Q: Compute the following: (i) $\quad\left(\begin{array}{cc}a & b \\ -b & a\end{array}\right)+\left(\begin{array}{ll}a & b \\ b & a\end{array}\right)$ (ii) $\left(\begin{array}{ll}a^{2}+b^{2} & b^{2}+c^{2} \\ a^{2}+c^{2} & a^{2}+b^{2}\end{array}\right)+\left(\begin{array}{cc}2 a b & 2 b c \\ -2 a c & -2 a b\end{array}\right)$ (iii) $\quad\left(\begin{array}{ccc}-1 & 4 & -6 \\ 8 & 5 & 16 \\ 2 & 8 & 5\end{array}\right)+\left(\begin{array}{ccc}12 & 7 & 6 \\ 8 & 0 & 5 \\ 3 & 2 & 4\end{array}\right)$ (iv) $\left(\begin{array}{ll}\cos ^{2} x & \sin ^{2} x \\ \sin ^{2} x & \cos ^{2} x\end{array}\right)+\left(\begin{array}{ll}\sin ^{2} x & \cos ^{2} x \\ \cos ^{2} x & \sin ^{2} x\end{array}\right)$

(i) $\quad\left(\begin{array}{cc}a & b \\ -b & a\end{array}\right)+\left(\begin{array}{ll}a & b \\ b & a\end{array}\right)$ $$ \begin{aligned} & \Rightarrow\left(\begin{array}{cc} a+a & b+b \\ -b+b & a+a \end{array}\right) \\ & \Rightarrow\left(\begin{array}{cc} 2 a & 2 b \\ 0 & 2 a \end{array}\right) \end{aligned} $$ (ii) $\left(\begin{array}{ll}a^{2}+b^{2} & b^{2}+c^{2} \\ a^{2}+c^{2} & a^{2}+b^{2}\end{array}\right)+\left(\begin{array}{cc}2 a b & 2 b c \\ -2 a c & -2 a b\end{array}\right)$ $$ \begin{aligned} & \Rightarrow\left(\begin{array}{ll} a^{2}+b^{2}+2 a b & b^{2}+c^{2}+2 b c \\ a^{2}+c^{2}-2 a c & a^{2}+b^{2}-2 a b \end{array}\right) \\ & \Rightarrow\left(\begin{array}{ll} (a+b)^{2} & (b+c)^{2} \\ (a-c)^{2} & (a-b)^{2} \end{array}\right) \end{aligned} $$ (iii) $\quad\left(\begin{array}{ccc}-1 & 4 & -6 \\ 8 & 5 & 16 \\ 2 & 8 & 5\end{array}\right)+\left(\begin{array}{ccc}12 & 7 & 6 \\ 8 & 0 & 5 \\ 3 & 2 & 4\end{array}\right)$ $$ \begin{aligned} & \Rightarrow\left(\begin{array}{ccc} -1+12 & 4+7 & -6+6 \\ 8+8 & 5+0 & 16+5 \\ 2+3 & 8+2 & 5+4 \end{array}\right) \\ & \Rightarrow\left(\begin{array}{ccc} 11 & 11 & 0 \\ 16 & 5 & 21 \\ 5 & 10 & 9 \end{array}\right) \end{aligned} $$ (iv) $\left(\begin{array}{cc}\cos ^{2} x & \sin ^{2} x \\ \sin ^{2} x & \cos ^{2} x\end{array}\right)+\left(\begin{array}{cc}\sin ^{2} x & \cos ^{2} x \\ \cos ^{2} x & \sin ^{2} x\end{array}\right)$ $$ \begin{aligned} & \Rightarrow\left(\begin{array}{ll} \cos ^{2} x+\sin ^{2} x & \sin ^{2} x+\cos ^{2} x \\ \sin ^{2} x+\cos ^{2} x & \cos ^{2} x+\sin ^{2} x \end{array}\right) \\ & \Rightarrow\left(\begin{array}{ll} 1 & 1 \\ 1 & 1 \end{array}\right) \end{aligned} $$

Q: Simplify $\cos \theta\left(\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta\end{array}\right)+\sin \theta\left(\begin{array}{cc}\sin \theta & -\cos \theta \\ \cos \theta & \sin \theta\end{array}\right)$.

$$ \begin{aligned} \cos \theta\left(\begin{array}{cc} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right)+\sin \theta\left(\begin{array}{cc} \sin \theta & -\cos \theta \\ \cos \theta & \sin \theta \end{array}\right) & \\ & \Rightarrow\left(\begin{array}{cc} \cos ^{2} \theta & \cos \theta \sin \theta \\ -\sin \theta \cos \theta & \cos ^{2} \theta \end{array}\right)+\left(\begin{array}{cc} \sin ^{2} \theta & -\sin \theta \cos \theta \\ \sin \theta \cos \theta & \sin ^{2} \theta \end{array}\right) \\ & \Rightarrow\left(\begin{array}{cc} \cos ^{2} \theta+\sin ^{2} \theta & \sin \theta \cos \theta-\sin \theta \cos \theta \\ -\sin \theta \cos \theta+\sin \theta \cos \theta & \cos ^{2} \theta+\sin ^{2} \theta \end{array}\right) \\ & \Rightarrow\left(\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right) \end{aligned} $$

Q: Find $X$, if $Y=\left(\begin{array}{ll}3 & 2 \\ 1 & 4\end{array}\right)$ and $2 X+Y=\left(\begin{array}{cc}1 & 0 \\ -3 & 2\end{array}\right)$.

$2 X+Y=\left(\begin{array}{cc}1 & 0 \\ -3 & 2\end{array}\right)$ $\Rightarrow 2 X+\left(\begin{array}{ll}3 & 2 \\ 1 & 4\end{array}\right)=\left(\begin{array}{cc}1 & 0 \\ -3 & 2\end{array}\right)$ $\Rightarrow 2 X=\left(\begin{array}{cc}1 & 0 \\ -3 & 2\end{array}\right)-\left(\begin{array}{ll}3 & 2 \\ 1 & 4\end{array}\right)$ $\Rightarrow 2 X=\left(\begin{array}{ll}-2 & -2 \\ -4 & -2\end{array}\right)$ $\Rightarrow X=\frac{1}{2}\left(\begin{array}{ll}-2 & -2 \\ -4 & -2\end{array}\right)$ $\Rightarrow X=\left(\begin{array}{ll}-1 & -1 \\ -2 & -1\end{array}\right)$

Question 1

Let $A=\left[\begin{array}{ll}2 & 4 \ 3 & 2\end{array}ight], B=\left[\begin{array}{cc}1 & 3 \ -2 & 5\end{array}ight], C=\left[\begin{array}{cc}-2 & 5 \ 3 & 4\end{array}ight]$. Find each of the following:

(i) $A+B$

(ii) $\quad A-B$

(iii) $3 A-C$

(iv) $A B$

(v) $B A$

Accepted Answer

(i) $A+B$

$$

\begin{aligned}

& \Rightarrow\left(\begin{array}{ll}

2 & 4 \

3 & 2

\end{array}ight)+\left(\begin{array}{cc}

1 & 3 \

-2 & 5

\end{array}ight) \

& \Rightarrow\left(\begin{array}{ll}

2+1 & 4+3 \

3-2 & 2+5

\end{array}ight) \

& \Rightarrow\left(\begin{array}{ll}

3 & 7 \

1 & 7

\end{array}ight)

\end{aligned}

$$

(ii) $A-B$

$$

\begin{aligned}

& \Rightarrow\left(\begin{array}{cc}

2 & 4 \

3 & 2

\end{array}ight)-\left(\begin{array}{cc}

1 & 3 \

-2 & 5

\end{array}ight) \

& \Rightarrow\left(\begin{array}{cc}

2-1 & 4-3 \

3+2 & 2-5

\end{array}ight) \

& \Rightarrow\left(\begin{array}{cc}

1 & 1 \

5 & -3

\end{array}ight)

