Question ID: #887
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 $|Q|=2^{10}$. Then the value of $|adj(adj~P)|$ is:
- (1) 32
- (2) 16
- (3) 81
- (4) 124
Solution:
$$Q = \begin{bmatrix} 2^{1}p_{11} & 2^{2}p_{12} & 2^{3}p_{13} \\ 2^{2}p_{21} & 2^{3}p_{22} & 2^{4}p_{23} \\ 2^{3}p_{31} & 2^{4}p_{32} & 2^{5}p_{33} \end{bmatrix}$$
Taking out common factors from rows:
Row 1: $2^1$, Row 2: $2^2$, Row 3: $2^3$.
$$|Q| = 2^1 \cdot 2^2 \cdot 2^3 \begin{vmatrix} p_{11} & 2p_{12} & 4p_{13} \\ p_{21} & 2p_{22} & 4p_{23} \\ p_{31} & 2p_{32} & 4p_{33} \end{vmatrix}$$
Now take out common factors from columns of the remaining determinant:
Col 1: $1$, Col 2: $2$, Col 3: $4=2^2$.
$$|Q| = 2^6 \cdot (1 \cdot 2 \cdot 2^2) |P| = 2^6 \cdot 2^3 |P| = 2^9 |P|$$
Given $|Q| = 2^{10}$.
$$2^{9} |P| = 2^{10} \Rightarrow |P| = 2$$
We need to find $|adj(adj~P)|$.
Formula: $|adj(adj~P)| = |P|^{(n-1)^2}$.
Here $n=3$, so exponent is $(3-1)^2 = 4$.
$$|adj(adj~P)| = |P|^4 = 2^4 = 16$$
Ans. (2)
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