Question

If $A=\left[\begin{array}{c}12\\ -13\end{array}\right]B=\left[\begin{array}{c}40\\ 15\end{array}\right]C=\left[\begin{array}{c}20\\ 1-2\end{array}\right]$ a = 4, and b = - 2, then show that

(vi) $(bA{)}^T$ = b $A^T$

Answer 1

We have,$A=\left[\begin{array}{c}12\\ -13\end{array}\right]B=\left[\begin{array}{c}40\\ 15\end{array}\right]$

$C=\left[\begin{array}{c}20\\ 1-2\end{array}\right]$ and a = 4, b = - 2

(vi) $(bA{)}^T{\left[\begin{array}{c}-2-4\\ 2-6\end{array}\right]}^T$ [$\because$ b = - 2]

$=\left[\begin{array}{c}-2-4\\ 2-6\end{array}\right]$

and $A^T=\left[\begin{array}{c}1-1\\ 23\end{array}\right]$

$\therefore b{A}^T=\left[\begin{array}{c}-22\\ -4-6\end{array}\right]=(bA{)}^T$