Let $\mathrm{a} \in \mathbf{R}$ and A be a matrix of order $3 \times 3$ such that $\operatorname{det}(A)=-4$ and $A+I=\left[\begin{array}{lll}1 & a & 1 \\ 2 & 1 & 0 \\ a & 1 & 2\end{array}\right]$, where $I$ is the identity matrix of order $3 \times 3$.
If $\operatorname{det}((a+1) \operatorname{adj}((a-1) A))$ is $2^m 3^n, m, n \in$ $\{0,1,2, \ldots .20\}$, then $\mathrm{m}+\mathrm{n}$ is equal to :

Question

Aicharya · Accepted Answer

$\mathrm{A}=\left[\begin{array}{lll}1 & \mathrm{a} & 1 \\ 2 & 1 & 0 \\ \mathrm{a} & 1 & 2\end{array}\right]-\mathrm{I}=\left[\begin{array}{lll}0 & \mathrm{a} & 1 \\ 2 & 0 & 0 \\ \mathrm{a} & 1 & 1\end{array}\right]$
$\begin{aligned} & |\mathrm{A}|=-4 \Rightarrow 2-2 \mathrm{a}=-4 \Rightarrow \mathrm{a}=3 \\ & |(\mathrm{a}+1) \operatorname{adj}(\mathrm{a}-1) \mathrm{A}|=|4 \operatorname{adj} 3 \mathrm{~A}| \\ & =4^3|\operatorname{adj} 3 \mathrm{~A}| \\ & =4^3 \times|3 \mathrm{~A}|^{3-1}=64|3 \mathrm{~A}|^2 \\ & =64 \times\left(3^3\right)^2|\mathrm{~A}|^2 \\ & =2^6 \times 3^6 \times 16 \\ & 2^{\mathrm{m}} \times 3^{\mathrm{n}}=2^{10} \times 3^6 \\ & \therefore \mathrm{~m}=10, \mathrm{n}=6 \\ & \Rightarrow \mathrm{~m}+\mathrm{n}=16\end{aligned}$

Let $\mathrm{a} \in \mathbf{R}$ and A be a matrix of order $3 \times 3$ such that $\operatorname{det}(A)=-4$…

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