Let A = ( 1, - 1, 0; 0, 1, - 1; 0, 0, 1 ) and B = 7A²⁰ - 20A⁷ + 2I, where I is an identity matrix of order 3 × 3. If B = [bᵢⱼ], then b₁₃is equal to _____________.

Solution

Maths sir · 4 steps

Split A into I plus a nilpotent N, and the binomial theorem gives any power of A in one line.

1. Split A. Write A = I + N with N = (0, -1, 0; 0, 0, -1; 0, 0, 0). Squaring gives N² = (0, 0, 1; 0, 0, 0; 0, 0, 0), and N³ = O.
2. Binomial expansion. I commutes with N, so the binomial theorem applies. Every term from N³ onwards is zero, which leaves only three terms.

   Aⁿ = I + nN + n(n-1)/2N²

3. Read off entry (1,3). I and N both have 0 in position (1,3), and N² has 1 there. So the (1,3) entry of Aⁿ is n(n-1)/2. That gives 190 for A²⁰ and 21 for A⁷.
4. So. 2I changes only the diagonal, so it adds 0 to b₁₃. Then b₁₃ = 7 × 190 - 20 × 21 = 1330 - 420 = 910.

The idea to keep. When a matrix is I plus a strictly upper triangular N, the powers of N die quickly. Expand (I+N)ⁿ binomially and keep only the first few terms.

Answer from the NTA final answer key. Solution by PrepWiser.