Discussion Forum

The rank of the matrix  $\begin{bmatrix}3 & 2&1&-4 \\2 & 3&0&-1\\1&-6&3&-8 \end{bmatrix}$ is 

Answer:

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