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Chapter 9

Matrices — Exercise 9(B)

Class - 10 Concise Mathematics Selina

Exercise 9(B)

Question 1(a)

If 4[5x]−5[y−2]=[1022]4\begin{bmatrix*}[r] 5 & x \end{bmatrix*} - 5\begin{bmatrix*}[r] y & -2 \end{bmatrix*} = \begin{bmatrix*}[r] 10 & 22 \end{bmatrix*}, the values of x and y are :

  1. x = 2 and y = 3

  2. x = 3 and y = 2

  3. x = -3 and y = 2

  4. x = 3 and y = -2

Answer

Given,

⇒4[5x]−5[y−2]=[1022]⇒[204x]−[5y−10]=[1022]⇒[20−5y4x−(−10)]=[1022]⇒[20−5y4x+10]=[1022]\Rightarrow 4\begin{bmatrix*}[r] 5 & x \end{bmatrix*} - 5\begin{bmatrix*}[r] y & -2 \end{bmatrix*} = \begin{bmatrix*}[r] 10 & 22 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 20 & 4x \end{bmatrix*} - \begin{bmatrix*}[r] 5y & -10 \end{bmatrix*} = \begin{bmatrix*}[r] 10 & 22 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 20 - 5y & 4x - (-10) \end{bmatrix*} = \begin{bmatrix*}[r] 10 & 22 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 20 - 5y & 4x + 10 \end{bmatrix*} = \begin{bmatrix*}[r] 10 & 22 \end{bmatrix*}

∴ 20 - 5y = 10 and 4x + 10 = 22

⇒ 5y = 20 - 10 and 4x = 22 - 10

⇒ 5y = 10 and 4x = 12

⇒ y = 2 and x = 3.

Hence, Option 2 is the correct option.

Question 1(b)

If A = [−3−70−8] and A−B=[64−30]\begin{bmatrix*}[r] -3 & -7 \\ 0 & -8 \end{bmatrix*}\text{ and } A - B = \begin{bmatrix*}[r] 6 & 4 \\ -3 & 0 \end{bmatrix*}, then matrix B is :

  1. [911−318]\begin{bmatrix*}[r] 9 & 11 \\ -3 & 18 \end{bmatrix*}

  2. [−9−1138]\begin{bmatrix*}[r] -9 & -11 \\ 3 & 8 \end{bmatrix*}

  3. [9−11−38]\begin{bmatrix*}[r] 9 & -11 \\ -3 & 8 \end{bmatrix*}

  4. [−9−11−3−8]\begin{bmatrix*}[r] -9 & -11 \\ -3 & -8 \end{bmatrix*}

Answer

Given,

A - B = [64−30]\begin{bmatrix*}[r] 6 & 4 \\ -3 & 0 \end{bmatrix*}

Substituting value of A in above equation we get :

⇒[−3−70−8]−B=[64−30]⇒B=[−3−70−8]−[64−30]⇒B=[−3−6−7−40−(−3)−8−0]⇒B=[−9−113−8]\Rightarrow \begin{bmatrix*}[r] -3 & -7 \\ 0 & -8 \end{bmatrix*} - B = \begin{bmatrix*}[r] 6 & 4 \\ -3 & 0 \end{bmatrix*} \\[1em] \Rightarrow B = \begin{bmatrix*}[r] -3 & -7 \\ 0 & -8 \end{bmatrix*} - \begin{bmatrix*}[r] 6 & 4 \\ -3 & 0 \end{bmatrix*} \\[1em] \Rightarrow B = \begin{bmatrix*}[r] -3 - 6 & -7 - 4 \\ 0 - (-3) & -8 - 0 \end{bmatrix*} \\[1em] \Rightarrow B = \begin{bmatrix*}[r] -9 & -11 \\ 3 & -8 \end{bmatrix*}

Hence, Option 4 is the correct option.

Question 1(c)

If I is a unit matrix of order 2 and M + 4I = [8−342]\begin{bmatrix*}[r] 8 & -3 \\ 4 & 2 \end{bmatrix*}, then matrix M is :

  1. [434−2]\begin{bmatrix*}[r] 4 & 3 \\ 4 & -2 \end{bmatrix*}

  2. [4342]\begin{bmatrix*}[r] 4 & 3 \\ 4 & 2 \end{bmatrix*}

  3. [4−3−42]\begin{bmatrix*}[r] 4 & -3 \\ -4 & 2 \end{bmatrix*}

  4. [4−34−2]\begin{bmatrix*}[r] 4 & -3 \\ 4 & -2 \end{bmatrix*}

Answer

As, I is a unit matrix of order 2.

