Let $a\in R$ and A be a matrix of order $3\times 3$ such that $\det (A)=-4$ and $A+I=\left[\begin{matrix}1 & a & 1 \ 2 & 1 & 0 \ a & 1 & 2\end{matrix}\right]$, where $I$ is the identity matrix of order $3\times 3$. If $\det ((\mathrm{a}+1)adj((\mathrm{a})\mathrm{A}))$ is ${2}^{\mathrm{m}}{3}^{\mathrm{n}},\mathrm{m},\mathrm{n}\in$ ${0,1,2,\ldots ..20}$, then $\mathrm{m}+\mathrm{n}$ is equal to : [JEE Main 2025, 2 Apr (Shift 1)]

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YESPYQ · Accepted Answer

Correct answer: d) $16$. Since $A+I=[\begin{matrix}1 & a & 1 \ 2 & 1 & 0 \ a & 1 & 2\end{matrix}]$, we subtract the identity matrix $I$ to get $A$: $A=[\begin{matrix}1 & a & 1 \ 2 & 1 & 0 \ a & 1 & 2\end{matrix}]−[\begin{matrix}1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1\end{matrix}]=[\begin{matrix}0 & a & 1 \ 2 & 0 & 0 \ a & 1 & 1\end{matrix}]$ We compute the determinant of $A$: $\det ⁡(A)=∣\begin{matrix}0 & a & 1 \ 2 & 0 & 0 \ a & 1 & 1\end{matrix}∣$ $−2a+2=−4\text{ }⟹\text{ }−2a=−6\text{ }⟹\text{ }a=3$ $\det ⁡((a+1)\text{adj}(aI)A)={[(a+1){a}^{2}]}^{3}\det ⁡(A)$ Substitute $a=3$ and $\det ⁡(A)=−4$: $(a+1){a}^{2}=(3+1)⋅{3}^{2}=4⋅9=36$ ${[36]}^{3}⋅(−4)={36}^{3}⋅(−4)$ ${36}^{3}⋅(−4)={2}^{6}⋅{3}^{6}⋅(−{2}^{2})=−{2}^{8}⋅{3}^{6}$ $\mathrm{∣}\det ⁡((a+1)\text{adj}(aI)A)\mathrm{∣}={2}^{8}⋅{3}^{6}$ Hence, $m=8$ and $n=6$.

Let \(a\in R\) and A be a matrix of order \(3\times 3\) such that \(\det (A)=-4\) and \(A+I=\left[\begin{matrix}1 & a & …

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