JEE Main Mathematics Question (2026) — Solution

Step-by-step solution

Let C_1, C_2, C_3 be the columns of A and R_1, R_2, R_3 be the rows of A. From the given equations, we have: A bmatrix 0\\0\\1 bmatrix = C_3 = bmatrix 1\\3\\1 bmatrix A bmatrix 1\\0\\1 bmatrix = C_1 + C_3 = bmatrix 3\\4\\4 bmatrix C_1 = bmatrix 2\\1\\3 bmatrix Similarly, for the transpose A^T, the columns correspond to the rows of A: A^T bmatrix 0\\0\\1 bmatrix = R_3^T = bmatrix 3\\1\\1 bmatrix R_3 = bmatrix 3 & 1 & 1 bmatrix A^T bmatrix 1\\0\\1 bmatrix = R_1^T + R_3^T = bmatrix 5\\2\\2 bmatrix R_1^T = bmatrix 2\\1\\1 bmatrix R_1 = bmatrix 2 & 1 & 1 bmatrix Using the elements of C_1, C_3, R_1, and R_3, we can construct the matrix A with an unknown central element b: A = bmatrix 2 & 1 & 1 \\ 1 & b & 3 \\ 3 & 1 & 1 bmatrix We are given that (A) = 1. Expanding along the first row: (A) = 2(b - 3) - 1(1 - 9) + 1(1 - 3b) = 1 2b - 6 + 8 + 1 - 3b = 1 3 - b = 1 b = 2 Thus, the matrix A is: A = bmatrix 2 & 1 & 1 \\ 1 & 2 & 3 \\ 3 & 1 & 1 bmatrix We need to find ( adj (A^2 + A)). Using the properties of determinants and adjoints: ( adj (A^2 + A)) = ( (A^2 + A))^2 = ( (A) (A + I))^2 First, find A + I: A + I = bmatrix 3 & 1 & 1 \\ 1 & 3 & 3 \\ 3 & 1 & 2 bmatrix Now, calculate (A + I): (A + I) = 3(6 - 3) - 1(2 - 9) + 1(1 - 9) (A + I) = 3(3) - 1(-7) + 1(-8) = 9 + 7 - 8 = 8 Since (A) = 1, we have: (A^2 + A) = 1 8 = 8 Finally, for a 3 3 matrix: ( adj (A^2 + A)) = 8^ 3-1 = 8^2 = 64 Answer: 64