Question
Given matrix $B=\left[\begin{array}{ll}1 & 1 \\ 8 & 3\end{array}\right]$ Find the matrix $X$ if, $X=B^2-4 B$. Hence, solve for $a$ and $b$ given $X\left[\begin{array}{l}a \\ b\end{array}\right]=\left[\begin{array}{c}5 \\ 50\end{array}\right]$

Answer

$B^2=B \times B=\left[\begin{array}{ll}1 & 1 \\ 8 & 3\end{array}\right]\left[\begin{array}{ll}1 & 1 \\ 8 & 3\end{array}\right]$
$=\left[\begin{array}{ll}1 \times 1+1 \times 8 & 1 \times 1+1 \times 3 \\ 8 \times 1+3 \times 8 & 8 \times 1+3 \times 3\end{array}\right]$
$=\left[\begin{array}{cc}9 & 4 \\ 32 & 17\end{array}\right]$
$4 B=4\left[\begin{array}{ll}1 & 1 \\ 8 & 3\end{array}\right]=\left[\begin{array}{cc}4 & 4 \\ 32 & 12\end{array}\right]$
Given : $X=B^2-4 B$
$\Rightarrow X=\left[\begin{array}{cc}9 & 4 \\ 32 & 17\end{array}\right]-\left[\begin{array}{cc}4 & 4 \\ 32 & 12\end{array}\right]$
$=\left[\begin{array}{ll}5 & 0 \\ 0 & 5\end{array}\right]$
To find: $a$ and $b \times\left[\begin{array}{l}a \\ b\end{array}\right]=\left[\begin{array}{c}5 \\ 50\end{array}\right] .......$ given 
$ \Rightarrow\left[\begin{array}{ll}5 & 0 \\ 0 & 5\end{array}\right]\left[\begin{array}{l}a \\ b\end{array}\right]=\left[\begin{array}{c}5 \\ 50\end{array}\right]  $
$ \Rightarrow\left[\begin{array}{l}5 a \\ 5 b\end{array}\right]=\left[\begin{array}{c}5 \\ 50\end{array}\right] $
$ \Rightarrow 5\left[\begin{array}{l}a \\ b\end{array}\right]=5\left[\begin{array}{c}1 \\ 10\end{array}\right]$
$\Rightarrow a = 1$ and $b = 10$

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