Solution

(i) $\quad A+B= \begin{bmatrix} 2 & 4 \\ 3 & 2\end{bmatrix} + \begin{bmatrix} 1 & 3 \\ -2 & 5\end{bmatrix} = \begin{bmatrix} 2+1 & 4+3 \\ 3-2 & 2+5\end{bmatrix} = \begin{bmatrix} 3 & 7 \\ 1 & 7 \end{bmatrix} $

(ii) $A-B= \begin{bmatrix} 2 & 4 \\ 3 & 2\end{bmatrix} - \begin{bmatrix} 1 & 3 \\ -2 & 5 \end{bmatrix} =$ (iii) $3 A-C=3 \begin{bmatrix} 2 & 4 \\ 3 & 2\end{bmatrix} - \begin{bmatrix} -2 & 5 \\ 3 & 4 \end{bmatrix} $

$= \begin{bmatrix} 3 \times 2 & 3 \times 4 \\ 3 \times 3 & 3 \times 2\end{bmatrix} - \begin{bmatrix} -2 & 5 \\ 3 & 4 \end{bmatrix} $

$= \begin{bmatrix} 6 & 12 \\ 9 & 6\end{bmatrix} - \begin{bmatrix} -2 & 5 \\ 3 & 4 \end{bmatrix} $

$= \begin{bmatrix} 6+2 & 12-5 \\ 9-3 & 6-4 \end{bmatrix} $

$= \begin{bmatrix} 8 & 7 \\ 6 & 2 \end{bmatrix} $

(iv) Matrix $A$ has 2 columns. This number is equal to the number of rows in matrix $B$.

Therefore, $A B$ is defined as:

$ \begin{aligned} A B & = \begin{bmatrix} 2 & 4 \\ 3 & 2 \end{bmatrix} \begin{bmatrix} 1 & 3 \\ -2 & 5 \end{bmatrix} = \begin{bmatrix} 2(1)+4(-2) & 2(3)+4(5) \\ 3(1)+2(-2) & 3(3)+2(5) \end{bmatrix} \\ & = \begin{bmatrix} 2-8 & 6+20 \\ 3-4 & 9+10 \end{bmatrix} = \begin{bmatrix} -6 & 26 \\ -1 & 19 \end{bmatrix} \end{aligned} $

(v) Matrix $B$ has 2 columns. This number is equal to the number of rows in matrix $A$. Therefore, $B A$ is defined as:

$ \begin{aligned} B A & = \begin{bmatrix} 1 & 3 \\ -2 & 5 \end{bmatrix} \begin{bmatrix} 2 & 4 \\ 3 & 2 \end{bmatrix} = \begin{bmatrix} 1(2)+3(3) & 1(4)+3(2) \\ -2(2)+5(3) & -2(4)+5(2) \end{bmatrix} \\ & = \begin{bmatrix} 2+9 & 4+6 \\ -4+15 & -8+10 \end{bmatrix} = \begin{bmatrix} 11 & 10 \\ 11 & 2 \end{bmatrix} \end{aligned} $

Solution

(i) $ \begin{bmatrix} a & b \\ -b & a\end{bmatrix} + \begin{bmatrix} a & b \\ b & a\end{bmatrix} = \begin{bmatrix} a+a & b+b \\ -b+b & a+a\end{bmatrix} = \begin{bmatrix} 2 a & 2 b \\ 0 & 2 a \end{bmatrix} $

(ii) $ \begin{bmatrix} a^{2}+b^{2} & b^{2}+c^{2} \\ a^{2}+c^{2} & a^{2}+b^{2}\end{bmatrix} + \begin{bmatrix} 2 a b & 2 b c \\ -2 a c & -2 a b \end{bmatrix} $

$ \begin{aligned} & = \begin{bmatrix} 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{bmatrix} \\ & = \begin{bmatrix} (a+b)^{2} & (b+c)^{2} \\ (a-c)^{2} & (a-b)^{2} \end{bmatrix} \end{aligned} $

(iii) $ \begin{bmatrix} -1 & 4 & -6 \\ 8 & 5 & 16 \\ 2 & 8 & 5\end{bmatrix} + \begin{bmatrix} 12 & 7 & 6 \\ 8 & 0 & 5 \\ 3 & 2 & 4 \end{bmatrix} $

$= \begin{bmatrix} -1+12 & 4+7 & -6+6 \\ 8+8 & 5+0 & 16+5 \\ 2+3 & 8+2 & 5+4 \end{bmatrix} $

$= \begin{bmatrix} 11 & 11 & 0 \\ 16 & 5 & 21 \\ 5 & 10 & 9 \end{bmatrix} $

(iv) $ \begin{bmatrix} \cos ^{2} x & \sin ^{2} x \\ \sin ^{2} x & \cos ^{2} x\end{bmatrix} + \begin{bmatrix} \sin ^{2} x & \cos ^{2} x \\ \cos ^{2} x & \sin ^{2} x \end{bmatrix} $

