If the matrix A = ( 0, 2; K, - 1 ) satisfies A(A³ + 3I) = 2I, then the value of K is :

Solution

Maths sir · 4 steps

Use Cayley–Hamilton to bring every power of A down to a combination of A and I, then compare.

1. Characteristic equation. For a 2 × 2 matrix, A² - (trace)A + (det A)I = O. Here the trace is 0 + (-1) = -1 and det A = 0 - 2K = -2K. So A² + A - 2KI = O, that is, A² = -A + 2KI.
2. Climb the powers. Multiply by A each time and replace every A² with -A + 2KI. This gives A³ = (1+2K)A - 2KI and A⁴ = -(1+4K)A + (4K²+2K)I.

   A(A³ + 3I) = A⁴ + 3A = (2-4K)A + (4K²+2K)I

3. Compare with 2I. A has a non-zero off-diagonal entry 2, so it is not a multiple of I. The coefficient of A must therefore vanish: 2 - 4K = 0, which gives K = 1/2. Check the constant part: 4 × 1/4 + 2 × 1/2 = 2, as required.
4. So. K = 1/2, which is option (a).

The idea to keep. For a 2 × 2 matrix, Cayley–Hamilton turns any power into pA + qI. If A is not a scalar matrix, equate the coefficients of A and I separately.

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