Let A be a 3 times 3 matrix such that |operat | vedclass

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(D) Given $|\operatorname{adj}(\operatorname{adj}(\operatorname{adj} A ))|=81$. Since $|\operatorname{adj} M| = |M|^{n-1}$ for an $n 	imes n$ matrix,

Vedclass · Accepted Answer

$732$

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(D) Given $|\operatorname{adj}(\operatorname{adj}(\operatorname{adj} A ))|=81$. Since $|\operatorname{adj} M| = |M|^{n-1}$ for an $n 	imes n$ matrix,here $n=3$,so $|\operatorname{adj} M| = |M|^2$. $|\operatorname{adj}(\operatorname{adj}(\operatorname{adj} A ))| = (|\operatorname{adj}(\operatorname{adj} A)|)^2 = ((|\operatorname{adj} A|)^2)^2 = (|\operatorname{adj} A|)^4 = (|A|^2)^4 = |A|^8$. Thus,$|A|^8 = 81 = 3^4$,which implies $|A|^2 = \sqrt{3^4} = 3^2 = 9$ is incorrect; let's re-evaluate: $|A|^8 = 3^4 \Rightarrow |A| = 3^{4/8} = 3^{1/2} = \sqrt{3}$. So,$|A|^2 = 3$. Now,$|\operatorname{adj}(\operatorname{adj} A)| = (|\operatorname{adj} A|)^2 = (|A|^2)^2 = |A|^4$. The given equation is $(|A|^4)^{\frac{(n-1)^2}{2}} = |A|^{(3n^2-5n-4)}$. $|A|^{2(n-1)^2} = |A|^{(3n^2-5n-4)}$. Equating exponents: $2(n^2-2n+1) = 3n^2-5n-4$. $2n^2-4n+2 = 3n^2-5n-4$. $n^2-n-6 = 0$. $(n-3)(n+2) = 0$,so $n=3$ or $n=-2$. We need to calculate $\sum_{n \in S} |A|^{n^2+n}$. For $n=3$,$|A|^{3^2+3} = |A|^{12} = (|A|^2)^6 = 3^6 = 729$. For $n=-2$,$|A|^{(-2)^2+(-2)} = |A|^{4-2} = |A|^2 = 3$. Sum $= 729 + 3 = 732$.

Let $A$ be a $3 \times 3$ matrix such that $|\operatorname{adj}(\operatorname{adj}(\operatorname{adj} A ))|=81$. If $S =\{ n \in \mathbb{Z} :(|\operatorname{adj}(\operatorname{adj} A)|)^{\frac{(n-1)^2}{2}}=|A|^{(3n^2-5n-4)}\}$,then $\sum_{n \in S}|A^{(n^2+n)}|$ is equal to

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