\end{aligned}

$$

(iii) $3 A-C$

$$

\begin{aligned}

& \Rightarrow 3\left(\begin{array}{ll}

2 & 4 \

3 & 2

\end{array}ight)-\left(\begin{array}{cc}

-2 & 5 \

3 & 4

\end{array}ight) \

& \Rightarrow\left(\begin{array}{cc}

3 	imes 2 & 3 	imes 4 \

3 	imes 3 & 3 	imes 2

\end{array}ight)-\left(\begin{array}{cc}

-2 & 5 \

3 & 4

\end{array}ight) \

& \Rightarrow\left(\begin{array}{cc}

6+2 & 12-5 \

9-3 & 6-4

\end{array}ight) \

& \Rightarrow\left(\begin{array}{ll}

8 & 7 \

6 & 2

\end{array}ight)

\end{aligned}

$$

(iv) $A B$

$$

\begin{aligned}

& \Rightarrow\left(\begin{array}{ll}

2 & 4 \

3 & 2

\end{array}ight)\left(\begin{array}{cc}

1 & 3 \

-2 & 5

\end{array}ight) \

& \Rightarrow\left(\begin{array}{ll}

2(1)+4(-2) & 2(3)+4(5) \

3(1)+2(-2) & 3(3)+2(5)

\end{array}ight) \

& \Rightarrow\left(\begin{array}{cc}

2-8 & 6+20 \

3-4 & 9+10

\end{array}ight) \

& \Rightarrow\left(\begin{array}{cc}

-6 & 26 \

-1 & 19

\end{array}ight)

\end{aligned}

$$

(v) $B A$

$$

\begin{aligned}

& \Rightarrow\left(\begin{array}{cc}

1 & 3 \

-2 & 5

\end{array}ight)\left(\begin{array}{ll}

2 & 4 \

3 & 2

\end{array}ight) \

& \Rightarrow\left(\begin{array}{cc}

1(2)+3(3) & 1(4)+3(2) \

-2(2)+5(3) & -2(4)+5(2)

\end{array}ight) \

& \Rightarrow\left(\begin{array}{cc}

2+9 & 4+6 \

-4+15 & -8+10

\end{array}ight) \

& \Rightarrow\left(\begin{array}{cc}

11 & 10 \

11 & 2

\end{array}ight)

\end{aligned}

$$

Question 2

Compute the following:

(i) $\quad\left(\begin{array}{cc}a & b \ -b & a\end{array}ight)+\left(\begin{array}{ll}a & b \ b & a\end{array}ight)$

(ii) $\left(\begin{array}{ll}a^{2}+b^{2} & b^{2}+c^{2} \ a^{2}+c^{2} & a^{2}+b^{2}\end{array}ight)+\left(\begin{array}{cc}2 a b & 2 b c \ -2 a c & -2 a b\end{array}ight)$

(iii) $\quad\left(\begin{array}{ccc}-1 & 4 & -6 \ 8 & 5 & 16 \ 2 & 8 & 5\end{array}ight)+\left(\begin{array}{ccc}12 & 7 & 6 \ 8 & 0 & 5 \ 3 & 2 & 4\end{array}ight)$

(iv) $\left(\begin{array}{ll}\cos ^{2} x & \sin ^{2} x \ \sin ^{2} x & \cos ^{2} x\end{array}ight)+\left(\begin{array}{ll}\sin ^{2} x & \cos ^{2} x \ \cos ^{2} x & \sin ^{2} x\end{array}ight)$

Accepted Answer

(i) $\quad\left(\begin{array}{cc}a & b \ -b & a\end{array}ight)+\left(\begin{array}{ll}a & b \ b & a\end{array}ight)$

$$

\begin{aligned}

& \Rightarrow\left(\begin{array}{cc}

a+a & b+b \

-b+b & a+a

\end{array}ight) \

& \Rightarrow\left(\begin{array}{cc}

2 a & 2 b \

0 & 2 a

\end{array}ight)

\end{aligned}

$$

(ii) $\left(\begin{array}{ll}a^{2}+b^{2} & b^{2}+c^{2} \ a^{2}+c^{2} & a^{2}+b^{2}\end{array}ight)+\left(\begin{array}{cc}2 a b & 2 b c \ -2 a c & -2 a b\end{array}ight)$

$$

\begin{aligned}

& \Rightarrow\left(\begin{array}{ll}

a^{2}+b^{2}+2 a b & b^{2}+c^{2}+2 b c \

a^{2}+c^{2}-2 a c & a^{2}+b^{2}-2 a b

\end{array}ight) \

& \Rightarrow\left(\begin{array}{ll}

(a+b)^{2} & (b+c)^{2} \

(a-c)^{2} & (a-b)^{2}

\end{array}ight)

\end{aligned}

$$

(iii) $\quad\left(\begin{array}{ccc}-1 & 4 & -6 \ 8 & 5 & 16 \ 2 & 8 & 5\end{array}ight)+\left(\begin{array}{ccc}12 & 7 & 6 \ 8 & 0 & 5 \ 3 & 2 & 4\end{array}ight)$

$$

\begin{aligned}

& \Rightarrow\left(\begin{array}{ccc}

-1+12 & 4+7 & -6+6 \

8+8 & 5+0 & 16+5 \

2+3 & 8+2 & 5+4

\end{array}ight) \

& \Rightarrow\left(\begin{array}{ccc}

11 & 11 & 0 \

16 & 5 & 21 \

5 & 10 & 9

\end{array}ight)

\end{aligned}

$$

(iv) $\left(\begin{array}{cc}\cos ^{2} x & \sin ^{2} x \ \sin ^{2} x & \cos ^{2} x\end{array}ight)+\left(\begin{array}{cc}\sin ^{2} x & \cos ^{2} x \ \cos ^{2} x & \sin ^{2} x\end{array}ight)$

$$

\begin{aligned}

& \Rightarrow\left(\begin{array}{ll}

\cos ^{2} x+\sin ^{2} x & \sin ^{2} x+\cos ^{2} x \

\sin ^{2} x+\cos ^{2} x & \cos ^{2} x+\sin ^{2} x

\end{array}ight) \

& \Rightarrow\left(\begin{array}{ll}

1 & 1 \

1 & 1

\end{array}ight)

\end{aligned}

$$

Question 3

Compute the indicated products:

(i) $\quad\left(\begin{array}{cc}a & b \ -b & a\end{array}ight)\left(\begin{array}{cc}a & -b \ b & a\end{array}ight)$

(ii) $\left(\begin{array}{l}1 \ 2 \ 3\end{array}ight)\left(\begin{array}{lll}2 & 3 & 4\end{array}ight)$

(iii) $\quad\left(\begin{array}{cc}1 & -2 \ 2 & 3\end{array}ight)\left(\begin{array}{lll}1 & 2 & 3 \ 2 & 3 & 1\end{array}ight)$

(iv) $\left(\begin{array}{lll}2 & 3 & 4 \ 3 & 4 & 5 \ 4 & 5 & 6\end{array}ight)\left(\begin{array}{ccc}1 & -3 & 5 \ 0 & 2 & 4 \ 3 & 0 & 5\end{array}ight)$

(v) $\quad\left(\begin{array}{cc}2 & 1 \ 3 & 2 \ -1 & 1\end{array}ight)\left(\begin{array}{ccc}1 & 0 & 1 \ -1 & 2 & 1\end{array}ight)$

(vi) $\quad\left(\begin{array}{ll}3 & 1\end{array}ight)$

Solution:

(i) $\left(\begin{array}{cc}a & b \ -b & a\end{array}ight)\left(\begin{array}{cc}a & -b \ b & a\end{array}ight)$

$$

\begin{aligned}

& \Rightarrow\left(\begin{array}{cc}

a(a)+b(b) & a(-b)+b(a) \

-b(a)+a(b) & -b(-b)+a(a)

\end{array}ight) \

& \Rightarrow\left(\begin{array}{cc}

a^{2}+b^{2} & -a b+a b \

-a b+a b & b^{2}+a^{2}

\end{array}ight) \

& \Rightarrow\left(\begin{array}{cc}

a^{2}+b^{2} & 0 \

0 & a^{2}+b^{2}

\end{array}ight)

\end{aligned}

$$

(ii) $\left(\begin{array}{l}1 \ 2 \ 3\end{array}ight)\left(\begin{array}{lll}2 & 3 & 4\end{array}ight)$

$$

\begin{aligned}

& \Rightarrow\left(\begin{array}{lll}

1(2) & 1(3) & 1(4) \

2(2) & 2(3) & 2(4) \

3(2) & 3(3) & 3(4)

