Let α β ≠ 0 and A=[β, α, 3; α, α, β; -β, α, 2 α]. If B=[3 α, -9, 3 α; -α, 7, -2 α; -2 α, 5, -2 β] is the matrix of cofactors of the elements of A, then det(A B) is equal to :

Solution

Maths sir · 4 steps

Match three cofactors from row 1 to find α and β, then use det(B) = |A|² for a 3×3 matrix.

1. Row 1 cofactors. C₁₁ = 2α² - αβ must equal 3α, so 2α - β = 3. C₁₃ = α² + αβ must equal 3α, so α + β = 3. We can divide by α because αβ ≠ 0.
2. Find α and β. Adding the two equations gives 3α = 6, so α = 2 and β = 1. Check with C₁₂ = -(2α² + β²) = -(8 + 1) = -9, which matches B.
3. Find det A. Expand along row 1 using these cofactors: |A| = β C₁₁ + α C₁₂ + 3C₁₃ = 1(6) + 2(-9) + 3(6) = 6.

   det(B) = |A|ⁿ⁻¹ = |A|² = 36

4. So. det(AB) = |A| × det(B) = 6 × 36 = 216, which is option (d).

The idea to keep. The cofactor matrix is the transpose of adj A, so its determinant is |A|ⁿ⁻¹. For a 3×3 matrix, det(A × cofactor matrix) is always |A|³.

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