We have, $A=\left[\begin{array}{cc}1 & 5 \\ 7 & 12\end{array}\right]$ and $B=\left[\begin{array}{ll}9 & 1 \\ 7 & 8\end{array}\right]$

Let $$C=\left[\begin{array}{ll} a & b \\ c & d \end{array}\right]$$

$$\begin{aligned} & \therefore \quad 3 A+5 B+2 C=0 \\ & \Rightarrow\left[\begin{array}{cc} 3 & 15 \\ 21 & 36 \end{array}\right]+\left[\begin{array}{cc} 45 & 5 \\ 35 & 40 \end{array}\right]+\left[\begin{array}{cc} 2 \mathrm{a} & 2 \mathrm{~b} \\ 2 \mathrm{c} & 2 \mathrm{~d} \end{array}\right]=\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right] \\ & \Rightarrow \quad\left[\begin{array}{ll} 48+2 a & 20+2 b \\ 56+2 c & 76+2 d \end{array}\right]=\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right] \\ & \Rightarrow \quad 2 a+48=0 \Rightarrow a=-24 \\ & \text { Also, } \quad 20+2 b=0 \Rightarrow b=-10 \\ & 56+2 c=0 \Rightarrow c=-28 \\ & \text { and } \quad 76+2 d=0 \Rightarrow d=-38 \\ & \therefore\quad C=\left[\begin{array}{ll} -24 & -10 \\ -28 & -38 \end{array}\right] \end{aligned} $$

$$\begin{aligned} \text{We have,}\quad A & =\left[\begin{array}{cc} 3 & -5 \\ -4 & 2 \end{array}\right] \quad\text{.... (i)}\\ \therefore\quad A^2 & =A \cdot A=\left[\begin{array}{cc} 3 & -5 \\ -4 & 2 \end{array}\right]\left[\begin{array}{cc} 3 & -5 \\ -4 & 2 \end{array}\right] \\ & =\left[\begin{array}{cc} 29 & -25 \\ -20 & 24 \end{array}\right]\quad \text{.... (ii)} \end{aligned}$$

$$\begin{aligned} \therefore \quad A^2-5 A-14 I & =\left[\begin{array}{cc} 29 & -25 \\ -20 & 24 \end{array}\right]-\left[\begin{array}{cc} 15 & -25 \\ -20 & 10 \end{array}\right]-\left[\begin{array}{cc} 14 & 0 \\ 0 & 14 \end{array}\right] \\ & =\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right] \end{aligned}$$

$$\begin{aligned} \text { Now, } & A^2-5 A-14 I =0 \\ \Rightarrow & \quad A \cdot A^2-5 A \cdot A-14 A I =0 \\ \Rightarrow & \quad A^3-5 A^2-14 A =0 \quad [\because AI=A]\\ \Rightarrow & \quad A^3 =5 A^2=14 A \end{aligned}$$

$$\begin{aligned} & =5\left[\begin{array}{cc} 29 & -25 \\ -20 & 24 \end{array}\right]+14\left[\begin{array}{cc} 3 & -5 \\ -4 & 2 \end{array}\right] \quad \text { [using Eqs. (i) and (ii)] } \\ & =\left[\begin{array}{cc} 145 & -125 \\ -100 & 120 \end{array}\right]+\left[\begin{array}{cc} 42 & -70 \\ -56 & 28 \end{array}\right] \\ & =\left[\begin{array}{cc} 187 & -195 \\ -156 & 148 \end{array}\right] \end{aligned}$$

