Question

For a system \(AX = b\) of linear equations to have infinitely many solutions, which one of the following options must be true?

• A. Determinant of the matrix A is non-zero.
• B. The rank of the matrix A is less than the rank of the augmented matrix \((A|b)\).
• C. The rank of the matrix A is equal to the rank of the augmented matrix \((A|b)\), but less than the number of variables.
• D. The system must be homogeneous.

Hint:

Apply the rank theorem: infinitely many solutions need equal ranks that are less than the number of unknowns.

Answer 1

The Correct Option is C

Step 1: The rank (Rouche-Capelli) theorem governs the solvability of a linear system \(AX = b\). The system is consistent, meaning it has at least one solution, if and only if the rank of the coefficient matrix \(A\) equals the rank of the augmented matrix \((A|b)\).

Step 2: Once consistency holds, compare the common rank to the number of unknowns \(n\). If rank\((A)\) = rank\((A|b)\) = \(n\), the system has exactly one solution. If rank\((A)\) = rank\((A|b)\) \(< n\), the system has infinitely many solutions, since \(n - \text{rank}(A)\) variables remain free.

Step 3: Eliminate the incorrect options. (A) A non-zero determinant means \(A\) is invertible, which gives a unique solution, not infinitely many, so (A) is wrong. (B) If rank\((A) <\) rank\((A|b)\), the augmented matrix carries information not present in \(A\), which makes the system inconsistent, meaning no solution exists, so (B) is wrong. (D) A homogeneous system always has at least the trivial solution and can still have a unique solution when rank\((A) = n\), so homogeneity alone does not guarantee infinitely many solutions, so (D) is wrong.

Infinitely many solutions occur exactly when rank\((A)\) = rank\((A|b)\) but this common value is less than the number of variables.

\[\boxed{\text{Option (C)}}\]