$= \begin{bmatrix} \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{bmatrix} $

$= \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} \quad(\because \sin ^{2} x+\cos ^{2} x=1)$

Solution

(i) $ \begin{bmatrix} a & b \\ -b & a\end{bmatrix} \begin{bmatrix} a & -b \\ b & a \end{bmatrix} $

$= \begin{bmatrix} a(a)+b(b) & a(-b)+b(a) \\ -b(a)+a(b) & -b(-b)+a(a) \end{bmatrix} $

$= \begin{bmatrix} a^{2}+b^{2} & -a b+a b \\ -a b+a b & b^{2}+a^{2}\end{bmatrix} = \begin{bmatrix} a^{2}+b^{2} & 0 \\ 0 & a^{2}+b^{2} \end{bmatrix} $

(ii) $ \begin{bmatrix} 1 \\ 2 \\ 3\end{bmatrix} \begin{bmatrix} 2 & 3 & 4\end{bmatrix} = \begin{bmatrix} 1(2) & 1(3) & 1(4) \\ 2(2) & 2(3) & 2(4) \\ 3(2) & 3(3) & 3(4)\end{bmatrix} = \begin{bmatrix} 2 & 3 & 4 \\ 4 & 6 & 8 \\ 6 & 9 & 12 \end{bmatrix} $

(iii) $ \begin{bmatrix} 1 & -2 \\ 2 & 3\end{bmatrix} \begin{bmatrix} 1 & 2 & 3 \\ 2 & 3 & 1 \end{bmatrix} $

$ \begin{aligned} & = \begin{bmatrix} 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{bmatrix} \\ & = \begin{bmatrix} 1-4 & 2-6 & 3-2 \\ 2+6 & 4+9 & 6+3 \end{bmatrix} = \begin{bmatrix} -3 & -4 & 1 \\ 8 & 13 & 9 \end{bmatrix} \end{aligned} $

(iv) $ \begin{bmatrix} 2 & 3 & 4 \\ 3 & 4 & 5 \\ 4 & 5 & 6\end{bmatrix} \begin{bmatrix} 1 & -3 & 5 \\ 0 & 2 & 4 \\ 3 & 0 & 5 \end{bmatrix} $

$= \begin{bmatrix} 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{bmatrix} $

$= \begin{bmatrix} 2+0+12 & -6+6+0 & 10+12+20 \\ 3+0+15 & -9+8+0 & 15+16+25 \\ 4+0+18 & -12+10+0 & 20+20+30\end{bmatrix} = \begin{bmatrix} 14 & 0 & 42 \\ 18 & -1 & 56 \\ 22 & -2 & 70 \end{bmatrix} $

(v) $ \begin{bmatrix} 2 & 1 \\ 3 & 2 \\ -1 & 1\end{bmatrix} \begin{bmatrix} 1 & 0 & 1 \\ -1 & 2 & 1 \end{bmatrix} $

$= \begin{bmatrix} 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{bmatrix} $

$= \begin{bmatrix} 2-1 & 0+2 & 2+1 \\ 3-2 & 0+4 & 3+2 \\ -1-1 & 0+2 & -1+1\end{bmatrix} = \begin{bmatrix} 1 & 2 & 3 \\ 1 & 4 & 5 \\ -2 & 2 & 0 \end{bmatrix} $

(vi) $ \begin{bmatrix} 3 & -1 & 3 \\ -1 & 0 & 2\end{bmatrix} \begin{bmatrix} 2 & -3 \\ 1 & 0 \\ 3 & 1 \end{bmatrix} $

$ \begin{aligned} & = \begin{bmatrix} 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{bmatrix} \\ & = \begin{bmatrix} 6-1+9 & -9-0+3 \\ -2+0+6 & 3+0+2 \end{bmatrix} = \begin{bmatrix} 14 & -6 \\ 4 & 5 \end{bmatrix} \end{aligned} $

Solution

$ \begin{aligned} A+B & = \begin{bmatrix} 1 & 2 & -3 \\ 5 & 0 & 2 \\ 1 & -1 & 1 \end{bmatrix} + \begin{bmatrix} 3 & -1 & 2 \\ 4 & 2 & 5 \\ 2 & 0 & 3 \end{bmatrix} \\ & = \begin{bmatrix} 1+3 & 2-1 & -3+2 \\ 5+4 & 0+2 & 2+5 \\ 1+2 & -1+0 & 1+3 \end{bmatrix} = \begin{bmatrix} 4 & 1 & -1 \\ 9 & 2 & 7 \\ 3 & -1 & 4 \end{bmatrix} \\ B-C & = \begin{bmatrix} 3 & -1 & 2 \\ 4 & 2 & 5 \\ 2 & 0 & 3 \end{bmatrix} - \begin{bmatrix} 4 & 1 & 2 \\ 0 & 3 & 2 \\ 1 & -2 & 3 \end{bmatrix} \\ & = \begin{bmatrix} 3-4 & -1-1 & 2-2 \\ 4-0 & 2-3 & 5-2 \\ 2-1 & 0-(-2) & 3-3 \end{bmatrix} = \begin{bmatrix} -1 & -2 & 0 \\ 4 & -1 & 3 \\ 1 & 2 & 0 \end{bmatrix} \end{aligned} $

