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Examine whether the vectors $X_1 = [3,1,1] , X_2=[2,0,-1] X_3 = [4,2,1]$ are Linearly independent
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Consider the matrix equation

$k_1 X_1+k_2 X_2+k_3 X_3=0 \\ ∴k_1 [3,1,1]+k_2 [2,0,-1]+k_3 [4,2,1]=[0,0,0] \\ ∴3k_1+2k_2+4k_3=0 k_1+0k_2+2k_3=0 \\ k_1-k_2+k_3=0$

It can be written as

$\begin{vmatrix} 3&2&4\\1&0&2\\1&-1&1\\ \end{vmatrix} \begin{vmatrix} k_1\\k_2\\ k_3 \\ \end{vmatrix}= \begin{vmatrix} 0\\ 0\\ 0 \\ \end {vmatrix}$

By $R_{13}$

$\begin{vmatrix} 1&-1&1\\1&0&2\\3&2&4\\ \end{vmatrix} \begin{vmatrix} k_1\\k_2\\ k_3 \\ \end{vmatrix}= \begin{vmatrix} 0\\ 0\\ 0 \\ \end {vmatrix}$

By $R_2-R_1$ and $R_3-3R_1$

$\begin{vmatrix} 1&-1&1\\0&1&1\\0&5&1\\ \end{vmatrix} \begin{vmatrix} k_1\\k_2\\ k_3 \\ \end{vmatrix}= \begin{vmatrix} 0\\ 0\\ 0 \\ \end {vmatrix}$

By $R_3-R_2$

$\begin{vmatrix} 1&-1&1\\0&1&1\\0&4&0\\ \end{vmatrix} \begin{vmatrix} k_1\\k_2\\ k_3 \\ \end{vmatrix}= \begin{vmatrix} 0\\ 0\\ 0 \\ \end {vmatrix}$

$∴k_1-k_2+k_3=0 ……………. (1) \\ k_2+k_3=0 ∴ k_2=-k_3 ………..(2) \\ 4k_2=0 ∴ k_2=0 \\ ∴ k_3=0 \space\space\space\space from (2) ……. (3) \\ k_1=0 \space\space\space\space from (1), (2) and (3)$

Since all $k_1,k_2,k_3$ are zero the vectors are Linearly Independent.

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