Let A = \begin{pmatrix} 2 & 3 \\ a & 0 \end{pmatrix}, a \in \mathbb{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:
Step-by-Step Solution
Key Concept: Any square matrix A can be uniquely written as A = P + Q, where P = (A + A^T)/2 is symmetric and Q = (A - A^T)/2 is skew-symmetric. Calculate P and Q, use the condition det(Q) = 9 to find 'a', then calculate det(P) for those values of 'a'.
Given A = [[2, 3], [a, 0]]. Then A^T = [[2, a], [3, 0]]. P = (A + A^T)/2 = [[2, (3+a)/2], [(3+a)/2, 0]]. Q = (A - A^T)/2 = [[0, (3-a)/2], [(a-3)/2, 0]]. det(Q) = 0 - ((a-3)/2 * (3-a)/2) = -(-(a-3)^2 / 4) = (a-3)^2 / 4. Given det(Q) = 9, so (a-3)^2 / 4 = 9 => (a-3)^2 = 36 => a-3 = \pm6. So a = 9 or a = -3. det(P) = 2*0 - ((3+a)/2)^2 = -(3+a)^2 / 4. If a = 9, det(P) = -(12)^2 / 4 = -144/4 = -36. If a = -3, det(P) = -(0)^2 / 4 = 0. Sum of all possible values of det(P) = -36 + 0 = -36. Modulus of the sum = |-36| = 36.
Correct Answer: 1