Question

State true/false:

If $A=\left[\begin{array}{c}-1\\ 2\\ 3\end{array}\right]$ and $B=[-2-1-4],$ then $(AB{)}^t\ne B^t{A}^t$.

• A. True
• B. False

Answer 1

The correct option is B False

Given,

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

Now,

$AB=\left[\begin{array}{c}-1\\ 2\\ 3\end{array}\right]\times \left[\begin{array}{ccc}-2& -1& -4\end{array}\right]=\left[\begin{array}{ccc}2& 1& 4\\ -4& -2& -8\\ -6& -3& -12\end{array}\right]\therefore {\left(AB\right)}^t=\left[\begin{array}{ccc}2& -4& -6\\ 1& -2& -3\\ 4& -8& -12\end{array}\right]Consider,B^t\times A^t=\left[\begin{array}{c}-2\\ -1\\ -4\end{array}\right]\times \left[\begin{array}{ccc}-1& 2& 3\end{array}\right]=\left[\begin{array}{ccc}2& -4& -6\\ 1& -2& -3\\ 4& -8& -12\end{array}\right]\therefore {\left(AB\right)}^t=B^t\times A^t$

Hence, given statement is false.