\end{array}ight) \

& \Rightarrow\left(\begin{array}{ccc}

2 & 3 & 4 \

4 & 6 & 8 \

6 & 9 & 12

\end{array}ight)

\end{aligned}

$$

(iii) $\left(\begin{array}{cc}1 & -2 \ 2 & 3\end{array}ight)\left(\begin{array}{lll}1 & 2 & 3 \ 2 & 3 & 1\end{array}ight)$

$$

\begin{aligned}

& \Rightarrow\left(\begin{array}{lll}

1(1)-2(2) & 1(2)-2(3) & 1(3)-2(1) \

2(1)+3(2) & 2(2)+3(3) & 2(3)+3(1)

\end{array}ight) \

& \Rightarrow\left(\begin{array}{ccc}

-3 & -4 & 1 \

8 & 13 & 9

\end{array}ight)

\end{aligned}

$$

(iv) $\left(\begin{array}{lll}2 & 3 & 4 \ 3 & 4 & 5 \ 4 & 5 & 6\end{array}ight)\left(\begin{array}{ccc}1 & -3 & 5 \ 0 & 2 & 4 \ 3 & 0 & 5\end{array}ight)$

$$

\begin{aligned}

& \Rightarrow\left(\begin{array}{lll}

2(1)+3(0)+4(3) & 2(-3)+3(2)+4(0) & 2(5)+3(4)+4(5) \

3(1)+4(0)+5(3) & 3(-3)+4(2)+5(0) & 3(5)+4(4)+5(5) \

4(1)+5(0)+6(3) & 4(-3)+5(2)+6(0) & 4(5)+5(4)+6(5)

\end{array}ight) \

& \Rightarrow\left(\begin{array}{ccc}

14 & 0 & 42 \

18 & -1 & 56 \

22 & -2 & 70

\end{array}ight)

\end{aligned}

$$

(v) $\left(\begin{array}{cc}2 & 1 \ 3 & 2 \ -1 & 1\end{array}ight)\left(\begin{array}{ccc}1 & 0 & 1 \ -1 & 2 & 1\end{array}ight)$

$$

\begin{aligned}

& \Rightarrow\left(\begin{array}{ccc}

2(1)+1(-1) & 2(0)+1(2) & 2(1)+1(1) \

3(1)+2(-1) & 3(0)+2(2) & 3(1)+2(1) \

-1(1)+1(-1) & -1(0)+1(2) & -1(1)+1(1)

\end{array}ight) \

& \Rightarrow\left(\begin{array}{ccc}

1 & 2 & 3 \

1 & 4 & 5 \

-2 & 2 & 0

\end{array}ight)

\end{aligned}

$$

(vi)

$$

\begin{aligned}

& \left(\begin{array}{ccc}

3 & -1 & 3 \

-1 & 0 & 2

\end{array}ight)\left(\begin{array}{cc}

2 & -3 \

1 & 0 \

3 & 1

\end{array}ight) \

& \quad \Rightarrow\left(\begin{array}{cc}

3(2)-1(1)+3(3) & 3(-3)-1(0)+3(1) \

-1(2)+0(1)+2(3) & -1(-3)+0(0)+2(1)

\end{array}ight) \

& \quad \Rightarrow\left(\begin{array}{cc}

14 & -6 \

4 & 5

\end{array}ight)

\end{aligned}

$$

Accepted Answer

(No solution provided.)

Question 4

If $A=\left(\begin{array}{ccc}1 & 2 & -3 \ 5 & 0 & 2 \ 1 & -1 & 1\end{array}ight), B=\left(\begin{array}{ccc}3 & -1 & 2 \ 4 & 2 & 5 \ 2 & 0 & 3\end{array}ight), C=\left(\begin{array}{ccc}4 & 1 & 2 \ 0 & 3 & 2 \ 1 & -2 & 3\end{array}ight)$, then compute $(A+B)$ and $(B-C)$. Also, verify that $A+(B-C)=(A+B)-C$.

Solution:

$$

\begin{aligned}

(A+B) & =\left(\begin{array}{ccc}

1 & 2 & -3 \

5 & 0 & 2 \

1 & -1 & 1

\end{array}ight)+\left(\begin{array}{ccc}

3 & -1 & 2 \

4 & 2 & 5 \

2 & 0 & 3

\end{array}ight) \

& =\left(\begin{array}{ccc}

4 & 1 & -1 \

9 & 2 & 7 \

3 & -1 & 4

\end{array}ight) \

(B-C) & =\left(\begin{array}{ccc}

3 & -1 & 2 \

4 & 2 & 5 \

2 & 0 & 3

\end{array}ight)-\left(\begin{array}{ccc}

4 & 1 & 2 \

0 & 3 & 2 \

1 & -2 & 3

\end{array}ight) \

& =\left(\begin{array}{ccc}

-1 & -2 & 0 \

4 & -1 & 3 \

1 & 2 & 0

\end{array}ight)

\end{aligned}

$$

Now,

$$

\begin{aligned}

A+(B-C) & =\left(\begin{array}{ccc}

1 & 2 & -3 \

5 & 0 & 2 \

1 & -1 & 1

\end{array}ight)+\left(\begin{array}{ccc}

-1 & -2 & 0 \

4 & -1 & 3 \

1 & 2 & 0

\end{array}ight) \

& =\left(\begin{array}{ccc}

0 & 0 & -3 \

9 & -1 & 5 \

2 & 1 & 1

\end{array}ight) \

(A+B)-C & =\left(\begin{array}{ccc}

4 & 1 & -1 \

9 & 2 & 7 \

3 & -1 & 4

\end{array}ight)-\left(\begin{array}{ccc}

4 & 1 & 2 \

0 & 3 & 2 \

1 & -2 & 3

\end{array}ight) \

& =\left(\begin{array}{ccc}

0 & 0 & -3 \

9 & -1 & 5 \

2 & 1 & 1

\end{array}ight)

\end{aligned}

$$

Hence, $A+(B-C)=(A+B)-C$.

Accepted Answer

(No solution provided.)

Question 5

$A=\left(\begin{array}{ccc}\frac{2}{3} & 1 & \frac{5}{3} \ \frac{1}{3} & \frac{2}{3} & \frac{4}{3} \ \frac{7}{3} & 2 & \frac{2}{3}\end{array}ight)$ and $B=\left(\begin{array}{ccc}\frac{2}{5} & \frac{3}{5} & 1 \ \frac{1}{5} & \frac{2}{5} & \frac{4}{5} \ \frac{7}{5} & \frac{6}{5} & \frac{2}{5}\end{array}ight)$, then compute $3 A-5 B$.

Accepted Answer

$$

\begin{aligned}

3 A-5 B & =3\left(\begin{array}{ccc}

\frac{2}{3} & 1 & \frac{5}{3} \

\frac{1}{3} & \frac{2}{3} & \frac{4}{3} \

\frac{7}{3} & 2 & \frac{2}{3}

\end{array}ight)-5\left(\begin{array}{ccc}

\frac{2}{5} & \frac{3}{5} & 1 \

\frac{1}{5} & \frac{2}{5} & \frac{4}{5} \

\frac{7}{5} & \frac{6}{5} & \frac{2}{5}

\end{array}ight) \

& =\left(\begin{array}{lll}

2 & 3 & 5 \

1 & 2 & 4 \

7 & 6 & 2

\end{array}ight)-\left(\begin{array}{ccc}

2 & 3 & 5 \

1 & 2 & 4 \

7 & 6 & 2

\end{array}ight)=\left(\begin{array}{lll}

0 & 0 & 0 \

0 & 0 & 0

\end{array}ight)

\end{aligned}

$$

Question 6

Simplify $\cos 	heta\left(\begin{array}{cc}\cos 	heta & \sin 	heta \ -\sin 	heta & \cos 	heta\end{array}ight)+\sin 	heta\left(\begin{array}{cc}\sin 	heta & -\cos 	heta \ \cos 	heta & \sin 	heta\end{array}ight)$.

