Question

If $A=\left[\begin{array}{c}-1\\ 2\\ 3\end{array}\right]$ and $B=A=\left[\begin{array}{ccc}-2& -1& -4\end{array}\right]$, verify that $(AB{)}^T=B^T{A}^T$.

Answer 1

Given

$A=\left[\begin{array}{c}-1\\ 2\\ 3\end{array}\right]B=\left[\begin{array}{ccc}-2& -1& -4\end{array}\right]$

To verify ${\left(AB\right)}^T=B^T{A}^T$

Calculating AB

$\left[\begin{array}{c}-1\\ 2\\ 3\end{array}\right]\left[\begin{array}{ccc}-2& -1& -4\end{array}\right]=\left[\begin{array}{ccc}-2\times -1& -1\times -1& -1\times -4\\ 2\times -2& 2\times -1& 2\times -4\\ 3\times -2& 3\times -1& 3\times -4\end{array}\right]$

$AB=\left[\begin{array}{ccc}2& 1& 4\\ -4& -2& -8\\ -6& -3& -12\end{array}\right]$

${\left(AB\right)}^T=\left[\begin{array}{ccc}2& -4& -6\\ 1& -2& -3\\ 4& -8& -12\end{array}\right]⟶1$

Calculating $B^T{A}^T$

$B^T{A}^T={\left[\begin{array}{ccc}-2& -1& -4\end{array}\right]}^T{\left[\begin{array}{c}-1\\ 2\\ 3\end{array}\right]}^T=\left[\begin{array}{c}-2\\ -1\\ -4\end{array}\right]\left[\begin{array}{ccc}-1& 2& 3\end{array}\right]$

$B^T{A}^T=\left[\begin{array}{ccc}-2\times -1& -2\times 2& -2\times 3\\ -1\times -1& -1\times 2& -1\times 3\\ -4\times -1& -4\times 2& -4\times 3\end{array}\right]$

$B^T{A}^T=\left[\begin{array}{ccc}2& -4& -6\\ -2& -2& -3\\ 4& -8& -12\end{array}\right]⟶2$

From equation $1$ and $2$ we see that ${\left(AB\right)}^T=B^T{A}^T$ Proved.