∴ I = [1001]\begin{bmatrix*}[r] 1 & 0 \\ 0 & 1 \end{bmatrix*}

Given,

⇒M+4I=[8−342]⇒M+4[1001]=[8−342]⇒M+[4004]=[8−342]⇒M=[8−342]−[4004]⇒M=[8−4−3−04−02−4]⇒M=[4−34−2].\Rightarrow M + 4I = \begin{bmatrix*}[r] 8 & -3 \\ 4 & 2 \end{bmatrix*} \\[1em] \Rightarrow M + 4\begin{bmatrix*}[r] 1 & 0 \\ 0 & 1 \end{bmatrix*} = \begin{bmatrix*}[r] 8 & -3 \\ 4 & 2 \end{bmatrix*} \\[1em] \Rightarrow M + \begin{bmatrix*}[r] 4 & 0 \\ 0 & 4 \end{bmatrix*} = \begin{bmatrix*}[r] 8 & -3 \\ 4 & 2 \end{bmatrix*} \\[1em] \Rightarrow M = \begin{bmatrix*}[r] 8 & -3 \\ 4 & 2 \end{bmatrix*} - \begin{bmatrix*}[r] 4 & 0 \\ 0 & 4 \end{bmatrix*} \\[1em] \Rightarrow M = \begin{bmatrix*}[r] 8 - 4 & -3 - 0 \\ 4 - 0 & 2 - 4 \end{bmatrix*} \\[1em] \Rightarrow M = \begin{bmatrix*}[r] 4 & -3 \\ 4 & -2 \end{bmatrix*}.

Hence, Option 4 is the correct option.

Question 1(d)

If 2[3x01]+3[13y2]=[z−7158]2\begin{bmatrix*}[r] 3 & x \\ 0 & 1 \end{bmatrix*} + 3\begin{bmatrix*}[r] 1 & 3 \\ y & 2 \end{bmatrix*} = \begin{bmatrix*}[r] z & -7 \\ 15 & 8 \end{bmatrix*}, the values of x, y and z are :

  1. x = 8, y = -5 and z = 9

  2. x = -8, y = 5 and z = 9

  3. x = -8, y = -5 and z = -9

  4. x = -8, y = 5 and z = -9

Answer

Given,

⇒2[3x01]+3[13y2]=[z−7158]⇒[62x02]+[393y6]=[z−7158]⇒[6+32x+90+3y2+6]=[z−7158]⇒[92x+93y8]=[z−7158]\Rightarrow 2\begin{bmatrix*}[r] 3 & x \\ 0 & 1 \end{bmatrix*} + 3\begin{bmatrix*}[r] 1 & 3 \\ y & 2 \end{bmatrix*} = \begin{bmatrix*}[r] z & -7 \\ 15 & 8 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 6 & 2x \\ 0 & 2 \end{bmatrix*} + \begin{bmatrix*}[r] 3 & 9 \\ 3y & 6 \end{bmatrix*} = \begin{bmatrix*}[r] z & -7 \\ 15 & 8 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 6 + 3 & 2x + 9 \\ 0 + 3y & 2 + 6 \end{bmatrix*} = \begin{bmatrix*}[r] z & -7 \\ 15 & 8 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 9 & 2x + 9 \\ 3y & 8 \end{bmatrix*} = \begin{bmatrix*}[r] z & -7 \\ 15 & 8 \end{bmatrix*} \\[1em]

From above equation we get :

⇒ z = 9, 3y = 15 and 2x + 9 = -7

⇒ z = 9, y = 153\dfrac{15}{3} and 2x = -7 - 9

⇒ z = 9, y = 5 and 2x = -16

⇒ z = 9, y = 5 and x = −162-\dfrac{16}{2}

⇒ z = 9, y = 5 and x = -8.

Hence, Option 2 is the correct option.

