Question
If $A=\left[\begin{array}{cc}2 & -4 \\ 3 & -2 \\ 0 & 1\end{array}\right], B=\left[\begin{array}{ccc}1 & -1 & 2 \\ -2 & 1 & 0\end{array}\right]$, show that $(A B)^{\top}=B^{\top} A^{\top}$.
$=\left[\begin{array}{ccc}10 & -6 & 4 \\ 7 & -5 & 6 \\ -2 & 1 & 0\end{array}\right]$
$\therefore \quad(A B)^T=\left[\begin{array}{ccc}10 & 7 & -2 \\ -6 & -5 & 1 \\ 4 & 6 & 0\end{array}\right]$
$\ldots$...(i)
Now, $A^{\mathrm{T}}=\left[\begin{array}{ccc}2 & 3 & 0 \\ -4 & -2 & 1\end{array}\right]$ and $B^{\mathrm{T}}=\left[\begin{array}{cc}1 & -2 \\ -1 & 1 \\ 2 & 0\end{array}\right]$
$\begin{aligned} \therefore \quad \mathbf{B}^{\top} \mathbf{A}^{\top} & =\left[\begin{array}{cc}1 & -2 \\ -1 & 1 \\ 2 & 0\end{array}\right]\left[\begin{array}{ccc}2 & 3 & 0 \\ -4 & -2 & 1\end{array}\right] \\ & =\left[\begin{array}{ccc}2+8 & 3+4 & 0-2 \\ -2-4 & -3-2 & 0+1 \\ 4-0 & 6-0 & 0+0\end{array}\right]\end{aligned}$
$\therefore \quad \mathbf{B}^{\mathrm{T}} \mathbf{A}^{\mathrm{T}}=\left[\begin{array}{ccc}10 & 7 & -2 \\ -6 & -5 & 1 \\ 4 & 6 & 0\end{array}\right]$
...(ii)
From (i) and (ii), we get
$(A B)^{\mathrm{T}}=\mathrm{B}^{\mathrm{T}} \mathrm{A}^{\mathrm{T}}$
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