$ \begin{aligned} A+(B-C) & = \begin{bmatrix} 1 & 2 & -3 \\ 5 & 0 & 2 \\ 1 & -1 & 1 \end{bmatrix} + \begin{bmatrix} -1 & -2 & 0 \\ 4 & -1 & 3 \\ 1 & 2 & 0 \end{bmatrix} \\ & = \begin{bmatrix} 1+(-1) & 2+(-2) & -3+0 \\ 5+4 & 0+(-1) & 2+3 \\ 1+1 & -1+2 & 1+0 \end{bmatrix} = \begin{bmatrix} 0 & 0 & -3 \\ 9 & -1 & 5 \\ 2 & 1 & 1 \end{bmatrix} \\ (A+B)-C & = \begin{bmatrix} 4 & 1 & -1 \\ 9 & 2 & 7 \\ 3 & -1 & 4 \end{bmatrix} - \begin{bmatrix} 4 & 1 & 2 \\ 0 & 3 & 2 \\ 1 & -2 & 3 \end{bmatrix} \\ & = \begin{bmatrix} 4-4 & 1-1 \\ 9-0 & 2-3 \\ 3-1 & -1-(-2) & 4-3 \end{bmatrix} \begin{bmatrix} 0 & 0 & -3 \\ 9 & -1 & 5 \\ 2 & 1 & 1 \end{bmatrix} \end{aligned} $

Hence, we have verified that $A+(B-C)=(A+B)-C$.

Solution

$ \begin{aligned} 3 A-5 B & =3 \begin{bmatrix} \frac{2}{3} & 1 & \frac{5}{3} \\ \frac{1}{3} & \frac{2}{3} & \frac{4}{3} \\ \frac{7}{3} & 2 & \frac{2}{3} \end{bmatrix} -5 \begin{bmatrix} \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{bmatrix} \\ & = \begin{bmatrix} 2 & 3 & 5 \\ 1 & 2 & 4 \\ 7 & 6 & 2 \end{bmatrix} - \begin{bmatrix} 2 & 3 & 5 \\ 1 & 2 & 4 \\ 7 & 6 & 2 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \end{aligned} $

Solution

$ \begin{aligned} & \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} \cos ^{2} \theta & \cos \theta \sin \theta \\ -\sin \theta \cos \theta & \cos ^{2} \theta \end{bmatrix} + \begin{bmatrix} \sin ^{2} \theta & -\sin \theta \cos \theta \\ \sin \theta \cos \theta & \sin ^{2} \theta \end{bmatrix} \\ & = \begin{bmatrix} \cos ^{2} \theta+\sin ^{2} \theta & \cos \theta \sin \theta-\sin \theta \cos \theta \\ -\sin \theta \cos \theta+\sin \theta \cos \theta & \cos ^{2} \theta+\sin ^{2} \theta \end{bmatrix} \\ & = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \quad(\because \cos ^{2} \theta+\sin ^{2} \theta=1) \end{aligned} $

Solution

$ X-Y= \begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix} $

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

$ \begin{aligned} & 2 X= \begin{bmatrix} 7 & 0 \\ 2 & 5 \end{bmatrix} + \begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix} = \begin{bmatrix} 7+3 & 0+0 \\ 2+0 & 5+3 \end{bmatrix} = \begin{bmatrix} 10 & 0 \\ 2 & 8 \end{bmatrix} \\ & \therefore X=\frac{1}{2} \begin{bmatrix} 10 & 0 \\ 2 & 8 \end{bmatrix} = \begin{bmatrix} 5 & 0 \\ 1 & 4 \end{bmatrix} \end{aligned} $

Now, $X+Y= \begin{bmatrix} 7 & 0 \\ 2 & 5 \end{bmatrix} $

$\Rightarrow \begin{bmatrix} 5 & 0 \\ 1 & 4\end{bmatrix} +Y= \begin{bmatrix} 7 & 0 \\ 2 & 5 \end{bmatrix} $

$\Rightarrow Y= \begin{bmatrix} 7 & 0 \\ 2 & 5\end{bmatrix} - \begin{bmatrix} 5 & 0 \\ 1 & 4 \end{bmatrix} $