Question 1(e)

Given A = [473−2] and B=[12−14]\begin{bmatrix*}[r] 4 & 7 \\ 3 & -2 \end{bmatrix*} \text{ and } B = \begin{bmatrix*}[r] 1 & 2 \\ -1 & 4 \end{bmatrix*}, then A - 2B is :

  1. [−235−10]\begin{bmatrix*}[r] -2 & 3 \\ 5 & -10 \end{bmatrix*}

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

  3. [235−10]\begin{bmatrix*}[r] 2 & 3 \\ 5 & -10 \end{bmatrix*}

  4. [23510]\begin{bmatrix*}[r] 2 & 3 \\ 5 & 10 \end{bmatrix*}

Answer

Substituting values of A and B in A - 2B, we get :

⇒A−2B=[473−2]−2[12−14]=[473−2]−[24−28]=[4−27−43−(−2)−2−8]=[233+2−10]=[235−10].\Rightarrow A - 2B = \begin{bmatrix*}[r] 4 & 7 \\ 3 & -2 \end{bmatrix*} - 2\begin{bmatrix*}[r] 1 & 2 \\ -1 & 4 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 4 & 7 \\ 3 & -2 \end{bmatrix*} - \begin{bmatrix*}[r] 2 & 4 \\ -2 & 8 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 4 - 2 & 7 - 4 \\ 3 - (-2) & -2 - 8 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 2 & 3 \\ 3 + 2 & -10 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 2 & 3 \\ 5 & -10 \end{bmatrix*}.

Hence, Option 3 is the correct option.

Question 2

Find x and y if :

(i) 3[4x]+2[y−3]=[100]3\begin{bmatrix*}[r] 4 & x \end{bmatrix*} + 2\begin{bmatrix*}[r] y & -3 \end{bmatrix*} = \begin{bmatrix*}[r] 10 & 0 \end{bmatrix*}

(ii) x[−12]−4[−2y]=[7−8]x\begin{bmatrix*}[r] -1 \\ 2 \end{bmatrix*} - 4\begin{bmatrix*}[r] -2 \\ y \end{bmatrix*} = \begin{bmatrix*}[r] 7 \\ -8 \end{bmatrix*}

Answer

(i) Given,

⇒3[4x]+2[y−3]=[100]⇒[123x]+[2y−6]=[100]⇒[12+2y3x+(−6)]=[100]⇒[12+2y3x−6]=[100]\Rightarrow 3\begin{bmatrix*}[r] 4 & x \end{bmatrix*} + 2\begin{bmatrix*}[r] y & -3 \end{bmatrix*} = \begin{bmatrix*}[r] 10 & 0 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 12 & 3x \end{bmatrix*} + \begin{bmatrix*}[r] 2y & -6 \end{bmatrix*} = \begin{bmatrix*}[r] 10 & 0 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 12 + 2y & 3x + (-6) \end{bmatrix*} = \begin{bmatrix*}[r] 10 & 0 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 12 + 2y & 3x - 6 \end{bmatrix*} = \begin{bmatrix*}[r] 10 & 0 \end{bmatrix*}

By equality of matrices we get,

12 + 2y = 10
⇒ 2y = 10 - 12
⇒ 2y = -2
⇒ y = -1.

3x - 6 = 0
⇒ 3x = 6
⇒ x = 2

Hence, x = 2 and y = -1.

(ii) Given,

⇒x[−12]−4[−2y]=[7−8]⇒[−x2x]−[−84y]=[7−8]⇒[−x−(−8)2x−4y]=[7−8]⇒[−x+82x−4y]=[7−8]\Rightarrow x\begin{bmatrix*}[r] -1 \\ 2 \end{bmatrix*} - 4\begin{bmatrix*}[r] -2 \\ y \end{bmatrix*} = \begin{bmatrix*}[r] 7 \\ -8 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] -x \\ 2x \end{bmatrix*} - \begin{bmatrix*}[r] -8 \\ 4y \end{bmatrix*} = \begin{bmatrix*}[r] 7 \\ -8 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] -x -(-8) \\ 2x - 4y \end{bmatrix*} = \begin{bmatrix*}[r] 7 \\ -8 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] -x + 8 \\ 2x - 4y \end{bmatrix*} = \begin{bmatrix*}[r] 7 \\ -8 \end{bmatrix*}

By definition of equality of matrices we get,

-x + 8 = 7 ........(i)

2x - 4y = -8 ......(ii)

Solving eq. (i) we get,

⇒ x = 8 - 7 = 1.

Substituting x = 1 in eq. (ii) we get,

⇒ 2x - 4y = -8
⇒ 2(1) - 4y = -8
⇒ 2 - 4y = -8
⇒ 4y = 2 + 8
⇒ 4y = 10
⇒ y = 104=52=2.5\dfrac{10}{4} = \dfrac{5}{2} = 2.5.

