Question

If $A^{\prime }=\left[\begin{array}{cc}-2& 3\\ 1& 2\end{array}\right]$ and $B=\left[\begin{array}{cc}-1& 0\\ 1& 2\end{array}\right]$, then find ${(A+2B)}^{\prime }$.

Answer 1

We know that $A={\left(A^{\prime }\right)}^{\prime }$
$\therefore A+2B=\left[\begin{array}{cc}-2& 1\\ 3& 2\end{array}\right]+2\left[\begin{array}{cc}-1& 0\\ 1& 2\end{array}\right]=\left[\begin{array}{cc}-2& 1\\ 3& 2\end{array}\right]+\left[\begin{array}{cc}-2& 0\\ 2& 4\end{array}\right]=\left[\begin{array}{cc}-4& 1\\ 5& 6\end{array}\right]$
$\therefore {(A+2B)}^{\prime }=\left[\begin{array}{cc}-4& 5\\ 1& 6\end{array}\right]$