Consider a system of linear equations \(AX = B\), where A is a square matrix. Which of the following statements are correct?
(A) If \(|A| \ne 0\), the system is consistent and has a unique solution \(X = A^{-1}B\).
(B) If \(|A| = 0\) and \((\mathrm{adj}\,A)B \ne O\), the system has no solution.
(C) If \(|A| = 0\) and \((\mathrm{adj}\,A)B = O\), the system always has infinitely many solutions.
(D) A consistent system always has a unique solution.
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A) and (B) only
A system is consistent if it has one or more solutions, and inconsistent if it has none. The rules for \(AX = B\) with A square are:
\(|A| \ne 0\): unique solution \(X = A^{-1}B\)
\(|A| = 0,\ (\mathrm{adj}\,A)B \ne O\): no solution
\(|A| = 0,\ (\mathrm{adj}\,A)B = O\): may be consistent or inconsistent
An example of the last case with no solution is
\(x + y + z = 1\)
\(x + y + z = 2\)
\(x + y + z = 3\)
Here every cofactor of A is 0, so \(\mathrm{adj}\,A = O\) and \((\mathrm{adj}\,A)B = O\), yet \(x + y + z\) cannot be 1, 2 and 3 at once.
So the test \((\mathrm{adj}\,A)B = O\) decides nothing on its own, while a consistent system can have one solution or infinitely many.
Checking the statements
(A) If \(|A| \ne 0\), the system is consistent and has a unique solution \(X = A^{-1}B\). — Correct. \(A^{-1}\) exists, and multiplying by it gives the only solution.
(B) If \(|A| = 0\) and \((\mathrm{adj}\,A)B \ne O\), the system has no solution. — Correct. This is the rule for an inconsistent system.
(C) If \(|A| = 0\) and \((\mathrm{adj}\,A)B = O\), the system always has infinitely many solutions. — Incorrect. The system \(x + y + z = 1, 2, 3\) satisfies both conditions and has no solution.
(D) A consistent system always has a unique solution. — Incorrect. \(x + y = 3,\ 2x + 2y = 6\) is consistent with infinitely many solutions.