Question:

Consider a homogeneous system of three linear equations in three unknowns represented by $AX=O$. If $X = \begin{pmatrix}l \\ m \\ 0 \end{pmatrix}, l \neq 0, m \neq 0, l, m \in \mathbb{R}$ represents an infinite number of solutions of this system, then rank of A is

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If the solution space of $AX=0$ contains an entire plane or line, directly think in terms of nullity. Then use the relation: rank + nullity = number of columns.
Updated On: Mar 30, 2026
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The Correct Option is C

Solution and Explanation

Step 1: Interpreting the given solution form:
It is given that $X = [l, m, 0]^T$ is a solution of $AX = O$ for all non-zero $l, m$.
This implies that every vector lying in the xy-plane (i.e., vectors with $z = 0$) satisfies the equation.
Hence, the entire xy-plane is contained in the null space (kernel) of matrix $A$.

Step 2: Finding the nullity of $A$:
The xy-plane is a two-dimensional subspace, and vectors such as \[ \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}, \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} \] form its basis.
Therefore, the null space has dimension $2$, i.e., \[ \text{nullity}(A) = 2. \] Since $A$ is a $3 \times 3$ matrix, the nullity cannot exceed $3$. If it were $3$, all vectors would be solutions, which contradicts the given condition.

Step 3: Applying the Rank-Nullity Theorem:
For a matrix with $n$ columns, \[ \text{rank}(A) + \text{nullity}(A) = n. \] Here, $n = 3$, so \[ \text{rank}(A) + 2 = 3 \Rightarrow \text{rank}(A) = 1. \]
Step 4: Verification using column interpretation:
Since every vector of the form $[l, m, 0]^T$ is a solution, we have \[ A \begin{pmatrix} l \\ m \\ 0 \end{pmatrix} = l \cdot C_1 + m \cdot C_2 = 0, \] where $C_1$ and $C_2$ are the first two columns of $A$.
This is possible only if both $C_1$ and $C_2$ are zero vectors.
Thus, $A$ must be of the form \[ \begin{pmatrix} 0 & 0 & c_1 \\ 0 & 0 & c_2 \\ 0 & 0 & c_3 \end{pmatrix}, \] which clearly has rank $1$ (provided it is not the zero matrix).
Final Conclusion:
Therefore, the rank of matrix $A$ is \[ \boxed{1}. \]
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