Question:
Consider the linear system of equations \[ \begin{bmatrix} 3 & -1 & 4 \\ 6 & 3 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \end{bmatrix}. \] In this system of equations, if \(x\) is always a fixed constant, then the system has:& nbsp;
Consider the linear system of equations \[ \begin{bmatrix} 3 & -1 & 4 \\ 6 & 3 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \end{bmatrix}. \] In this system of equations, if \(x\) is always a fixed constant, then the system has:& nbsp;
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For a consistent system with fewer equations than unknowns, at least one free variable survives. Such systems generally admit infinitely many solutions.
Updated On: Jun 25, 2026
- No solution
- Infinite solutions
- Unique solution
- Exactly two solutions
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The Correct Option is B
Solution and Explanation
Concept:
The system contains two equations in three unknowns. Such systems generally possess infinitely many solutions whenever they are consistent.
A unique solution is impossible because the number of variables exceeds the number of independent equations.
Step 1: Write the equations explicitly.
\[ 3x-y+4z=1 \] \[ 6x+3y+z=1. \]
Step 2: Eliminate one variable.
Multiply the first equation by \(3\), \[ 9x-3y+12z=3. \] Adding to the second equation, \[ 15x+13z=4. \] Hence, \[ x=\frac{4-13z}{15}. \]
Step 3: Use the given condition that \(x\) is fixed.
For \(x\) to remain constant, \(z\) can still vary while corresponding values of \(y\) adjust accordingly. Since one free parameter remains, the solution set represents a line in \(\mathbb R^3\). Therefore the system possesses infinitely many solutions. \[ \boxed{\text{Infinite solutions}} \]
Step 1: Write the equations explicitly.
\[ 3x-y+4z=1 \] \[ 6x+3y+z=1. \]
Step 2: Eliminate one variable.
Multiply the first equation by \(3\), \[ 9x-3y+12z=3. \] Adding to the second equation, \[ 15x+13z=4. \] Hence, \[ x=\frac{4-13z}{15}. \]
Step 3: Use the given condition that \(x\) is fixed.
For \(x\) to remain constant, \(z\) can still vary while corresponding values of \(y\) adjust accordingly. Since one free parameter remains, the solution set represents a line in \(\mathbb R^3\). Therefore the system possesses infinitely many solutions. \[ \boxed{\text{Infinite solutions}} \]
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