Let A = { aᵢⱼ} be a 3 × 3 matrix,
where aᵢⱼ = { ( - 1)ʲ⁻ⁱ, if, i < j,; 2, if, i = j,; ( - 1)ⁱ⁺ʲ, if, i > j
then det (3Adj(2A⁻¹)) is equal to _____________.

Solution

Maths sir · 4 steps

Build A, find its determinant, then peel off the 3, the adjoint and the inverse one by one.

1. Build the matrix. Above the diagonal a₁₂ = (-1)¹ = -1, a₁₃ = (-1)² = 1, a₂₃ = -1. Below it a₂₁ = (-1)³ = -1, a₃₁ = (-1)⁴ = 1, a₃₂ = (-1)⁵ = -1. So A = (2, -1, 1; -1, 2, -1; 1, -1, 2).
2. Find det A. Expand along the first row: |A| = 2(4 - 1) + 1(-2 + 1) + 1(1 - 2) = 6 - 1 - 1 = 4.
3. Use the rules. For an n × n matrix, |kM| = kⁿ |M| and |adj M| = |M|ⁿ⁻¹. Here n = 3, so |2A⁻¹| = 2³ × 1/4 = 2.

   |3 adj(2A⁻¹)| = 3³ |2A⁻¹|² = 27 × 2²

4. So. det(3 adj(2A⁻¹)) = 27 × 4 = 108.

The common slip: 12. Taking |3M| = 3|M| instead of 3³|M| gives 3 × 4 = 12. A scalar multiplies every row, so it comes out as 3³.

The idea to keep. Pulling a scalar out of a determinant raises it to the power n, and |adj M| = |M|ⁿ⁻¹. Work from the outside in, one layer at a time.

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