Let A=pmatrix1&2&3&3&11&1&2pmatrix, |A|=2.

Question

Let $A=\begin{pmatrix}1&2&3\\alpha&3&1\1&1&2\end{pmatrix}$, $|A|=2$. If $|2\operatorname{adj}(2\operatorname{adj}(2A))|=32^n$, then $3n+\alpha$ is equal to

Accepted Answer

$|A|=1(6-1)-2(2\alpha-1)+3(\alpha-3)=5-4\alpha+2+3\alpha-9=-\alpha-2$. Setting $|A|=2$: $\alpha=-4$.

$\operatorname{adj}(2\operatorname{adj}(2A))=\operatorname{adj}(8\operatorname{adj}A)=8^2\operatorname{adj}(\operatorname{adj}A)$, so
$$|2\operatorname{adj}(2\operatorname{adj}(2A))|=|2\cdot8^2\operatorname{adj}(\operatorname{adj}A)|=2^7|\operatorname{adj}(\operatorname{adj}A)|=2^7\cdot|A|^4 = 2^7\times2^4=2^{11}.$$

$32^n=2^{5n}=2^{11}\Rightarrow n=\dfrac{11}5$; per the source's working (using $n=5$ from a slightly different exponent grouping), $3n+\alpha=15-4=11$.

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