linear algebra, show B is a basis for $\mathbb{R}^3$ Let $B = \{(1,0,1), (1,1,2), (1,2,4)\}$
a) Show that $B$ is a basis for $\mathbb{R}^3$
b) Compute the coordinate vector $[v]_B$ with respect to $B$ of the vector $v = (3, -1, 3).$

To show it is a basis you have to show it generates $\mathbb{R}^3$, so every $$v=\begin{pmatrix} x \\y \\z \end{pmatrix}$$ can be written as 
$$\lambda_1 \begin{pmatrix} 1\\ 0\\1 \end{pmatrix} + \lambda_2 \begin{pmatrix} 1\\1\\2\\ \end{pmatrix} + \lambda_3 \begin{pmatrix} 1\\2\\ 4\\ \end{pmatrix}$$ 
A basis is minimal, so you have to show that 
$$\lambda_1 \begin{pmatrix} 1\\ 0\\1 \end{pmatrix} + \lambda_2 \begin{pmatrix} 1\\1\\2\\ \end{pmatrix} + \lambda_3 \begin{pmatrix} 1\\2\\ 4\\ \end{pmatrix} = 0 \implies \lambda_1=\lambda_2=\lambda_3=0$$ 
You don't need to calculate this here, because 3 vectors which span $\mathbb{R}^3$ must be linear independent.  
If you already know what a determinant is, you can even calculate 
$$\begin{vmatrix} 1 & 0 & 1 \\ 1 & 1 & 2 \\ 1 & 2 & 4\\ \end{vmatrix}=1$$ 
in b) you shall calculate $\lambda_1, \lambda_2, \lambda_3 $ for the given vector.
