 28

If x[21]+y[35]+[−8−11]x\begin{bmatrix} 2 \\ 1 \end{bmatrix} + y\begin{bmatrix} 3 \\ 5 \end{bmatrix} + \begin{bmatrix} -8 \\ -11 \end{bmatrix} = 0, then the value of (2x – 3y) is:

  1. 1

  2. -4

  3. 3

  4. -3

Answer

⇒x[21]+y[35]+[−8−11]=[00]⇒[2xx]+[3y5y]+[−8−11]=[00]⇒[2x+3y−8x+5y−11]=[00].\Rightarrow x\begin{bmatrix} 2 \\ 1 \end{bmatrix} + y\begin{bmatrix} 3 \\ 5 \end{bmatrix} + \begin{bmatrix} -8 \\ -11 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 2x \\ x \end{bmatrix} + \begin{bmatrix} 3y \\ 5y \end{bmatrix} + \begin{bmatrix} -8 \\ -11 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 2x + 3y - 8 \\ x + 5y - 11 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}.

∴ 2x + 3y - 8 = 0

⇒ 2x + 3y = 8 ....(1)

∴ x + 5y - 11 = 0

⇒ x = 11 - 5y ....(2)

Substituting value of x from equation (2) in (1), we get :

⇒ 2(11 - 5y) + 3y = 8

⇒ 22 - 10y + 3y = 8

⇒ -7y = 8 - 22

⇒ -7y = -14

⇒ y = −14−7\dfrac{-14}{-7}

⇒ y = 2.

Substituting value of y in equation (2) we get:

⇒ x = 11 - 5(2)

⇒ x = 11 - 10

⇒ x = 1.

∴ 2x - 3y = 2(1) - 3(2) = 2 - 6 = -4.

Hence, option 2 is the correct option.

Question 29

If [23−41]+2A=4[−1234]\begin{bmatrix} 2 & 3 \\ -4 & 1 \end{bmatrix} + 2A = 4 \begin{bmatrix} -1 & 2 \\ 3 & 4 \end{bmatrix}, then the matrix A is:

  1. [−35815]\begin{bmatrix} -3 & 5 \\ 8 & 15 \end{bmatrix}

  2. [5−3158]\begin{bmatrix} 5 & -3 \\ 15 & 8 \end{bmatrix}

  3. [−3528152]\begin{bmatrix} -3 & \dfrac{5}{2} \\ 8 & \dfrac{15}{2} \end{bmatrix}

  4. [52−31528]\begin{bmatrix} \dfrac{5}{2} & -3 \\ \dfrac{15}{2} & 8 \end{bmatrix}

Answer

Given,

⇒[23−41]+2A=4[−1234]⇒2A=4[−1234]−[23−41]⇒A=12(4[−1234]−[23−41])⇒A=12([−481216]−[23−41])⇒A=12([−4−28−312−(−4)16−1])⇒12[−651615]⇒[−3528152].\Rightarrow \begin{bmatrix} 2 & 3 \\ -4 & 1 \end{bmatrix} + 2A = 4 \begin{bmatrix} -1 & 2 \\ 3 & 4 \end{bmatrix} \\[1em] \Rightarrow 2A = 4 \begin{bmatrix} -1 & 2 \\ 3 & 4 \end{bmatrix} - \begin{bmatrix} 2 & 3 \\ -4 & 1 \end{bmatrix} \\[1em] \Rightarrow A = \dfrac{1}{2} \Big(4\begin{bmatrix} -1 & 2 \\ 3 & 4 \end{bmatrix} - \begin{bmatrix} 2 & 3 \\ -4 & 1 \end{bmatrix} \Big) \\[1em] \Rightarrow A = \dfrac{1}{2} \Big(\begin{bmatrix} -4 & 8 \\ 12 & 16 \end{bmatrix} - \begin{bmatrix} 2 & 3 \\ -4 & 1 \end{bmatrix} \Big) \\[1em] \Rightarrow A = \dfrac{1}{2} \Big(\begin{bmatrix} -4 - 2 & 8 - 3 \\ 12 - (-4) & 16 - 1 \end{bmatrix}\Big) \\[1em] \Rightarrow \dfrac{1}{2}\begin{bmatrix} -6 & 5 \\ 16 & 15 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} -3 & \dfrac{5}{2} \\ 8 & \dfrac{15}{2} \end{bmatrix}.

Hence, option 3 is the correct option.

Question 30

The product AB of two matrices A and B is possible if:

  1. A and B have the same number of rows.

  2. The number of columns of A is equal to the number of rows of B.

  3. The number of rows of A is equal to the number of columns of B.

  4. A and B have the same number of columns.

Answer

For matrix multiplication A × B to be defined, the inner dimensions must match.

If matrix A has the order m × n.And matrix B has the order p × q.The product AB is only possible if n = p.

