Let P=[p_{ij}] and Q=[q_{ij}] be two square m | vedclass

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(B) Given $q_{ij} = 2^{(i+j-1)}p_{ij}$. The matrix $Q$ can be written as: $Q = \begin{bmatrix} 2^1 p_{11} & 2^2 p_{12} & 2^3 p_{13} \ 2^2 p_{21} & 2^

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$16$

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(B) Given $q_{ij} = 2^{(i+j-1)}p_{ij}$. The matrix $Q$ can be written as: $Q = \begin{bmatrix} 2^1 p_{11} & 2^2 p_{12} & 2^3 p_{13} \ 2^2 p_{21} & 2^3 p_{22} & 2^4 p_{23} \ 2^3 p_{31} & 2^4 p_{32} & 2^5 p_{33} \end{bmatrix}$ Taking common factors from each row: $\det(Q) = (2^1 \cdot 2^2 \cdot 2^3) \begin{vmatrix} p_{11} & p_{12} & p_{13} \ 2 p_{21} & 2 p_{22} & 2 p_{23} \ 2^2 p_{31} & 2^2 p_{32} & 2^2 p_{33} \end{vmatrix} = 2^6 \cdot (1 \cdot 2 \cdot 2^2) \det(P) = 2^6 \cdot 2^3 \det(P) = 2^9 \det(P)$ Given $\det(Q) = 2^{10}$,so $2^9 \det(P) = 2^{10} \implies \det(P) = 2$. We know that $\det(	ext{adj}(	ext{adj } P)) = \det(P)^{(n-1)^2}$,where $n=3$. $\det(	ext{adj}(	ext{adj } P)) = \det(P)^{(3-1)^2} = \det(P)^4 = 2^4 = 16$.

Let $P=[p_{ij}]$ and $Q=[q_{ij}]$ be two square matrices of order $3$ such that $q_{ij}=2^{(i+j-1)}p_{ij}$ and $\det(Q)=2^{10}$. Then the value of $\det(\text{adj}(\text{adj } P))$ is:

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