Linear combinations of four-dimensional vectors

Question

Let $v_1  = begin{pmatrix} 1 \ 1 \ 0 \ 0 end{pmatrix}$ $v_2  = begin{pmatrix} 0 \ 1 \ 1 \ 0 end{pmatrix}$ $v_3  =  begin{pmatrix} 0 \ 0 \ 1 \ 1 end{pmatrix}$ $v_4  =  begin{pmatrix} 2 \ 0 \ 0 \ 1 end{pmatrix}$ $mathbb{R}^4$ vectors.
Show that every $v in mathbb{R}^{4times1}$  can be written as vectors $(v_1,v_2,v_3,v_4)$ linear combination.
My attempt:
$left[begin{array}{cccc|l}
1 & 0 & 0 & 2  &v_1\
1 & 1 & 0 & 0  & v_2\
0 & 1 & 1 & 0  &v_3\
0 & 0 & 1 & 1  &v_4\
end{array}
right]$
Where do I go from here? Every input is appreciated.

Raffaele · Accepted Answer

As $det A=-1ne 0$, given any vector $B=(b_1,b_2,b_3,b_4)$, we solve the linear system $AX=B$ with
$$A=left(
begin{array}{cccc}
 1 & 0 & 0 & 2 \
 1 & 1 & 0 & 0 \
 0 & 1 & 1 & 0 \
 0 & 0 & 1 & 1 \
end{array}
right);;X=left(
x_1,x_2,x_3,x_4
right)$$
and thank to Cramer's theorem we get as unique solution  the components $X$ of $B$ in the base $left(v_1,v_2,v_3,v_4right)$.

J. W. Tanner · Answer

$begin{vmatrix}1&0&0&2\1&1&0&0\0&1&1&0\0&0&1&1end{vmatrix}=begin{vmatrix}1&0&0\1&1&0\0&1&1end{vmatrix}-begin{vmatrix}0&0&2\1&1&0\0&1&1end{vmatrix}=1+begin{vmatrix}0&2\1&1end{vmatrix}=-1ne0$,
so the four vectors are linearly independent, so they span $mathbb R^4$,
so any vector in $mathbb R^4$ can be expressed as a linear combination of them.

Linear combinations of four-dimensional vectors

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