Question
If $A=\left[\begin{array}{cc}2 & -4 \\ 3 & -2 \\ 0 & 1\end{array}\right], B=\left[\begin{array}{ccc}1 & -1 & 2 \\ -2 & 1 & 0\end{array}\right]$ then showthat $(A B)^{\top}=B^{\top} A^{\top}$.

Answer

$
\begin{aligned}
& AB =\left[\begin{array}{cc}
2 & -4 \\
3 & -2 \\
0 & 1
\end{array}\right]\left[\begin{array}{ccc}
1 & -1 & 2 \\
-2 & 1 & 0
\end{array}\right] \\
& =\left[\begin{array}{ccc}
2+8 & -2-4 & 4+0 \\
3+4 & -3-2 & 6-0 \\
0-2 & 0+1 & 0+0
\end{array}\right] \\
& =\left[\begin{array}{ccc}
10 & -6 & 4 \\
7 & -5 & 6 \\
-2 & 1 & 0
\end{array}\right] \\
& \therefore( AB )^{\top}=\left[\begin{array}{ccc}
10 & 7 & -2 \\
-6 & -5 & 1 \\
4 & 6 & 0
\end{array}\right] \\
& \text { Now,AT }=\left[\begin{array}{ccc}
2 & 3 & 0 \\
-4 & -2 & 1
\end{array}\right] \text { and } B^{\top}=\left[\begin{array}{cc}
1 & -2 \\
-1 & 1 \\
2 & 0
\end{array}\right] \\
& \therefore B ^{\top} A ^{\top}=\left[\begin{array}{cc}
1 & -2 \\
-1 & 1 \\
2 & 0
\end{array}\right]\left[\begin{array}{ccc}
2 & 3 & 0 \\
-4 & -2 & 1
\end{array}\right] \\
& =\left[\begin{array}{ccc}
2+8 & 3+4 & 0-2 \\
-2-4 & -3-2 & 0+1 \\
4-0 & 6-0 & 0+0
\end{array}\right] \\
&
\end{aligned}
$
$
\therefore B ^{\top} A ^{\top}=\left[\begin{array}{ccc}
10 & 7 & -2 \\
-6 & -5 & 1 \\
4 & 6 & 0
\end{array}\right]
$
From (i) and (ii), we get $(A B)^{\top}=B^{\top} A^{\top}$.

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