Accepted Answer

$$

\begin{aligned}

\cos 	heta\left(\begin{array}{cc}

\cos 	heta & \sin 	heta \

-\sin 	heta & \cos 	heta

\end{array}ight)+\sin 	heta\left(\begin{array}{cc}

\sin 	heta & -\cos 	heta \

\cos 	heta & \sin 	heta

\end{array}ight) & \

& \Rightarrow\left(\begin{array}{cc}

\cos ^{2} 	heta & \cos 	heta \sin 	heta \

-\sin 	heta \cos 	heta & \cos ^{2} 	heta

\end{array}ight)+\left(\begin{array}{cc}

\sin ^{2} 	heta & -\sin 	heta \cos 	heta \

\sin 	heta \cos 	heta & \sin ^{2} 	heta

\end{array}ight) \

& \Rightarrow\left(\begin{array}{cc}

\cos ^{2} 	heta+\sin ^{2} 	heta & \sin 	heta \cos 	heta-\sin 	heta \cos 	heta \

-\sin 	heta \cos 	heta+\sin 	heta \cos 	heta & \cos ^{2} 	heta+\sin ^{2} 	heta

\end{array}ight) \

& \Rightarrow\left(\begin{array}{ll}

1 & 0 \

0 & 1

\end{array}ight)

\end{aligned}

$$

Question 7

Find $X$ and $Y$, if

(i) $\quad X+Y=\left(\begin{array}{ll}7 & 0 \ 2 & 5\end{array}ight)$ and $X-Y=\left(\begin{array}{ll}3 & 0 \ 0 & 3\end{array}ight)$

(ii) $\quad 2 X+3 Y=\left(\begin{array}{ll}2 & 3 \ 4 & 0\end{array}ight)$ and $3 X+2 Y=\left(\begin{array}{cc}2 & -2 \ -1 & 5\end{array}ight)$

Solution:

(i) $\quad X+Y=\left(\begin{array}{ll}7 & 0 \ 2 & 5\end{array}ight)$

$$

X-Y=\left(\begin{array}{ll}3 & 0  	ag{1}\ 0 & 3\end{array}ight)

$$

Adding equations (1) and (2),

$$

\begin{aligned}

2 X & =\left(\begin{array}{ll}

7 & 0 \

2 & 5

\end{array}ight)+\left(\begin{array}{ll}

3 & 0 \

0 & 3

\end{array}ight) \

& =\left(\begin{array}{cc}

10 & 0 \

2 & 8

\end{array}ight) \

X & =\frac{1}{2}\left(\begin{array}{cc}

10 & 0 \

2 & 8

\end{array}ight) \

& =\left(\begin{array}{ll}

5 & 0 \

1 & 4

\end{array}ight)

\end{aligned}

$$

Now,

$$

\begin{aligned}

& \Rightarrow X+Y=\left(\begin{array}{ll}

7 & 0 \

2 & 5

\end{array}ight) \

& \Rightarrow Y=\left(\begin{array}{ll}

7 & 0 \

2 & 5

\end{array}ight)-\left(\begin{array}{ll}

5 & 0 \

1 & 4

\end{array}ight) \

& \Rightarrow Y=\left(\begin{array}{ll}

2 & 0 \

1 & 1

\end{array}ight)

\end{aligned}

$$

(ii) $\quad 2 X+3 Y=\left(\begin{array}{ll}2 & 3 \ 4 & 0\end{array}ight)$

$$

3 X+2 Y=\left(\begin{array}{cc}

2 & -2  	ag{1}\

-1 & 5

\end{array}ight)

$$

Multiplying equation ( 1 ) by 2,

$$

\begin{align*}

& 2(2 X+3 Y)=2\left(\begin{array}{ll}

2 & 3 \

4 & 0

\end{array}ight) \

& 4 X+6 Y=\left(\begin{array}{ll}

4 & 6 \

8 & 0

\end{array}ight) 	ag{3}

\end{align*}

$$

Multiplying equation (2) by 3,

$$

\begin{align*}

& 3(3 X+2 Y)=3\left(\begin{array}{cc}

2 & -2 \

-1 & 5

\end{array}ight) \

& 9 X+6 Y=\left(\begin{array}{cc}

6 & -6 \

-3 & 15

\end{array}ight) 	ag{4}

\end{align*}

$$

From (3) and (4),

$$

\begin{aligned}

(4 X & +6 Y)-(9 X+6 Y)=\left(\begin{array}{ll}

4 & 6 \

8 & 0

\end{array}ight)-\left(\begin{array}{cc}

6 & -6 \

-3 & 15

\end{array}ight) \

-5 X & =\left(\begin{array}{cc}

4-6 & 6+6 \

8+3 & 0-15

\end{array}ight) \

-5 X & =\left(\begin{array}{cc}

-2 & 12 \

11 & -15

\end{array}ight) \

X & =\frac{-1}{5}\left(\begin{array}{cc}

-2 & 12 \

11 & -15

\end{array}ight) \

& =\left(\begin{array}{cc}

\frac{2}{5} & \frac{-12}{5} \

\frac{-11}{5} & 3

\end{array}ight)

\end{aligned}

$$

Now

$$

\begin{aligned}

& \Rightarrow 2 X+3 Y=\left(\begin{array}{ll}

2 & 3 \

4 & 0

\end{array}ight) \

& \Rightarrow 2\left(\begin{array}{cc}

\frac{2}{5} & \frac{-12}{5} \

\frac{-11}{5} & 3

\end{array}ight)+3 Y=\left(\begin{array}{ll}

2 & 3 \

4 & 0

\end{array}ight) \

& \Rightarrow\left(\begin{array}{cc}

\frac{4}{5} & \frac{-24}{5} \

\frac{-22}{5} & 6

\end{array}ight)+3 Y=\left(\begin{array}{ll}

2 & 3 \

4 & 0

\end{array}ight) \

& \Rightarrow 3 Y=\left(\begin{array}{ll}

2 & 3 \

4 & 0

\end{array}ight)-\left(\begin{array}{cc}

\frac{4}{5} & \frac{-24}{5} \

\frac{-22}{5} & 6

\end{array}ight) \

& \Rightarrow 3 Y=\left(\begin{array}{cc}

\frac{6}{5} & \frac{39}{5} \

\frac{42}{5} & -6

\end{array}ight) \

& \Rightarrow Y=\frac{1}{3}\left(\begin{array}{cc}

\frac{6}{5} & \frac{39}{5} \

\frac{42}{5} & -6

\end{array}ight) \

& \Rightarrow Y=\left(\begin{array}{cc}

\frac{2}{5} & \frac{13}{5} \

\frac{14}{5} & -2

\end{array}ight)

\end{aligned}

$$

Accepted Answer

(No solution provided.)

Question 8

Find $X$, if $Y=\left(\begin{array}{ll}3 & 2 \ 1 & 4\end{array}ight)$ and $2 X+Y=\left(\begin{array}{cc}1 & 0 \ -3 & 2\end{array}ight)$.

Accepted Answer

$2 X+Y=\left(\begin{array}{cc}1 & 0 \ -3 & 2\end{array}ight)$

$\Rightarrow 2 X+\left(\begin{array}{ll}3 & 2 \ 1 & 4\end{array}ight)=\left(\begin{array}{cc}1 & 0 \ -3 & 2\end{array}ight)$

$\Rightarrow 2 X=\left(\begin{array}{cc}1 & 0 \ -3 & 2\end{array}ight)-\left(\begin{array}{ll}3 & 2 \ 1 & 4\end{array}ight)$

$\Rightarrow 2 X=\left(\begin{array}{ll}-2 & -2 \ -4 & -2\end{array}ight)$

$\Rightarrow X=\frac{1}{2}\left(\begin{array}{ll}-2 & -2 \ -4 & -2\end{array}ight)$

$\Rightarrow X=\left(\begin{array}{ll}-1 & -1 \ -2 & -1\end{array}ight)$

Question 9

Find $x$ and $y$, if ${ }^{2}\left(\begin{array}{ll}1 & 3 \ 0 & x\end{array}ight)+\left(\begin{array}{ll}y & 0 \ 1 & 2\end{array}ight)=\left(\begin{array}{ll}5 & 6 \ 1 & 8\end{array}ight)$.

