Let A=pmatrix&1&22&3&00&4&5pmatrix and B=pmatrix1&0&00&-5&00&4&-2pmatrix+adj(A).

Question

Let $A=\begin{pmatrix}\alpha&1&2\2&3&0\0&4&5\end{pmatrix}$ and $B=\begin{pmatrix}1&0&0\0&-5\alpha&0\0&4\alpha&-2\alpha\end{pmatrix}+\operatorname{adj}(A)$. If $\det(B)=66$, then $\det(\operatorname{adj}(A))$ equals:

Accepted Answer

$|A|=15\alpha-10+16=15\alpha+6$.

$\operatorname{adj}(A)=\begin{pmatrix}15&3&-6\-10&5\alpha&4\8&-4\alpha&3\alpha-2\end{pmatrix}$.

So $B=\begin{pmatrix}16&3&-6\-10&0&4\8&0&\alpha-2\end{pmatrix}$, and expanding along the middle column, $\det(B)=6(5\alpha+6)$.

Given $\det(B)=66$: $6(5\alpha+6)=66\Rightarrow5\alpha+6=11\Rightarrow\alpha=1$, so $|A|=21$.

For a $3	imes3$ matrix, $\det(\operatorname{adj}A)=|A|^2=21^2=441$.

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