Question

Show that the set 𝑆 = {(1, 0, 1)(1,1, 0)(−1, 0, −1)} is linearly dependent in 𝑉3(𝑅)

Answer 1

Let $\vec v_1=\begin{pmatrix} 1\\ 0 \\ 1 \end{pmatrix}, \vec v_2=\begin{pmatrix} 1\\ 1 \\ 0 \end{pmatrix}$ and $\vec v_3=\begin{pmatrix} -1\\ 0 \\ -1 \end{pmatrix}.$

We have to determine whether or not we can find real numbers $a_1, a_2, a_3$ which are not all zero, such that

$$a_1\vec v_1+a_2\vec v_2+a_3\vec v_3=\vec 0$$

Substitute

$$a_1\begin{pmatrix} 1\\ 0 \\ 1 \end{pmatrix}+a_2\begin{pmatrix} 1\\ 1 \\ 0 \end{pmatrix}+a_3\begin{pmatrix} -1\\ 0 \\ -1 \end{pmatrix}=\begin{pmatrix} 0\\ 0 \\ 0 \end{pmatrix}$$

Augmented matrix

$$A=\begin{pmatrix} 1 & 1 & -1 & & 0 \\ 0 & 1 & 0 & & 0 \\ 1 & 0 & -1 & & 0 \\ \end{pmatrix}$$

$R_3=R_3-R_1$

$$\begin{pmatrix} 1 & 1 & -1 & & 0 \\ 0 & 1 & 0 & & 0 \\ 0 & -1 & 0 & & 0 \\ \end{pmatrix}$$

$R_1=R_1-R_2$

$$\begin{pmatrix} 1 & 0 & -1 & & 0 \\ 0 & 1 & 0 & & 0 \\ 0 & -1 & 0 & & 0 \\ \end{pmatrix}$$

$R_3=R_3+R_2$

$$\begin{pmatrix} 1 & 0 & -1 & & 0 \\ 0 & 1 & 0 & & 0 \\ 0 & 0 & 0 & & 0 \\ \end{pmatrix}$$

If we take $a_3=1,$ then the solution is $(a_1, a_2, a_3)=(1, 0, 1)$

So this set of equations has a non–zero solution.

Therefore, set $𝑆 = \{(1, 0, 1)(1,1, 0)(−1, 0, −1)\}$ is a linearly dependent in $V^3.$