Let P = [ 3, - 1, - 2; 2, 0, α; 3, - 5, 0 ], where α ∈ R. Suppose Q = [ qᵢⱼ] is a matrix satisfying PQ = kl₃ for some non-zero k ∈ R.
If q₂₃ = - k/8 and |Q| = k²/2, then a² + k² is equal to ______.

Solution

Maths sir · 4 steps

PQ equals k times the identity, so Q is just k times the inverse of P. Everything follows from that.

1. Write Q. From PQ = kI₃ we get Q = kP⁻¹ = k/|P| adj P. First find |P| by expanding along row 1: |P| = 3(0 + 5α) + 1(0 - 3α) - 2(-10 - 0) = 12α + 20.
2. Use q23. Entry (2,3) of adj P is the cofactor C₃₂ of P, because the adjoint is the transpose of the cofactor matrix. Deleting row 3 and column 2 leaves |3, -2; 2, α| = 3α + 4, so C₃₂ = -(3α + 4).

   q₂₃ = -k(3α+4)/(12α+20) = -k/8 ⇒ 24α + 32 = 12α + 20 ⇒ α = -1

3. Use |Q|. With α = -1, |P| = 8. For a 3 × 3 matrix, |kP⁻¹| = k³ · 1/|P|, so k³/8 = k²/2. Since k ≠ 0, this gives k = 4.
4. So. α² + k² = 1 + 16 = 17.

The common slip: 1.0625. Taking |Q| = k |P⁻¹| instead of k³|P⁻¹| gives k = 1/4, so the sum comes out as 1.0625, not 17.

The idea to keep. When PQ = kI, write Q = kP⁻¹ straight away. Remember that entry (i,j) of adj P is the cofactor Cⱼᵢ, and that the scalar k comes out of an n × n determinant as kⁿ.

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