Let A = [ 2, 3; a, 0 ], a∈R be written as P + Q where P is a symmetric matrix and Q is skew symmetric matrix. If det(Q) = 9, then the modulus of the sum of all possible values of determinant of P is equal to :

Solution

Maths sir · 4 steps

Split A into its symmetric and skew halves. det Q fixes a, and then det P follows.

1. Split the matrix. Every square matrix splits uniquely as P = 1/2(A + Aᵀ) (symmetric) plus Q = 1/2(A - Aᵀ) (skew-symmetric). Here Aᵀ = (2, a; 3, 0).

   P = (2, (3+a)/2; (3+a)/2, 0), Q = (0, (3-a)/2; (a-3)/2, 0)

2. Use det Q. det Q = 0 - (3-a)/2 × (a-3)/2 = (3-a)²/4. Setting this equal to 9 gives (3-a)² = 36, so a = 9 or a = -3.
3. Find det P. det P = 2 × 0 - ((3+a)/2)² = -(3+a)²/4. For a = 9 this is -144/4 = -36. For a = -3 it is 0.
4. So. The sum of the possible values is -36 + 0 = -36, and its modulus is 36, option (a).

The idea to keep. Whenever a matrix is written as symmetric plus skew-symmetric, go straight to P = 1/2(A + Aᵀ) and Q = 1/2(A - Aᵀ). The split is unique, so there is nothing to guess.

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