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Let $I$ be a unit matrix of order 6 . Let $A=\left(a_{i j}\right)$ be a square matrix of order 6 such that $a_{i j}=\left\{\begin{array}{l}1, \text { if } i+j=7 \\ 0, \text { if } i+j \neq 7\end{array}\right.$ then $\left(A(\operatorname{adj} A) A^{-1}\right) A^2=$
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The correct answer is:
$-A$
Given that,
$a_{i j}= \begin{cases}1, & \text { if } i+j=7 \\ 0, & \text { if } i+j \neq 7\end{cases}$
Now, $\left(A(\operatorname{adj} A) A^{-1}\right) A^2=\left(A(\operatorname{Adj} A) A^{-1}\right) A^2$
$=\left((|A| I) A^{-1}\right) A^2 \quad[\because A$ adj $A=|A| I]$
$=-1\left(A^{-1} A\right) A \quad[\because|A|=-1]$
$=-I A \quad\left[\because A^{-1} A=I\right]$
$=-A$
$a_{i j}= \begin{cases}1, & \text { if } i+j=7 \\ 0, & \text { if } i+j \neq 7\end{cases}$
Now, $\left(A(\operatorname{adj} A) A^{-1}\right) A^2=\left(A(\operatorname{Adj} A) A^{-1}\right) A^2$
$=\left((|A| I) A^{-1}\right) A^2 \quad[\because A$ adj $A=|A| I]$
$=-1\left(A^{-1} A\right) A \quad[\because|A|=-1]$
$=-I A \quad\left[\because A^{-1} A=I\right]$
$=-A$
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