Question

If $A=\left[\begin{array}{cc}1& 2\\ -2& 3\end{array}\right],B=\left[\begin{array}{cc}2& 1\\ 2& 3\end{array}\right]$ and $C=\left[\begin{array}{cc}-3& 1\\ 2& 0\end{array}\right]$. check whether

$\left(AB\right)C=A\left(BC\right)$ and $A(B+C)=BA+CA$ ?

Answer 1

$AB=\left[\begin{array}{cc}1& 2\\ -2& 3\end{array}\right]\times \left[\begin{array}{cc}2& 1\\ 2& 3\end{array}\right]$
$=\left[\begin{array}{cc}1.2+2.2& 1.1+2.3\\ (-2).2+3.2& (-2).1+3.3\end{array}\right]$
$=\left[\begin{array}{cc}6& 7\\ 2& 7\end{array}\right]$
$BC=\left[\begin{array}{cc}2& 1\\ 2& 3\end{array}\right]\times \left[\begin{array}{cc}-3& 1\\ 2& 0\end{array}\right]$
$=\left[\begin{array}{cc}-6+2& 2+0\\ -6+6& 2+0\end{array}\right]$
$=\left[\begin{array}{cc}-4& 2\\ 0& 2\end{array}\right]$
$AC=\left[\begin{array}{cc}1& 2\\ -2& 3\end{array}\right]\times \left[\begin{array}{cc}-3& 1\\ 2& 0\end{array}\right]$
$=\left[\begin{array}{cc}-3+4& 1+0\\ 6+6& -2+0\end{array}\right]$
$=\left[\begin{array}{cc}1& 1\\ 12& -2\end{array}\right]$
$B+C=\left[\begin{array}{cc}2& 1\\ 2& 3\end{array}\right]+\left[\begin{array}{cc}-3& 1\\ 2& 0\end{array}\right]$
$=\left[\begin{array}{cc}2-3& 1+1\\ 2+2& 3+0\end{array}\right]=\left[\begin{array}{cc}-1& 2\\ 4& 3\end{array}\right]$
$\left(i\right)\left(AB\right)C=\left[\begin{array}{cc}6& 7\\ 2& 7\end{array}\right]\times \left[\begin{array}{cc}-3& 1\\ 2& 0\end{array}\right]$
$=\left[\begin{array}{cc}-18+14& 6+0\\ -6+14& 2+0\end{array}\right]=\left[\begin{array}{cc}-4& 6\\ 8& 2\end{array}\right]...\left(1\right)$
$A\left(BC\right)=\left[\begin{array}{cc}1& 2\\ -2& 3\end{array}\right]\times \left[\begin{array}{cc}-4& 2\\ 0& 2\end{array}\right]$
$=\left[\begin{array}{cc}-4+0& 2+4\\ 8+0& -4+6\end{array}\right]=\left[\begin{array}{cc}-4& 6\\ 8& 2\end{array}\right]...\left(2\right)$
From $\left(1\right)$ and $\left(2\right)$, we have $\left(AB\right)C=A\left(BC\right)$
$\left(ii\right)A(B+C)=\left[\begin{array}{cc}1& 2\\ -2& 3\end{array}\right]\times \left[\begin{array}{cc}-1& 2\\ 4& 3\end{array}\right]$
$=\left[\begin{array}{cc}-1+8& 2+6\\ 2+12& -4+9\end{array}\right]=\left[\begin{array}{cc}7& 8\\ 14& 5\end{array}\right]...\left(3\right)$
and $AB+AC=\left[\begin{array}{cc}6& 7\\ 2& 7\end{array}\right]+\left[\begin{array}{cc}1& 1\\ 12& -2\end{array}\right]$
$=\left[\begin{array}{cc}6+1& 7+1\\ 2+12& 7-2\end{array}\right]=\left[\begin{array}{cc}7& 8\\ 14& 5\end{array}\right]...\left(4\right)$
From $\left(3\right)$ and $\left(4\right)$, we have $A(B+C)=AB+AC$