Question

If $A=\left[\begin{array}{cc}2& 3\\ 0& 1\end{array}\right]B=\left[\begin{array}{cc}3& 4\\ 1& 1\end{array}\right]$. Prove that $(AB{)}^{\prime }=B^{\prime }{A}^{\prime }$.

Answer 1

$AB=\left[\begin{array}{cc}2& 3\\ 0& 1\end{array}\right]\left[\begin{array}{cc}3& 4\\ 1& 1\end{array}\right]=\left[\begin{array}{cc}2\times 3+3\times 1& 2\times 4+3\times 1\\ 0\times 3+1\times 1& 0\times 4+1\times 1\end{array}\right]=\left[\begin{array}{cc}9& 11\\ 1& 1\end{array}\right]$

$\therefore \text{LHS}=(AB{)}^{\prime }={\left[\begin{array}{cc}9& 11\\ 1& 1\end{array}\right]}^{\prime }=\left[\begin{array}{cc}9& 1\\ 11& 1\end{array}\right]$

Now, consider

$\therefore \text{RHS}=B^{\prime }{A}^{\prime }={\left[\begin{array}{cc}3& 4\\ 1& 1\end{array}\right]}^{\prime }{\left[\begin{array}{cc}2& 3\\ 0& 1\end{array}\right]}^{\prime }=\left[\begin{array}{cc}3& 1\\ 4& 1\end{array}\right]\left[\begin{array}{cc}2& 0\\ 3& 1\end{array}\right]$

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

$\therefore \text{RHS}=\left[\begin{array}{cc}9& 1\\ 11& 1\end{array}\right]$

$\therefore \text{RHS}=\text{LHS}$

Hence, Proved