Let A be a 33 matrix such that (A)=-4, and A+I=pmatrix1&a&12&1&0&1&2pmatrix, where I is th…

Question

Let $A$ be a $3	imes3$ matrix such that $\det(A)=-4$, and $A+I=\begin{pmatrix}1&a&1\2&1&0\a&1&2\end{pmatrix}$, where $I$ is the identity matrix of order 3. If $\det\big((a+1)\operatorname{adj}((a-1)A)\big)=2^m3^n$, $m,n\in\{0,1,\dots,20\}$, then $m+n$ is equal to:

Accepted Answer

$A=(A+I)-I=\begin{pmatrix}0&a&1\2&0&0\a&1&1\end{pmatrix}$.

$\det(A)=0-a(2\cdot1-0\cdot a)+1(2\cdot1-0\cdot a)=-2a+2$.

Setting $-2a+2=-4$ gives $a=3$.

With $a=3$: $(a+1)\operatorname{adj}((a-1)A)=4\operatorname{adj}(3A)$.

$\det\big(4\operatorname{adj}(3A)\big)=4^3\det(\operatorname{adj}(3A))=4^3\cdot\det(3A)^2$ (using $\det(\operatorname{adj}M)=\det(M)^{n-1}=\det(M)^2$ for $n=3$).

$\det(3A)=3^3\det(A)=27\times(-4)=-108$, so $\det(3A)^2=108^2$.

$$\det\big(4\operatorname{adj}(3A)\big)=4^3\times108^2=2^6\times(2^2\cdot3^3)^2=2^6\times2^4\times3^6=2^{10}\times3^6.$$

So $m=10,\ n=6$, and $m+n=16$.

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