Given matrix B = $\begin{bmatrix$}[r] 1 & 1 \ 8 & 3 \end{bmatrix}. Find the matrix X if X = B² - 4B. Hence, solve for a and b given $X\begin{bmatrix$}[r] a \ b \end{bmatrix} = \begin{bmatrix}[r] 5 \ 50 \end{bmatrix}.

Given,

$X = \begin{bmatrix$}[r] 1 & 1 \ 8 & 3 \end{bmatrix}\begin{bmatrix}[r] 1 & 1 \ 8 & 3 \end{bmatrix} - 4\begin{bmatrix}[r] 1 & 1 \ 8 & 3 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] 1 \times 1 + 1 \times 8 & 1 \times 1 + 1 \times 3 \ 8 \times 1 + 3 \times 8 & 8 \times 1 + 3 \times 3 \end{bmatrix} - \begin{bmatrix}[r] 4 & 4 \ 32 & 12 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] 1 + 8 & 1 + 3 \ 8 + 24 & 8 + 9 \end{bmatrix} - \begin{bmatrix}[r] 4 & 4 \ 32 & 12 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] 9 & 4 \ 32 & 17 \end{bmatrix} - \begin{bmatrix}[r] 4 & 4 \ 32 & 12 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] 9 - 4 & 4 - 4 \ 32 - 32 & 17 - 12 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] 5 & 0 \ 0 & 5 \end{bmatrix}.

Substituting value of X in $X\begin{bmatrix$}[r] a \ b \end{bmatrix} = \begin{bmatrix}[r] 5 \ 50 \end{bmatrix} we get,

$\Rightarrow \begin{bmatrix$}[r] 5 & 0 \ 0 & 5 \end{bmatrix}\begin{bmatrix}[r] a \ b \end{bmatrix} = \begin{bmatrix}[r] 5 \ 50 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix}[r] 5 \times a + 0 \times b \ 0 \times a + 5 \times b \end{bmatrix} = \begin{bmatrix}[r] 5 \ 50 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix}[r] 5a \ 5b \end{bmatrix} = \begin{bmatrix}[r] 5 \ 50 \end{bmatrix}

By definition of equality matrices we get,

5a = 5
⇒ a = 1.

5b = 50
⇒ b = 10.

Hence, X = $\begin{bmatrix$}[r] 5 & 0 \ 0 & 5 \end{bmatrix} and a = 1, b = 10.