Question

If $A=\left[\begin{array}{cc}2& -1\\ 3& 2\end{array}\right]$, $B=\left[\begin{array}{cc}0& 4\\ -1& 7\end{array}\right]$. Find $3A^2-2B+1$.

Answer 1

Given $A=\left[\begin{array}{cc}2& -1\\ 3& 2\end{array}\right]$ and $B=\left[\begin{array}{cc}0& 4\\ -1& 7\end{array}\right]$

Now $A^2=A\times A$ $=\left[\begin{array}{cc}2& -1\\ 3& 2\end{array}\right]\times \left[\begin{array}{cc}2& -1\\ 3& 2\end{array}\right]$.
$=\left[\begin{array}{cc}4-3& -2-2\\ 6+6& -3+4\end{array}\right]$
$=\left[\begin{array}{cc}1& -4\\ 12& 1\end{array}\right]$

$\therefore 3A^2-2B+I=3\left[\begin{array}{cc}1& -4\\ 12& 1\end{array}\right]-\left[\begin{array}{cc}0& 8\\ -2& 14\end{array}\right]+\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$
$=\left[\begin{array}{cc}3+1& -12-8\\ 36+2& 3-14+1\end{array}\right]$
$=\left[\begin{array}{cc}4& -20\\ 38& -10\end{array}\right]$