The number of columns of A is equal to the number of rows of B.

Hence, option 2 is the correct option.

Question 31

If the orders of matrices A and B are m × n and p × q respectively, then which of the following is true for the product BA?

  1. m = p

  2. m = q

  3. n = p

  4. n = q

Answer

Given,

Matrix A: m × n

Matrix B: p × q

The product BA is defined if number of columns in B = number of rows in A, m = q .

Hence, option 2 is the correct option.

Question 32

If P and Q are two different matrices of order 3 × n and n × p, then the order of the matrix PQ is:

  1. 3 × p

  2. p × 3

  3. n × n

  4. 3 × 3

Answer

Given,

Matrix P: 3 × n

Matrix Q: n × p

The product PQ is defined if number of columns in P × number of rows in Q, 3 × p .

Hence, option 1 is the correct option.

Question 33

If [2432]×P=[68]\begin{bmatrix} 2 & 4 \\ 3 & 2 \end{bmatrix} \times P = \begin{bmatrix} 6 \\ 8 \end{bmatrix}, then the order of matrix P is:

  1. 2 × 2

  2. 2 × 1

  3. 1 × 2

  4. cannot be determined

Answer

Let A = [2432]\begin{bmatrix} 2 & 4 \\ 3 & 2 \end{bmatrix}

Let B = [68]\begin{bmatrix} 6 \\ 8 \end{bmatrix}

Let order of P be m × n:

Thus, AP = B

For matrix multiplication:

The number of columns in A must equal the number of rows in P.

⇒ m = 2

The order of the product B = Rows of A × Columns of P.

⇒ 2 × 1 = 2 × n

⇒ n = 1.

Order of matrix P = 2 × 1.

Hence, option 2 is the correct option.

Question 34

If A = [3−6]\begin{bmatrix} 3 & -6 \end{bmatrix} and B = [3−143]\begin{bmatrix} 3 & -1 \\ 4 & 3 \end{bmatrix}, then the order of matrix AB is:

  1. 1 × 2

  2. 2 × 1

  3. 2 × 2

  4. 1 × 3

Answer

A = [3−6]\begin{bmatrix} 3 & -6 \end{bmatrix}

Order of matrix A = 1 × 2

B = [3−143]\begin{bmatrix} 3 & -1 \\ 4 & 3 \end{bmatrix}

Order of matrix B = 2 × 2

The resulting matrix AB will have the number of rows from the first matrix A and the number of columns from the second matrix B.

Order of matrix AB = 1 × 2

Hence, option 1 is the correct option.

Question 35

If [2004][xy]\begin{bmatrix} 2 & 0 \\ 0 & 4 \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix} = [2−8]\begin{bmatrix} 2 \\ -8 \end{bmatrix}, the values of x and y respectively are:

  1. 1, -2

  2. -2, 1

  3. 1, 2

  4. -2, -1

Answer

Given,

[2004][xy]\begin{bmatrix} 2 & 0 \\ 0 & 4 \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix} = [2−8]\begin{bmatrix} 2 \\ -8 \end{bmatrix}

Solving,

⇒[2(x)+0(y)0(x)+4(y)]=[2−8]⇒[2x4y]=[2−8]\Rightarrow \begin{bmatrix} 2(x) + 0(y) \\ 0(x) + 4(y) \end{bmatrix}= \begin{bmatrix} 2 \\ -8 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 2x \\ 4y \end{bmatrix}= \begin{bmatrix} 2 \\ -8 \end{bmatrix}

∴ 2x = 2

⇒ x = 22\dfrac{2}{2}

⇒ x = 1.

∴ 4y = -8

⇒ y = −84\dfrac{-8}{4}

⇒ y = -2.

Hence, option 1 is the correct option.

Question 36

If [x1][10−20]=0\begin{bmatrix} x & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ -2 & 0 \end{bmatrix} = 0, then the value of x is :

  1. 0

  2. 1

  3. 2

  4. -2

Answer

Given,

[x1][10−20]=0\begin{bmatrix} x & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ -2 & 0 \end{bmatrix} = 0

Solving,

⇒[x(1)+(1)(−2)(x)(0)+(1)(0)]=0⇒[x−20]=[00]\Rightarrow \begin{bmatrix} x(1) + (1)(-2) & (x)(0) + (1)(0) \end{bmatrix} = 0 \\[1em] \Rightarrow \begin{bmatrix} x - 2 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 \end{bmatrix}

∴ x - 2 = 0

⇒ x = 2

Hence, option 3 is the correct option.

