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Chapter 9

Matrices — Exercise 9(A)

Class - 10 Concise Mathematics Selina

Exercise 9(A)

Question 1(a)

If [x+27y+3a−2]=[4b−343]\begin{bmatrix*}[r] x + 2 & 7 \\ y + 3 & a - 2 \end{bmatrix*} = \begin{bmatrix*}[r] 4 & b - 3 \\ 4 & 3 \end{bmatrix*}, the value of x, y, a and b are :

  1. x = 2, y = 1, a = 5 and b = 10

  2. x = -2, y = 1, a = 5 and b = 10

  3. x = 2, y = -1, a = 5 and b = 10

  4. x = 2, y = 1, a = -5 and b = 10

Answer

Given,

[x+27y+3a−2]=[4b−343]\begin{bmatrix*}[r] x + 2 & 7 \\ y + 3 & a - 2 \end{bmatrix*} = \begin{bmatrix*}[r] 4 & b - 3 \\ 4 & 3 \end{bmatrix*}.

∴ x + 2 = 4

⇒ x = 4 - 2 = 2.

∴ y + 3 = 4

⇒ y = 4 - 3 = 1.

∴ b - 3 = 7

⇒ b = 7 + 3 = 10.

∴ a - 2 = 3

⇒ a = 3 + 2 = 5.

Hence, Option 1 is the correct option.

Question 1(b)

If A = [5−53−3] and B=[−55−33]\begin{bmatrix*}[r] 5 & -5 \\ 3 & -3 \end{bmatrix*} \text{ and } B = \begin{bmatrix*}[r] -5 & 5 \\ -3 & 3 \end{bmatrix*}; the value of matrix (A - B) is :

  1. [0000]\begin{bmatrix*}[r] 0 & 0 \\ 0 & 0 \end{bmatrix*}

  2. [10−106−6]\begin{bmatrix*}[r] 10 & -10 \\ 6 & -6 \end{bmatrix*}

  3. [10−10−66]\begin{bmatrix*}[r] 10 & -10 \\ -6 & 6 \end{bmatrix*}

  4. [−1010−66]\begin{bmatrix*}[r] -10 & 10 \\ -6 & 6 \end{bmatrix*}

Answer

Given,

A = [5−53−3] and B=[−55−33]\begin{bmatrix*}[r] 5 & -5 \\ 3 & -3 \end{bmatrix*} \text{ and } B = \begin{bmatrix*}[r] -5 & 5 \\ -3 & 3 \end{bmatrix*}

A−B=[5−53−3]−[−55−33]=[5−(−5)−5−53−(−3)−3−3]=[10−106−6]A -B = \begin{bmatrix*}[r] 5 & -5 \\ 3 & -3 \end{bmatrix*} - \begin{bmatrix*}[r] -5 & 5 \\ -3 & 3 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 5 - (-5) & -5 - 5 \\ 3 - (-3) & -3 - 3 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 10 & -10 \\ 6 & -6 \end{bmatrix*}

Hence, Option 2 is the correct option.

Question 1(c)

If A = [5540],B=[3214] and C=[−2321]\begin{bmatrix*}[r] 5 & 5 \\ 4 & 0 \end{bmatrix*}, B = \begin{bmatrix*}[r] 3 & 2 \\ 1 & 4 \end{bmatrix*} \text{ and } C = \begin{bmatrix*}[r] -2 & 3 \\ 2 & 1 \end{bmatrix*} then matrix (A + B - C) is :

