Question:medium

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

Show Hint

Apply the rank theorem: infinitely many solutions need equal ranks that are less than the number of unknowns.
Updated On: Jul 3, 2026
  • Determinant of the matrix A is non-zero.
  • The rank of the matrix A is less than the rank of the augmented matrix \((A|b)\).
  • The rank of the matrix A is equal to the rank of the augmented matrix \((A|b)\), but less than the number of variables.
  • The system must be homogeneous.
Show Solution

The Correct Option is C

Solution and Explanation

Method: Think in terms of the solution set structure rather than quoting the rank theorem directly.

Any solution of $AX = b$ can be written as $X = X_p + X_h$, where $X_p$ is one particular solution and $X_h$ is any vector in the null space of $A$, i.e. any solution of $AX = 0$.

For a particular solution $X_p$ to exist at all, the system must be consistent. By the fundamental consistency criterion, this happens exactly when adding the column $b$ to $A$ does not increase the rank, that is, rank$(A) = $ rank$(A|b)$.

Given consistency, the number of solutions depends entirely on the size of the null space. By the rank-nullity theorem, if $A$ has $n$ columns (unknowns):

\[ \dim(\text{null space of } A) = n - \text{rank}(A) \]

If rank$(A) = n$, the null space contains only the zero vector, so $X = X_p$ is the unique solution.

If rank$(A) < n$, the null space has dimension at least $1$, so infinitely many vectors $X_h$ can be added to $X_p$, producing infinitely many solutions.

Combining both requirements, infinitely many solutions occur precisely when the system is consistent, rank$(A) = $ rank$(A|b)$, and this rank is strictly less than the number of variables $n$.

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