Let I be the identity matrix of order 3 × 3 and for the matrix A=[λ, 2, 3; 4, 5, 6; 7, -1, 2],|A|=-1. Let B be the inverse of the matrix adj(Aadj(A²)). Then |(λ B+I)| is equal to______

Solution

Maths sir · 4 steps

Skip computing adjoints: the rules collapse the whole matrix to -A, then one determinant finishes it.

1. Find lambda. Expand along the first row: |A| = λ(10+6) - 2(8-42) + 3(-4-35) = 16λ - 49. Setting 16λ - 49 = -1 gives λ = 3.
2. Collapse the matrix. adj(A²) = (adjA)², so A adj(A²) = (A adjA)adjA = |A|adjA = -adjA. For a 3 × 3 matrix, adj(kM) = k²adjM and adj(adjA) = |A| A, so adj(-adjA) = |A| A = -A.
3. Rewrite the target. So B = (-A)⁻¹ = -A⁻¹, and λ B + I = I - 3A⁻¹ = A⁻¹(A - 3I). Since |A⁻¹| = 1/|A| = -1, the determinant becomes -|A - 3I|.

   |A - 3I| = |0, 2, 3; 4, 2, 6; 7, -1, -1| = 0 - 2(-4-42) + 3(-4-14) = 92 - 54 = 38

4. So. |λ B + I| = -1 × 38 = -38.

The common slip: 38. Forgetting the factor |A⁻¹| = -1 when writing I - 3A⁻¹ = A⁻¹(A - 3I) gives 38 instead of -38.

The idea to keep. For 3 × 3: A adjA = |A|I, adj(kA) = k²adjA, adj(adjA) = |A|A. Shrink nested adjoints first, compute last.

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