X1= [3,1,1]

X2=[2,0,-1] and

X3=[4,2,1]

• Consider the matrix equation where$k_1, k_2, k_3$ are scalars such that$k_1 X_1 + k_2X_2+k_3X_3=0$
• After substitution we get,

$k_1 [3,1,1]+k_2[2,0,-1]+k_3[4,2,1]=[0,0,0] $

• Hence

$3k_1+2k_2+4k_3=0\\ 1k_1+0k_2+2k_3=0\\ 1k_1-1k_2+1k_3=0$

• In matrix form, it can be written as:

$\begin{bmatrix} 3 & 2 & 4 \\[0.3em] 1 &0 & 2 \\[0.3em] 1 & -1 & 1 \end{bmatrix}$$\begin{bmatrix} k_1 \[0.3em] k_2 \[0.3em] k_3 \end{bmatrix}$= $\begin{bmatrix} 0 \[0.3em] 0 \[0.3em] 0 \end{bmatrix}$   * By R13;      $\begin{bmatrix} 1 & -1 & 1 \[0.3em] 1 &0 & 2 \[0.3em] 3 & 2 & 4 \end{bmatrix}$$\begin{bmatrix} k_1 \[0.3em] k_2 \[0.3em] k_3 \end{bmatrix}$=$\begin{bmatrix} 0 \\[0.3em] 0 \\[0.3em] 0 \end{bmatrix}$

• By R2 - R1; R3 - 3R1;

$\begin{bmatrix} 1 & -1 & 1 \\[0.3em] 0 &1 & 1 \\[0.3em] 0 & 5 & 1 \end{bmatrix}$$\begin{bmatrix} k_1 \[0.3em] k_2 \[0.3em] k_3 \end{bmatrix}$= $\begin{bmatrix} 0 \[0.3em] 0 \[0.3em] 0 \end{bmatrix}$   * By R3 - R2; $\begin{bmatrix} 1 & -1 & 1 \[0.3em] 0 &1 & 1 \[0.3em] 0 & 4 & 0 \end{bmatrix}$$\begin{bmatrix} k_1 \[0.3em] k_2 \[0.3em] k_3 \end{bmatrix}$= $\begin{bmatrix} 0 \[0.3em] 0 \[0.3em] 0 \end{bmatrix}$   * Hence $k_1 - k_2 + k_3 = 0; $ $k_2+k_3=0;$ $4k_2=0 $ * Thus$k_2=0, k_3=0, $ hence$k_1=0$

As all k1, k2, k3 are zero, vectors X1, X2 and X3  are linearly independent.