JEE Main Mathematics Question (2026) — Solution

Step-by-step solution

Given A = bmatrix 2 & -2 \\ 4 & -2 bmatrix , we find its determinant: |A| = (2)(-2) - (-2)(4) = 4 Since |A| 0, A is invertible. The inverse of A is: A^ -1 = 1 4 bmatrix -2 & 2 \\ -4 & 2 bmatrix = bmatrix -1/2 & 1/2 \\ -1 & 1/2 bmatrix We are given PA = B P = BA^ -1 and AQ = B Q = A^ -1 B. Using the cyclic property of trace Tr (XY) = Tr (YX), we have: Tr (P) = Tr (BA^ -1 ) = Tr (A^ -1 B) = Tr (Q) Let us calculate Tr (P) by finding the diagonal elements of P = BA^ -1 : P = bmatrix 3 & 9 \\ 1 & 3 bmatrix bmatrix -1/2 & 1/2 \\ -1 & 1/2 bmatrix The diagonal elements of P are: P_ 11 = 3 (- 1 2 ) + 9(-1) = - 21 2 P_ 22 = 1 ( 1 2 ) + 3 ( 1 2 ) = 2 Tr (P) = P_ 11 + P_ 22 = - 21 2 + 2 = - 17 2 Since Tr (P) = Tr (Q) = - 17 2 , the sum of the diagonal elements of 2(P + Q) is: Tr (2(P + Q)) = 2( Tr (P) + Tr (Q)) = 2 (- 17 2 - 17 2 ) = 2(-17) = -34 The absolute value is |-34| = 34. Answer: 34