  1. [104−33]\begin{bmatrix*}[r] 10 & 4 \\ -3 & 3 \end{bmatrix*}

  2. [−1043−3]\begin{bmatrix*}[r] -10 & 4 \\ 3 & -3 \end{bmatrix*}

  3. [10433]\begin{bmatrix*}[r] 10 & 4 \\ 3 & 3 \end{bmatrix*}

  4. [10−433]\begin{bmatrix*}[r] 10 & -4 \\ 3 & 3 \end{bmatrix*}

Answer

Given,

A = [5540],B=[3214] and C=[−2321]\begin{bmatrix*}[r] 5 & 5 \\ 4 & 0 \end{bmatrix*}, B = \begin{bmatrix*}[r] 3 & 2 \\ 1 & 4 \end{bmatrix*} \text{ and } C = \begin{bmatrix*}[r] -2 & 3 \\ 2 & 1 \end{bmatrix*}

(A+B−C)=[5540]+[3214]−[−2321]=[5+3−(−2)5+2−34+1−20+4−1]=[8+27−35−24−1]=[10433].(A + B - C) = \begin{bmatrix*}[r] 5 & 5 \\ 4 & 0 \end{bmatrix*} + \begin{bmatrix*}[r] 3 & 2 \\ 1 & 4 \end{bmatrix*} - \begin{bmatrix*}[r] -2 & 3 \\ 2 & 1 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 5 + 3 - (-2) & 5 + 2 - 3 \\ 4 + 1 - 2 & 0 + 4 - 1 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 8 + 2 & 7 - 3 \\ 5 - 2 & 4 - 1 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 10 & 4 \\ 3 & 3 \end{bmatrix*}.

Hence, Option 3 is the correct option.

Question 1(d)

If A = [75−33] and B=[−2510]\begin{bmatrix*}[r] 7 & 5 \\ -3 & 3 \end{bmatrix*} \text{ and B} = \begin{bmatrix*}[r] -2 & 5 \\ 1 & 0 \end{bmatrix*}, then the matrix P (such that A + P = B) is :

  1. [409−3]\begin{bmatrix*}[r] 4 & 0 \\ 9 & -3 \end{bmatrix*}

  2. [904−2]\begin{bmatrix*}[r] 9 & 0 \\ 4 & -2 \end{bmatrix*}

  3. [−9043]\begin{bmatrix*}[r] -9 & 0 \\ 4 & 3 \end{bmatrix*}

  4. [−904−3]\begin{bmatrix*}[r] -9 & 0 \\ 4 & -3 \end{bmatrix*}

Answer

Given,

⇒ A + P = B

⇒ P = B - A

Substituting values we get :

⇒P=[−2510]−[75−33]⇒P=[−2−75−51−(−3)0−3]⇒P=[−904−3].\Rightarrow P = \begin{bmatrix*}[r] -2 & 5 \\ 1 & 0 \end{bmatrix*} - \begin{bmatrix*}[r] 7 & 5 \\ -3 & 3 \end{bmatrix*} \\[1em] \Rightarrow P = \begin{bmatrix*}[r] -2 - 7 & 5 - 5 \\ 1 - (-3) & 0 - 3 \end{bmatrix*} \\[1em] \Rightarrow P = \begin{bmatrix*}[r] -9 & 0 \\ 4 & -3 \end{bmatrix*}.

Hence, Option 4 is the correct option.

Question 1(e)

The additive inverse of matrix A + B, where

A = [427−2] and B=[−213−4]\begin{bmatrix*}[r] 4 & 2 \\ 7 & -2 \end{bmatrix*} \text{ and } B = \begin{bmatrix*}[r] -2 & 1 \\ 3 & -4 \end{bmatrix*} is :

  1. [−2−3−106]\begin{bmatrix*}[r] -2 & -3 \\ -10 & 6 \end{bmatrix*}

  2. [23−10−6]\begin{bmatrix*}[r] 2 & 3 \\ -10 & -6 \end{bmatrix*}

  3. [−2−3−10−6]\begin{bmatrix*}[r] -2 & -3 \\ -10 & -6 \end{bmatrix*}

  4. [−2310−6]\begin{bmatrix*}[r] -2 & 3 \\ 10 & -6 \end{bmatrix*}

Answer

Additive inverse of a matrix M is given by -M.

