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Q. If $A=\begin{bmatrix}1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9\end{bmatrix}$, then $\left(A A'\right)'=$

TS EAMCET 2018

Solution:

We have,
$A=\begin{bmatrix}1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9\end{bmatrix}$
$\Rightarrow \,A'=\begin{bmatrix}1 & 4 & 7 \\ 2 & 5 & 8 \\ 3 & 6 & 9\end{bmatrix}$
Now, $A \cdot A'=\begin{bmatrix}1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9\end{bmatrix} \begin{bmatrix}1 & 4 & 7 \\ 2 & 5 & 8 \\ 3 & 6 & 9\end{bmatrix}$
$=\begin{bmatrix}1+4+9 & 4+10+18 & 7+16+27 \\ 4+10+18 & 16+25+36 & 28+40+54 \\ 7+16+27 & 28+40+54 & 49+64+81\end{bmatrix}$
So, $A A'=\begin{bmatrix}14 & 32 & 50 \\ 32 & 77 & 122 \\ 50 & 122 & 194\end{bmatrix}\left(A A'\right)'=\begin{bmatrix}14 & 32 & 50 \\ 32 & 77 & 122 \\ 50 & 122 & 194\end{bmatrix}$