If A = ( 2, 2; 9, 4 ) and I = ( 1, 0; 0, 1 ) then 10A –1 is equal to :

Solution

Maths sir · 4 steps

Use Cayley–Hamilton: trace and determinant give A's equation, then multiply through by A⁻¹.

1. Trace and determinant. For A, the trace is 2 + 4 = 6. The determinant is |A| = 2 × 4 - 2 × 9 = 8 - 18 = -10. It is not zero, so A⁻¹ exists.
2. Cayley–Hamilton. Every 2 × 2 matrix satisfies A² - (trace)A + |A| I = O. Here that gives A² - 6A - 10I = O. Multiply every term by A⁻¹.

   A² - 6A - 10I = O ⇒ A - 6I - 10A⁻¹ = O

3. Check directly. A⁻¹ = 1/-10(4, -2; -9, 2), so 10A⁻¹ = (-4, 2; 9, -2). This equals A - 6I = (2-6, 2; 9, 4-6).
4. So. 10A⁻¹ = A - 6I, which is option (c).

The idea to keep. For any 2 × 2 matrix, A² - (trA)A + |A|I = O. Multiply it by A⁻¹ and you get the inverse in terms of A and I, with no adjoint needed.

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