So, additive inverse of (A + B) = -(A + B).

Substituting values we get :

⇒−(A+B)=−([427−2]+[−213−4])=−([4+(−2)2+17+3−2+(−4)])=−([2310−6])=[−2−3−106]\Rightarrow -(A + B) = -\Big(\begin{bmatrix*}[r] 4 & 2 \\ 7 & -2 \end{bmatrix*} + \begin{bmatrix*}[r] -2 & 1 \\ 3 & -4 \end{bmatrix*}\Big) \\[1em] = -\Big(\begin{bmatrix*}[r] 4 + (-2) & 2 + 1 \\ 7 + 3 & -2 + (-4) \end{bmatrix*}\Big) \\[1em] = -\Big(\begin{bmatrix*}[r] 2 & 3 \\ 10 & -6 \end{bmatrix*}\Big) \\[1em] = \begin{bmatrix*}[r] -2 & -3 \\ -10 & 6 \end{bmatrix*}

Hence, Option 1 is the correct option.

Question 2

State, whether the following statements are true or false. If false, give a reason.

(i) If A and B are two matrices of orders 3 × 2 and 2 × 3 respectively; then their sum A + B is possible.

(ii) The matrices A2 × 3 and B2 × 3 are conformable for subtraction.

(iii) Transpose of a 2 × 1 matrix is a 2 × 1 matrix.

(iv) Transpose of a square matrix is a square matrix.

(v) A column matrix has many columns and only one row.

Answer

(i) False

Reason — For addition the order of both the matrices must be same.

Here, A and B have different orders.

Hence, the statement is false because the orders of both matrices are different.

(ii) True

Reason — For subtraction the order of both the matrices must be same.

Here, A and B have same orders.

Hence, the statement is true.

(iii) False

Reason — On transposing, the no. of rows and columns gets interchanged.

Hence, transpose of a 2 × 1 matrix will be of order 1 × 2.

Hence, the statement is false.

(iv) True

Reason — The transpose of a square matrix is also a square matrix.

Hence, the statement is true.

(v) False

Reason — A column matrix has only one column and can have many rows.

Hence, the statement is false.

Question 3(i)

Solve for a, b and c; if :

[−4a+532]=[b+423c−1]\begin{bmatrix} -4 & a + 5 \\ 3 & 2 \end{bmatrix} = \begin{bmatrix} b + 4 & 2 \\ 3 & c - 1 \end{bmatrix}

Answer

Given, [−4a+532]=[b+423c−1]\begin{bmatrix} -4 & a + 5 \\ 3 & 2 \end{bmatrix} = \begin{bmatrix} b + 4 & 2 \\ 3 & c - 1 \end{bmatrix}

By definition of equality of matrices we get,

-4 = b + 4 ⇒ b = -4 - 4 = -8,

a + 5 = 2 ⇒ a = 2 - 5 = -3,

2 = c - 1 ⇒ c = 2 + 1 = 3.

Hence, a = -3, b = -8 and c = 3.

Question 3(ii)

Solve for a, b and c; if :

[aa−bb+c0]=[3−120]\begin{bmatrix} a & a - b \\ b + c & 0 \end{bmatrix} = \begin{bmatrix} 3 & -1 \\ 2 & 0 \end{bmatrix}

Answer

Given, [aa−bb+c0]=[3−120]\begin{bmatrix} a & a - b \\ b + c & 0 \end{bmatrix} = \begin{bmatrix} 3 & -1 \\ 2 & 0 \end{bmatrix}

By definition of equality of matrices we get,

a = 3,

a - b = -1
⇒ 3 - b = -1 [∵ a = 3]
⇒ b = 3 + 1 = 4

b + c = 2
⇒ 4 + c = 2
⇒ c = -2.

Hence, a = 3, b = 4 and c = -2.

Question 4

Wherever possible, write each of the following as a single matrix.

