Question

If $A=\left[\begin{array}{cc}1& 4\\ 2& 1\end{array}\right],B=\left[\begin{array}{cc}-3& 2\\ 4& 0\end{array}\right],C=\left[\begin{array}{cc}1& 0\\ 0& 2\end{array}\right]$; find $A^2+BC$

Answer 1

Given that:

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

To Find:

$A^2+BC=?$

Solution:

$A^2=\left[\begin{array}{cc}1& 4\\ 2& 1\end{array}\right]\times \left[\begin{array}{cc}1& 4\\ 2& 1\end{array}\right]$

$=\left[\begin{array}{cc}1\times 1+4\times 2& 1\times 4+4\times 1\\ 2\times 1+1\times 2& 2\times 4+1\times 1\end{array}\right]$

$A^2=\left[\begin{array}{cc}9& 8\\ 4& 9\end{array}\right]$

$BC=\left[\begin{array}{cc}-3& 2\\ 4& 0\end{array}\right]\times \left[\begin{array}{cc}1& 0\\ 0& 2\end{array}\right]$

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

$BC=\left[\begin{array}{cc}-3& 4\\ 4& 0\end{array}\right]$

$A^2+BC=\left[\begin{array}{cc}9& 8\\ 4& 9\end{array}\right]+\left[\begin{array}{cc}-3& 4\\ 4& 0\end{array}\right]$

$A^2+BC=\left[\begin{array}{cc}6& 12\\ 8& 9\end{array}\right]$