Let A = [ x, 1; 1, 0 ], x ∈ R and A⁴ = [aᵢⱼ].
If a₁₁ = 109, then a₂₂ is equal to _______ .

Solution

Maths sir · 4 steps

Square A once, then square the result: A⁴ depends only on x², so solve for x² first.

1. Square A. Multiply row by column: A² = (x²+1, x; x, 1). Getting A⁴ as (A²)² needs just one more multiplication.

   A⁴ = (x²+1, x; x, 1)² = ((x²+1)² + x², x(x²+2); x(x²+2), x²+1)

2. Use a11 = 109. Put t = x². Then (t+1)² + t = 109, which gives t² + 3t - 108 = 0, that is (t+12)(t-9) = 0.
3. Pick the valid root. Since x ∈ ℝ, t = x² cannot be negative. So t = -12 is rejected and x² = 9.
4. So. a₂₂ = x² + 1 = 9 + 1 = 10.

The common slip: 9. Stopping at x² = 9 and reporting it as a₂₂ gives 9. The entry is x² + 1, so add the 1.

The idea to keep. For a high power like A⁴, square twice instead of multiplying four times. When only x² shows up, solve for x² directly and reject any negative root.

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