Question

If $A=\left[\begin{array}{ccc}1& -2& 3\\ -4& 2& 5\end{array}\right]$ and $B=\left[\begin{array}{cc}1& 3\\ -1& 0\\ 2& 4\end{array}\right]$. Show that $(AB{)}^{\prime }=?$

• A. $B^{\prime }{A}^{\prime }$
• B. $A^{\prime }{B}^{\prime }$
• C. $A{B}^{\prime }$
• D. $A^{\prime }B$

Answer 1

The correct option is A $B^{\prime }{A}^{\prime }$

We have,
$A^{\prime }=\left[\begin{array}{cc}1& -4\\ -2& 2\\ 3& 5\end{array}\right]$
and $B^{\prime }=\left[\begin{array}{ccc}1& -1& 2\\ 3& 0& 4\end{array}\right]$
$\therefore AB=\left[\begin{array}{ccc}1& -2& 3\\ -4& 2& 5\end{array}\right]\times \left[\begin{array}{cc}1& 3\\ -1& 0\\ 2& 4\end{array}\right]$
$=\left[\begin{array}{cc}1+2+6& 3+0+12\\ -4-2+10& -12+0+20\end{array}\right]=\left[\begin{array}{cc}9& 15\\ 4& 8\end{array}\right]$
$\therefore (AB{)}^{\prime }=\left[\begin{array}{cc}9& 4\\ 15& 8\end{array}\right]...(i)$
and
$B^{\prime }{A}^{\prime }=\left[\begin{array}{ccc}1& -1& 2\\ 3& 0& 4\end{array}\right]\times \left[\begin{array}{cc}1& -4\\ -2& 2\\ 3& 5\end{array}\right]$
$=\left[\begin{array}{cc}1+2+6& -4-2+10\\ 3+0+12& -12+0+20\end{array}\right]$
$=\left[\begin{array}{cc}9& 4\\ 15& 8\end{array}\right]...\left(ii\right)$
From $\left(i\right)$ and $\left(ii\right)$, we get
$(AB{)}^{\prime }=B^{\prime }{A}^{\prime }$