Solution:

$\Rightarrow 2\left(\begin{array}{ll}1 & 3 \ 0 & x\end{array}ight)+\left(\begin{array}{ll}y & 0 \ 1 & 2\end{array}ight)=\left(\begin{array}{ll}5 & 6 \ 1 & 8\end{array}ight)$

$\Rightarrow\left(\begin{array}{cc}2 & 6 \ 0 & 2 x\end{array}ight)+\left(\begin{array}{ll}y & 0 \ 1 & 2\end{array}ight)=\left(\begin{array}{ll}5 & 6 \ 1 & 8\end{array}ight)$

$\Rightarrow\left(\begin{array}{cc}2+y & 6 \ 1 & 2 x+2\end{array}ight)=\left(\begin{array}{ll}5 & 6 \ 1 & 8\end{array}ight)$

Comparing the corresponding elements of these two matrices,

$$

\begin{aligned}

& 2+y=5 \

& \Rightarrow y=3 \

& 2 x+2=8 \

& \Rightarrow x=3

\end{aligned}

$$

Therefore, $x=3$ and $y=3$.

Accepted Answer

(No solution provided.)

Question 10

Solve the equation for $x, y, z$ and $t$ if $2\left(\begin{array}{cc}x & z \ y & t\end{array}ight)+3\left(\begin{array}{cc}1 & -1 \ 0 & 2\end{array}ight)=3\left(\begin{array}{cc}3 & 5 \ 4 & 6\end{array}ight)$.

Solution:

$\Rightarrow 2\left(\begin{array}{ll}x & z \ y & t\end{array}ight)+3\left(\begin{array}{cc}1 & -1 \ 0 & 2\end{array}ight)=3\left(\begin{array}{ll}3 & 5 \ 4 & 6\end{array}ight)$

$\Rightarrow\left(\begin{array}{ll}2 x & 2 z \ 2 y & 2 t\end{array}ight)+\left(\begin{array}{cc}3 & -3 \ 0 & 6\end{array}ight)=\left(\begin{array}{cc}9 & 15 \ 12 & 18\end{array}ight)$

$\Rightarrow\left(\begin{array}{cc}2 x+3 & 2 z-3 \ 2 y & 2 t+6\end{array}ight)=\left(\begin{array}{cc}9 & 15 \ 12 & 18\end{array}ight)$

Comparing the corresponding elements of these two matrices,

$$

\begin{aligned}

& 2 x+3=9 \

& \Rightarrow 2 x=6 \

& \Rightarrow x=3 \

& 2 y=12 \

& \Rightarrow y=6 \

& 2 z-3=15 \

& \Rightarrow 2 z=18 \

& \Rightarrow z=9 \

& 2 t+6=18 \

& \Rightarrow 2 t=12 \

& \Rightarrow t=6

\end{aligned}

$$

Therefore, $x=3, y=6, z=9$ and $t=6$.

Accepted Answer

(No solution provided.)

Question 11

If $x\binom{2}{3}+y\binom{-1}{1}=\binom{10}{5}$, find values of $x$ and $y$.

Accepted Answer

$\Rightarrow x\binom{2}{3}+y\binom{-1}{1}=\binom{10}{5}$

$\Rightarrow\binom{2 x}{3 x}+\binom{-y}{y}=\binom{10}{5}$

$\Rightarrow\binom{2 x-y}{3 x+y}=\binom{10}{5}$

Comparing the corresponding elements of these two matrices,

$$

\begin{align*}

& 2 x-y=10  	ag{1}\

& 3 x+y=5 	ag{2}

\end{align*}

$$

By adding these two equations, we get

$$

\begin{aligned}

& 5 x=15 \

& \Rightarrow x=3

\end{aligned}

$$

Now, putting this value in (2)

$$

\begin{aligned}

& \Rightarrow 3 x+y=5 \

& \Rightarrow y=5-3 x \

& \Rightarrow y=5-3(3) \

& \Rightarrow y=5-9 \

& \Rightarrow y=-4

\end{aligned}

$$

Therefore, $x=3$ and $y=4$.

Question 12

Given $3\left(\begin{array}{cc}x & y \ z & w\end{array}ight)=\left(\begin{array}{cc}x & 6 \ -1 & 2 w\end{array}ight)+\left(\begin{array}{cc}4 & x+y \ z+w & 3\end{array}ight)$, find values of $w, x, y$ and $z$.

Accepted Answer

$$

\begin{aligned}

& \Rightarrow 3\left(\begin{array}{ll}

x & y \

z & w

\end{array}ight)=\left(\begin{array}{cc}

x & 6 \

-1 & 2 w

\end{array}ight)+\left(\begin{array}{cc}

4 & x+y \

z+w & 3

\end{array}ight) \

& \Rightarrow\left(\begin{array}{ll}

3 x & 3 y \

3 z & 3 w

\end{array}ight)=\left(\begin{array}{cc}

x+4 & 6+x+y \

-1+z+w & 2 w+3

\end{array}ight)

\end{aligned}

$$

Comparing the corresponding elements of these two matrices,

$$

\begin{aligned}

& \Rightarrow 3 x=x+4 \

& \Rightarrow 2 x=4 \

& \Rightarrow x=2 \

& \Rightarrow 3 y=6+x+y \

& \Rightarrow 2 y=6+x \

& \Rightarrow 2 y=6+2 \

& \Rightarrow 2 y=8 \

& \Rightarrow y=4 \

& \Rightarrow 3 w=2 w+3 \

& \Rightarrow w=3 \

& \Rightarrow 3 z=-1+z+w \

& \Rightarrow 2 z=w-1 \

& \Rightarrow 2 z=3-1 \

& \Rightarrow 2 z=2 \

& \Rightarrow z=1

\end{aligned}

$$

Therefore, $x=2, y=4, z=1$ and $w=3$

Question 13

$\begin{aligned} F(x) & =\left(\begin{array}{ccc}\cos x & -\sin x & 0 \ \sin x & \cos x & 0 \ 0 & 0 & 1\end{array}ight) 	ext {, show that } F(x) F(y)=F(x+y) .\end{aligned}$

Accepted Answer

It is given that $F(x)=\left(\begin{array}{ccc}\cos x & -\sin x & 0 \ \sin x & \cos x & 0 \ 0 & 0 & 1\end{array}ight)$

Then, $\quad F(y)=\left(\begin{array}{ccc}\cos y & -\sin y & 0 \ \sin y & \cos y & 0 \ 0 & 0 & 1\end{array}ight)$

Now,

$F(x+y)=\left(\begin{array}{ccc}\cos (x+y) & -\sin (x+y) & 0 \ \sin (x+y) & \cos (x+y) & 0 \ 0 & 0 & 1\end{array}ight)$

$$

\begin{aligned}

F(x) F(y) & =\left(\begin{array}{ccc}

\cos x & -\sin x & 0 \

\sin x & \cos x & 0 \

0 & 0 & 1

\end{array}ight)\left(\begin{array}{ccc}

\cos y & -\sin y & 0 \

\sin y & \cos y & 0 \

0 & 0 & 1

\end{array}ight) \

& =\left(\begin{array}{ccc}

\cos x \cos y-\sin x \sin y+0 & -\cos x \sin y-\sin x \cos y+0 & 0 \

\sin x \cos y+\cos x \sin y+0 & -\sin x \sin y+\cos x \cos y+0 & 0 \

0 & 0 & 0

\end{array}ight) \

& =\left(\begin{array}{ccc}

\cos (x+y) & -\sin (x+y) & 0 \

\sin (x+y) & \cos (x+y) & 0 \

0 & 0 & 1

\end{array}ight) \

& =F(x+y)

\end{aligned}

$$

Therefore, $F(x) F(y)=F(x+y)$

Question 14

\section*{Show that}

(i) $\quad\left(\begin{array}{cc}5 & -1 \ 6 & 7\end{array}ight)\left(\begin{array}{ll}2 & 1 \ 3 & 4\end{array}ight) 
eq\left(\begin{array}{ll}2 & 1 \ 3 & 4\end{array}ight)\left(\begin{array}{cc}5 & -1 \ 6 & 7\end{array}ight)$

