JEE Main Mathematics Question (2026) — Solution

Step-by-step solution

Let A = I + C, where I is the identity matrix and C = bmatrix 0 & 0 & 0 \\ 3 & 0 & 0 \\ 9 & 3 & 0 bmatrix . Calculating the powers of C, we get: C^2 = bmatrix 0 & 0 & 0 \\ 3 & 0 & 0 \\ 9 & 3 & 0 bmatrix bmatrix 0 & 0 & 0 \\ 3 & 0 & 0 \\ 9 & 3 & 0 bmatrix = bmatrix 0 & 0 & 0 \\ 0 & 0 & 0 \\ 9 & 0 & 0 bmatrix C^3 = bmatrix 0 & 0 & 0 \\ 0 & 0 & 0 \\ 9 & 0 & 0 bmatrix bmatrix 0 & 0 & 0 \\ 3 & 0 & 0 \\ 9 & 3 & 0 bmatrix = bmatrix 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 bmatrix Since C^3 = 0, all higher powers of C are also zero matrices. Using the binomial expansion for A^ 99 : A^ 99 = (I + C)^ 99 = I + 99C + ^ 99 C_ 2 C^2 Given B = A^ 99 - I, we have: B = 99C + 4851C^2 Substituting the matrices C and C^2: B = 99 bmatrix 0 & 0 & 0 \\ 3 & 0 & 0 \\ 9 & 3 & 0 bmatrix + 4851 bmatrix 0 & 0 & 0 \\ 0 & 0 & 0 \\ 9 & 0 & 0 bmatrix From this, we can extract the required elements of B: b_ 31 = 99 9 + 4851 9 = 9(99 + 4851) = 9 4950 = 44550 b_ 21 = 99 3 = 297 b_ 32 = 99 3 = 297 Now, substituting these values into the given expression: b_ 31 - b_ 21 b_ 32 = 44550 - 297 297 = 44550 297 - 1 b_ 31 - b_ 21 b_ 32 = 150 - 1 = 149 Answer: 149