Question

If $A=\left[\begin{array}{cc}2& 3\\ 5& 7\end{array}\right],B=\left[\begin{array}{cc}0& 4\\ -1& 7\end{array}\right]$ and $C=\left[\begin{array}{cc}1& 0\\ -1& 4\end{array}\right]$, find $AC+B^2-10C$.

Answer 1

$A=\left[\begin{array}{cc}2& 3\\ 5& 7\end{array}\right],B=\left[\begin{array}{cc}0& 4\\ -1& 7\end{array}\right]$ and $C=\left[\begin{array}{cc}1& 0\\ -1& 4\end{array}\right]$,

Upon multiplying, we find:

$AC=\left[\begin{array}{cc}2& 3\\ 5& 7\end{array}\right]\times \left[\begin{array}{cc}1& 0\\ -1& 4\end{array}\right]=\left[\begin{array}{cc}2-3& 12\\ 5-7& 0+28\end{array}\right]=\left[\begin{array}{cc}-1& 12\\ -2& 28\end{array}\right]$

$B^2=\left[\begin{array}{cc}0& 4\\ -1& 7\end{array}\right]\times =\left[\begin{array}{cc}0& 4\\ -1& 7\end{array}\right]=\left[\begin{array}{cc}-4& 28\\ -7& 45\end{array}\right]$

$10C=\left[\begin{array}{cc}10& 0\\ -10& 40\end{array}\right]$

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

$\therefore AC+B^2-10C=\left[\begin{array}{cc}-15& 40\\ 1& 33\end{array}\right]$