Let I be a unit matrix of order 6. Let A = (a | vedclass

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(C) Given that $a_{ij} = 1$ if $i+j=7$ and $0$ otherwise. This represents an anti-diagonal matrix where all elements on the anti-diagonal are $1$.For

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$-A$

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(C) Given that $a_{ij} = 1$ if $i+j=7$ and $0$ otherwise. This represents an anti-diagonal matrix where all elements on the anti-diagonal are $1$.For a matrix $A$ of order $n$,the determinant $|A|$ is $(-1)^{n(n-1)/2}$. Here $n=6$,so $|A| = (-1)^{6(5)/2} = (-1)^{15} = -1$.We know that $A(	ext{adj } A) = |A| I$.Substituting this into the expression: $(A(	ext{adj } A) A^{-1}) A^2 = (|A| I) A^{-1} A^2$.Since $|A| = -1$,this becomes $(-I) A^{-1} A^2 = -I (A^{-1} A) A$.Since $A^{-1} A = I$,we have $-I (I) A = -A$.

Let $I$ be a unit matrix of order $6$. Let $A = (a_{ij})$ be a square matrix of order $6$ such that $a_{ij} = \begin{cases} 1, & \text{if } i+j=7 \\ 0, & \text{if } i+j \neq 7 \end{cases}$. Then $(A(\text{adj } A) A^{-1}) A^2 = $

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