If the matrix A = [ 1, 0, 0; 0, 2, 0; 3, 0, - 1 ] satisfies the equation
A²⁰ + α A¹⁹ + β A = [ 1, 0, 0; 0, 4, 0; 0, 0, 1 ] for some real numbers α and β, then β - α is equal to ___________.

Solution

Maths sir · 4 steps

Square A first: A² turns out diagonal, so every power of A follows an even and odd pattern.

1. Find the pattern. Multiply A by itself. The third row gives (3 × 1 + (-1) × 3, 0, (-1)(-1)) = (0, 0, 1), so A² = diag(1, 4, 1). Every even power is therefore diag(1, 2ⁿ, 1), and every odd power looks like A with 2ⁿ in the middle.

   A²⁰ = (1, 0, 0; 0, 2²⁰, 0; 0, 0, 1), A¹⁹ = (1, 0, 0; 0, 2¹⁹, 0; 3, 0, -1)

2. Compare entries. The (1,1) entry gives 1 + α + β = 1, so β = -α. The (3,1) and (3,3) entries give the same condition, so they add nothing new.
3. Middle entry. The (2,2) entry gives 2²⁰ + 2¹⁹α + 2β = 4. Put β = -α: α(2¹⁹ - 2) = -(2²⁰ - 4) = -2(2¹⁹ - 2), so α = -2 and β = 2.
4. So. β - α = 2 - (-2) = 4.

The common slip: -4. Computing α - β instead of β - α gives -4. Read the order asked for before the last line.

The idea to keep. For a high power of a matrix, compute A² first. If it comes out diagonal or equal to I, the powers repeat in a short cycle and the big exponent stops being scary.

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