Answer to Question #290592 in Linear Algebra for abed

Question

 Study whether the vectors v1(1, 1, 2), v2(2, 3, 0), v3(0, 1, 2) in R 3 are linearly independent.

Answer 1

A sequence of vectors $\vec v_1, \vec v_2,..., \vec v_k$  from a vector space $V$ is said to be linearly dependent, if there exist scalars $a_1, a_2, ..., a_k$ not all zero, such that

$$a_1\vec v_1+a_2\vec v_1+...+a_k\vec v_k=\vec 0$$

where $\vec 0$ denotes the zero vector.

Consider the set of vectors $\vec v_1=(1,1,2), \vec v_2=(2,3,0), \vec v_3=(0,1,2)$ then the condition for linear dependence seeks a set of non-zero scalars, such that

$$\begin{bmatrix} 1 & 2 & 0 \\ 1 & 3 & 1 \\ 2 & 0 & 2 \\ \end{bmatrix}\begin{bmatrix} a_1 \\ a_2 \\ a_3 \end{bmatrix}=\begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$$

Augmented matrix

$$\begin{bmatrix} 1 & 2 & 0 & \ 0 \\ 1 & 3 & 1 & \ 0 \\ 2 & 0 & 2 & \ 0 \\ \end{bmatrix}$$

$R_2=R_2-R_1$

$$\begin{bmatrix} 1 & 2 & 0 & \ 0 \\ 0 & 1 & 1 & \ 0 \\ 2 & 0 & 2 & \ 0 \\ \end{bmatrix}$$

$R_3=R_3-2R_1$

$$\begin{bmatrix} 1 & 2 & 0 & \ 0 \\ 0 & 1 & 1 & \ 0 \\ 0 & -4 & 2 & \ 0 \\ \end{bmatrix}$$

$R_1=R_1-2R_2$

$$\begin{bmatrix} 1 & 0 & -2 & \ 0 \\ 0 & 1 & 1 & \ 0 \\ 0 & -4 & 2 & \ 0 \\ \end{bmatrix}$$

$R_3=R_3+4R_2$

$$\begin{bmatrix} 1 & 0 & -2 & \ 0 \\ 0 & 1 & 1 & \ 0 \\ 0 & 0 & 6 & \ 0 \\ \end{bmatrix}$$

$R_3=R_3/6$

$$\begin{bmatrix} 1 & 0 & -2 & \ 0 \\ 0 & 1 & 1 & \ 0 \\ 0 & 0 & 1 & \ 0 \\ \end{bmatrix}$$

$R_1=R_1+2R_3$

$$\begin{bmatrix} 1 & 0 & 0 & \ 0 \\ 0 & 1 & 1 & \ 0 \\ 0 & 0 & 1 & \ 0 \\ \end{bmatrix}$$

$R_2=R_2-R_3$

$$\begin{bmatrix} 1 & 0 & 0 & \ 0 \\ 0 & 1 & 0 & \ 0 \\ 0 & 0 & 1 & \ 0 \\ \end{bmatrix}$$

Then $a_1=a_2=a_3=0.$

Therefore the vectors $\vec v_1=(1,1,2), \vec v_2=(2,3,0), \vec v_3=(0,1,2)$ in $R ^3$ are linearly independent.