Let B=[1, 3; 1, 5] and A be a 2 × 2 matrix such that A B⁻¹=A⁻¹. If B C B⁻¹=A and C⁴+α C²+β I=O, then 2 β-α is equal to

Solution

Maths sir · 5 steps

First show A² = B. Then C² = B, and Cayley–Hamilton on B gives α and β at once.

1. Find A squared. Multiply AB⁻¹ = A⁻¹ on the left by A to get A²B⁻¹ = I. Now multiply on the right by B, which gives A² = B.
2. Find C squared. From BCB⁻¹ = A we get C = B⁻¹AB. Then C² = B⁻¹A²B = B⁻¹BB = B, so C⁴ = B².
3. Rewrite the condition. So C⁴ + α C² + β I = O becomes B² + α B + β I = O. For B, the trace is 1 + 5 = 6 and the determinant is 1 × 5 - 3 × 1 = 2. Cayley–Hamilton then gives:

   B² - 6B + 2I = O

4. Compare. B is not a multiple of I, so this is the only equation of the form B² + α B + β I = O that B satisfies. That gives α = -6 and β = 2.
5. So. 2β - α = 4 + 6 = 10, which is option (b).

The idea to keep. When a matrix is sandwiched as B⁻¹AB, every power is sandwiched too: (B⁻¹AB)ⁿ = B⁻¹AⁿB. For any 2 × 2 matrix M, M² - (tr M)M + (det M)I = O.

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