$\Rightarrow Y= \begin{bmatrix} 7-5 & 0-0 \\ 2-1 & 5-4 \end{bmatrix} $

$\therefore Y= \begin{bmatrix} 2 & 0 \\ 1 & 1 \end{bmatrix} $

(ii) $2 X+3 Y= \begin{bmatrix} 2 & 3 \\ 4 & 0 \end{bmatrix} $

$ 3 X+2 Y= \begin{bmatrix} 2 & -2 \\ -1 & 5 \end{bmatrix} $

Multiplying equation (3) with (2), we get:

$$ \begin{align} & 2(2 X+3 Y)=2 \begin{bmatrix} 2 & 3 \\ 4 & 0 \\ \end{bmatrix} \\ & \Rightarrow 4 X+6 Y= \begin{bmatrix} 4 & 6 \\ 8 & 0 \\ \end{bmatrix} \end{align} $$

Multiplying equation (4) with (3), we get:

$$ \begin{align*} & 3(3 X+2 Y)=3 \begin{bmatrix} 2 & -2 \\ -1 & 5 \\ \end{bmatrix} \\ & \Rightarrow 9 X+6 Y= \begin{bmatrix} 6 & -6 \\ -3 & 15 \\ \end{bmatrix} \tag{6} \end{align*} $$

From (5) and (6), we have:

$ \begin{aligned} & (4 X+6 Y)-(9 X+6 Y)= \begin{bmatrix} 4 & 6 \\ 8 & 0 \end{bmatrix} - \begin{bmatrix} 6 & -6 \\ -3 & 15 \end{bmatrix} \\ & \Rightarrow-5 X= \begin{bmatrix} 4-6 & 6-(-6) \\ 8-(-3) & 0-15 \end{bmatrix} = \begin{bmatrix} -2 & 12 \\ 11 & -15 \end{bmatrix} \\ & \therefore X=-\frac{1}{5} \begin{bmatrix} -2 & 12 \\ 11 & -15 \end{bmatrix} = \begin{bmatrix} \frac{2}{5} & -\frac{12}{5} \\ -\frac{11}{5} & 3 \end{bmatrix} \end{aligned} $

Now, $2 X+3 Y= \begin{bmatrix} 2 & 3 \\ 4 & 0 \end{bmatrix} $ $\Rightarrow 2 \begin{bmatrix} \frac{2}{5} & -\frac{12}{5} \\ -\frac{11}{5} & 3\end{bmatrix} +3 Y= \begin{bmatrix} 2 & 3 \\ 4 & 0 \end{bmatrix} $

$\Rightarrow \begin{bmatrix} \frac{4}{5} & -\frac{24}{5} \\ -\frac{22}{5} & 6\end{bmatrix} +3 Y= \begin{bmatrix} 2 & 3 \\ 4 & 0 \end{bmatrix} $

$\Rightarrow 3 Y= \begin{bmatrix} 2 & 3 \\ 4 & 0\end{bmatrix} - \begin{bmatrix} \frac{4}{5} & -\frac{24}{5} \\ -\frac{22}{5} & 6 \end{bmatrix} $

$\Rightarrow 3 Y= \begin{bmatrix} 2-\frac{4}{5} & 3+\frac{24}{5} \\ 4+\frac{22}{5} & 0-6\end{bmatrix} = \begin{bmatrix} \frac{6}{5} & \frac{39}{5} \\ \frac{42}{5} & -6 \end{bmatrix} $

$\therefore Y=\frac{1}{3} \begin{bmatrix} \frac{6}{5} & \frac{39}{5} \\ \frac{42}{5} & -6\end{bmatrix} = \begin{bmatrix} \frac{2}{5} & \frac{13}{5} \\ \frac{14}{5} & -2 \end{bmatrix} $

Solution

$ \begin{aligned} & 2 X+Y= \begin{bmatrix} 1 & 0 \\ -3 & 2 \end{bmatrix} \\ & \Rightarrow 2 X+ \begin{bmatrix} 3 & 2 \\ 1 & 4 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ -3 & 2 \end{bmatrix} \\ & \Rightarrow 2 X= \begin{bmatrix} 1 & 0 \\ -3 & 2 \end{bmatrix} - \begin{bmatrix} 3 & 2 \\ 1 & 4 \end{bmatrix} = \begin{bmatrix} 1-3 & 0-2 \\ -3-1 & 2-4 \end{bmatrix} \\ & \Rightarrow 2 X= \begin{bmatrix} -2 & -2 \\ -4 & -2 \end{bmatrix} \\ & \therefore X=\frac{1}{2} \begin{bmatrix} -2 & -2 \\ -4 & -2 \end{bmatrix} = \begin{bmatrix} -1 & -1 \\ -2 & -1 \end{bmatrix} \end{aligned} $

