If P is a 3 × 3 real matrix such that Pᵀ=a P+(a-1) I, where a>1, then :

Solution

Maths sir · 5 steps

Take the transpose of both sides and put the given equation back in; P comes out as -I.

1. Transpose both sides. Use (Pᵀ)ᵀ = P and Iᵀ = I. This gives P = aPᵀ + (a-1)I.
2. Substitute for P transpose. Replace Pᵀ with aP + (a-1)I from the question and expand.

   P = a(aP + (a-1)I) + (a-1)I = a²P + (a²-1)I

3. Solve for P. Rearranging gives (1-a²)P = (a²-1)I. Since a>1, a²-1 ≠ 0, so you can divide by it to get P = -I. Check: aP + (a-1)I = -aI + aI - I = -I = Pᵀ.
4. Determinants. |P| = (-1)³ = -1, so P is non-singular. For a 3 × 3 matrix, |adj P| = |P|³⁻¹ = (-1)² = 1.
5. So. |adj P| = 1, which is option (a).

The idea to keep. When a matrix equation involves Pᵀ, transpose it once more and substitute back. Then remember |adj P| = |P|ⁿ⁻¹ for an n × n matrix.

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