Hence, x = 1 and y = 2.5

Question 3

Given A = [2130],B=[1152] and C=[−3−100]\begin{bmatrix*}[r] 2 & 1 \\ 3 & 0 \end{bmatrix*}, B = \begin{bmatrix*}[r] 1 & 1 \\ 5 & 2 \end{bmatrix*} \text{ and } C = \begin{bmatrix*}[r] -3 & -1 \\ 0 & 0 \end{bmatrix*}; find :

(i) 2A - 3B + C

(ii) A + 2C - B

Answer

(i) Given,

2A - 3B + C

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

⇒2A−3B+C=2[2130]−3[1152]+[−3−100]=[4260]−[33156]+[−3−100]=[4−3+(−3)2−3+(−1)6−15+00−6+0]=[−2−2−9−6].\Rightarrow 2A - 3B + C = 2\begin{bmatrix*}[r] 2 & 1 \\ 3 & 0 \end{bmatrix*} - 3\begin{bmatrix*}[r] 1 & 1 \\ 5 & 2 \end{bmatrix*} + \begin{bmatrix*}[r] -3 & -1 \\ 0 & 0 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 4 & 2 \\ 6 & 0 \end{bmatrix*} - \begin{bmatrix*}[r] 3 & 3 \\ 15 & 6 \end{bmatrix*} + \begin{bmatrix*}[r] -3 & -1 \\ 0 & 0 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 4 - 3 + (-3) & 2 - 3 + (-1) \\ 6 - 15 + 0 & 0 - 6 + 0 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -2 & -2 \\ -9 & -6 \end{bmatrix*}.

Hence 2A - 3B + C = [−2−2−9−6].\begin{bmatrix*}[r] -2 & -2 \\ -9 & -6 \end{bmatrix*}.

(ii) Given,

A + 2C - B

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

⇒[2130]+2[−3−100]−[1152]=[2130]+[−6−200]−[1152]=[2+(−6)−11+(−2)−13+0−50+0−2]=[−5−2−2−2].\Rightarrow \begin{bmatrix*}[r] 2 & 1 \\ 3 & 0 \end{bmatrix*} + 2\begin{bmatrix*}[r] -3 & -1 \\ 0 & 0 \end{bmatrix*} - \begin{bmatrix*}[r] 1 & 1 \\ 5 & 2 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 2 & 1 \\ 3 & 0 \end{bmatrix*} + \begin{bmatrix*}[r] -6 & -2 \\ 0 & 0 \end{bmatrix*} - \begin{bmatrix*}[r] 1 & 1 \\ 5 & 2 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 2 + (-6) - 1 & 1 + (-2) - 1 \\ 3 + 0 - 5 & 0 + 0 - 2 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -5 & -2 \\ -2 & -2 \end{bmatrix*}.

Hence, A + 2C - B = [−5−2−2−2].\begin{bmatrix*}[r] -5 & -2 \\ -2 & -2 \end{bmatrix*}.

Question 4

If [4−240]+3A=[−2−21−3]\begin{bmatrix*}[r] 4 & -2 \\ 4 & 0 \end{bmatrix*} + 3A = \begin{bmatrix*}[r] -2 & -2 \\ 1 & -3 \end{bmatrix*}; find A.

Answer

Given,

⇒[4−240]+3A=[−2−21−3]⇒3A=[−2−21−3]−[4−240]⇒3A=[−2−4−2−(−2)1−4−3−0]⇒3A=[−60−3−3]⇒A=13[−60−3−3]⇒A=[−20−1−1].\Rightarrow \begin{bmatrix*}[r] 4 & -2 \\ 4 & 0 \end{bmatrix*} + 3A = \begin{bmatrix*}[r] -2 & -2 \\ 1 & -3 \end{bmatrix*} \\[1em] \Rightarrow 3A = \begin{bmatrix*}[r] -2 & -2 \\ 1 & -3 \end{bmatrix*} - \begin{bmatrix*}[r] 4 & -2 \\ 4 & 0 \end{bmatrix*} \\[1em] \Rightarrow 3A = \begin{bmatrix*}[r] -2 - 4 & -2 - (-2) \\ 1 - 4 & -3 - 0 \end{bmatrix*} \\[1em] \Rightarrow 3A = \begin{bmatrix*}[r] -6 & 0 \\ -3 & -3 \end{bmatrix*} \\[1em] \Rightarrow A = \dfrac{1}{3}\begin{bmatrix*}[r] -6 & 0 \\ -3 & -3 \end{bmatrix*} \\[1em] \Rightarrow A = \begin{bmatrix*}[r] -2 & 0 \\ -1 & -1 \end{bmatrix*}.

