Question:
Let \(m>1\) and \(n>1\) be integers.
Let \(A\) be an \(m \times n\) matrix such that for some \(m \times 1\) matrix \(b_1,\)
the equation \(A x = b_1\) has infinitely many solutions.
Let \(b_2\) denote an \(m \times 1\) matrix different from \(b_1.\)
Then \(A x = b_2\) has
Let \(m>1\) and \(n>1\) be integers.
Let \(A\) be an \(m \times n\) matrix such that for some \(m \times 1\) matrix \(b_1,\)
the equation \(A x = b_1\) has infinitely many solutions.
Let \(b_2\) denote an \(m \times 1\) matrix different from \(b_1.\)
Then \(A x = b_2\) has
Show Hint
A consistent system with infinitely many solutions means rank deficiency in \(A.\)
That guarantees existence of some vectors \(b_2\) outside the column space,
for which the system becomes inconsistent.
Updated On: Dec 6, 2025
- infinitely many solutions for some \(b_2.\)
- a unique solution for some \(b_2.\)
- no solution for some \(b_2.\)
- finitely many solutions for some \(b_2.\)
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The Correct Option is A, C
Solution and Explanation
Step 1: Given condition.
The system \(A x = b_1\) has infinitely many solutions \(\Rightarrow\) system is consistent and \(\operatorname{rank}(A)<n.\)
Step 2: Implication for other right-hand sides.
The system \(A x = b_2\) is consistent \(\Leftrightarrow b_2 \in \mathcal{R}(A)\) (range of \(A\)). Since \(\mathcal{R}(A)\) is a proper subspace of \(\mathbb{R}^m\) (because rank \(< m\)), there exists some \(b_2 \notin \mathcal{R}(A)\). For such \(b_2,\) no solution exists.
Step 3: Conclusion.
Hence, (C) is correct.
The system \(A x = b_1\) has infinitely many solutions \(\Rightarrow\) system is consistent and \(\operatorname{rank}(A)<n.\)
Step 2: Implication for other right-hand sides.
The system \(A x = b_2\) is consistent \(\Leftrightarrow b_2 \in \mathcal{R}(A)\) (range of \(A\)). Since \(\mathcal{R}(A)\) is a proper subspace of \(\mathbb{R}^m\) (because rank \(< m\)), there exists some \(b_2 \notin \mathcal{R}(A)\). For such \(b_2,\) no solution exists.
Step 3: Conclusion.
Hence, (C) is correct.
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