Let $a \in R$ 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 $2^{\mathrm{m}} 3^{\mathrm{n}}, \mathrm{m}$, $\mathrm{n} \in\{0,1,2, \ldots, 20\}$, then $\mathrm{m}+\mathrm{n}$ is equal to :

<p>Given</p> <p>$$A+I=\begin{bmatrix}1 & a & 1\\ 2 & 1 & 0\\ a & 1 & 2\end{bmatrix}$$</p> <p>So,</p> <p>$$A=\begin{bmatrix}1-1 & a & 1\\ 2 & 1-1 & 0\\ a & 1 & 2-1\end{bmatrix} =\begin{bmatrix}0 & a & 1\\ 2 & 0 & 0\\ a & 1 & 1\end{bmatrix}$$</p> <h3>1) Find $a$ using $\det(A)=-4$</h3> <p>Compute $\det(A)$ by expanding along the first row:</p> <p>$ \det(A)=0\cdot C_{11}+a\cdot C_{12}+1\cdot C_{13} $</p> <p>Now,</p> <p>$ C_{12}=(-1)^{1+2}\begin{vmatrix}2&0\\ a&1\end{vmatrix} =-\,(2\cdot 1-0\cdot a)=-2 $</p> <p>$ C_{13}=(-1)^{1+3}\begin{vmatrix}2&0\\ a&1\end{vmatrix} =+\,(2\cdot 1-0\cdot a)=2 $</p> <p>So,</p> <p>$ \det(A)=a(-2)+1(2)=2-2a=2(1-a) $</p> <p>Given $\det(A)=-4$:</p> <p>$ 2(1-a)=-4 \implies 1-a=-2 \implies a=3 $</p> <h3>2) Evaluate $\det\left((a+1)\,\operatorname{adj}((a-1)A)\right)$</h3> <p>Use properties for a $3\times 3$ matrix:</p> <ul> <li><p>$\det(kM)=k^3\det(M)$</p></li> <li><p>$\det(\operatorname{adj}(M))=(\det M)^{3-1}=(\det M)^2$</p></li> <li><p>$\det((a-1)A)=(a-1)^3\det(A)$</p></li> </ul> <p>Let $B=(a-1)A$. Then</p> <p>$ \det\left((a+1)\operatorname{adj}(B)\right)=(a+1)^3\det(\operatorname{adj}(B)) =(a+1)^3(\det B)^2 $</p> <p>Now,</p> <p>$ \det B=((a-1)^3)(\det A)=(a-1)^3(-4) $</p> <p>So,</p> <p>$ \det\left((a+1)\operatorname{adj}((a-1)A)\right) =(a+1)^3\left[(-4)(a-1)^3\right]^2 =(a+1)^3\cdot 16\cdot (a-1)^6 $</p> <p>Put $a=3$:</p> <p>$ (a+1)^3=4^3=2^6,\quad (a-1)^6=2^6,\quad 16=2^4 $</p> <p>Hence,</p> <p>$ \text{Determinant}=2^6\cdot 2^4\cdot 2^6=2^{16}=2^m3^n $</p> <p>So $m=16,\; n=0 \implies m+n=16$.</p> <p><strong>Answer: 16 (Option D)</strong></p>