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] …