KnowledgeBoat Logo
|
OPEN IN APP

Chapter 9

Matrices — Exercise 9(B)

Class - 10 RS Aggarwal Mathematics Solutions

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.

PrevNext