(ii) $\left(\begin{array}{lll}1 & 2 & 3 \ 0 & 1 & 0 \ 1 & 1 & 0\end{array}ight)\left(\begin{array}{ccc}-1 & 1 & 0 \ 0 & -1 & 1 \ 2 & 3 & 4\end{array}ight) 
eq\left(\begin{array}{ccc}-1 & 1 & 0 \ 0 & -1 & 1 \ 2 & 3 & 4\end{array}ight)\left(\begin{array}{lll}1 & 2 & 3 \ 0 & 1 & 0 \ 1 & 1 & 0\end{array}ight)$

Accepted Answer

(i) $\quad\left(\begin{array}{cc}5 & -1 \ 6 & 7\end{array}ight)\left(\begin{array}{ll}2 & 1 \ 3 & 4\end{array}ight) 
eq\left(\begin{array}{ll}2 & 1 \ 3 & 4\end{array}ight)\left(\begin{array}{cc}5 & -1 \ 6 & 7\end{array}ight)$

$$

\begin{aligned}

\left(\begin{array}{cc}

5 & -1 \

6 & 7

\end{array}ight)\left(\begin{array}{ll}

2 & 1 \

3 & 4

\end{array}ight) & =\left(\begin{array}{cc}

5(2)-1(3) & 5(1)-1(4) \

6(2)+7(3) & 6(1)+7(4)

\end{array}ight) \

& =\left(\begin{array}{cc}

10-3 & 5-4 \

12+21 & 6+28

\end{array}ight) \

& =\left(\begin{array}{cc}

7 & 1 \

33 & 34

\end{array}ight)

\end{aligned}

$$

\begin{aligned}

\left(\begin{array}{ll}

2 & 1 \

3 & 4

\end{array}ight)\left(\begin{array}{cc}

5 & -1 \

6 & 7

\end{array}ight) & =\left(\begin{array}{cc}

2(5)+1(6) & 2(-1)+1(7) \

3(5)+4(6) & 3(-1)+4(7)

\end{array}ight) \

& =\left(\begin{array}{cc}

10+6 & -2+7 \

15+24 & -3+28

\end{array}ight) \

& =\left(\begin{array}{cc}

16 & 5 \

39 & 25

\end{array}ight)

\end{aligned}

$$

Thus, $\left(\begin{array}{cc}5 & -1 \ 6 & 7\end{array}ight)\left(\begin{array}{ll}2 & 1 \ 3 & 4\end{array}ight) 
eq\left(\begin{array}{ll}2 & 1 \ 3 & 4\end{array}ight)\left(\begin{array}{cc}5 & -1 \ 6 & 7\end{array}ight)$

(ii) $\quad\left(\begin{array}{lll}1 & 2 & 3 \ 0 & 1 & 0 \ 1 & 1 & 0\end{array}ight)\left(\begin{array}{ccc}-1 & 1 & 0 \ 0 & -1 & 1 \ 2 & 3 & 4\end{array}ight) 
eq\left(\begin{array}{ccc}-1 & 1 & 0 \ 0 & -1 & 1 \ 2 & 3 & 4\end{array}ight)\left(\begin{array}{lll}1 & 2 & 3 \ 0 & 1 & 0 \ 1 & 1 & 0\end{array}ight)$

$$

\begin{aligned}

&\left(\begin{array}{lll}

1 & 2 & 3 \

0 & 1 & 0 \

1 & 1 & 0

\end{array}ight)\left(\begin{array}{ccc}

-1 & 1 & 0 \

0 & -1 & 1 \

2 & 3 & 4

\end{array}ight)=\left(\begin{array}{ccc}

1(-1)+2(0)+3(2) & 1(1)+2(-1)+3(3) & 1(0)+2(1)+3(4) \

0(-1)+1(0)+0(2) & 0(1)+1(-1)+0(3) & 0(0)+1(1)+0(4) \

1(-1)+1(0)+0(2) & 1(1)+1(-1)+0(3) & 1(0)+1(1)+0(4)

\end{array}ight) \

&=\left(\begin{array}{ccc}

5 & 8 & 14 \

0 & -1 & 1 \

-1 & 0 & 1

\end{array}ight) \

&\left(\begin{array}{ccc}

-1 & 1 & 0 \

0 & -1 & 1 \

2 & 3 & 4

\end{array}ight)\left(\begin{array}{lll}

1 & 2 & 3 \

0 & 1 & 0 \

1 & 1 & 0

\end{array}ight)=\left(\begin{array}{ccc}

-1(1)+1(0)+0(1) & -1(2)+1(1)+0(1) & -1(3)+1(0)+0(0) \

0(1)+(-1)(0)+1(1) & 0(2)+(-1)(1)+1(1) & 0(3)+-1(0)+1(0) \

2(1)+3(0)+4(1) & 2(2)+3(1)+4(1) & 2(3)+3(0)+4(0)

\end{array}ight) \

&=\left(\begin{array}{ccc}

-1 & -1 & -3 \

1 & 0 & 0 \

6 & 11 & 6

\end{array}ight) \

& 	ext { Thus, }\left(\begin{array}{lll}

1 & 2 & 3 \

0 & 1 & 0 \

1 & 1 & 0

\end{array}ight)\left(\begin{array}{ccc}

-1 & 1 & 0 \

0 & -1 & 1 \

2 & 3 & 4

\end{array}ight) 
eq\left(\begin{array}{ccc}

-1 & 1 & 0 \

0 & -1 & 1 \

2 & 3 & 4

\end{array}ight)\left(\begin{array}{lll}

1 & 2 & 3 \

0 & 1 & 0 \

1 & 1 & 0

\end{array}ight)

\end{aligned}

$$

Question 15

Find $A^{2}-5 A+6 I$, if $A=\left(\begin{array}{ccc}2 & 0 & 1 \ 2 & 1 & 3 \ 1 & -1 & 0\end{array}ight)$

Solution:

$$

\begin{aligned}

A^{2} & =A . A \

& =\left(\begin{array}{ccc}

2 & 0 & 1 \

2 & 1 & 3 \

1 & -1 & 0

\end{array}ight)\left(\begin{array}{ccc}

2 & 0 & 1 \

2 & 1 & 3 \

1 & -1 & 0

\end{array}ight) \

& =\left(\begin{array}{ccc}

2(2)+0(2)+1(1) & 2(0)+0(1)+1(-1) & 2(1)+0(3)+1(0) \

2(2)+1(2)+3(1) & 2(0)+1(1)+1(1) & 2(1)+1(3)+3(0) \

1(2)+(-1)(2)+0(1) & 1(0)+(-1)(1)+0(-1) & 1(1)+(-1)(3)+0(0)

\end{array}ight) \

& =\left(\begin{array}{ccc}

4+0+1 & 0+0-1 & 2+0+0 \

4+2+3 & 0+1-3 & 2+3+0 \

2-2+0 & 0-1+0 & 1-3+0

\end{array}ight) \

& =\left(\begin{array}{ccc}

5 & -1 & 2 \

9 & -2 & 5 \

0 & -1 & -2

\end{array}ight)

\end{aligned}

$$

Therefore,

$$

\begin{aligned}

A^{2}-5 A+6 I & =\left(\begin{array}{ccc}

5 & -1 & 2 \

9 & -2 & 5 \

0 & -1 & -2

\end{array}ight)-5\left(\begin{array}{ccc}

2 & 0 & 1 \

2 & 1 & 3 \

1 & -1 & 0

\end{array}ight)+6\left(\begin{array}{ccc}

1 & 0 & 0 \

0 & 1 & 0 \

0 & 0 & 1

\end{array}ight) \

& =\left(\begin{array}{ccc}

5 & -1 & 2 \

9 & -2 & 5 \

0 & -1 & -2

\end{array}ight)-\left(\begin{array}{ccc}

10 & 0 & 5 \

10 & 5 & 15 \

5 & -5 & 0

\end{array}ight)+\left(\begin{array}{ccc}

6 & 0 & 0 \

0 & 6 & 0 \

0 & 0 & 6

\end{array}ight) \

& =\left(\begin{array}{ccc}

5-10 & -1-0 & 2-5 \

9-10 & -2-5 & 5-15 \

0-5 & -1+5 & -2-0

\end{array}ight)+\left(\begin{array}{lll}

6 & 0 & 0 \

0 & 6 & 0 \

0 & 0 & 6

\end{array}ight) \

& =\left(\begin{array}{ccc}

-5 & -1 & -3 \

-1 & -7 & -10 \

-5 & 4 & -2

\end{array}ight)+\left(\begin{array}{ccc}

6 & 0 & 0 \

0 & 6 & 0 \

0 & 0 & 6

\end{array}ight) \

& =\left(\begin{array}{ccc}

1 & -1 & -3 \

-1 & -1 & -10 \

-5 & 4 & 4

\end{array}ight)

\end{aligned}

$$

Accepted Answer

(No solution provided.)

