NDA Mathematics Question (2025) — Solution

Step-by-step solution

Given A = pmatrix 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 pmatrix First, calculating A^2: A^2 = pmatrix 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 pmatrix pmatrix 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 pmatrix A^2 = pmatrix 1+4+4 & 2+2+4 & 2+4+2 \\ 2+2+4 & 4+1+4 & 4+2+2 \\ 2+4+2 & 4+2+2 & 4+4+1 pmatrix = pmatrix 9 & 8 & 8 \\ 8 & 9 & 8 \\ 8 & 8 & 9 pmatrix Now, calculating 4A: 4A = 4 pmatrix 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 pmatrix = pmatrix 4 & 8 & 8 \\ 8 & 4 & 8 \\ 8 & 8 & 4 pmatrix Subtracting 4A from A^2: A^2 - 4A = pmatrix 9 & 8 & 8 \\ 8 & 9 & 8 \\ 8 & 8 & 9 pmatrix - pmatrix 4 & 8 & 8 \\ 8 & 4 & 8 \\ 8 & 8 & 4 pmatrix A^2 - 4A = pmatrix 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 pmatrix A^2 - 4A = 5 pmatrix 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 pmatrix = 5I_3 Answer: 5I_3