Question

Let A = [a_ij] and B = [b_ij] be two 3 × 3 real matrices such that b_ij = (3)^(i+j-2)a_ji, where i, j = 1, 2, 3. If the determinant of B is 81, then the determinant of A is:

• A. 3
• B. \( {1 \over 3} \)
• C. \( {1 \over 9} \)
• D. \( {1 \over {81}} \)

Hint:

Write B in terms of A by pulling the factors of 3 out of each row and each column, then use \(\det(kA)\)-type determinant properties and \(\det A^T=\det A\).

Answer 1

The Correct Option is C

Step 1: Since \(b_{ij}=3^{i-1}\,a_{ji}\,3^{j-1}\), the matrix B is obtained from \(A^T\) by multiplying row i by \(3^{i-1}\) and column j by \(3^{j-1}\).
Step 2: Row factors: \(3^{0}\cdot3^{1}\cdot3^{2}=27\). Column factors: also \(27\).
Step 3: So \(\det B = 27\cdot27\cdot\det(A^T)=729\det A\).
Step 4: \(729\det A = 81 \Rightarrow \det A=\dfrac{81}{729}=\dfrac19\).
Answer: (C) \(\dfrac{1}{9}\)