# For the matrices $A$ and $B$, verify that $(A B)^{prime}=B^{prime} A^{prime}$ where

Question:

For the matrices $A$ and $B$, verify that $(A B)^{\prime}=B^{\prime} A^{\prime}$ where

(i) $A=\left[\begin{array}{r}1 \\ -4 \\ 3\end{array}\right], B=\left[\begin{array}{lll}-1 & 2 & 1\end{array}\right]$

(ii) $A=\left[\begin{array}{l}0 \\ 1 \\ 2\end{array}\right]$,$B=\left[\begin{array}{lll}1 & 5 & 7\end{array}\right]$

Solution:

(i)

$A B=\left[\begin{array}{r}1 \\ -4 \\ 3\end{array}\right]\left[\begin{array}{lll}-1 & 2 & 1\end{array}\right]=\left[\begin{array}{rrr}-1 & 2 & 1 \\ 4 & -8 & -4 \\ -3 & 6 & 3\end{array}\right]$

$\therefore(A B)^{\prime}=\left[\begin{array}{rrr}-1 & 4 & -3 \\ 2 & -8 & 6 \\ 1 & -4 & 3\end{array}\right]$

Now, $A^{\prime}=\left[\begin{array}{lll}1 & -4 & 3\end{array}\right], B^{\prime}=\left[\begin{array}{c}-1 \\ 2 \\ 1\end{array}\right]$

$\therefore B^{\prime} A^{\prime}=\left[\begin{array}{c}-1 \\ 2 \\ 1\end{array}\right]\left[\begin{array}{lll}1 & -4 & 3\end{array}\right]=\left[\begin{array}{rrr}-1 & 4 & -3 \\ 2 & -8 & 6 \\ 1 & -4 & 3\end{array}\right]$

Hence, we have verified that $(A B)^{\prime}=B^{\prime} A^{\prime}$.

(ii)

$A B=\left[\begin{array}{l}0 \\ 1 \\ 2\end{array}\right]\left[\begin{array}{lll}1 & 5 & 7\end{array}\right]=\left[\begin{array}{ccc}0 & 0 & 0 \\ 1 & 5 & 7 \\ 2 & 10 & 14\end{array}\right]$

$\therefore(A B)^{\prime}=\left[\begin{array}{rrr}0 & 1 & 2 \\ 0 & 5 & 10 \\ 0 & 7 & 14\end{array}\right]$

Now, $A^{\prime}=\left[\begin{array}{lll}0 & 1 & 2\end{array}\right], B^{\prime}=\left[\begin{array}{l}1 \\ 5 \\ 7\end{array}\right]$

$\therefore B^{\prime} A^{\prime}=\left[\begin{array}{l}1 \\ 5 \\ 7\end{array}\right]\left[\begin{array}{lll}0 & 1 & 2\end{array}\right]=\left[\begin{array}{rrr}0 & 1 & 2 \\ 0 & 5 & 10 \\ 0 & 7 & 14\end{array}\right]$

Hence, we have verified that $(A B)^{\prime}=B^{\prime} A^{\prime}$.