JEE Main Mathematics Question (2026) — Solution

Step-by-step solution

From B = (I + A)^ -1 we get B(I + A) = I, so BA = I - B. From A + C = I: C = I - A. Calculating CB = C(I+A)^ -1 = (I-A)(I+A)^ -1 . Since B(I+A) = I, we have (I+A) = B^ -1 , so C = (I-A)B^ -1 . Multiplying by B: CB = (I-A). However, using the given BC = bmatrix 1 & -5 \\ -1 & 2 bmatrix and B(I+A) = I implies B - BA = B - (I-B) = 2B - I = BC, i.e. BC = 2B - I. Therefore B = 1 2 (BC + I) = 1 2 bmatrix 2 & -5 \\ -1 & 3 bmatrix = bmatrix 1 & -5/2 \\ -1/2 & 3/2 bmatrix . Also CB = 2B - I = bmatrix 1 & -5 \\ -1 & 2 bmatrix . From CB bmatrix x_1 \\ x_2 bmatrix = bmatrix 12 \\ -6 bmatrix : (CB)^ -1 = - 1 3 bmatrix 2 & 5 \\ 1 & 1 bmatrix . So bmatrix x_1 \\ x_2 bmatrix = - 1 3 bmatrix -6 \\ 6 bmatrix = bmatrix 2 \\ -2 bmatrix . Therefore x_1 + x_2 = 0