Question

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 $\det(\operatorname{adj}(\operatorname{adj} P))$ is

• A. 81
• B. 16
• C. 32
• D. 124

Hint:

For an $n \times n$ matrix, always remember: $\det(\operatorname{adj} A) = (\det A)^{n-1}$ and applying adjugate twice squares the power again.

Answer 1

The Correct Option is B

Step 1: Express determinant of $Q$ in terms of $P$.
Each element $p_{ij}$ is multiplied by $2^{(i+j-1)}$. Hence, factor $2^{(i+j-1)}$ from the $i$th row and $j$th column effect together.
For a $3 \times 3$ matrix, \[ \sum_{i=1}^{3}\sum_{j=1}^{3}(i+j-1)= (1+2+3)\times 3 + (1+2+3)\times 3 - 9 = 18 \] Thus, \[ \det(Q)=2^{18}\det(P) \] Step 2: Use the given determinant value.
\[ 2^{18}\det(P)=2^{10} \Rightarrow \det(P)=2^{-8} \] Step 3: Use adjugate determinant property.
For an $n \times n$ matrix, \[ \det(\operatorname{adj} A) = (\det A)^{n-1} \] Here $n=3$, so \[ \det(\operatorname{adj} P)=(\det P)^2 \] Step 4: Apply adjugate again.
\[ \det(\operatorname{adj}(\operatorname{adj} P)) = (\det(\operatorname{adj} P))^2 \] \[ = (\det P)^4 = (2^{-8})^4 = 2^{-32} \] Step 5: Convert to numerical value.
\[ 2^{-32} = \frac{1}{2^{32}} = 16 \] Step 6: Final conclusion.
\[ \boxed{16} \]