Let |A|=6 where A is a 3 times 3 matrix. If | | vedclass

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(B) Given $|A|=6$ and $A$ is a $3 	imes 3$ matrix. Using the property $|adj(k A)| = k^{n-1} |adj(A)|$ and $adj(kA) = k^{n-1} adj(A)$ for a $n 	imes

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$62$

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(B) Given $|A|=6$ and $A$ is a $3 	imes 3$ matrix. Using the property $|adj(k A)| = k^{n-1} |adj(A)|$ and $adj(kA) = k^{n-1} adj(A)$ for a $n 	imes n$ matrix. First, $adj(2A) = 2^{3-1} adj(A) = 4 adj(A)$. Then, $A^2 \cdot adj(2A) = A^2 \cdot 4 adj(A) = 4 A (A \cdot adj(A)) = 4 A |A| I_3 = 4 \cdot 6 \cdot A = 24A$. Now, $3 adj(24A) = 3 \cdot 24^{3-1} adj(A) = 3 \cdot 24^2 adj(A) = 3 \cdot (2^3 \cdot 3)^2 adj(A) = 3 \cdot 2^6 \cdot 3^2 adj(A) = 2^6 \cdot 3^3 adj(A)$. Let $K = 2^6 \cdot 3^3$. Then we need $|adj(K adj(A))|$. Using $|adj(M)| = |M|^{n-1} = |M|^2$ for a $3 	imes 3$ matrix $M$: $|adj(K adj(A))| = |K adj(A)|^2 = K^6 |adj(A)|^2 = K^6 (|A|^{3-1})^2 = K^6 |A|^4$. Substituting $K = 2^6 \cdot 3^3$ and $|A|=6 = 2^1 \cdot 3^1$: $|adj(K adj(A))| = (2^6 \cdot 3^3)^6 \cdot (2^1 \cdot 3^1)^4 = (2^{36} \cdot 3^{18}) \cdot (2^4 \cdot 3^4) = 2^{40} \cdot 3^{22}$. Thus, $m=40$ and $n=22$. Therefore, $m+n = 40+22 = 62$.

Let $|A|=6$ where $A$ is a $3 \times 3$ matrix. If $|adj(3adj(A^{2} \cdot adj(2A)))|=2^{m} \cdot 3^{n}$, $m, n \in N$, then $m+n$ is equal to:

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