The number of matrices A=(a, b; c, d), where a, b, c, d ∈{-1,0,1,2,3, … …, 10}, such that A=A⁻¹, is ___________.

Solution

Maths sir · 4 steps

Turn A = A⁻¹ into A² = I, then split on whether the trace a + d is zero.

1. Rewrite the condition. Multiply A = A⁻¹ by A to get A² = I. Square the matrix and match it entry by entry with I.

   A² = (a² + bc, b(a+d); c(a+d), bc + d²) = (1, 0; 0, 1)

2. Nonzero trace. If a + d ≠ 0, the off-diagonal entries force b = c = 0, so a² = d² = 1. Since a + d ≠ 0, we need a = d = 1 or a = d = -1. That gives 2 matrices.
3. Zero trace. If a + d = 0, then d = -a, and both must lie in the set, so a ∈ {-1, 0, 1}. The only condition left is a² + bc = 1.

   (a, d)	Need	Choices of (b, c)	Count
   (0, 0)	bc = 1	(1, 1) or (-1, -1)	2
   (1, -1)	bc = 0	b = 0 or c = 0: 12 + 12 - 1	23
   (-1, 1)	bc = 0	b = 0 or c = 0: 12 + 12 - 1	23

4. So. The set has 12 values, which gives the 23s above. The total is 2 + 2 + 23 + 23 = 50.

The common slip: 48. Counting only the trace-zero case and missing A = I and A = -I gives 48, not 50.

The idea to keep. For a 2 × 2 matrix with A² = I, the trace decides everything. Either A = ± I, or the trace is 0 and det A = -1. Always check both cases.

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