JEE Main Mathematics Question (2026) — Solution

Step-by-step solution

The determinant of matrix A is: |A| = -1(0 - 0) - 1(1 - 0) - 1(0 - 0) = -1 Using the properties of the adjoint of a 3 3 matrix: adj(adj(A)) = |A|^ 3-2 A = -A adj(A)(adj(adj(A))) = adj(A)(-A) = -|A|I = -(-1)I = I The given equation simplifies to: A^2 - A + I = bmatrix 2 & -2 & 2 \\ -2 & 0 & -1 \\ 0 & 0 & -1 bmatrix Calculating A^2: A^2 = bmatrix -1 & 1 & -1 \\ 1 & 0 & 1 \\ 0 & 0 & 1 bmatrix bmatrix -1 & 1 & -1 \\ 1 & 0 & 1 \\ 0 & 0 & 1 bmatrix = bmatrix 2 & -1 & 1 \\ -1 & 1 & 0 \\ 0 & 0 & 1 bmatrix Substituting A^2, A, and I into the simplified equation: bmatrix 2 & -1 & 1 \\ -1 & 1 & 0 \\ 0 & 0 & 1 bmatrix - bmatrix -1 & 1 & -1 \\ 1 & 0 & 1 \\ 0 & 0 & 1 bmatrix + bmatrix 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 bmatrix = bmatrix 2 & -2 & 2 \\ -2 & 0 & -1 \\ 0 & 0 & -1 bmatrix Comparing the corresponding elements on both sides: From the (1,2) element: -1 - = -2 = 1 From the (2,2) element: 1 + = 0 = -1 Therefore, the value of ( - )^2 is: (1 - (-1))^2 = 2^2 = 4 Answer: 4