Solution

$ \begin{aligned} & 2 \begin{bmatrix} 1 & 3 \\ 0 & x \end{bmatrix} + \begin{bmatrix} y & 0 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 5 & 6 \\ 1 & 8 \end{bmatrix} \\ & \Rightarrow \begin{bmatrix} 2 & 6 \\ 0 & 2 x \end{bmatrix} + \begin{bmatrix} y & 0 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 5 & 6 \\ 1 & 8 \end{bmatrix} \\ & \Rightarrow \begin{bmatrix} 2+y & 6 \\ 1 & 2 x+2 \end{bmatrix} = \begin{bmatrix} 5 & 6 \\ 1 & 8 \end{bmatrix} \end{aligned} $

Comparing the corresponding elements of these two matrices, we have:

$ \begin{aligned} & 2+y=5 \\ & \Rightarrow y=3 \\ & 2 x+2=8 \\ & \Rightarrow x=3 \\ & \therefore x=3 \text{ and } y=3 \end{aligned} $

Solution

$ \begin{aligned} & 2 \begin{bmatrix} x & z \\ y & t \end{bmatrix} +3 \begin{bmatrix} 1 & -1 \\ 0 & 2 \end{bmatrix} =3 \begin{bmatrix} 3 & 5 \\ 4 & 6 \end{bmatrix} \\ & \Rightarrow \begin{bmatrix} 2 x & 2 z \\ 2 y & 2 t \end{bmatrix} + \begin{bmatrix} 3 & -3 \\ 0 & 6 \end{bmatrix} = \begin{bmatrix} 9 & 15 \\ 12 & 18 \end{bmatrix} \\ & \Rightarrow \begin{bmatrix} 2 x+3 & 2 z-3 \\ 2 y & 2 t+6 \end{bmatrix} = \begin{bmatrix} 9 & 15 \\ 12 & 18 \end{bmatrix} \end{aligned} $

Comparing the corresponding elements of these two matrices, we get:

$ \begin{aligned} & 2 x+3=9 \\ & \Rightarrow 2 x=6 \\ & \Rightarrow x=3 \end{aligned} $

$2 y=12$

$\Rightarrow y=6$

$ \begin{aligned} & 2 z-3=15 \\ & \Rightarrow 2 z=18 \\ & \Rightarrow z=9 \end{aligned} $

$ \begin{aligned} & 2 t+6=18 \\ & \Rightarrow 2 t=12 \\ & \Rightarrow t=6 \end{aligned} $

$\therefore x=3, y=6, z=9$, and $t=6$

Solution

$ \begin{aligned} & x \begin{bmatrix} 2 \\ 3 \end{bmatrix} +y \begin{bmatrix} -1 \\ 1 \end{bmatrix} = \begin{bmatrix} 10 \\ 5 \end{bmatrix} \\ & \Rightarrow \begin{bmatrix} 2 x \\ 3 x \end{bmatrix} + \begin{bmatrix} -y \\ y \end{bmatrix} = \begin{bmatrix} 10 \\ 5 \end{bmatrix} \\ & \Rightarrow \begin{bmatrix} 2 x-y \\ 3 x+y \end{bmatrix} = \begin{bmatrix} 10 \\ 5 \end{bmatrix} \end{aligned} $

Comparing the corresponding elements of these two matrices, we get:

$2 x-y=10$ and $3 x+y=5$

Adding these two equations, we have:

$5 x=15$

$\Rightarrow x=3$

Now, $3 x+y=5$

$\Rightarrow y=5-3 x$

$\Rightarrow y=5-9=-4$

$\therefore x=3$ and $y=-4$

Solution

$ \begin{aligned} & 3 \begin{bmatrix} x & y \\ z & w \end{bmatrix} = \begin{bmatrix} x & 6 \\ -1 & 2 w \end{bmatrix} + \begin{bmatrix} 4 & x+y \\ z+w & 3 \end{bmatrix} \\ & \Rightarrow \begin{bmatrix} 3 x & 3 y \\ 3 z & 3 w \end{bmatrix} = \begin{bmatrix} x+4 & 6+x+y \\ -1+z+w & 2 w+3 \end{bmatrix} \end{aligned} $

Comparing the corresponding elements of these two matrices, we get:

$ \begin{aligned} & 3 x=x+4 \\ & \Rightarrow 2 x=4 \\ & \Rightarrow x=2 \end{aligned} $

