Answer:
Option B
Explanation:
$\begin{bmatrix}3 & 2&1&-4 \\2 & 3&0&-1\\1&-6&3&-8 \end{bmatrix}$
On applying $R_{2} \rightarrow R_{2}-\frac{2}{3} R_{1}$ and $R_{3} \rightarrow R_{3}-\frac{1}{3} R_{1}$
we get
A= $\begin{bmatrix}3 & 2&1&-4 \\0 &5/3&-2/3&5/3\\0&-20/3&8/3&-20/3 \end{bmatrix} $
On applying $R_{3} \rightarrow R_{3}+4R_{2}$, we get
$A=\begin{bmatrix}3 & 2&1&-4 \\0 &5/3&-2/3&5/3\\0&0&0&0 \end{bmatrix} $
$\therefore$ There are two linear independent row, i.e.
$R_{1} and R_{2}$
$\therefore$ Rank of matrix A=2
