Let AX = B be a system of three linear equations in three variables.…

The correct answer is Option (2) → (B), (C) and (D) only

Note: The given answer is as per the official NTA Answer Key. However, the answer given by NTA appears to be incorrect, and the correct answer should be Option (4), because statement (D) is not true in every case, as explained below.

For \(AX = B\) with a square matrix A, the standard results are:

    \(|A| \ne 0\): a unique solution, \(X = A^{-1}B\).

    \(|A| = 0\) and \((\mathrm{adj}\,A)B \ne O\): no solution (inconsistent).

    \(|A| = 0\) and \((\mathrm{adj}\,A)B = O\): the system may be consistent or inconsistent. It has infinitely many solutions or none.

Checking the statements

(A) a unique solutions if \(|A| = 0\) — Incorrect. When \(|A| = 0\), \(A^{-1}\) does not exist, so the solution is never unique.

(B) a unique solutions if \(|A| \ne 0\) — Correct. \(X = A^{-1}B\) is the only solution.

(C) no solutions if \(|A| = 0\) and \((\mathrm{adj}\,A)B \ne 0\) — Correct. This is the test for an inconsistent system.

(D) infinitely many solutions if \(|A| = 0\) and \((\mathrm{adj}\,A)B = 0\) — Incorrect. This is not always true. Take

    \(x + y + z = 1\)

    \(x + y + z = 2\)

    \(x + y + z = 3\)

    \( A = \begin{bmatrix}1&1&1\\1&1&1\\1&1&1\end{bmatrix},\quad |A| = 0 \)

    Every 2 × 2 minor of A is 0, so \(\mathrm{adj}\,A = O\) and \((\mathrm{adj}\,A)B = O\).

    But \(x + y + z\) cannot equal 1 and 2 at the same time, so there is no solution.

So only (B) and (C) are true in general.

Tip: \(|A| = 0\) with \((\mathrm{adj}\,A)B = O\) is not enough to decide. Solve or reduce the equations to see whether there are infinitely many solutions or none.

Advice for the exam: In the real exam, mark the answer that is mathematically correct. If NTA’s provisional answer key marks a different option, challenge it during the answer-key challenge window, giving proper reasoning. NTA does revise its key when a challenge is well justified, so a sound challenge can raise your marks.