$3 y=6+x+y$

$\Rightarrow 2 y=6+x=6+2=8$

$\Rightarrow y=4$

$3 w=2 w+3$

$\Rightarrow w=3$

$3 z=-1+z+w$

$\Rightarrow 2 z=-1+w=-1+3=2$

$\Rightarrow z=1$

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

Solution

$ \begin{aligned} & F(x)= \begin{bmatrix} \cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{bmatrix} , F(y)= \begin{bmatrix} \cos y & -\sin y & 0 \\ \sin y & \cos y & 0 \\ 0 & 0 & 1 \end{bmatrix} \\ & F(x+y)= \begin{bmatrix} \cos (x+y) & -\sin (x+y) & 0 \\ \sin (x+y) & \cos (x+y) & 0 \\ 0 & 0 & 1 \end{bmatrix} \\ & F(x) F(y) \\ & = \begin{bmatrix} \cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} \cos y & -\sin y & 0 \\ \sin y & \cos y & 0 \\ 0 & 0 & 1 \end{bmatrix} \\ & = \begin{bmatrix} \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{bmatrix} \\ & = \begin{bmatrix} \cos (x+y) & -\sin (x+y) & 0 \\ \sin (x+y) & \cos (x+y) & 0 \\ 0 & 0 & 1 \end{bmatrix} \\ & =F(x+y) \\ & \therefore F(x) F(y)=F(x+y) \end{aligned} $

Solution

(i) $ \begin{bmatrix} 5 & -1 \\ 6 & 7\end{bmatrix} \begin{bmatrix} 2 & 1 \\ 3 & 4 \end{bmatrix} $

$= \begin{bmatrix} 5(2)-1(3) & 5(1)-1(4) \\ 6(2)+7(3) & 6(1)+7(4) \end{bmatrix} $

$= \begin{bmatrix} 10-3 & 5-4 \\ 12+21 & 6+28\end{bmatrix} = \begin{bmatrix} 7 & 1 \\ 33 & 34 \end{bmatrix} $

$ \begin{bmatrix} 2 & 1 \\ 3 & 4\end{bmatrix} \begin{bmatrix} 5 & -1 \\ 6 & 7 \end{bmatrix} $

$= \begin{bmatrix} 2(5)+1(6) & 2(-1)+1(7) \\ 3(5)+4(6) & 3(-1)+4(7) \end{bmatrix} $

$= \begin{bmatrix} 10+6 & -2+7 \\ 15+24 & -3+28\end{bmatrix} = \begin{bmatrix} 16 & 5 \\ 39 & 25 \end{bmatrix} $

$\therefore \begin{bmatrix} 5 & -1 \\ 6 & 7\end{bmatrix} \begin{bmatrix} 2 & 1 \\ 3 & 4\end{bmatrix} \neq \begin{bmatrix} 2 & 1 \\ 3 & 4\end{bmatrix} \begin{bmatrix} 5 & -1 \\ 6 & 7 \end{bmatrix} $

(ii) $ \begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 0 \\ 1 & 1 & 0\end{bmatrix} \begin{bmatrix} -1 & 1 & 0 \\ 0 & -1 & 1 \\ 2 & 3 & 4 \end{bmatrix} $

$= \begin{bmatrix} 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{bmatrix} $

$= \begin{bmatrix} 5 & 8 & 14 \\ 0 & -1 & 1 \\ -1 & 0 & 1 \end{bmatrix} $

$ \begin{aligned} & { \begin{bmatrix} -1 & 1 & 0 \\ 0 & -1 & 1 \\ 2 & 3 & 4 \end{bmatrix} \begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \end{bmatrix} } \\ & = \begin{bmatrix} -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{bmatrix} \\ & = \begin{bmatrix} -1 & -1 & -3 \\ 1 & 0 & 0 \\ 6 & 11 & 6 \end{bmatrix} \\ & \therefore \begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \end{bmatrix} \begin{bmatrix} -1 & 1 & 0 \\ 0 & -1 & 1 \\ 2 & 3 & 4 \end{bmatrix} \neq \begin{bmatrix} -1 & 1 & 0 \\ 0 & -1 & 1 \\ 2 & 3 & 4 \end{bmatrix} \begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \end{bmatrix} \end{aligned} $

Solution

We have $A^{2}=A \times A$

$ \begin{aligned} & A^{2}=A A= \begin{bmatrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{bmatrix} \begin{bmatrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{bmatrix} \\ & = \begin{bmatrix} 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)+3(-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{bmatrix} \\ & = \begin{bmatrix} 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{bmatrix} \\ & = \begin{bmatrix} 5 & -1 & 2 \\ 9 & -2 & 5 \\ 0 & -1 & -2 \end{bmatrix} \\ & \therefore A^{2}-5 A+6 I \\ & = \begin{bmatrix} 5 & -1 & 2 \\ 9 & -2 & 5 \\ 0 & -1 & -2 \end{bmatrix} -5 \begin{bmatrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{bmatrix} +6 \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \\ & = \begin{bmatrix} 5 & -1 & 2 \\ 9 & -2 & 5 \\ 0 & -1 & -2 \end{bmatrix} - \begin{bmatrix} 10 & 0 & 5 \\ 10 & 5 & 15 \\ 5 & -5 & 0 \end{bmatrix} + \begin{bmatrix} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \end{bmatrix} \\ & = \begin{bmatrix} 5-10 & -1-0 & 2-5 \\ 9-10 & -2-5 & 5-15 \\ 0-5 & -1+5 & -2-0 \end{bmatrix} + \begin{bmatrix} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \end{bmatrix} \\ & = \begin{bmatrix} -5 & -1 & -3 \\ -1 & -7 & -10 \\ -5 & 4 & -2 \end{bmatrix} + \begin{bmatrix} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \end{bmatrix} \\ & = \begin{bmatrix} -5+6 & -1+0 & -3+0 \\ -1+0 & -7+6 & -10+0 \\ -5+0 & 4+0 & -2+6 \end{bmatrix} \\ & = \begin{bmatrix} 1 & -1 & -3 \\ -1 & -1 & -10 \\ -5 & 4 & 4 \end{bmatrix} \end{aligned} $

