Let A=begin{bmatrix} 2 & 2+p & 2+p+q 4 & 6+2 | vedclass

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(B) Given $A = \begin{bmatrix} 2 & 2+p & 2+p+q \ 4 & 6+2p & 8+3p+2q \ 6 & 12+3p & 20+6p+3q \end{bmatrix}$.Applying column operations: $C_2 ightarr

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$24$

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(B) Given $A = \begin{bmatrix} 2 & 2+p & 2+p+q \ 4 & 6+2p & 8+3p+2q \ 6 & 12+3p & 20+6p+3q \end{bmatrix}$.Applying column operations: $C_2 ightarrow C_2 - C_1$ and $C_3 ightarrow C_3 - C_2$,we simplify the determinant.$|A| = \begin{vmatrix} 2 & p & q \ 4 & 2+2p & 2+p+q \ 6 & 6+3p & 8+3p+q \end{vmatrix}$.Further simplifying,we find $|A| = 8 = 2^3$.We know that $\operatorname{det}(\operatorname{adj}(\operatorname{adj}(M))) = |M|^{(n-1)^2}$,where $n$ is the order of the matrix.Here $n=3$,so $\operatorname{det}(\operatorname{adj}(\operatorname{adj}(3A))) = |3A|^{(3-1)^2} = |3A|^4$.Since $|3A| = 3^3 |A| = 3^3 \cdot 2^3$,we have $|3A|^4 = (3^3 \cdot 2^3)^4 = 3^{12} \cdot 2^{12}$.Comparing with $2^m \cdot 3^n$,we get $m=12$ and $n=12$.Therefore,$m+n = 12+12 = 24$.

Let $A=\begin{bmatrix} 2 & 2+p & 2+p+q \\ 4 & 6+2p & 8+3p+2q \\ 6 & 12+3p & 20+6p+3q \end{bmatrix}$. If $\operatorname{det}(\operatorname{adj}(\operatorname{adj}(3A)))=2^m \cdot 3^n$,where $m, n \in N$,then $m+n$ is equal to:

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