Question 16

If $A=\left(\begin{array}{lll}1 & 0 & 2 \ 0 & 2 & 1 \ 2 & 0 & 3\end{array}ight)$, prove that $A^{3}-6 A^{2}+7 A+2 I=0$.

Solution:

$$

\begin{aligned}

A^{2} & =A \cdot A \

& =\left(\begin{array}{lll}

1 & 0 & 2 \

0 & 2 & 1 \

2 & 0 & 3

\end{array}ight)\left(\begin{array}{lll}

1 & 0 & 2 \

0 & 2 & 1 \

2 & 0 & 3

\end{array}ight) \

& =\left(\begin{array}{ccc}

1+0+4 & 0+0+0 & 2+0+6 \

0+0+2 & 0+4+0 & 0+2+3 \

2+0+6 & 0+0+0 & 4+0+9

\end{array}ight) \

& =\left(\begin{array}{ccc}

5 & 0 & 8 \

2 & 4 & 5 \

8 & 0 & 13

\end{array}ight)

\end{aligned}

$$

Now,

$$

\begin{aligned}

A^{3} & =A^{2} \cdot A \

& =\left(\begin{array}{ccc}

5 & 0 & 8 \

2 & 4 & 5 \

8 & 0 & 13

\end{array}ight)\left(\begin{array}{ccc}

1 & 0 & 2 \

0 & 2 & 1 \

2 & 0 & 3

\end{array}ight) \

& =\left(\begin{array}{ccc}

5+0+16 & 0+0+0 & 10+0+24 \

2+0+10 & 0+8+0 & 4+4+15 \

8+0+26 & 0+0+0 & 16+0+39

\end{array}ight) \

& =\left(\begin{array}{lll}

21 & 0 & 34 \

12 & 8 & 23 \

34 & 0 & 55

\end{array}ight)

\end{aligned}

$$

Therefore,

$$

\begin{aligned}

A^{3}-6 A^{2}+7 A+2 I & =\left(\begin{array}{lll}

21 & 0 & 34 \

12 & 8 & 23 \

34 & 0 & 55

\end{array}ight)-6\left(\begin{array}{ccc}

5 & 0 & 8 \

2 & 4 & 5 \

8 & 0 & 13

\end{array}ight)+7\left(\begin{array}{ccc}

1 & 0 & 2 \

0 & 2 & 1 \

2 & 0 & 3

\end{array}ight)+2\left(\begin{array}{ccc}

1 & 0 & 0 \

0 & 1 & 0 \

0 & 0 & 1

\end{array}ight) \

& =\left(\begin{array}{lll}

21 & 0 & 34 \

12 & 8 & 23 \

34 & 0 & 55

\end{array}ight)-\left(\begin{array}{ccc}

30 & 0 & 48 \

12 & 24 & 30 \

48 & 0 & 78

\end{array}ight)+\left(\begin{array}{ccc}

7 & 0 & 14 \

0 & 14 & 7 \

14 & 0 & 21

\end{array}ight)+\left(\begin{array}{lll}

2 & 0 & 0 \

0 & 2 & 0 \

0 & 0 & 2

\end{array}ight) \

& =\left(\begin{array}{ccc}

21+7+2 & 0+0+0 & 34+14+0 \

12+0+0 & 8+14+2 & 23+7+0 \

34+14+0 & 0+0+0 & 55+21+2

\end{array}ight)-\left(\begin{array}{ccc}

30 & 0 & 48 \

12 & 24 & 30 \

48 & 0 & 78

\end{array}ight)

\end{aligned}

$$

\begin{aligned}

& =\left(\begin{array}{ccc}

30 & 0 & 48 \

12 & 24 & 30 \

48 & 0 & 78

\end{array}ight)-\left(\begin{array}{ccc}

30 & 0 & 48 \

12 & 24 & 30 \

48 & 0 & 78

\end{array}ight) \

& =\left(\begin{array}{lll}

0 & 0 & 0 \

0 & 0 & 0

\end{array}ight)=0

\end{aligned}

$$

Hence, $A^{3}-6 A^{2}+7 A+2 I=0$.

Accepted Answer

(No solution provided.)

Question 17

If $A=\left(\begin{array}{ll}3 & -2 \ 4 & -2\end{array}ight)$ and $I=\left(\begin{array}{ll}1 & 0 \ 0 & 1\end{array}ight)$, find $k$ so that $A^{2}=k A-2 I$.

Solution:

$$

\begin{aligned}

A^{2} & =A \cdot A \

& =\left(\begin{array}{ll}

3 & -2 \

4 & -2

\end{array}ight)\left(\begin{array}{ll}

3 & -2 \

4 & -2

\end{array}ight) \

& =\left(\begin{array}{ll}

3(3)+(-2)(4) & 3(-2)+(-2)(-2) \

4(3)+(-2)(4) & 4(-2)+(-2)(-2)

\end{array}ight) \

& =\left(\begin{array}{ll}

1 & -2 \

4 & -4

\end{array}ight)

\end{aligned}

$$

Now,

$$

\begin{aligned}

& \Rightarrow A^{2}=k A-2 I \

& \Rightarrow\left(\begin{array}{ll}

1 & -2 \

4 & -4

\end{array}ight)=k\left(\begin{array}{ll}

3 & -2 \

4 & -2

\end{array}ight)-2\left(\begin{array}{ll}

1 & 0 \

0 & 1

\end{array}ight) \

& \Rightarrow\left(\begin{array}{ll}

1 & -2 \

4 & -4

\end{array}ight)=\left(\begin{array}{ll}

3 k & -2 k \

4 k & -2 k

\end{array}ight)-\left(\begin{array}{ll}

2 & 0 \

0 & 2

\end{array}ight) \

& \Rightarrow\left(\begin{array}{ll}

1 & -2 \

4 & -4

\end{array}ight)=\left(\begin{array}{cc}

3 k-2 & -2 k \

4 k & -2 k-2

\end{array}ight)

\end{aligned}

$$

Comparing the corresponding elements, we have:

$$

\begin{aligned}

& 3 k-2=1 \

& \Rightarrow 3 k=3 \

& \Rightarrow k=1

\end{aligned}

$$

Therefore, the value of $k=1$.

Accepted Answer

(No solution provided.)