(i) [1234]+[−1−21−7]\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} + \begin{bmatrix} -1 & -2 \\ 1 & -7 \end{bmatrix}

(ii) [234567]−[0236−10]\begin{bmatrix} 2 & 3 & 4 \\ 5 & 6 & 7 \end{bmatrix} - \begin{bmatrix} 0 & 2 & 3 \\ 6 & -1 & 0 \end{bmatrix}

(iii) [012467]+[3468]\begin{bmatrix} 0 & 1 & 2 \\ 4 & 6 & 7 \end{bmatrix} + \begin{bmatrix} 3 & 4 \\ 6 & 8 \end{bmatrix}

Answer

(i) Given,

⇒[1234]+[−1−21−7]⇒[1+(−1)2+(−2)3+14+(−7)]⇒[004−3].\phantom{\Rightarrow} \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} + \begin{bmatrix} -1 & -2 \\ 1 & -7 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 1 + (-1) & 2 + (-2) \\ 3 + 1 & 4 + (-7) \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 0 & 0 \\ 4 & -3 \end{bmatrix}.

Hence, resultant matrix = [004−3]\begin{bmatrix} 0 & 0 \\ 4 & -3 \end{bmatrix}.

(ii) Given,

⇒[234567]−[0236−10]⇒[2−03−24−35−66−(−1)7−0]⇒[211−177].\phantom{\Rightarrow} \begin{bmatrix} 2 & 3 & 4 \\ 5 & 6 & 7 \end{bmatrix} - \begin{bmatrix} 0 & 2 & 3 \\ 6 & -1 & 0 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 2 - 0 & 3 - 2 & 4 - 3 \\ 5 - 6 & 6 - (-1) & 7 - 0 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 2 & 1 & 1 \\ -1 & 7 & 7 \end{bmatrix}.

Hence, resultant matrix = [211−177]\begin{bmatrix} 2 & 1 & 1 \\ -1 & 7 & 7 \end{bmatrix}.

(iii) Given,

[012467]+[3468]\begin{bmatrix} 0 & 1 & 2 \\ 4 & 6 & 7 \end{bmatrix} + \begin{bmatrix} 3 & 4 \\ 6 & 8 \end{bmatrix}

The above calculation is not possible because for addition the order of both matrices must be equal.

Question 5(i)

Find, x and y from the following equations :

[52−1y−1]−[1x−12−3]=[47−32]\begin{bmatrix} 5 & 2 \\ -1 & y - 1 \end{bmatrix} - \begin{bmatrix} 1 & x - 1 \\ 2 & -3 \end{bmatrix} = \begin{bmatrix} 4 & 7 \\ -3 & 2 \end{bmatrix}

Answer

Given,

⇒[52−1y−1]−[1x−12−3]=[47−32]⇒[5−12−(x−1)−1−2y−1−(−3)]=[47−32]⇒[43−x−3y+2]=[47−32]\Rightarrow \begin{bmatrix} 5 & 2 \\ -1 & y - 1 \end{bmatrix} - \begin{bmatrix} 1 & x - 1 \\ 2 & -3 \end{bmatrix} = \begin{bmatrix} 4 & 7 \\ -3 & 2 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 5 - 1 & 2 - (x - 1) \\ -1 - 2 & y - 1 - (-3) \end{bmatrix} = \begin{bmatrix} 4 & 7 \\ -3 & 2 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 4 & 3 - x \\ -3 & y + 2 \end{bmatrix} = \begin{bmatrix} 4 & 7 \\ -3 & 2 \end{bmatrix} \\[1em]

By definition of equality of matrices we get,

⇒ 3 - x = 7
⇒ x = 3 - 7 = -4.

⇒ y + 2 = 2 ⇒ y = 0.

Hence, x = -4 and y = 0.

