JEE Main 2022 · 26 July · morning · Single correct · +4 / −1

Let A be a 2 × 2 matrix with det (A) = - 1 and det ((A + I) (Adj (A) + I)) = 4. Then the sum of the diagonal elements of A can be :

  1. (a) -1
  2. (b) 2
  3. (c) 1
  4. (d) - √2
Answer and solution

Answer (b) 2

Solution

Maths sir · 4 steps

Use A adjA = det(A) I and the whole product collapses to A + adjA.

  1. Expand the product. Multiply out: (A + I)(adjA + I) = A adjA + A + adjA + I. Since A adjA = det(A) I = -I, the -I and +I cancel, leaving A + adjA.
  2. Write the adjoint. For A = (a, b; c, d), the adjoint swaps the diagonal entries and negates the off-diagonal ones: adjA = (d, -b; -c, a). Adding the two cancels the off-diagonal entries.

    A + adjA = (a+d, 0; 0, a+d) = (a+d) I

  3. Take the determinant. For a 2 by 2 matrix, det((a+d) I) = (a+d)². Setting this equal to 4 gives a + d = ± 2.
  4. So. The sum of the diagonal elements can be 2 or -2. Among the options only 2 appears, so the answer is (b).

The idea to keep. For any 2 by 2 matrix, A + adjA = (tr A) I. And whenever you see A adjA, replace it with det(A) I straight away.

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

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Asked about 110 times in JEE Main since 2020. This question is from 2022.