Let A be a 2 × 2 real matrix with entries from {0,1} and |A| ≠ 0. Consider the following two statements:<br>(P) If A ≠ I<sub>2</sub>, then |A| = -1<br>(Q) If |A| = 1, then tr(A) = 2,<br>where I<sub>2</sub> denotes 2 × 2 identity matrix and tr(A) denotes the sum of the diagonal entries of A.<br>Then:
(1) (P) is true and (Q) is false
(2) Both (P) and (Q) are false
(3) Both (P) and (Q) are true
(4) (P) is false and (Q) is true
Step-by-Step Solution
Key Concept: Test all possible 2x2 matrices with entries 0 and 1 that have a non-zero determinant. For (P), check if there exists a matrix where |A| is not -1 when A is not identity. For (Q), check if |A|=1 implies tr(A)=2.
The possible 2x2 matrices with entries from {0,1} and |A| ≠ 0 are:<br>A1 = [[1,0],[0,1]] (I2), |A1|=1, tr(A1)=2<br>A2 = [[0,1],[1,0]], |A2|=-1, tr(A2)=0<br>A3 = [[1,1],[1,0]], |A3|=-1, tr(A3)=1<br>A4 = [[0,1],[1,1]], |A4|=-1, tr(A4)=1<br>A5 = [[1,0],[1,1]], |A5|=1, tr(A5)=2<br>A6 = [[1,1],[0,1]], |A6|=1, tr(A6)=2<br>Statement (P): If A ≠ I2, then |A| = -1. Counterexample: A5 or A6, where |A|=1. So (P) is false.<br>Statement (Q): If |A| = 1, then tr(A) = 2. For all matrices with |A|=1 (A1, A5, A6), tr(A)=2. So (Q) is true.
Correct Answer: 4