A) 4
B) 1
C) 2
D) 3
Correct Answer: D
Solution :
Given \[A=\left[ \begin{matrix} 1 & 3 & 1 \\ 2 & 1 & -1 \\ 3 & 0 & 1 \\ \end{matrix} \right]\] \[=\left[ \begin{matrix} 1 & 3 & 1 \\ 0 & -5 & -3 \\ 0 & -9 & -2 \\ \end{matrix} \right]\] \[[\text{applying }{{\text{R}}_{2}}\to {{R}_{2}}-2{{R}_{1}},{{R}_{3}}\to {{R}_{3}}-3{{R}_{1}}]\] \[\approx \left[ \begin{matrix} 1 & 3 & 1 \\ 0 & -5 & -3 \\ 0 & 0 & \frac{17}{5} \\ \end{matrix} \right]\] \[\left[ {{R}_{3}}\to {{R}_{3}}-\frac{9}{5}{{R}_{2}} \right]\]Hence, rank (A) = 3
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