Let A = [ x, y, z; y, z, x; z, x, y ], where x, y and z are real numbers such that x + y + z > 0 and xyz = 2. If A² = I₃, then the value of x³ + y³ + z³ is ____________.

Solution

Maths sir · 4 steps

Square A by multiplying rows into rows; A² = I then gives the sum of squares and the pair-products directly.

1. Read off A squared. A is symmetric, so each column of A equals the matching row. The entry (i, j) of A² is therefore row i dotted with row j.

   A² = (x²+y²+z², xy+yz+zx, xy+yz+zx; xy+yz+zx, x²+y²+z², xy+yz+zx; xy+yz+zx, xy+yz+zx, x²+y²+z²) = I₃

2. Compare with I. The diagonal gives x²+y²+z² = 1 and the off-diagonal gives xy+yz+zx = 0. Then (x+y+z)² = 1 + 2 × 0 = 1, and since x+y+z > 0, we get x+y+z = 1.
3. Use the cube identity. x³+y³+z³ - 3xyz = (x+y+z)(x²+y²+z² - xy - yz - zx) = 1 × (1 - 0) = 1.
4. So. x³+y³+z³ = 1 + 3xyz = 1 + 3 × 2 = 7.

The common slip: 5. Taking x+y+z = -1 gives 6 - 1 = 5. The condition x+y+z > 0 is there to rule out this negative root.

The idea to keep. Every row of this cyclic matrix holds the same three numbers. So A² = I turns into x²+y²+z² = 1 and xy+yz+zx = 0, and the identity for x³+y³+z³ - 3xyz finishes the job.

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