Discussion Forum

If A= $\begin{bmatrix}1 & 3&1 \\2 & 1&-1\\3&0&1 \end{bmatrix}$  , then rank(A) is equal to 

Answer:

Explanation:

$A= \begin{bmatrix}1 & 3&1 \\2 & 1&-1\\3&0&1 \end{bmatrix}=\begin{bmatrix}1 & 3&1 \\0 & -5&-3\\0&-9&-2 \end{bmatrix}$

[Applying R₂→  R₂ -2R₁   , R₃ →   R₃ -3R₁ ]

$\approx \begin{bmatrix}1 & 3&1 \\0 & -5&-3\\0&0&\frac{17}{5} \end{bmatrix}$

$[R_{3}\rightarrow R_{3}-\frac{9}{5}R_{2}]$

  $\therefore$  rank (A)=3