Let X = [ 0, 1, 0; 0, 0, 1; 0, 0, 0 ], Y = α I + β X + γ X² and Z = α ²I - α β X + (β ² - α γ )X², α, β, γ ∈ R. If Y⁻¹ = [ 1/5, (- 2)/5, 1/5; 0, 1/5, (- 2)/5; 0, 0, 1/5 ], then (α - β + γ)² is equal to ____________.

Solution

Maths sir · 4 steps

Check that YZ = α³ I, then match Z/α³ entry by entry with the given inverse.

1. Powers of X. X² has a single 1 in the top-right corner, and X³ = O. So Y is upper triangular, with α on the diagonal, β just above it, and γ in the corner.
2. Multiply Y and Z. Expand YZ and drop every X³ and X⁴ term. The X terms give -α²β + α²β = 0. The X² terms give α(β² - αγ) - αβ² + α²γ = 0. So YZ = α³ I, which means Y⁻¹ = Z/α³.

   Y⁻¹ = 1/αI - β/α²X + (β² - αγ)/α³X²

3. Match entries. Diagonal: 1/α = 1/5, so α = 5. Just above the diagonal: β/25 = 2/5, so β = 10. Corner: (100 - 5γ)/125 = 1/5, so 100 - 5γ = 25 and γ = 15.
4. So. α - β + γ = 5 - 10 + 15 = 10, and its square is 100.

The common slip: 0.64. Reading α, β, γ straight off Y⁻¹ as if it were Y gives α = 1/5, β = -2/5, γ = 1/5. That leads to (4/5)² = 0.64, but those are the entries of the inverse, not of Y.

The idea to keep. When X is nilpotent (X³ = O), any aI + bX + cX² with a ≠ 0 has an inverse of the same form. Multiply and force the X and X² terms to vanish.

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