Solution

$ \begin{aligned} A^{2}=A A & = \begin{bmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{bmatrix} \begin{bmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{bmatrix} \\ & = \begin{bmatrix} 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{bmatrix} = \begin{bmatrix} 5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13 \end{bmatrix} \end{aligned} $

Now $A^{3}=A^{2} \cdot A$

$ \begin{aligned} & = \begin{bmatrix} 5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13 \end{bmatrix} \begin{bmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{bmatrix} \\ & = \begin{bmatrix} 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{bmatrix} \\ & = \begin{bmatrix} 21 & 0 & 34 \\ 12 & 8 & 23 \\ 34 & 0 & 55 \end{bmatrix} \\ & \therefore A^{3}-6 A^{2}+7 A+2 I \\ & = \begin{bmatrix} 21 & 0 & 34 \\ 12 & 8 & 23 \\ 34 & 0 & 55 \end{bmatrix} -6 \begin{bmatrix} 5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13 \end{bmatrix} +7 \begin{bmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{bmatrix} +2 \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \end{aligned} $

$ \begin{aligned} & = \begin{bmatrix} 21 & 0 & 34 \\ 12 & 8 & 23 \\ 34 & 0 & 55 \end{bmatrix} - \begin{bmatrix} 30 & 0 & 48 \\ 12 & 24 & 30 \\ 48 & 0 & 78 \end{bmatrix} + \begin{bmatrix} 7 & 0 & 14 \\ 0 & 14 & 7 \\ 14 & 0 & 21 \end{bmatrix} + \begin{bmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{bmatrix} \\ & = \begin{bmatrix} 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{bmatrix} - \begin{bmatrix} 30 & 0 & 48 \\ 12 & 24 & 30 \\ 48 & 0 & 78 \end{bmatrix} \\ & = \begin{bmatrix} 30 & 0 & 48 \\ 12 & 24 & 30 \\ 48 & 0 & 78 \end{bmatrix} - \begin{bmatrix} 30 & 0 & 48 \\ 12 & 24 & 30 \\ 48 & 0 & 78 \end{bmatrix} \\ & = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} =O \\ & \therefore A^{3}-6 A^{2}+7 A+2 I=O \end{aligned} $

Solution

$ \begin{aligned} A^{2}=A \cdot A & = \begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix} \begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix} \\ & = \begin{bmatrix} 3(3)+(-2)(4) & 3(-2)+(-2)(-2) \\ 4(3)+(-2)(4) & 4(-2)+(-2)(-2) \end{bmatrix} = \begin{bmatrix} 1 & -2 \\ 4 & -4 \end{bmatrix} \end{aligned} $

Now $A^{2}=k A-2 I$

$ \begin{aligned} & \Rightarrow \begin{bmatrix} 1 & -2 \\ 4 & -4 \end{bmatrix} =k \begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix} -2 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \\ & \Rightarrow \begin{bmatrix} 1 & -2 \\ 4 & -4 \end{bmatrix} = \begin{bmatrix} 3 k & -2 k \\ 4 k & -2 k \end{bmatrix} - \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix} \\ & \Rightarrow \begin{bmatrix} 1 & -2 \\ 4 & -4 \end{bmatrix} = \begin{bmatrix} 3 k-2 & -2 k \\ 4 k & -2 k-2 \end{bmatrix} \end{aligned} $

Comparing the corresponding elements, we have:

$3 k-2=1$

$\Rightarrow 3 k=3$

$\Rightarrow k=1$

Thus, the value of $k$ is 1 .

Solution

On the L.H.S.

$I+A=\begin{bmatrix}1 & 0 \\ 0 & 1 \end{bmatrix} + \begin{bmatrix} 0 & -\tan \frac{\alpha}{2} \\ \tan \frac{\alpha}{2} & 0 \end{bmatrix}$

$=\begin{bmatrix} 0 & -\tan \frac{\alpha}{2} \\ \tan \frac{\alpha}{2} & 0 \end{bmatrix}\quad …(1)$

On the R.H.S.

