Given : A = $\left[\begin{array}{cc}3& \mathrm{x}\\ 0& 1\end{array}\right]$, B = $\left[\begin{array}{cc}9& 16\\ 0& -\mathrm{y}\end{array}\right]$ and A² = B

As A = A $\times$ A = $\left[\begin{array}{cc}3& \mathrm{x}\\ 0& 1\end{array}\right] \times \left[\begin{array}{cc}3& \mathrm{x}\\ 0& 1\end{array}\right] = \left[\begin{array}{cc}9& 3\mathrm{x} + \mathrm{x}\\ 0& 1\end{array}\right] = \left[\begin{array}{cc}9& 4\mathrm{x}\\ 0& 1\end{array}\right]$

We have A² = B

Two matrices are eqaul if each and every corrosponding element is equal.

Thus, $\left[\begin{array}{cc}9& 4\mathrm{x}\\ 0& 1\end{array}\right] = \left[\begin{array}{cc}9& 16\\ 0& -\mathrm{y}\end{array}\right]$

as 4x = 16 and 1 = -y

x = 4 and y = -1

Given:   B = $\left[ \begin{array}{cc}1& 1\\ 8& 3\end{array} \right]$    and     X =  B² - 4 B

Now,  B² = B x B

  $$\begin{gathered} = \left[ \begin{array}{cc}1 & 1\\ 8 & 3\end{array} \right] \mathrm{x} \left[ \begin{array}{cc}1 & 1\\ 8 & 3\end{array} \right] \\ = \left[ \begin{array}{cc}1 \mathrm{x} 1 + 1 \mathrm{x} 8& 1 \mathrm{x} 1 + 1 \mathrm{x} 3 \\ 8 \mathrm{x} 1 + 3 \mathrm{x} 8& 8 \mathrm{x} 1 + 3 \mathrm{x} 3 \end{array} \right] \\ = \left[ \begin{array}{cc}1 + 8& 1 + 3\\ 8 + 24& 8 + 9\end{array} \right] \\ = \left[ \begin{array}{cc}9& 4\\ 32& 17\end{array} \right] \end{gathered}$$

$$\begin{gathered} \mathrm{X} = \mathrm{B} - 4 \mathrm{B} = \left[ \begin{array}{cc}9& 4\\ 32& 17\end{array} \right] - 4 \left[ \begin{array}{cc}1& 1\\ 8& 3\end{array} \right] \\ = \left[ \begin{array}{cc}9& 4\\ 32& 17\end{array} \right] - \left[ \begin{array}{cc}4& 4\\ 32& 12\end{array} \right] \\ = \left[ \begin{array}{cc}5& 0\\ 0& 5\end{array} \right] \\ \mathrm{Now}, \mathrm{X} \left[ \begin{array}{c}\mathrm{a}\\ \mathrm{b}\end{array} \right] = \left[ \begin{array}{c}5\\ 50\end{array} \right] \end{gathered}$$

$$\begin{gathered} \Rightarrow \left[ \begin{array}{cc}5& 0\\ 0& 5\end{array} \right] \left[ \begin{array}{c}\mathrm{a}\\ \mathrm{b}\end{array} \right] = \left[ \begin{array}{c}5\\ 50\end{array} \right] \\ \Rightarrow \left[ \begin{array}{c} 5 \mathrm{a} + 0 \mathrm{b}\\ 0 \mathrm{a} + 5 \mathrm{b}\end{array} \right] = \left[ \begin{array}{c}5\\ 50\end{array} \right] \\ \Rightarrow \left[ \begin{array}{c}5 \mathrm{a}\\ 5 \mathrm{b}\end{array} \right] = \left[ \begin{array}{c}5\\ 50\end{array} \right] \\ \Rightarrow 5 \mathrm{a} = 5 \mathrm{and} 5 \mathrm{b} = 50 \\ \Rightarrow \mathrm{a} = 1 \mathrm{and} \mathrm{b} = 10 \end{gathered}$$

A² = 9 A + mI

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

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

$$\begin{gathered} \mathrm{Given}: \mathrm{A} = \left[ \begin{array}{cc}2 & 3\\ 5 & 7\end{array} \right], \mathrm{B} = \left[\begin{array}{cc}0& 4\\ - 1 & 7\end{array} \right] \mathrm{and} \mathrm{C} = \left[\begin{array}{cc}1& 0\\ - 1 & 4\end{array} \right] \\ \mathrm{Now}, \mathrm{AC} = \mathrm{A} \times \mathrm{C} = \left[ \begin{array}{cc}2 & 3\\ 5 & 7\end{array} \right] \times \left[\begin{array}{cc}1& 0\\ - 1 & 4\end{array} \right] \\ = \left[ \begin{array}{cc}2 \mathrm{x} 1 + 3 \mathrm{x} ( - 1 )& 2 \mathrm{x} 0 + 3 \mathrm{x} 4\\ 5 \mathrm{x} 1 + 7 \mathrm{x} ( - 1 )& 5 \mathrm{x} 0 + 7 \mathrm{x} 4\end{array} \right] \\ = \left[ \begin{array}{cc}2 - 3 & 0 + 12\\ 5 - 7 & 0 + 28\end{array} \right] \\ = \left[ \begin{array}{cc}- 1 & 12 \\ - 2 & 28 \end{array} \right] \end{gathered}$$

$$\begin{gathered} \mathrm{And}, {\mathrm{B}}^2 = \mathrm{B} \mathrm{x} \mathrm{B} = \left[\begin{array}{cc}0& 4\\ - 1 & 7\end{array} \right] \mathrm{x} \left[\begin{array}{cc}0& 4\\ - 1 & 7\end{array} \right] \\ = \left[\begin{array}{cc}0 \mathrm{x} 0 + 4 \mathrm{x} ( - 1 )& 0 \mathrm{x} 4 + 4 \mathrm{x} 7\\ - 1 \mathrm{x} 0 + 7 \mathrm{x} ( - 1 ) & - 1 \mathrm{x} 4 + 7 \mathrm{x} 7\end{array} \right] \\ = \left[ \begin{array}{cc}0 - 4& 0 + 28 \\ 0 - 7& - 4 + 49 \end{array}\right] \\ = \left[ \begin{array}{cc}- 4& 28\\ - 7& 45\end{array} \right] \end{gathered}$$

$$\begin{gathered} \mathrm{Now}, \mathrm{AC} + {\mathrm{B}}^2 - 10 \mathrm{C} = \left[\begin{array}{cc}- 1& 12\\ - 2& 28\end{array} \right] + \left[\begin{array}{cc}- 4& 28\\ - 7& 45\end{array} \right] - 10 \left[\begin{array}{cc} 1& 0\\ - 1& 4\end{array} \right] \\ = \left[\begin{array}{cc}- 1& 12\\ - 2& 28\end{array} \right] + \left[\begin{array}{cc}- 4& 28\\ - 7& 45\end{array} \right] - \left[\begin{array}{cc} 10& 0\\ - 10& 40\end{array} \right] \\ = \left[ \begin{array}{cc}- 1 - 4 - 10& 12 + 28 - 0\\ - 2 - 7 + 10& 28 + 45 - 40\end{array} \right] \\ = \left[ \begin{array}{cc}- 15& 40\\ 1& 33\end{array} \right] \end{gathered}$$

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

( 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

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