Let A=[0, 1, 2; a, 0, 3; 1, c, 0], where a, c ∈ ℝ. If A³=A and the positive value of a belongs to the interval (n-1, n], where n ∈ ℕ, then n is equal to ___________.

Solution

Maths sir · 5 steps

Write Cayley–Hamilton for A, set its A³ equal to A, and match the coefficients.

1. Three invariants. The trace is 0 + 0 + 0 = 0. The principal 2 × 2 minors add to (0 - a) + (0 - 2) + (0 - 3c) = -a - 2 - 3c. Expanding along the first row, det A = -1(0 - 3) + 2(ac - 0) = 3 + 2ac.
2. Cayley–Hamilton. Every 3 × 3 matrix satisfies A³ - (trA)A² + (sum of minors)A - (det A)I = O. With the values above, this gives the equation below.

   A³ = (a + 2 + 3c) A + (3 + 2ac) I

3. Apply the condition. Putting A³ = A gives (a + 1 + 3c)A + (3 + 2ac)I = O. A is not a multiple of I, because its diagonal is zero but it has nonzero entries. So both coefficients must be zero: a + 3c = -1 and ac = -3/2.
4. Solve for a. Put c = (-1 - a)/3 into ac = -3/2. This gives 2a² + 2a - 9 = 0, so a = (-1 ± √19)/2. The positive root is a = (√19 - 1)/2 ≈ (4.36 - 1)/2 ≈ 1.68.
5. So. 1.68 lies in (1, 2], which is (n-1, n] with n = 2. So n = 2.

The common slip: 1. Taking n as the whole-number part of a ≈ 1.68 gives 1. The interval (n-1, n] is closed at the top, so n is the integer just above a, which is 2.

The idea to keep. When a 3 × 3 matrix satisfies a polynomial like A³ = A, write its Cayley–Hamilton equation and compare coefficients of A and I. You only need the trace, the sum of principal minors and the determinant.

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