Let A be a 2 × 2 symmetric matrix such that A[1; 1]=[3; 7] and the determinant of A be 1 . If A⁻¹=α A+β I, where I is an identity matrix of order 2 × 2, then α+β equals _________.

Solution

Maths sir · 4 steps

Pin down A from the three conditions, then let Cayley-Hamilton write A⁻¹ for you.

1. Set up A. A symmetric 2 × 2 matrix has the form A = (a, b; b, c). The given product gives a + b = 3 and b + c = 7, so a = 3 - b and c = 7 - b.
2. Use the determinant. ac - b² = (3 - b)(7 - b) - b² = 21 - 10b = 1, so b = 2. Then a = 1, c = 5, and A = (1, 2; 2, 5) with trace 6.
3. Cayley-Hamilton. Every 2 × 2 matrix satisfies A² - (tr A)A + (det A)I = O, which here is A² - 6A + I = O. Multiply through by A⁻¹ to get the line below.

   A - 6I + A⁻¹ = O ⇒ A⁻¹ = -A + 6I

4. So. α = -1 and β = 6. As a check, 6I - A = (5, -2; -2, 1) times A gives I. So α + β = 5.

The common slip: -5. Moving the terms to the wrong side and writing A⁻¹ = A - 6I gives α = 1, β = -6 and a sum of -5. The correct form is A⁻¹ = 6I - A.

The idea to keep. For any 2 × 2 matrix, A² - (tr A)A + (det A)I = O. Multiply by A⁻¹ and you get the inverse as a mix of A and I straight away.

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