We have, $$A=\left[\begin{array}{cc} \sqrt{3} & 1 \\ 2 & 3 \end{array}\right]_{2 \times 2} \text { and } B=\left[\begin{array}{lll} x & y & z \\ a & b & 6 \end{array}\right]_{2 \times 3}$$

Here, $A$ and $B$ are of different orders. Also, we know that the addition of two matrices $A$ and $B$ is possible only if order of both the matrices $A$ and $B$ should be same. Hence, the sum of matrices $A$ and $B$ is not possible.

We have, $X=\left[\begin{array}{ccc}3 & 1 & -1 \\ 5 & -2 & -3\end{array}\right]_{2 \times 3}$ and $Y=\left[\begin{array}{ccc}2 & 1 & -1 \\ 7 & 2 & 4\end{array}\right]_{2 \times 3}$

(i) $X+Y=\left[\begin{array}{ccc}3+2 & 1+1 & -1-1 \\ 5+7 & -2+2 & -3+4\end{array}\right]=\left[\begin{array}{ccc}5 & 2 & -2 \\ 12 & 0 & 1\end{array}\right]$

$$\begin{aligned} \text{(ii) }\quad & \because 2 X=2\left[\begin{array}{lll} 3 & 1 & -1 \\ 5 & -2 & -3 \end{array}\right]=\left[\begin{array}{rrr} 6 & 2 & -2 \\ 10 & -4 & -6 \end{array}\right] \\ & \text { and } \\ & 3 Y=3\left[\begin{array}{ccc} 2 & 1 & -1 \\ 7 & 2 & 4 \end{array}\right]=\left[\begin{array}{ccc} 6 & 3 & -3 \\ 21 & 6 & 12 \end{array}\right] \\ & \therefore \quad 2 X-3 Y=\left[\begin{array}{ccc} 6-6 & 2-3 & -2+3 \\ 10-21 & -4-6 & -6-12 \end{array}\right]=\left[\begin{array}{ccc} 0 & -1 & 1 \\ -11 & -10 & -18 \end{array}\right] \end{aligned}$$

(iii) $X+Y=\left[\begin{array}{ccc}3+2 & 1+1 & -1-1 \\ 5+7 & -2+2 & -3+4\end{array}\right]=\left[\begin{array}{ccc}5 & 2 & -2 \\ 12 & 0+1\end{array}\right]$

Also, $$X+Y+Z=\left[\begin{array}{lll} 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right]$$

We see that $Z$ is the additive inverse of $(X+Y)$ or negative of $(X+Y)$.

$$\therefore \quad Z=\left[\begin{array}{ccc} -5 & -2 & 2 \\ -12 & 0 & -1 \end{array}\right] \quad [\because Z=-(X+Y)]$$

$$\begin{aligned} &\text { Given that, }\\ &\begin{array}{cc} & x\left[\begin{array}{cc} 2 x & 2 \\ 3 & x \end{array}\right]+2\left[\begin{array}{ll} 8 & 5 x \\ 4 & 4 x \end{array}\right]=2\left[\begin{array}{cc} \left(x^2+8\right) & 24 \\ 10 & 6 x \end{array}\right] \\ \Rightarrow & {\left[\begin{array}{cc} 2 x^2 & 2 x \\ 3 x & x^2 \end{array}\right]+\left[\begin{array}{cc} 16 & 10 x \\ 8 & 8 x \end{array}\right]=\left[\begin{array}{cc} 2 x^2+16 & 48 \\ 20 & 12 x \end{array}\right]} \\ \Rightarrow & {\left[\begin{array}{cc} 2 x^2+16 & 2 x+10 x \\ 3 x+8 & x^2+8 x \end{array}\right]=\left[\begin{array}{cc} 2 x^2+16 & 48 \\ 20 & 12 x \end{array}\right]} \\ \Rightarrow & 2 x+10 x=48 \\ \Rightarrow & 12 x=48 \\ \therefore & x=\frac{48}{12}=4 \end{array} \end{aligned}$$

$$\begin{aligned} \text{We have,}\quad A & =\left[\begin{array}{ll} 0 & 1 \\ 1 & 1 \end{array}\right] \text { and } B=\left[\begin{array}{cc} 0 & -1 \\ 1 & 0 \end{array}\right] \\ \therefore\quad(A+B) & =\left[\begin{array}{ll} 0+0 & 1-1 \\ 1+1 & 1+0 \end{array}\right]=\left[\begin{array}{ll} 0 & 0 \\ 2 & 1 \end{array}\right]_{2 \times 2} \\ \text{and}\quad(A-B) & =\left[\begin{array}{ll} 0-0 & 1+1 \\ 1-1 & 1-0 \end{array}\right]=\left[\begin{array}{ll} 0 & 2 \\ 0 & 1 \end{array}\right]_{2 \times 2} \end{aligned}$$

Since, $(A+B) \cdot(A-B)$ is defined, if the number of columns of $(A+B)$ is equal to the number of rows of $(A-B)$, so here multiplication of matrices $(A+B) \cdot(A-B)$ is possible.

