Let a ∈ R and A be a matrix of order 3 × 3 such that det(A)=-4 and A+I= [ … ], where I is the identity matrix of order 3 × 3. If det((a+1) adj((a-1)…

$$\mathrm{A}=\left[\begin{array}{lll}1 & \mathrm{a} & 1 \\ 2 & 1 & 0 \\ \mathrm{a} & 1 & 2\end{array}\right]-\mathrm{I}=\left[\begin{array}{lll}0 & \mathrm{a} & 1 \\ 2 & 0 & 0 \\ \mathrm{a} & 1 & 1\end{array}\right]$$A=​12a​a11​102​​−I=​02a​a01​101​​

$$\begin{aligned} & |\mathrm{A}|=-4 \Rightarrow 2-2 \mathrm{a}=-4 \Rightarrow \mathrm{a}=3 \\ & |(\mathrm{a}+1) \operatorname{adj}(\mathrm{a}-1) \mathrm{A}|=|4 \operatorname{adj} 3 \mathrm{~A}| \\ & =4^3|\operatorname{adj} 3 \mathrm{~A}| \\ & =4^3 \times|3 \mathrm{~A}|^{3-1}=64|3 \mathrm{~A}|^2 \\ & =64 \times\left(3^3\right)^2|\mathrm{~A}|^2 \\ & =2^6 \times 3^6 \times 16 \\ & 2^{\mathrm{m}} \times 3^{\mathrm{n}}=2^{10} \times 3^6 \\ & \therefore \mathrm{~m}=10, \mathrm{n}=6 \\ & \Rightarrow \mathrm{~m}+\mathrm{n}=16\end{aligned}$$​∣A∣=−4⇒2−2a=−4⇒a=3∣(a+1)adj(a−1)A∣=∣4adj3 A∣=43∣adj3 A∣=43×∣3 A∣3−1=64∣3 A∣2=64×(33)2∣ A∣2=26×36×162m×3n=210×36∴ m=10,n=6⇒ m+n=16​