Let a in R and A be a matrix of order 3 times | vedclass

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(D) Given $A+I = \begin{bmatrix} 1 & a & 1 \ 2 & 1 & 0 \ a & 1 & 2 \end{bmatrix}$.Then $A = \begin{bmatrix} 1 & a & 1 \ 2 & 1 & 0 \ a & 1 & 2 \end

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$16$

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(D) Given $A+I = \begin{bmatrix} 1 & a & 1 \ 2 & 1 & 0 \ a & 1 & 2 \end{bmatrix}$.Then $A = \begin{bmatrix} 1 & a & 1 \ 2 & 1 & 0 \ a & 1 & 2 \end{bmatrix} - I = \begin{bmatrix} 0 & a & 1 \ 2 & 0 & 0 \ a & 1 & 1 \end{bmatrix}$.Calculating the determinant of $A$: $\det(A) = 0(0) - a(2) + 1(2-0) = -2a + 2$.Given $\det(A) = -4$,so $-2a + 2 = -4 \Rightarrow -2a = -6 \Rightarrow a = 3$.Now,we need to find $\det((a+1) \operatorname{adj}((a-1)A))$.Substituting $a=3$: $\det((3+1) \operatorname{adj}((3-1)A)) = \det(4 \operatorname{adj}(2A))$.Since $A$ is a $3 	imes 3$ matrix,$\det(kM) = k^3 \det(M)$.Thus,$\det(4 \operatorname{adj}(2A)) = 4^3 \det(\operatorname{adj}(2A)) = 64 \det(\operatorname{adj}(2A))$.Using the property $\det(\operatorname{adj}(M)) = (\det(M))^{n-1}$,where $n=3$:$\det(\operatorname{adj}(2A)) = (\det(2A))^{3-1} = (\det(2A))^2$.Since $\det(2A) = 2^3 \det(A) = 8 	imes (-4) = -32$,we have $(\det(2A))^2 = (-32)^2 = 1024 = 2^{10}$.So,$\det(4 \operatorname{adj}(2A)) = 64 	imes 1024 = 2^6 	imes 2^{10} = 2^{16} = 2^{16} 	imes 3^0$.Comparing with $2^m 3^n$,we get $m=16$ and $n=0$.Therefore,$m+n = 16+0 = 16$.

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