Question 5(ii)

Find, x and y from the following equations :

[−8x]+[y−2]=[−32]\begin{bmatrix} -8 & x \end{bmatrix} + \begin{bmatrix} y & -2 \end{bmatrix} = \begin{bmatrix} -3 & 2 \end{bmatrix}

Answer

Given,

⇒[−8x]+[y−2]=[−32]⇒[−8+yx+(−2)]=[−32]\Rightarrow \begin{bmatrix} -8 & x \end{bmatrix} + \begin{bmatrix} y & -2 \end{bmatrix} = \begin{bmatrix} -3 & 2 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} -8 + y & x + (-2) \end{bmatrix} = \begin{bmatrix} -3 & 2 \end{bmatrix}

By definition of equality of matrices we get,

-8 + y = -3
⇒ y = -3 + 8 = 5.

x - 2 = 2
⇒ x = 2 + 2 = 4.

Hence, x = 4 and y = 5.

Question 6

Given : M = [5−3−24]\begin{bmatrix} 5 & -3 \\ -2 & 4 \end{bmatrix}, find its transpose matrix Mt. If possible, find :

(i) M + Mt

(ii) Mt - M

Answer

M = [5−3−24]\begin{bmatrix} 5 & -3 \\ -2 & 4 \end{bmatrix}

Mt = [5−2−34]\begin{bmatrix} 5 & -2 \\ -3 & 4 \end{bmatrix}

(i)

M+Mt=[5−3−24]+[5−2−34]=[5+5−3+(−2)−2+(−3)4+4]=[10−5−58].M + M^t = \begin{bmatrix} 5 & -3 \\ -2 & 4 \end{bmatrix} + \begin{bmatrix} 5 & -2 \\ -3 & 4 \end{bmatrix} \\[1em] = \begin{bmatrix} 5 + 5 & -3 + (-2) \\ -2 + (-3) & 4 + 4 \end{bmatrix} \\[1em] = \begin{bmatrix} 10 & -5 \\ -5 & 8 \end{bmatrix}.

Hence, M+Mt=[10−5−58].M + M^t = \begin{bmatrix} 10 & -5 \\ -5 & 8 \end{bmatrix}.

(ii)

Mt−M=[5−2−34]−[5−3−24]=[5−5−2−(−3)−3−(−2)4−4]=[01−10].M^t - M = \begin{bmatrix} 5 & -2 \\ -3 & 4 \end{bmatrix} - \begin{bmatrix} 5 & -3 \\ -2 & 4 \end{bmatrix} \\[1em] = \begin{bmatrix} 5 - 5 & -2 - (-3) \\ -3 - (-2) & 4 - 4 \end{bmatrix} \\[1em] = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}.

Hence, Mt - M = [01−10].\begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}.

Question 7

Given A = [2−3],B=[02] and C=[−14];\begin{bmatrix} 2 & -3 \end{bmatrix}, B = \begin{bmatrix} 0 & 2 \end{bmatrix} \text{ and C} = \begin{bmatrix} -1 & 4 \end{bmatrix}; find the matrix X in each of the following :

(i) X + B = C - A

(ii) A - X = B + C

Answer

(i) Given,

⇒ X + B = C - A

⇒ X = C - A - B

Substituting values of A, B and C in above equation we get,

⇒X=[−14]−[2−3]−[02]=[−1−2−04−(−3)−2]=[−35].\Rightarrow X = \begin{bmatrix} -1 & 4 \end{bmatrix} - \begin{bmatrix} 2 & -3 \end{bmatrix} - \begin{bmatrix} 0 & 2 \end{bmatrix} \\[1em] = \begin{bmatrix} -1 - 2 - 0 & 4 - (-3) - 2 \end{bmatrix} \\[1em] = \begin{bmatrix} -3 & 5 \end{bmatrix}.

Hence, X = [−35].\begin{bmatrix} -3 & 5 \end{bmatrix}.

