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

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

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

We have, $A=\left[\begin{array}{cc}\cos a & \sin a \\ -\sin a & \cos a\end{array}\right]$ and $A^{\prime}=\left[\begin{array}{cc}\cos a & -\sin a \\ \sin a & \cos a\end{array}\right]$

$$\begin{array}{lrl} \text { Also, } & A^{-1} & =A^{\prime} \\ \Rightarrow & A A^{-1} & =A A^{\prime} \end{array}$$

$\left.\begin{array}{lll}\Rightarrow & I & =\left[\begin{array}{cc}\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha\end{array}\right]\end{array} \begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha\end{array}\right]$

$\Rightarrow \quad\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]=\left[\begin{array}{lc}\cos ^2 \alpha+\sin ^2 \alpha & 0 \\ 0 & \sin ^2 \alpha+\cos ^2 \alpha\end{array}\right]$

By using equality of matrices, we get

$$\cos ^2 \alpha+\sin ^2 \alpha=1$$

which is true for all real values of $\alpha$.

Let $A=\left[\begin{array}{ccc}0 & a & 3 \\ 2 & b & -1 \\ c & 1 & 0\end{array}\right]$ Since, $A$ is skew-symmetric matrix.

$$\therefore \quad A^{\prime}=-A$$

$$\begin{aligned} &\begin{array}{ll} \Rightarrow \quad {\left[\begin{array}{ccc} 0 & 2 & c \\ a & b & 1 \\ 3 & -1 & 0 \end{array}\right]=-\left[\begin{array}{ccc} 0 & a & 3 \\ 2 & b & -1 \\ c & 1 & 0 \end{array}\right]} \\ \Rightarrow \quad\left[\begin{array}{ccc} 0 & 2 & c \\ a & b & 1 \\ 3 & -1 & 0 \end{array}\right]=\left[\begin{array}{ccc} 0 & -a & -3 \\ -2 & -b & +1 \\ -c & -1 & 0 \end{array}\right] \end{array}\\ &\text { By equality of matrices, we get }\\ &\begin{array}{ll} & a=-2, c=-3 \text { and } b=-b \Rightarrow b=0 \\ \therefore & a=-2, b=0 \text { and } c=-3 \end{array} \end{aligned}$$