Question:
If $(x \;\; 3 \;\; -1)\begin{pmatrix} 1 & 1 & 1 \\ -1 & 0 & 1 \\ 1 & 0 & -1 \end{pmatrix}\begin{pmatrix} 2 \\ 3 \\ 1 \end{pmatrix} = 0$, then the values of $x$ are:
If $(x \;\; 3 \;\; -1)\begin{pmatrix} 1 & 1 & 1 \\ -1 & 0 & 1 \\ 1 & 0 & -1 \end{pmatrix}\begin{pmatrix} 2 \\ 3 \\ 1 \end{pmatrix} = 0$, then the values of $x$ are:
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Always multiply matrices stepwise (column first, then row) to avoid mistakes.
Updated On: Apr 24, 2026
- $-2$
- $-\frac{1}{3}$
- $-3$
- $\frac{2}{3}$
- $-\frac{2}{3}$
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The Correct Option is D
Solution and Explanation
Concept:
• Matrix multiplication step-by-step
• Final result is scalar (since row × column)
Step 1: Multiply matrix and column vector
\[ \begin{pmatrix} 1 & 1 & 1 -1 & 0 & 1 1 & 0 & -1 \end{pmatrix} \begin{pmatrix} 2 3 1 \end{pmatrix} = \begin{pmatrix} 2+3+1 -2+0+1 2+0-1 \end{pmatrix} = \begin{pmatrix} 6 -1 1 \end{pmatrix} \]
Step 2: Multiply with row matrix
\[ (x \;\; 3 \;\; -1) \begin{pmatrix} 6 -1 1 \end{pmatrix} = 6x - 3 - 1 \] \[ = 6x - 4 \]
Step 3: Solve equation
\[ 6x - 4 = 0 \Rightarrow 6x = 4 \Rightarrow x = \frac{2}{3} \] Final Conclusion:
\[ x = \frac{2}{3} \]
• Matrix multiplication step-by-step
• Final result is scalar (since row × column)
Step 1: Multiply matrix and column vector
\[ \begin{pmatrix} 1 & 1 & 1 -1 & 0 & 1 1 & 0 & -1 \end{pmatrix} \begin{pmatrix} 2 3 1 \end{pmatrix} = \begin{pmatrix} 2+3+1 -2+0+1 2+0-1 \end{pmatrix} = \begin{pmatrix} 6 -1 1 \end{pmatrix} \]
Step 2: Multiply with row matrix
\[ (x \;\; 3 \;\; -1) \begin{pmatrix} 6 -1 1 \end{pmatrix} = 6x - 3 - 1 \] \[ = 6x - 4 \]
Step 3: Solve equation
\[ 6x - 4 = 0 \Rightarrow 6x = 4 \Rightarrow x = \frac{2}{3} \] Final Conclusion:
\[ x = \frac{2}{3} \]
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