Hence, A = [−20−1−1]\begin{bmatrix*}[r] -2 & 0 \\ -1 & -1 \end{bmatrix*}.

Question 5

Given A = [1423] and B=[−4−1−3−2]\begin{bmatrix*}[r] 1 & 4 \\ 2 & 3 \end{bmatrix*} \text{ and } B = \begin{bmatrix*}[r] -4 & -1 \\ -3 & -2 \end{bmatrix*}

(i) find the matrix 2A + B

(ii) find a matrix C such that :

C + B = [0000]\begin{bmatrix*}[r] 0 & 0 \\ 0 & 0 \end{bmatrix*}

Answer

(i)

2A+B=2[1423]+[−4−1−3−2]=[2846]+[−4−1−3−2]=[2+(−4)8+(−1)4+(−3)6+(−2)]=[−2714].2A + B = 2\begin{bmatrix*}[r] 1 & 4 \\ 2 & 3 \end{bmatrix*} + \begin{bmatrix*}[r] -4 & -1 \\ -3 & -2 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 2 & 8 \\ 4 & 6 \end{bmatrix*} + \begin{bmatrix*}[r] -4 & -1 \\ -3 & -2 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 2 + (-4) & 8 + (-1) \\ 4 + (-3) & 6 + (-2) \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -2 & 7 \\ 1 & 4 \end{bmatrix*}.

Hence, 2A + B = [−2714]\begin{bmatrix*}[r] -2 & 7 \\ 1 & 4 \end{bmatrix*}.

(ii) Given,

⇒C+B=[0000]⇒C=[0000]−B⇒C=[0000]−[−4−1−3−2]⇒C=[0−(−4)0−(−1)0−(−3)0−(−2)]⇒C=[4132].\Rightarrow C + B = \begin{bmatrix*}[r] 0 & 0 \\ 0 & 0 \end{bmatrix*} \\[1em] \Rightarrow C = \begin{bmatrix*}[r] 0 & 0 \\ 0 & 0 \end{bmatrix*} - B \\[1em] \Rightarrow C = \begin{bmatrix*}[r] 0 & 0 \\ 0 & 0 \end{bmatrix*} - \begin{bmatrix*}[r] -4 & -1 \\ -3 & -2 \end{bmatrix*} \\[1em] \Rightarrow C = \begin{bmatrix*}[r] 0 - (-4) & 0 - (-1) \\ 0 - (-3) & 0 - (-2) \end{bmatrix*} \\[1em] \Rightarrow C = \begin{bmatrix*}[r] 4 & 1 \\ 3 & 2 \end{bmatrix*}.

Hence, C = [4132]\begin{bmatrix*}[r] 4 & 1 \\ 3 & 2 \end{bmatrix*}.

Question 6

If 2[3x01]+3[13y2]=[z−7158]2\begin{bmatrix*}[r] 3 & x \\ 0 & 1 \end{bmatrix*} + 3\begin{bmatrix*}[r] 1 & 3 \\ y & 2 \end{bmatrix*} = \begin{bmatrix*}[r] z & -7 \\ 15 & 8 \end{bmatrix*}; find the values of x, y and z.

Answer

Given,

⇒2[3x01]+3[13y2]=[z−7158]⇒[62x02]+[393y6]=[z−7158]⇒[6+32x+90+3y2+6]=[z−7158]⇒[92x+93y8]=[z−7158]\Rightarrow 2\begin{bmatrix*}[r] 3 & x \\ 0 & 1 \end{bmatrix*} + 3\begin{bmatrix*}[r] 1 & 3 \\ y & 2 \end{bmatrix*} = \begin{bmatrix*}[r] z & -7 \\ 15 & 8 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 6 & 2x \\ 0 & 2 \end{bmatrix*} + \begin{bmatrix*}[r] 3 & 9 \\ 3y & 6 \end{bmatrix*} = \begin{bmatrix*}[r] z & -7 \\ 15 & 8 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 6 + 3 & 2x + 9 \\ 0 + 3y & 2 + 6 \end{bmatrix*} = \begin{bmatrix*}[r] z & -7 \\ 15 & 8 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 9 & 2x + 9 \\ 3y & 8 \end{bmatrix*} = \begin{bmatrix*}[r] z & -7 \\ 15 & 8 \end{bmatrix*}

By definition of equality of matrices we get,

z = 9,

2x + 9 = -7
⇒ 2x = -7 - 9
⇒ 2x = -16
⇒ x = -8,

3y = 15
⇒ y = 5.

