Let A be a 2 × 2 matrix with real entries such that A'=α A+I, where α ∈ ℝ-{-1,1}. If det(A²-A)=4, then the sum of all possible values of α is equal to :

Solution

Maths sir · 5 steps

Take the transpose of the given relation once more, and A is forced to be a multiple of I.

1. Transpose again. Taking the transpose of A' = α A + I gives A = α A' + I. Substitute A' = α A + I into this: A = α(α A + I) + I = α² A + (α + 1)I.
2. Find A. So (1 - α²)A = (1 + α)I. Since α ≠ ± 1, divide by (1-α)(1+α) to get A = kI with k = 1/(1-α).
3. Use the determinant. Now A² - A = (k² - k)I. For a 2 × 2 matrix, det(cI) = c², so (k² - k)² = 4 and k² - k = ± 2.

   k² - k - 2 = 0 ⇒ k = 2 or k = -1

4. Back to alpha. The other case, k² - k + 2 = 0, has discriminant 1 - 8 < 0, so it gives no real k. From α = 1 - 1/k: k = 2 gives α = 1/2, and k = -1 gives α = 2.
5. So. Both values satisfy α ≠ ± 1, and their sum is 1/2 + 2 = 5/2, which is option (d).

The idea to keep. When a relation links A' and A, transpose it once more and substitute. That usually pins A down to a scalar matrix kI, and then every condition becomes plain algebra in k.

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