Show that $(1,0,1), (-2,1,4), (0,3,1)$ forms the basis for $\in \mathbb {R^3}$.

Question

I want to show that $\begin{bmatrix}1 \\ 0 \\ 1 \end{bmatrix}, \begin{bmatrix}-2 \\ 1 \\ 4 \end{bmatrix}, \begin{bmatrix}0 \\ 3 \\ 1 \end{bmatrix}$ forms the basis for $\in \mathbb {R^3}$. First I tried to show that the vectors are linearly independent

$$α_1\begin{bmatrix}1 \\ 0 \\ 1 \end{bmatrix} + α_2\begin{bmatrix}-2 \\ 1 \\ 4 \end{bmatrix} + α_3\begin{bmatrix}0 \\ 3 \\ 1 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$$

But I tried to continue from this on wards but I get that $α_3=18α_3$, which doesn't make sense so I'm a bit lost.

$$α_1\begin{bmatrix}1 \\ 0 \\ 1 \end{bmatrix} + α_2\begin{bmatrix}-2 \\ 1 \\ 4 \end{bmatrix} + α_3\begin{bmatrix}0 \\ 3 \\ 1 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$$

Answer 1

$$α_1\begin{bmatrix}1 \\ 0 \\ 1 \end{bmatrix} + α_2\begin{bmatrix}-2 \\ 1 \\ 4 \end{bmatrix} + α_3\begin{bmatrix}0 \\ 3 \\ 1 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$$ You got a system of equations: $$\begin{cases}\alpha_1-2\alpha_2&=0\\\alpha_2+3\alpha_3&=0\\\alpha_1+4\alpha_2+\alpha_3&=0 \end{cases} \implies\begin{cases}\alpha_1=2\alpha_2\\\alpha_3=-\frac13\alpha_2\\ 2\alpha_2+4\alpha_2-\frac13\alpha_2=0 \implies\alpha_2=\alpha_1=\alpha_3=0\end{cases}$$

Answer 2

Another equivalent way to @Lorenzo is to show that the determinant of \begin{equation} \begin{bmatrix} 1 & -2 & 0 \\ 0 & 1 & 3 \\ 1 & 4 & 1 \end{bmatrix} \end{equation} is not zero. Doing this will give you: \begin{equation} \det \begin{bmatrix} 1 & -2 & 0 \\ 0 & 1 & 3 \\ 1 & 4 & 1 \end{bmatrix} = 1(1 - 12) + 1(-6 + 0) = -11 - 6 = -17 \neq 0 \end{equation}