Hence, x = -8, y = 5 and z = 9.

Question 7

Given A = [−360−9]\begin{bmatrix*}[r] -3 & 6 \\ 0 & -9 \end{bmatrix*} and At is its transpose matrix. Find :

(i) 2A + 3At

(ii) 2At - 3A

(iii) 12A−13At\dfrac{1}{2}A - \dfrac{1}{3}A^t

(iv) At−13AA^t - \dfrac{1}{3}A

Answer

A=[−360−9] and At=[−306−9]A = \begin{bmatrix*}[r] -3 & 6 \\ 0 & -9 \end{bmatrix*} \text{ and } A^t = \begin{bmatrix*}[r] -3 & 0 \\ 6 & -9 \end{bmatrix*}.

(i) Substituting value of A and At in 2A + 3At we get,

⇒2A+3At=2[−360−9]+3[−306−9]=[−6120−18]+[−9018−27]=[−6+(−9)12+00+18−18+(−27)]=[−151218−45].\Rightarrow 2A + 3A^t = 2\begin{bmatrix*}[r] -3 & 6 \\ 0 & -9 \end{bmatrix*} + 3\begin{bmatrix*}[r] -3 & 0 \\ 6 & -9 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -6 & 12 \\ 0 & -18 \end{bmatrix*} + \begin{bmatrix*}[r] -9 & 0 \\ 18 & -27 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -6 + (-9) & 12 + 0 \\ 0 + 18 & -18 + (-27) \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -15 & 12 \\ 18 & -45 \end{bmatrix*}.

Hence, 2A + 3At = [−151218−45].\begin{bmatrix*}[r] -15 & 12 \\ 18 & -45 \end{bmatrix*}.

(ii) Substituting value of A and At in 2At - 3A we get,

⇒2At−3A=2[−306−9]−3[−360−9]=[−6012−18]−[−9180−27]=[−6−(−9)0−1812−0−18−(−27)]=[3−18129].\Rightarrow 2A^t - 3A = 2\begin{bmatrix*}[r] -3 & 0 \\ 6 & -9 \end{bmatrix*} - 3\begin{bmatrix*}[r] -3 & 6 \\ 0 & -9 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -6 & 0 \\ 12 & -18 \end{bmatrix*} - \begin{bmatrix*}[r] -9 & 18 \\ 0 & -27 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -6 - (-9) & 0 - 18 \\ 12 - 0 & -18 - (-27) \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 3 & -18 \\ 12 & 9 \end{bmatrix*}.

Hence, 2At−3A=[3−18129].2A^t - 3A = \begin{bmatrix*}[r] 3 & -18 \\ 12 & 9 \end{bmatrix*}.

(iii) Substituting value of A and At in 12A−13At\dfrac{1}{2}A - \dfrac{1}{3}A^t we get,

⇒12A−13At=12[−360−9]−13[−306−9]=[−3230−92]−[−102−3]=[−32−(−1)3−00−2−92−(−3)]=[−32+13−2−92+3]=[−123−2−32]\Rightarrow \dfrac{1}{2}A - \dfrac{1}{3}A^t = \dfrac{1}{2}\begin{bmatrix*}[r] -3 & 6 \\ 0 & -9 \end{bmatrix*} - \dfrac{1}{3}\begin{bmatrix*}[r] -3 & 0 \\ 6 & -9 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -\dfrac{3}{2} & 3 \\ 0 & -\dfrac{9}{2} \end{bmatrix*} - \begin{bmatrix*}[r] -1 & 0 \\ 2 & -3 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -\dfrac{3}{2} - (-1) & 3 - 0 \\ 0 - 2 & -\dfrac{9}{2} - (-3) \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -\dfrac{3}{2} + 1 & 3 \\ -2 & -\dfrac{9}{2} + 3 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -\dfrac{1}{2} & 3 \\ -2 & -\dfrac{3}{2} \end{bmatrix*}

Hence, 12A−13At=[−123−2−32].\dfrac{1}{2}A - \dfrac{1}{3}A^t = \begin{bmatrix*}[r] -\dfrac{1}{2} & 3 \\ -2 & -\dfrac{3}{2} \end{bmatrix*}.

