Let A=(m, n; p, q), d=|A| ≠ 0 and |A-d(A d j A)|=0. Then

Solution

Maths sir · 4 steps

Write out adj A, take the determinant, and replace np by mq − d so everything collapses.

1. Build the matrix. For a 2 × 2 matrix, swap the diagonal entries and change the sign of the other two: adj A = (q, -n; -p, m). So A - d (adj A) = (m - dq, n(1+d); p(1+d), q - dm).
2. Expand the determinant. Setting it to zero gives (m - dq)(q - dm) - np(1+d)² = 0. The first product is mq - dm² - dq² + d² mq.
3. Remove np. Since d = mq - np, put np = mq - d. Then np(1+d)² = mq(1+d)² - d(1+d)² = mq + 2dmq + d² mq - d(1+d)².

   mq - dm² - dq² + d²mq - mq - 2dmq - d²mq + d(1+d)² = d[(1+d)² - (m+q)²]

4. So. The condition becomes d[(1+d)² - (m+q)²] = 0. Since d ≠ 0, (1+d)² = (m+q)², which is option (d).

The idea to keep. In 2 × 2 determinant questions, write adj A by swapping the diagonal and changing the off-diagonal signs. Then use np = mq - d to remove the off-diagonal product, and the expression usually factorises.

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