$ (I-A) \begin{bmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{bmatrix} $

$=\begin{bmatrix}\begin{bmatrix} 1 & 0 \\0 & 1\end{bmatrix} - \begin{bmatrix} 0 & -\tan \frac{\alpha}{2} \\ \tan \frac{\alpha}{2} & 0 \end{bmatrix} \end{bmatrix} \begin{bmatrix} \cos \alpha & -\sin \beta \\ \sin \alpha & \cos \alpha \end{bmatrix}$

$=\begin{bmatrix} 1 & -\tan \frac{\alpha}{2} \\ \tan \frac{\alpha}{2} & 1 \end{bmatrix} \begin{bmatrix} \cos \alpha & -\sin \beta \\ \sin \alpha & \cos \alpha \end{bmatrix}$

$=\left[\begin{array}{ll}\cos \alpha+\sin \alpha \tan \frac{\alpha}{2} & -\sin \alpha+\cos \alpha \tan \frac{\alpha}{2} \\ -\cos \alpha \tan \frac{\alpha}{2}+\sin \alpha & \sin \alpha \tan \frac{\alpha}{2}+\cos \alpha\end{array}\right]$

$\begin{aligned} & =\left[\begin{array}{lc}1-2 \sin ^2 \frac{\alpha}{2}+2 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2} \tan \frac{\alpha}{2} & -2 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2}+\left(2 \cos ^2 \frac{\alpha}{2}-1\right) \tan \frac{\alpha}{2} \\ -\left(2 \cos ^2 \frac{\alpha}{2}-1\right) \tan \frac{\alpha}{2}+2 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2} & 2 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2} \tan \frac{\alpha}{2}+1-2 \sin ^2 \frac{\alpha}{2}\end{array}\right] \\ & =\left[\begin{array}{lc}1-2 \sin ^2 \frac{\alpha}{2}+2 \sin ^2 \frac{\alpha}{2} & -2 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2}+2 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2}-\tan \frac{\alpha}{2} \\ -2 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2}+\tan \frac{\alpha}{2}+2 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2} & 2 \sin ^2 \frac{\alpha}{2}+1-2 \sin ^2 \frac{\alpha}{2}\end{array}\right] \\ & =\left[\begin{array}{cc}1 & -\tan \frac{\alpha}{2} \\ \tan \frac{\alpha}{2} & 1\end{array}\right]\end{aligned}$

Thus, from (1) and (2), we get L.H.S. = R.H.S.

Solution

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

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 Rs 1800, we have:

$ \begin{aligned} & { \begin{bmatrix} x & (30000-x) \end{bmatrix} \begin{bmatrix} \frac{5}{100} \\ \frac{7}{100} \end{bmatrix} =1800 \quad[\text{ S.I. for 1 year }=\frac{\text{ Principal } \times \text{ Rate }}{100}]} \\ & \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 Rs 1800 , the trust fund should invest Rs 15000 in the first bond and the remaining Rs 15000 in the second bond.

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

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

$ \begin{bmatrix} x & (30000-x) \end{bmatrix} \begin{bmatrix} \frac{5}{100} \\ \frac{7}{100} \end{bmatrix} =2000 $

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

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

$\Rightarrow 210000-2 x=200000$

$\Rightarrow 2 x=210000-200000$

$\Rightarrow 2 x=10000$

$\Rightarrow x=5000$

Thus, in order to obtain an annual total interest of Rs 2000 , the trust fund should invest Rs 5000 in the first bond and the remaining Rs 25000 in the second bond.

Solution

The bookshop has 10 dozen chemistry books, 8 dozen physics books, and 10 dozen economics books.

The selling prices of a chemistry book, a physics book, and an economics book are respectively given as Rs 80, Rs 60, and Rs 40.

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 \begin{bmatrix} 10 & 8 & 10 \end{bmatrix} \begin{bmatrix} 80 \\ 60 \\ 40 \end{bmatrix} \\ & =12[10 \times 80+8 \times 60+10 \times 40] \\ & =12(800+480+400) \\ & =12(1680) \\ & =20160 \end{aligned} $

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

Solution

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

Therefore, matrix $P Y$ will be defined if $k=3$. Consequently, $P Y$ will be of the order $p \times k$. Matrices $W$ and $Y$ are of the orders $n \times 3$ and $3 \times k$ respectively.

Since the number of columns in $W$ is equal to the number of rows in $Y$, matrix $W Y$ is well-defined and is of the order $n \times k$.

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

However, $P Y$ is of the order $p \times k$ and $W Y$ is of the order $n \times k$. 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.

Solution

The correct answer is $B$.

Matrix $X$ is of the order $2 \times n$.

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

Matrix $Z$ is of the order $2 \times p$, i.e., $2 \times n$ [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 $2 \times n$.

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