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

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

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

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

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

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

We have, $$ \begin{aligned} A & =\left[\begin{array}{cc} 5 & 3 \\ -1 & -2 \end{array}\right] \\ \therefore\quad A^2 & =A \cdot A=\left[\begin{array}{cc} 5 & 3 \\ -1 & -2 \end{array}\right] \cdot\left[\begin{array}{cc} 5 & 3 \\ -1 & -2 \end{array}\right] \end{aligned}$$

$$\begin{aligned} & =\left[\begin{array}{ll} 25-3 & 15-6 \\ -5+2 & -3+4 \end{array}\right]=\left[\begin{array}{ll} 22 & 9 \\ -3 & 1 \end{array}\right] \\ 3 A & =3\left[\begin{array}{cc} 5 & 3 \\ -1 & -2 \end{array}\right]=\left[\begin{array}{cc} 15 & 9 \\ -3 & -6 \end{array}\right] \\ \text{and}\quad 7 I & =7\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]=\left[\begin{array}{ll} 7 & 0 \\ 0 & 7 \end{array}\right] \end{aligned}$$

$$\begin{aligned} \therefore \quad A^2-3 A-7 I & =\left[\begin{array}{cc} 22 & 9 \\ -3 & 1 \end{array}\right]-\left[\begin{array}{cc} 15 & 9 \\ -3 & -6 \end{array}\right]-\left[\begin{array}{ll} 7 & 0 \\ 0 & 7 \end{array}\right] \\ & =\left[\begin{array}{ll} 22-15-7 & 9-9-0 \\ -3+3-0 & 1+6-7 \end{array}\right] \\ & =\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right] \\ & =0\quad\text{Hence proved.} \end{aligned}$$

$$\begin{array}{l} \text { Since, } & A^2-3 A-7 I =0 \\ \Rightarrow & A^{-1}\left[\left(A^2\right)-3 A-7 I\right] =A^{-1} 0 \\ \Rightarrow & A^{-1} A \cdot A-3 A^{-1} A-7 A^{-1} I=0 \quad \left[\because A^{-1} 0=0\right]\\ \Rightarrow & I A-3 I-7 A^{-1}=0 \quad \left[\because A^{-1} A=I\right]\\ \Rightarrow & A-3 I-7 A^{-1}=0\quad \left[\because A^{-1} I=A^{-1}\right] \end{array}$$

$$\begin{aligned} \Rightarrow \quad -7 A^{-1} & =-A+3 I \\ & =\left[\begin{array}{cc} -5 & -3 \\ 1 & 2 \end{array}\right]+\left[\begin{array}{ll} 3 & 0 \\ 0 & 3 \end{array}\right]=\left[\begin{array}{cc} -2 & -3 \\ 1 & 5 \end{array}\right] \end{aligned}$$

$\therefore \quad A^{-1}=\frac{-1}{7}\left[\begin{array}{cc}-2 & -3 \\ 1 & 5\end{array}\right]$

We have, $$ \left[\begin{array}{ll} 2 & 1 \\ 3 & 2 \end{array}\right]_{2 \times 2} A \cdot\left[\begin{array}{cc} -3 & 2 \\ 5 & -3 \end{array}\right]_{2 \times 2}=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]_{2 \times 2} $$

Let $A=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]_{2 \times 2}$

$$\begin{array}{ll} \therefore & {\left[\begin{array}{ll} 2 & 1 \\ 3 & 2 \end{array}\right]\left[\begin{array}{ll} a & b \\ c & d \end{array}\right]\left[\begin{array}{cc} -3 & 2 \\ 5 & -3 \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]} \\ \Rightarrow & {\left[\begin{array}{cc} 2 a+c & 2 b+d \\ 3 a+2 c & 3 b+2 d \end{array}\right]\left[\begin{array}{cc} -3 & 2 \\ 5 & -3 \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]} \\ \Rightarrow & {\left[\begin{array}{cc} -6 a-3 c+10 b+5 d & 4 a+2 c-6 b-3 d \\ -9 a-6 c+15 b+10 d & 6 a+4 c-9 b-6 d \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]} \\ \Rightarrow & -6 a-3 c+10 b+5 d=1\quad\text{.... (i)} \end{array}$$

$$\begin{array}{lr} \Rightarrow & 4 a+2 c-6 b-3 d=0 \quad\text{.... (ii)}\\ \Rightarrow & -9 a-6 c+15 b+10 d=0 \quad\text{.... (iii)}\\ \Rightarrow & 6 a+4 c-9 b-6 d=1\quad\text{.... (iv)} \end{array}$$

On adding Eqs. (i) and (iv), we get

$$c+b-d=2 \Rightarrow d=c+b-2\quad\text{.... (v)}$$

On adding Eqs. (ii) and (iii), we get

$$-5 a-4 c+9 b+7 d=0\quad\text{.... (vi)}$$

On adding Eqs. (vi) and (iv), we get

$$a+0+0+d=1 \Rightarrow d=1-a\quad\text{.... (vii)}$$

From Eqs. (v) and (vii),

$$c+b-2=1-a \Rightarrow a+b+c=3\quad\text{.... (viii)}$$

$$\Rightarrow \quad a=3-b-c$$

$$\begin{aligned} &\text { Now, using the values of a and } d \text { in Eq. (iii), we get }\\ &\begin{array}{l} & -9(3-b-c)-6 c+15 b+10(-2+b+c) =0 \\ \Rightarrow & -27+9 b+9 c-6 c+15 b-20+10 b+10 c =0 \\ \Rightarrow & 34 b+13 c =47\quad\text{.... (ix)} \end{array} \end{aligned}$$

$$\begin{aligned} &\text { Now, using the values of a and } d \text { in Eq. (ii), we get }\\ &\begin{array}{lrl} & 4(3-b-c)+2 c-6 b-3(b+c-2) \\ \Rightarrow & 12-4 b-4 c+2 c-6 b-3 b-3 c+6 =0 \\ \Rightarrow & -13 b-5 c =-18\quad\text{.... (x)} \end{array} \end{aligned}$$

On multiplying Eq. (ix) by 5 and Eq. (x) by 13, then adding, we get

$\begin{gathered}-169 b-65 c=-234 \\ \frac{170 b+65 c=235}{b=1}\end{gathered}$

$$\begin{array}{ll} \Rightarrow & -13 \times 1-5 c=-18 \quad\text{[from Eq. (x)]}\\ \Rightarrow & -5 c=-18+13=-5 \Rightarrow c=1 \\ \therefore & a=3-1-1=1 \text { and } d=1-1=0 \\ \therefore & A=\left[\begin{array}{ll} 1 & 1 \\ 1 & 0 \end{array}\right] \end{array}$$