We have, $$ 3\left[\begin{array}{ll} a & b \\ c & d \end{array}\right]=\left[\begin{array}{cc} a & 6 \\ -1 & 2 d \end{array}\right]+\left[\begin{array}{cc} 4 & a+b \\ c+d & 3 \end{array}\right] $$

$$\begin{array}{rlrl} \Rightarrow & {\left[\begin{array}{cc} 3 a & 3 b \\ 3 c & 3 d \end{array}\right]} =\left[\begin{array}{cc} a+4 & 6+a+b \\ c+d-1 & 3+2 d \end{array}\right] \\ \Rightarrow & 3 a =a+4 \Rightarrow a=2 ; \\ \Rightarrow & 3 b =6+a+b \\ \Rightarrow & 3 b-b =8 \Rightarrow b=4 ; \\ & 3 d =3+2 d \Rightarrow d=3 \end{array}$$

$$\begin{array}{ll} \text { and } \Rightarrow & 3 c=c+d-1 \\ \Rightarrow & 2 c=3-1 c=1 \\ \therefore & a=2, b=4, c=1 \text { and } d=3 \end{array}$$

We have, $$\left[\begin{array}{cc} 2 & -1 \\ 1 & 0 \\ -3 & 4 \end{array}\right]_{3 \times 2} \quad A=\left[\begin{array}{ccc} -1 & -8 & -10 \\ 1 & -2 & -5 \\ 9 & 22 & 15 \end{array}\right]_{3 \times 3}$$

From the given equation, it is clear that order of $A$ should be $2 \times 3$.

Let $$A=\left[\begin{array}{lll} a & b & c \\ d & e & f \end{array}\right]$$

$$\begin{array}{ll} \therefore & {\left[\begin{array}{cc} 2 & -1 \\ 1 & 0 \\ -3 & 4 \end{array}\right]\left[\begin{array}{lll} a & b & c \\ d & e & f \end{array}\right]=\left[\begin{array}{ccc} -1 & -8 & -10 \\ 1 & -2 & -5 \\ 9 & 22 & 15 \end{array}\right]} \\ \Rightarrow & {\left[\begin{array}{ccc} 2 a-d & 2 b-e & 2 c-f \\ a+0 d & b+0 \cdot e & c+0 \cdot f \\ -3 a+4 d & -3 b+4 e & -3 c+4 f \end{array}\right]=\left[\begin{array}{ccc} -1 & -8 & -10 \\ 1 & -2 & -5 \\ 9 & 22 & 15 \end{array}\right]} \\ \Rightarrow & {\left[\begin{array}{ccc} 2 a-d & 2 b-e & 2 c-f \\ a & b & c \\ -3 a+4 d & -3 b+4 e & -3 c+4 f \end{array}\right]=\left[\begin{array}{ccc} -1 & -8 & -10 \\ 1 & -2 & -5 \\ 9 & 22 & 15 \end{array}\right]} \end{array}$$

$$\begin{aligned} & \text { By equality of matrices, we get } \\ & a=1, b=-2, c=-5 \\ \text { and } \quad & 2 a-d=-1 \Rightarrow d=2 a+1=3 \text {; } \\ \Rightarrow\quad & 2 b-e=-8 \Rightarrow e=2(-2)+8=4 \\ & 2 c-f=-10 \Rightarrow f=2 c+10=0 \\ & \therefore \quad A=\left[\begin{array}{ccc} 1 & -2 & -5 \\ 3 & 4 & 0 \end{array}\right] \end{aligned}$$

We have, $A=\left[\begin{array}{ll}1 & 2 \\ 4 & 1\end{array}\right]$

$\therefore \quad A^2=\left[\begin{array}{ll}1 & 2 \\ 4 & 1\end{array}\right]\left[\begin{array}{ll}1 & 2 \\ 4 & 1\end{array}\right] \quad\left[\because A^2=A \cdot A\right]$

$$ \begin{aligned} & =\left[\begin{array}{lll} 1+8 & 2+2 \\ 4+4 & 8+1 \end{array}\right]=\left[\begin{array}{ll} 9 & 4 \\ 8 & 9 \end{array}\right] \\ \therefore\quad A^2+2 A+7 I & =\left[\begin{array}{ll} 9 & 4 \\ 8 & 9 \end{array}\right]+\left[\begin{array}{ll} 2 & 4 \\ 8 & 2 \end{array}\right]+\left[\begin{array}{ll} 7 & 0 \\ 0 & 7 \end{array}\right]=\left[\begin{array}{cc} 18 & 8 \\ 16 & 18 \end{array}\right] \end{aligned}$$