(iv) Substituting value of A and At in At−13AA^t - \dfrac{1}{3}A we get,

⇒At−13A=[−306−9]−13[−360−9]=[−306−9]−[−120−3]=[−3−(−1)0−26−0−9−(−3)]=[−2−26−6].\Rightarrow A^t - \dfrac{1}{3}A = \begin{bmatrix*}[r] -3 & 0 \\ 6 & -9 \end{bmatrix*} - \dfrac{1}{3}\begin{bmatrix*}[r] -3 & 6 \\ 0 & -9 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -3 & 0 \\ 6 & -9 \end{bmatrix*} - \begin{bmatrix*}[r] -1 & 2 \\ 0 & -3 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -3 - (-1) & 0 - 2 \\ 6 - 0 & -9 - (-3) \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -2 & -2 \\ 6 & -6 \end{bmatrix*}.

Hence, At−13A=[−2−26−6].A^t - \dfrac{1}{3}A = \begin{bmatrix*}[r] -2 & -2 \\ 6 & -6 \end{bmatrix*}.

Question 8

Given A = [11−20] and B=[2−111]\begin{bmatrix*}[r] 1 & 1 \\ -2 & 0 \end{bmatrix*} \text{ and } B = \begin{bmatrix*}[r] 2 & -1 \\ 1 & 1 \end{bmatrix*}.

Solve for matrix X :

(i) X + 2A = B

(ii) 3x + B + 2A = 0

(iii) 3A - 2X = X - 2B.

Answer

(i) Given,

⇒X+2A=B⇒X=B−2A⇒X=[2−111]−2[11−20]⇒X=[2−111]−[22−40]⇒X=[2−2−1−21−(−4)1−0]⇒X=[0−351]\Rightarrow X + 2A = B \\[1em] \Rightarrow X = B - 2A \\[1em] \Rightarrow X = \begin{bmatrix*}[r] 2 & -1 \\ 1 & 1 \end{bmatrix*} - 2\begin{bmatrix*}[r] 1 & 1 \\ -2 & 0 \end{bmatrix*} \\[1em] \Rightarrow X = \begin{bmatrix*}[r] 2 & -1 \\ 1 & 1 \end{bmatrix*} - \begin{bmatrix*}[r] 2 & 2 \\ -4 & 0 \end{bmatrix*} \\[1em] \Rightarrow X = \begin{bmatrix*}[r] 2 - 2 & -1 - 2 \\ 1 - (-4) & 1 - 0 \end{bmatrix*} \\[1em] \Rightarrow X = \begin{bmatrix*}[r] 0 & -3 \\ 5 & 1 \end{bmatrix*}

Hence, X = [0−351].\begin{bmatrix*}[r] 0 & -3 \\ 5 & 1 \end{bmatrix*}.

(ii) Given,

⇒3X+B+2A=0⇒3X=−(B+2A)⇒3X=−([2−111]+2[11−20])⇒3X=−([2−111]+[22−40])⇒3X=−([2+2−1+21+(−4)1+0])⇒3X=−([41−31])⇒X=−13([41−31])⇒X=[−43−131−13]\Rightarrow 3X + B + 2A = 0 \\[1em] \Rightarrow 3X = -(B + 2A) \\[1em] \Rightarrow 3X = -\Big(\begin{bmatrix*}[r] 2 & -1 \\ 1 & 1 \end{bmatrix*} + 2\begin{bmatrix*}[r] 1 & 1 \\ -2 & 0 \end{bmatrix*}\Big) \\[1em] \Rightarrow 3X = -\Big(\begin{bmatrix*}[r] 2 & -1 \\ 1 & 1 \end{bmatrix*} + \begin{bmatrix*}[r] 2 & 2 \\ -4 & 0 \end{bmatrix*}\Big) \\[1em] \Rightarrow 3X = -\Big(\begin{bmatrix*}[r] 2 + 2 & -1 + 2 \\ 1 + (-4) & 1 + 0 \end{bmatrix*}\Big) \\[1em] \Rightarrow 3X = -\Big(\begin{bmatrix*}[r] 4 & 1 \\ -3 & 1 \end{bmatrix*}\Big) \\[1em] \Rightarrow X = -\dfrac{1}{3}\Big(\begin{bmatrix*}[r] 4 & 1 \\ -3 & 1 \end{bmatrix*}\Big) \\[1em] \Rightarrow X = \begin{bmatrix*}[r] -\dfrac{4}{3} & -\dfrac{1}{3} \\ 1 & -\dfrac{1}{3} \end{bmatrix*}

Hence, X = [−43−131−13]\begin{bmatrix*}[r] -\dfrac{4}{3} & -\dfrac{1}{3} \\ 1 & -\dfrac{1}{3} \end{bmatrix*}.

