# if

Question:

If $A=\left[\begin{array}{l}3 \\ 5 \\ 2\end{array}\right]$ and $B=\left[\begin{array}{lll}1 & 0 & 4\end{array}\right]$, verify that $(A B)^{\top}=B^{T} A^{T}$

Solution:

Given : $A=\left[\begin{array}{l}3 \\ 5 \\ 2\end{array}\right]$

$A^{T}=\left[\begin{array}{lll}3 & 5 & 2\end{array}\right]$

$B=\left[\begin{array}{lll}1 & 0 & 4\end{array}\right]$

$B^{T}=\left[\begin{array}{l}1 \\ 0 \\ 4\end{array}\right]$

Now,

$A B=\left[\begin{array}{l}3 \\ 5 \\ 2\end{array}\right]\left[\begin{array}{lll}1 & 0 & 4\end{array}\right]$

$\Rightarrow A B=\left[\begin{array}{lll}3 & 0 & 12 \\ 5 & 0 & 20 \\ 2 & 0 & 8\end{array}\right]$

$\Rightarrow(A B)^{T}=\left[\begin{array}{ccc}3 & 5 & 2 \\ 0 & 0 & 0 \\ 12 & 20 & 8\end{array}\right]$       ...(1)

$B^{T} A^{T}=\left[\begin{array}{l}1 \\ 0 \\ 4\end{array}\right]\left[\begin{array}{lll}3 & 5 & 2\end{array}\right]$

$\Rightarrow B^{T} A^{T}=\left[\begin{array}{ccc}3 & 5 & 2 \\ 0 & 0 & 0 \\ 12 & 20 & 8\end{array}\right]$      ...(2)

$\Rightarrow(A B)^{T}=B^{T} A^{T}$                [From eqs. (1) and (2)]