Question 37

If A = [2−2−22]\begin{bmatrix} 2 & -2 \\ -2 & 2 \end{bmatrix} and A2 = kA, then the value of k is:

  1. 2

  2. 8

  3. -4

  4. 4

Answer

Given,

A = [2−2−22]\begin{bmatrix} 2 & -2 \\ -2 & 2 \end{bmatrix}

Solving:

⇒A2=[2−2−22]×[2−2−22]=[(2)(2)+(−2)(−2)(2)(−2)+(−2)(2)(−2)(2)+(2)(−2)(−2)(−2)+(2)(2)]=[4+4−4−4−4−44+4]=[8−8−88].\Rightarrow A^2 = \begin{bmatrix} 2 & -2 \\ -2 & 2 \end{bmatrix} \times \begin{bmatrix} 2 & -2 \\ -2 & 2 \end{bmatrix} \\[1em] = \begin{bmatrix} (2)(2) + (-2)(-2) & (2)(-2) + (-2)(2) \\ (-2)(2) + (2)(-2) & (-2)(-2) + (2)(2) \end{bmatrix} \\[1em] = \begin{bmatrix} 4 + 4 & -4 - 4 \\ -4 - 4 & 4 + 4 \end{bmatrix} \\[1em] = \begin{bmatrix} 8 & -8 \\ -8 & 8 \end{bmatrix}.

Given,

A2 = kA

⇒[8−8−88]=k[2−2−22]⇒[8−8−88]=[2k−2k−2k2k].\Rightarrow \begin{bmatrix} 8 & -8 \\ -8 & 8 \end{bmatrix} = k\begin{bmatrix} 2 & -2 \\ -2 & 2 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 8 & -8 \\ -8 & 8 \end{bmatrix} = \begin{bmatrix} 2k & -2k \\ -2k & 2k \end{bmatrix}.

∴ 2k = 8

⇒ k = 82\dfrac{8}{2}

⇒ k = 4.

Hence, option 4 is the correct option.

Question 38

If, x[23]+y[−11]=[105]x \begin{bmatrix} 2 \\ 3 \end{bmatrix} + y \begin{bmatrix} -1 \\ 1 \end{bmatrix} = \begin{bmatrix} 10 \\ 5 \end{bmatrix}, then the value of xy is:

  1. 3

  2. -12

  3. -6

  4. 15

Answer

Solving for x and y:

⇒x[23]+y[−11]=[105]⇒[2x3x]+[−yy]=[105]⇒[2x−y3x+y]=[105]\Rightarrow x \begin{bmatrix} 2 \\ 3 \end{bmatrix} + y \begin{bmatrix} -1 \\ 1 \end{bmatrix} = \begin{bmatrix} 10 \\ 5 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 2x \\ 3x \end{bmatrix} + \begin{bmatrix} -y \\ y \end{bmatrix} = \begin{bmatrix} 10 \\ 5 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 2x - y \\ 3x + y \end{bmatrix} = \begin{bmatrix} 10 \\ 5 \end{bmatrix} \\[1em]

∴ 2x - y = 10 ....(1)

∴ 3x + y = 5 .....(2)

Adding equations (1) and (2), we get :

⇒ 2x - y + 3x + y = 10 + 5

⇒ 2x + 3x = 15

⇒ 5x = 15

⇒ x = 155\dfrac{15}{5}

⇒ x = 3.

Substituting value of x in 2x - y = 10, we get :

⇒ 2(3) - y = 10

⇒ 6 - y = 10

⇒ y = 6 - 10

⇒ y = -4.

∴ xy = (3)(-4) = -12.

Hence, option 2 is the correct option.

Question 39

If, A = [4312]\begin{bmatrix} 4 & 3 \\ 1 & 2 \end{bmatrix} and B = [−43]\begin{bmatrix} -4 \\ 3 \end{bmatrix}, then the matrix AB is :

  1. [−72]\begin{bmatrix} -7 \\ 2 \end{bmatrix}

  2. [−72]\begin{bmatrix} -7 & 2 \end{bmatrix}

  3. [−7−2]\begin{bmatrix} -7 \\ -2 \end{bmatrix}

  4. [−7−2]\begin{bmatrix} -7 & -2 \end{bmatrix}

Answer

Given,

A = [4312]\begin{bmatrix} 4 & 3 \\ 1 & 2 \end{bmatrix} and B = [−43]\begin{bmatrix} -4 \\ 3 \end{bmatrix}

Solving for AB:

⇒[4312]×[−43]⇒[(4)(−4)+(3)(3)(1)(−4)+(2)(3)]⇒[−16+9−4+6]⇒[−72].\Rightarrow \begin{bmatrix} 4 & 3 \\ 1 & 2 \end{bmatrix} \times \begin{bmatrix} -4 \\ 3 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} (4)(-4) + (3)(3) \\ (1)(-4) + (2)(3) \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} -16 + 9 \\ -4 + 6 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} -7 \\ 2 \end{bmatrix}.

Hence, option 1 is the correct option.