(iii) Given,

⇒3A−2X=X−2B⇒X+2X=3A+2B⇒3X=3A+2B⇒3X=3[11−20]+2[2−111]⇒3X=[33−60]+[4−222]⇒3X=[3+43+(−2)−6+20+2]⇒3X=[71−42]⇒X=13[71−42]⇒X=[7313−4323].\Rightarrow 3A - 2X = X - 2B \\[1em] \Rightarrow X + 2X = 3A + 2B \\[1em] \Rightarrow 3X = 3A + 2B \\[1em] \Rightarrow 3X = 3\begin{bmatrix*}[r] 1 & 1 \\ -2 & 0 \end{bmatrix*} + 2\begin{bmatrix*}[r] 2 & -1 \\ 1 & 1 \end{bmatrix*} \\[1em] \Rightarrow 3X = \begin{bmatrix*}[r] 3 & 3 \\ -6 & 0 \end{bmatrix*} + \begin{bmatrix*}[r] 4 & -2 \\ 2 & 2 \end{bmatrix*} \\[1em] \Rightarrow 3X = \begin{bmatrix*}[r] 3 + 4 & 3 + (-2) \\ -6 + 2 & 0 + 2 \end{bmatrix*} \\[1em] \Rightarrow 3X = \begin{bmatrix*}[r] 7 & 1 \\ -4 & 2 \end{bmatrix*} \\[1em] \Rightarrow X = \dfrac{1}{3}\begin{bmatrix*}[r] 7 & 1 \\ -4 & 2 \end{bmatrix*} \\[1em] \Rightarrow X = \begin{bmatrix*}[r] \dfrac{7}{3} & \dfrac{1}{3} \\ -\dfrac{4}{3} & \dfrac{2}{3} \end{bmatrix*}.

Hence, X = [7313−4323].\begin{bmatrix*}[r] \dfrac{7}{3} & \dfrac{1}{3} \\ -\dfrac{4}{3} & \dfrac{2}{3} \end{bmatrix*}.

Question 9

If I is the unit matrix of order 2 × 2; find the matrix M such that :

5M + 3I = 4[2−50−3]4\begin{bmatrix*}[r] 2 & -5 \\ 0 & -3 \end{bmatrix*}

Answer

I = [1001]\begin{bmatrix*}[r] 1 & 0 \\ 0 & 1 \end{bmatrix*}

Given,

⇒5M+3I=4[2−50−3]⇒5M+3[1001]=[8−200−12]⇒5M+[3003]=[8−200−12]⇒5M=[8−200−12]−[3003]⇒5M=[8−3−20−00−0−12−3]⇒5M=[5−200−15]⇒M=15[5−200−15]⇒M=[1−40−3].\Rightarrow 5M + 3I = 4\begin{bmatrix*}[r] 2 & -5 \\ 0 & -3 \end{bmatrix*} \\[1em] \Rightarrow 5M + 3\begin{bmatrix*}[r] 1 & 0 \\ 0 & 1 \end{bmatrix*} = \begin{bmatrix*}[r] 8 & -20 \\ 0 & -12 \end{bmatrix*} \\[1em] \Rightarrow 5M + \begin{bmatrix*}[r] 3 & 0 \\ 0 & 3 \end{bmatrix*} = \begin{bmatrix*}[r] 8 & -20 \\ 0 & -12 \end{bmatrix*} \\[1em] \Rightarrow 5M = \begin{bmatrix*}[r] 8 & -20 \\ 0 & -12 \end{bmatrix*} - \begin{bmatrix*}[r] 3 & 0 \\ 0 & 3 \end{bmatrix*} \\[1em] \\[1em] \Rightarrow 5M = \begin{bmatrix*}[r] 8 - 3 & -20 - 0 \\ 0 - 0 & -12 - 3 \end{bmatrix*} \\[1em] \Rightarrow 5M = \begin{bmatrix*}[r] 5 & -20 \\ 0 & -15 \end{bmatrix*} \\[1em] \Rightarrow M = \dfrac{1}{5}\begin{bmatrix*}[r] 5 & -20 \\ 0 & -15 \end{bmatrix*} \\[1em] \Rightarrow M = \begin{bmatrix*}[r] 1 & -4 \\ 0 & -3 \end{bmatrix*}.

Hence, M = [1−40−3].\begin{bmatrix*}[r] 1 & -4 \\ 0 & -3 \end{bmatrix*}.

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