Now, $$(A+B)_{2 \times 2} \cdot(A-B)_{2 \times 2}=\left[\begin{array}{ll} 0+0 & 0+0 \\ 0+0 & 4+1 \end{array}\right]=\left[\begin{array}{ll} 0 & 0 \\ 0 & 5 \end{array}\right]\quad\text{.... (i)}$$

$$\begin{aligned} &\text { Also, }\\ &\begin{aligned} A^2 & =A \cdot A \\ & =\left[\begin{array}{ll} 0 & 1 \\ 1 & 1 \end{array}\right] \cdot\left[\begin{array}{ll} 0 & 1 \\ 1 & 1 \end{array}\right] \\ & =\left[\begin{array}{ll} 0+1 & 0+1 \\ 0+1 & 1+1 \end{array}\right]=\left[\begin{array}{ll} 1 & 1 \\ 1 & 2 \end{array}\right] \end{aligned} \end{aligned}$$

$$\begin{aligned} \text { and } B^2=B \cdot B & =\left[\begin{array}{cc} 0 & -1 \\ 1 & 0 \end{array}\right]\left[\begin{array}{cc} 0 & -1 \\ 1 & 0 \end{array}\right] \\ & =\left[\begin{array}{cc} 0-1 & 0+0 \\ 0+0 & -1+0 \end{array}\right]=\left[\begin{array}{cc} -1 & 0 \\ 0 & -1 \end{array}\right] \end{aligned}$$

$\therefore \quad A^2-B^2=\left[\begin{array}{ll}1 & 1 \\ 1 & 2\end{array}\right]-\left[\begin{array}{cc}-1 & 0 \\ 0 & -1\end{array}\right]=\left[\begin{array}{ll}2 & 1 \\ 1 & 3\end{array}\right]\quad\text{.... (ii)}$

Thus, we see that

$$\begin{aligned} (A+B) \cdot(A-B) & \neq A^2-B^2 \quad\text{using Eqs. (i) and (ii)]}\\ \Rightarrow {\left[\begin{array}{ll} 0 & 0 \\ 0 & 5 \end{array}\right] } & \neq\left[\begin{array}{ll} 2 & 1 \\ 1 & 3 \end{array}\right] \end{aligned}$$

Hence proved.

We have, $$ \left[\begin{array}{lll} x & 1 \end{array}\right]_{1 \times 3}\left[\begin{array}{ccc} 1 & 3 & 2 \\ 2 & 5 & 1 \\ 15 & 3 & 2 \end{array}\right]_{3 \times 3}\left[\begin{array}{l} 1 \\ 2 \\ x \end{array}\right]_{3 \times 1}=0 $$

$\Rightarrow\left[\begin{array}{lll}1+2 x+15 & 3+5 x+3 & 2+x+2\end{array}\right]_{1 \times 3} \left[\begin{array}{l} \\ 1 \\ 2 \\ x \end{array}\right]_{3 \times 1}=0$

$\Rightarrow \quad[16+2 x \quad 5 x+6 \quad x+4]_{1 \times 3}\left[\begin{array}{c} 1 \\ 2 \\ x \end{array}\right]_{3 \times 1}=0$

$$\begin{aligned} \Rightarrow \quad& {[16+2 x+(5 x+6) \cdot 2+(x+4) \cdot x]_{1 \times 1} } =0 \\ \Rightarrow\quad& {\left[16+2 x+10 x+12+x^2+4 x\right] } =0 \\ \Rightarrow \quad& {\left[x^2+16 x+28\right] } =0 \\ \Rightarrow \quad& {\left[x^2+2 x+14 x+28\right] } =0 \\ \Rightarrow \quad& (x+2)(x+14) =0 \\ \therefore \quad& x =-2,-14 \end{aligned}$$