(ii) Given,

⇒ A - X = B + C

⇒ X = A - (B + C)

Substituting values of A, B and C in above equation we get,

⇒X=[2−3]−([02]+[−14])=[2−3]−([0+(−1)2+4])=[2−3]−[−16]=[2−(−1)−3−6]=[3−9].\Rightarrow X = \begin{bmatrix} 2 & -3 \end{bmatrix} - \Big(\begin{bmatrix} 0 & 2 \end{bmatrix} + \begin{bmatrix} -1 & 4 \end{bmatrix}\Big) \\[1em] = \begin{bmatrix} 2 & -3 \end{bmatrix} - \Big(\begin{bmatrix} 0 + (-1) & 2 + 4 \end{bmatrix}\Big) \\[1em] = \begin{bmatrix} 2 & -3 \end{bmatrix} - \begin{bmatrix} -1 & 6 \end{bmatrix} \\[1em] = \begin{bmatrix} 2 - (-1) & -3 - 6 \end{bmatrix} \\[1em] = \begin{bmatrix} 3 & -9 \end{bmatrix}.

Hence, X = [3−9]\begin{bmatrix} 3 & -9 \end{bmatrix}.

Question 8

Given A = [−102−4] and B=[3−3−20]\begin{bmatrix} -1 & 0 \\ 2 & -4 \end{bmatrix} \text{ and } B = \begin{bmatrix} 3 & -3 \\ -2 & 0 \end{bmatrix}; find the matrix X in each of the following :

(i) A + X = B

(ii) A - X = B

(iii) X - B = A

Answer

(i) Given,

⇒ A + X = B

⇒ X = B - A

⇒X=[3−3−20]−[−102−4]=[3−(−1)−3−0−2−20−(−4)]=[4−3−44].\Rightarrow X = \begin{bmatrix} 3 & -3 \\ -2 & 0 \end{bmatrix} - \begin{bmatrix} -1 & 0 \\ 2 & -4 \end{bmatrix} \\[1em] = \begin{bmatrix} 3 - (-1) & -3 - 0 \\ -2 - 2 & 0 - (-4) \end{bmatrix} \\[1em] = \begin{bmatrix} 4 & -3 \\ -4 & 4 \end{bmatrix}.

Hence, X = [4−3−44].\begin{bmatrix} 4 & -3 \\ -4 & 4 \end{bmatrix}.

(ii) Given,

⇒ A - X = B

⇒ X = A - B

⇒X=[−102−4]−[3−3−20]=[−1−30−(−3)2−(−2)−4−0]=[−434−4].\Rightarrow X = \begin{bmatrix} -1 & 0 \\ 2 & -4 \end{bmatrix} - \begin{bmatrix} 3 & -3 \\ -2 & 0 \end{bmatrix} \\[1em] = \begin{bmatrix} -1 - 3 & 0 - (-3) \\ 2 - (-2) & -4 - 0 \end{bmatrix} \\[1em] = \begin{bmatrix} -4 & 3 \\ 4 & -4 \end{bmatrix}.

Hence, X = [−434−4].\begin{bmatrix} -4 & 3 \\ 4 & -4 \end{bmatrix}.

(iii) Given,

⇒ X - B = A

⇒ X = A + B

⇒X=[−102−4]+[3−3−20]=[−1+30+(−3)2+(−2)−4+0]=[2−30−4].\Rightarrow X = \begin{bmatrix} -1 & 0 \\ 2 & -4 \end{bmatrix} + \begin{bmatrix} 3 & -3 \\ -2 & 0 \end{bmatrix} \\[1em] = \begin{bmatrix} -1 + 3 & 0 + (-3) \\ 2 + (-2) & -4 + 0 \end{bmatrix} \\[1em] = \begin{bmatrix} 2 & -3 \\ 0 & -4 \end{bmatrix}.

Hence, X = [2−30−4].\begin{bmatrix} 2 & -3 \\ 0 & -4 \end{bmatrix}.

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