Question 40

If A is a matrix of order (m × n) such that AB and BA are both defined, then B is a:

  1. m × n matrix

  2. n × m matrix

  3. m × m matrix

  4. n × n matrix

Answer

If A is of order m × n:

For matrix multiplication AB to exist, the number of columns of A must equal the number of rows of B.

Number of rows in B = n

For matrix multiplication BA to exist, the number of columns of B must be equal to the number of rows of A.

Number of columns in B = m.

B is of order = n × m.

Hence, option 2 is the correct option.

Question 41

If A = [5103−4]\begin{bmatrix} 5 & 10 \\ 3 & -4 \end{bmatrix} and I = [1001]\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}, then AI is equal to:

  1. [1001]\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}

  2. [510−34]\begin{bmatrix} 5 & 10 \\ -3 & 4 \end{bmatrix}

  3. [5103−4]\begin{bmatrix} 5 & 10 \\ 3 & -4 \end{bmatrix}

  4. [1515−1−1]\begin{bmatrix} 15 & 15 \\ -1 & -1 \end{bmatrix}

Answer

Given,

A = [5103−4]\begin{bmatrix} 5 & 10 \\ 3 & -4 \end{bmatrix} and I = [1001]\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}

Solving for AI:

⇒[5103−4]×[1001]⇒[5(1)+10(0)5(0)+10(1)3(1)+(−4)(0)3(0)+(−4)(1)]⇒[5+00+103+00+(−4)]⇒[5103−4].\Rightarrow \begin{bmatrix} 5 & 10 \\ 3 & -4 \end{bmatrix} \times \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 5(1) + 10(0) & 5(0) + 10(1) \\ 3(1) + (-4)(0) & 3(0) + (-4)(1) \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 5 + 0 & 0 + 10 \\ 3 + 0 & 0 + (-4) \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 5 & 10 \\ 3 & -4 \end{bmatrix}.

Hence, option 3 is the correct option.

Assertion Reason Questions

Question 1

Assertion (A): The product of a row matrix and a column matrix is possible.

Reason (R): Two matrices of the same order can be multiplied.

  1. A is true, R is false.

  2. A is false, R is true.

  3. Both A and R are true.

  4. Both A and R are false.

Answer

The product of a row matrix and a column matrix is possible.

A row matrix has order 1 × n and a column matrix has order m × 1.

If n = m, so the product is defined and gives a 1 × 1 matrix.

So, assertion (A) is true.

Two matrices of the same order cannot be multiplied.

If the matrix are of the same order and not square, example 2 × 3 and 2 × 3, then (2 × 3) × (2 × 3) is not defined because the number of columns of the first matrix is not equal to the number of rows of the second matrix.

So the statement “Two matrices of the same order can be multiplied” is false.

Reason (R) is false.

A is true, R is false.

Hence, option 1 is the correct option.

Question 2

Assertion (A): A rectangular matrix can be a diagonal matrix.

Reason (R): A matrix in which every non-diagonal element is 0 is called a diagonal matrix.

  1. A is true, R is false.

  2. A is false, R is true.

  3. Both A and R are true.

  4. Both A and R are false.

Answer

In a rectangular matrix number of rows ≠ number of columns.

A diagonal matrix is defined only for square matrices, where all non-diagonal elements are 0 and diagonal elements may be nonzero.

Hence, a rectangular matrix can never be diagonal, because diagonal position exists only when rows = columns.

So, assertion (A) is false.

A matrix in which every non-diagonal element is 0 is called a diagonal matrix.

This is the correct definition of a diagonal matrix.

So, reason (R) is true.

A is false, R is true.

Hence, option 2 is the correct option.

Question 3

Assertion (A): For any two matrices A and B, A + B = B + A.

Reason (R): We can add only two matrices of same order.

  1. A is true, R is false.

  2. A is false, R is true.

  3. Both A and R are true.

  4. Both A and R are false.

Answer

Matrix addition follows the commutative law — but only when A and B are of the same order.

If that is the condition, A + B = B + A always holds true.

So, Assertion (A) is true.

Matrix addition is defined only when both matrices have the same order.

So, Reason (R) is true.

Both A and R are true.

Hence, option 3 is the correct option.

Question 4

Assertion (A): For any two square matrices A and B of same order, AB and BA both exist.

Reason (R): For any two matrices A and B, the product AB exists only when number of rows in A = number of columns in B.

  1. A is true, R is false.

  2. A is false, R is true.

  3. Both A and R are true.

  4. Both A and R are false.

Answer

If A and B are square matrices of the same order (say n × n), then :

Thus, for AB, no. of columns in A = no. of rows in B.

Thus, AB is possible.

For BA, no. of columns in B = no. of rows in A.

Thus, BA is possible.

Assertion (A) is true.

The rule for matrix multiplication is : AB exists if the number of columns of A = number of rows of B.

Reason (R) is false.

A is true, R is false.

Hence, option 1 is the correct option.

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