Question 18

If $A=\left(\begin{array}{cc}0 & -	an \frac{\alpha}{2} \ 	an \frac{\alpha}{2} & 0\end{array}ight)$ and $I$ is the identity matrix of order 2, show that $I+A=(I-A)\left(\begin{array}{cc}\cos \alpha & -\sin \alpha \ \sin \alpha & \cos \alpha\end{array}ight)$

Accepted Answer

LHS $=I+A$

$$

\begin{align*}

& =\left(\begin{array}{ll}

1 & 0 \

0 & 1

\end{array}ight)+\left(\begin{array}{cc}

0 & -	an \frac{\alpha}{2} \

an \frac{\alpha}{2} & 0

\end{array}ight) \

& =\left(\begin{array}{cc}

1 & -	an \frac{\alpha}{2} \

an \frac{\alpha}{2} & 1

\end{array}ight) 	ag{1}

\end{align*}

$$

\begin{align*}

R H S & =(I-A)\left(\begin{array}{cc}

\cos \alpha & -\sin \alpha \

\sin \alpha & \cos \alpha

\end{array}ight) \

& =\left(\left(\begin{array}{ll}

1 & 0 \

0 & 1

\end{array}ight)-\left(\begin{array}{cc}

0 & -	an \frac{\alpha}{2} \

an \frac{\alpha}{2} & 0

\end{array}ight)ight)\left(\begin{array}{cc}

\cos \alpha & -\sin \alpha \

\sin \alpha & \cos \alpha

\end{array}ight) \

& =\left(\begin{array}{cc}

1 & 	an \frac{\alpha}{2} \

-	an \frac{\alpha}{2} & 1

\end{array}ight)\left(\begin{array}{cc}

\cos \alpha & -\sin \alpha \

\sin \alpha & \cos \alpha

\end{array}ight) \

& =\left(\begin{array}{cc}

\cos \alpha+\sin \alpha 	an \frac{\alpha}{2} & -\sin \alpha+\cos \alpha 	an \frac{\alpha}{2} \

-\cos \alpha 	an \frac{\alpha}{2}+\sin \alpha & \sin \alpha 	an \frac{\alpha}{2}+\cos \alpha

\end{array}ight) \

& =\left(\begin{array}{cc}

1-2 \sin ^{2} \frac{\alpha}{2}+2 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2} 	an \frac{\alpha}{2} & -2 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2}+\left(2 \cos ^{2} \frac{\alpha}{2}-1ight) 	an \frac{\alpha}{2} \

-\left(2 \cos ^{2} \frac{\alpha}{2}-1ight) 	an \frac{\alpha}{2}+2 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2} & 2 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2} 	an \frac{\alpha}{2}+1-2 \sin ^{2} \frac{\alpha}{2}

\end{array}ight) \

& =\left(\begin{array}{cc}

1-2 \sin ^{2} \frac{\alpha}{2}+2 \sin ^{2} \frac{\alpha}{2} & -2 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2}+2 \sin ^{2} \frac{\alpha}{2} \cos \frac{\alpha}{2}-	an \frac{\alpha}{2} \

-2 \sin ^{2} \frac{\alpha}{2} \cos \frac{\alpha}{2}+	an \frac{\alpha}{2}+2 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2} & 2 \sin ^{2} \frac{\alpha}{2}+1-2 \sin ^{2} \frac{\alpha}{2}

\end{array}ight) \

& =\left(\begin{array}{cc}

1 & -	an \frac{\alpha}{2} \

an \frac{\alpha}{2} & 1

\end{array}ight) \quad \ldots(2) 	ag{2}

\end{align*}

$$

Thus, from (1) and (2), we get

$I+A=(I-A)\left(\begin{array}{cc}\cos \alpha & -\sin \alpha \ \sin \alpha & \cos \alpha\end{array}ight)$

Question 19

A trust fund has ₹ 30000 that must be invested in two different types of bonds. The first bond pays $5 \%$ interest per year, and the second bond pays $7 \%$ interest per year. Using matrix multiplication, determine how to divide ₹ 30000 among the two types of bonds. If the trust fund must obtain an annual total interest of:

(i) ₹ 1800

(ii) ₹ 2000

(i) ₹ 1800

(ii) ₹ 2000

Accepted Answer

(i) Let $₹ x$ be invested in the first bond. Then, the sum of money invested in the second bond will be $₹(30000-x)$.

It is given that the first bond pays $5 \%$ interest per year and the second bond pays $7 \%$ interest per year.

Therefore, in order to obtain an annual total interest of ₹1800, we have:

$$

\begin{aligned}

& {\left[\begin{array}{ll}

x & (30000-x)

\end{array}ight]\left[\begin{array}{c}

\frac{5}{100} \

\frac{7}{100}

\end{array}ight]=1800 \quad\left[	ext { S.I for 1 year }=\frac{	ext { Principal } 	imes 	ext { Rate }}{100}ight]} \

& \Rightarrow \frac{5 x}{100}+\frac{7(30000-x)}{100}=1800 \

& \Rightarrow 5 x+210000-7 x=180000 \

& \Rightarrow 210000-2 x=180000 \

& \Rightarrow 2 x=210000-180000 \

& \Rightarrow 2 x=30000 \

& \Rightarrow x=15000

\end{aligned}

$$

Thus, in order to obtain an annual total interest of ₹1800, the trust fund should invest ₹ 15000 in the first bond and the remaining ₹ 15000 in the second bond.

(ii) Let $₹ x$ be invested in the first bond. Then, the sum of money invested in the second bond will be $₹(30000-x)$.

Therefore, in order to obtain an annual total interest of ₹2000, we have:

$$

\begin{aligned}

& {\left[\begin{array}{ll}

x & (30000-x)

\end{array}ight]\left[\begin{array}{c}

\frac{5}{100} \

\frac{7}{100}

\end{array}ight]=2000} \

& \Rightarrow \frac{5 x}{100}+\frac{7(30000-x)}{100}=2000 \

& \Rightarrow 5 x+210000-7 x=200000 \

& \Rightarrow 210000-2 x=200000 \

& \Rightarrow 2 x=210000-200000 \

& \Rightarrow 2 x=10000 \

& \Rightarrow x=5000

\end{aligned}

$$

Thus, in order to obtain an annual total interest of ₹1800, the trust fund should invest ₹5000 in the first bond and the remaining ₹25000 in the second bond.

Question 20

The bookshop of a particular school has 10 dozen chemistry books, 8 dozen physics books, 10 dozen economics books. Their selling prices are ₹ 80 , ₹ 60 and ₹ 40 each respectively. Find the total amount the bookshop will receive from selling all the books using matrix algebra.

Accepted Answer

The total amount of money that will be received from the sale of all these books can be represented in the form of a matrix as:

$$

\begin{aligned}

12\left[\begin{array}{lll}

10 & 8 & 10

\end{array}ight]\left[\begin{array}{l}

80 \

60 \

40

\end{array}ight] & =12[10(80)+8(60)+10(40)] \

& =12(800+480+400) \

& =12(1680) \

& =20160

\end{aligned}

$$

Thus, the bookshop will receive ₹ 20160 from the sale of all these books.

Question 21

Assume $X, Y, Z, W$ and $P$ are matrices of order $2 	imes n, 3 	imes k, 2 	imes p, n 	imes 3$ and $p 	imes k$ respectively. The restriction on $n, k$ and $p$ so that $P Y+W Y$ will be defined are:

(A) $k=3, p=n$

(B) $k$ is arbitrary, $p=2$

(C) $p$ is arbitrary, $k=3$

(D) $k=2, p=3$

Accepted Answer

Matrices $P$ and $Y$ are of the orders $p 	imes k$ and $3 	imes k$ respectively.

Therefore, matrix $P Y$ will be defined if $k=3$.

Consequently, $P Y$ will be of the order $p 	imes k$.

Matrices $W$ and $Y$ are of the orders $n 	imes 3$ and $3 	imes k$ respectively.

Since the number of columns in $W$ is equal to the number of rows in $Y$, matrix $W Y$ is welldefined and is of the order $n 	imes k$.

Matrices $P Y$ and $W Y$ can be added only when their orders are the same.

However, $P Y$ is of the order $p 	imes k$ and $W Y$ is of the order $n 	imes k$. Therefore, we must have $p=n$.

Thus, $k=3$ and $p=n$ are the restrictions on $n, k$ and $p$ so that $P Y+W Y$ will be defined.

The correct option is A .

Question 22

Assume $X, Y, Z, W$ and $P$ are matrices of order $2 	imes n, 3 	imes k, 2 	imes p, n 	imes 3$ and $p 	imes k$ respectively. If $n=p$, then the order of the matrix $7 X-5 Z$ is:

(A) $p 	imes 2$

(B) $2 	imes n$

(C) $n 	imes 3$

(D) $p 	imes n$

Accepted Answer

Matrix $X$ is of the order $2 	imes n$.

Therefore, matrix $7 X$ is also of the same order.

Matrix $Z$ is of the order $2 	imes p$, i.e., $2 	imes n$ [Since $n=p$ ]

Therefore, matrix $5 Z$ is also of the same order.

Now, both the matrices $7 X$ and 5 Z are of the order $2 	imes n$.

Thus, matrix $7 X-5 Z$ is well-defined and is of the order $2 	imes n$.

The correct option is B.

\section*{EXERCISE 3.3}