Let A = [aᵢⱼ] and B = [bᵢⱼ] be two 3 × 3 real matrices such that bᵢⱼ = (3)⁽ⁱ⁺ʲ⁻²⁾aⱼᵢ, where i, j = 1, 2, 3. If the determinant of B is 81, then the determinant of A is:

Solution

Maths sir · 3 steps

Each row of B carries a power of 3 and so does each column: pull both out and you are left with A transpose.

1. Split the power. Write 3ⁱ⁺ʲ⁻² = 3ⁱ⁻¹ × 3ʲ⁻¹. So bᵢⱼ = 3ⁱ⁻¹ × 3ʲ⁻¹ × aⱼᵢ, and aⱼᵢ is the (i, j) entry of Aᵀ.
2. Take out factors. Row i of B has a common factor 3ⁱ⁻¹, which is 1, 3, 9 for the three rows. Column j has a common factor 3ʲ⁻¹, again 1, 3, 9. Take out every one of these factors and the determinant left behind is |Aᵀ| = |A|.

   |B| = (1 × 3 × 9) × (1 × 3 × 9) × |Aᵀ| = 27 × 27 × |A| = 729 |A|

3. So. 729 |A| = 81 gives |A| = 81/729 = 1/9, which is option (c).

The idea to keep. When an entry is a power that depends on both i and j, split it into a row part and a column part. Each part comes out of the determinant separately, and transposing does not change it.

Answer from the NTA final answer key. Solution by PrepWiser.