JEE Main 2020 · 2 September · morning · Single correct · +4 / −1

Let A be a 2 × 2 real matrix with entries from 0, 1 and |A| ≠ 0. Consider the following two statements :
(P) If A ≠ I₂ , then |A| = –1
(Q) If |A| = 1, then tr(A) = 2,
where I₂ denotes 2 × 2 identity matrix and tr(A) denotes the sum of the diagonal entries of A. Then :

  1. (a) (P) is true and (Q) is false
  2. (b) Both (P) and (Q) are false
  3. (c) Both (P) and (Q) are true
  4. (d) (P) is false and (Q) is true
Answer and solution

Answer (d) (P) is false and (Q) is true

Solution

Maths sir · 4 steps

With only 0s and 1s, ad - bc is just 1 or -1. List every such matrix and test both statements.

  1. Set up. Write A = (a, b; c, d), so |A| = ad - bc. Each product ad and bc is either 0 or 1, so a non-zero determinant can only be 1 or -1.
  2. Two cases. |A| = 1 needs ad = 1 and bc = 0, so a = d = 1. |A| = -1 needs bc = 1 and ad = 0, so b = c = 1. That gives exactly six matrices.
    Matrixad - bc|A|tr(A)
    (1, 0; 0, 1)1 - 012
    (1, 1; 0, 1)1 - 012
    (1, 0; 1, 1)1 - 012
    (0, 1; 1, 0)0 - 1-10
    (1, 1; 1, 0)0 - 1-11
    (0, 1; 1, 1)0 - 1-11
  3. Check P. (1, 1; 0, 1) is not I₂, yet its determinant is 1, not -1. One counterexample is enough, so (P) is false.
  4. So. Every matrix with |A| = 1 has a = d = 1, so its trace is always 2 and (Q) is true. The answer is (d).

The idea to keep. When entries come from a tiny set like {0, 1}, list every case. Here |A| = 1 forces a = d = 1, and one counterexample is enough to kill a general claim.

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

Study Matrices

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