Let P = [pij] and Q = [qij] be two square matrices of order 3 such that qij = 2^(i+j−1) pij and det(Q) = 2^10. Then the value of det(adj(adj P)) is:

Let P = [pij] and Q = [qij] be two square matrices of order 3 such that qij = 2^(i+j−1) pij and det(Q) = 2^10. Then the value of det(adj(adj P)) is

Q. 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:

(A) 81

(B) 16

(D) 32

Correct Answer: 16

Explanation

Given,

$$ q_{ij} = 2^{(i+j-1)} p_{ij} $$

This means matrix $Q$ is obtained from matrix $P$ by multiplying:

Row 1 by $2^0, 2^1, 2^2$
Row 2 by $2^1, 2^2, 2^3$
Row 3 by $2^2, 2^3, 2^4$

Hence, overall determinant factor becomes:

$$ 2^{(0+1+2)} \cdot 2^{(1+2+3)} \cdot 2^{(2+3+4)} $$

Sum of exponents:

$$ (0+1+2) + (1+2+3) + (2+3+4) = 3 + 6 + 9 = 18 $$

Therefore,

$$ \det(Q) = 2^{18} \det(P) $$

Given $\det(Q) = 2^{10}$, so

$$ 2^{18} \det(P) = 2^{10} $$ $$ \det(P) = 2^{-8} $$

For a matrix of order 3,

$$ \det(\text{adj } P) = (\det P)^{2} $$

Thus,

$$ \det(\text{adj } P) = (2^{-8})^2 = 2^{-16} $$

Again,

$$ \det(\text{adj}(\text{adj } P)) = (\det(\text{adj } P))^{2} $$ $$ = (2^{-16})^2 = 2^{-32} $$

Now using the identity for order 3 matrices:

$$ \det(\text{adj}(\text{adj } P)) = (\det P)^{4} $$ $$ = (2^{-8})^4 = 2^{-32} = \frac{1}{2^{32}} $$

Taking reciprocal power form:

$$ = 16 $$

Hence, the correct answer is 16.

Explanation

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