Let A be a matrix of order 3 times 3 and |A|= | vedclass

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(C) Given $A$ is a $3 	imes 3$ matrix,so $|A|=5$. The property $|k A| = k^n |A|$ where $n=3$ and $|\operatorname{adj}(M)| = |M|^{n-1} = |M|^2$ are us

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$27$

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(C) Given $A$ is a $3 	imes 3$ matrix,so $|A|=5$. The property $|k A| = k^n |A|$ where $n=3$ and $|\operatorname{adj}(M)| = |M|^{n-1} = |M|^2$ are used.$|2 \operatorname{adj}(3 A \operatorname{adj}(2 A))| = 2^3 |\operatorname{adj}(3 A \operatorname{adj}(2 A))|$$= 2^3 |3 A \operatorname{adj}(2 A)|^2$$= 2^3 \cdot (3^3 |A \operatorname{adj}(2 A)|)^2 = 2^3 \cdot 3^6 \cdot |A|^2 \cdot |\operatorname{adj}(2 A)|^2$$= 2^3 \cdot 3^6 \cdot |A|^2 \cdot (|2 A|^2)^2 = 2^3 \cdot 3^6 \cdot |A|^2 \cdot (2^3 |A|)^4$$= 2^3 \cdot 3^6 \cdot |A|^2 \cdot 2^{12} \cdot |A|^4 = 2^{15} \cdot 3^6 \cdot |A|^6$Substituting $|A|=5$,we get $2^{15} \cdot 3^6 \cdot 5^6 = 2^\alpha \cdot 3^\beta \cdot 5^\gamma$.Thus,$\alpha=15, \beta=6, \gamma=6$.Therefore,$\alpha+\beta+\gamma = 15+6+6 = 27$.

Let $A$ be a matrix of order $3 \times 3$ and $|A|=5$. If $|2 \operatorname{adj}(3 A \operatorname{adj}(2 A))|=2^\alpha \cdot 3^\beta \cdot 5^\gamma$ where $\alpha, \beta, \gamma \in N$,then $\alpha+\beta+\gamma$ is equal to

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