A² = 9 A + mI

$$\begin{gathered} \Rightarrow {\mathrm{A}}^2 - 9 \mathrm{A} = \mathrm{mI} ............( \mathrm{i} ) \\ {\mathrm{A}}^2 = \mathrm{A} \mathrm{A} \\ = \left[\begin{array}{cc} 2 & 0\\ - 1 & 7\end{array} \right] \left[\begin{array}{cc} 2 & 0\\ - 1 & 7\end{array} \right] \\ = \left[ \begin{array}{cc} 4 & 0\\ - 9& 49\end{array} \right] \\ \mathrm{Substitute} {\mathrm{A}}^2 \mathrm{in} ( \mathrm{i} ) \\ {\mathrm{A}}^2 - 9 \mathrm{A} = \mathrm{mI} \end{gathered}$$

$$\begin{gathered} \Rightarrow \left[ \begin{array}{cc} 4 & 0\\ - 9& 49\end{array} \right] - 9 \left[\begin{array}{cc} 2& 0\\ - 1& 7\end{array} \right] = \mathrm{m} \left[ \begin{array}{cc}1& 0\\ 0& 1\end{array} \right] \\ \Rightarrow \left[ \begin{array}{cc} 4 & 0\\ - 9& 49\end{array} \right] - \left[\begin{array}{cc} 18& 0\\ - 9& 63\end{array} \right] = \left[ \begin{array}{cc}\mathrm{m}& 0\\ 0& \mathrm{m}\end{array} \right] \\ \Rightarrow \left[ \begin{array}{cc}- 14& 0 \\ 0 & - 14\end{array} \right] = \left[ \begin{array}{cc}\mathrm{m}& 0\\ 0& \mathrm{m}\end{array} \right] \\ \Rightarrow \mathrm{m} = - 14 \end{gathered}$$

$\mathrm{A} = \left[ \begin{array}{cc}4 \sin 30°& \cos 0°\\ \cos 0° & 4 \sin 30°\end{array} \right] \mathrm{and} \mathrm{B} = \left[\begin{array}{c}4\\ 5\end{array}\right]$

( i ) Let the order of matrix  X = m x n

Order of matrix  A = 2 x 2

Order of matrix  B = 2 x 1

Now,  A X = B 

$\Rightarrow$ Order of matrix  X = m x n = 2 x 2

( ii ) Let the matrix  X = $\left[ \begin{array}{c}\mathrm{x}\\ \mathrm{y}\end{array} \right]$

AX = B

$$\begin{gathered} \Rightarrow \left[ \begin{array}{cc}4 \sin 30°& \cos 0°\\ \cos 0° & 4 \sin 30°\end{array} \right] \left[ \begin{array}{c}\mathrm{x}\\ \mathrm{y}\end{array} \right] = \left[\begin{array}{c}4\\ 5\end{array}\right] \\ \Rightarrow \left[ \begin{array}{cc}4 \left( \frac{1}{2} \right)& 1\\ 1& 4 \left( \frac{1}{2} \right)\end{array} \right] \left[ \begin{array}{c}\mathrm{x}\\ \mathrm{y}\end{array} \right] = \left[\begin{array}{c}4\\ 5\end{array}\right] \\ \Rightarrow \left[ \begin{array}{cc}2 & 1\\ 1 & 2\end{array} \right] \left[ \begin{array}{c}\mathrm{x}\\ \mathrm{y}\end{array} \right] = \left[\begin{array}{c}4\\ 5\end{array}\right] \\ \Rightarrow \left[ \begin{array}{c}2 \mathrm{x} + \mathrm{y}\\ \mathrm{x} + 2 \mathrm{y}\end{array} \right]= \left[\begin{array}{c}4\\ 5\end{array}\right] \\ \Rightarrow 2 \mathrm{x} + \mathrm{y} = 4 ..........( \mathrm{i} ) \\ \Rightarrow \mathrm{x} + 2 \mathrm{y} = 5 ..........( \mathrm{ii} ) \end{gathered}$$

Multiplying  ( i )  by 2,   we get

 4 x + 2 y = 8         ........( iii )

Subtracting  ( ii ) from  ( iii ),  we get

3 x = 3

$\Rightarrow$X = 1

Substitute  X in  ( i ),  we get

2 x 1 + y = 4

$$\begin{gathered} \Rightarrow \mathrm{y} = 2 \\ \mathrm{Hence}, \mathrm{the} \mathrm{matrix} \mathrm{X} = \left[ \begin{array}{c}\mathrm{x}\\ \mathrm{y}\end{array} \right] = \left[ \begin{array}{c}1\\ 2\end{array} \right] \end{gathered}$$