Editing Linear independence (section)

=== Vectors in R<sup>4</sup> ===
In order to determine if the three vectors in <math>\mathbb{R}^4,</math> 
:<math>\mathbf{v}_1= \begin{bmatrix}1\\4\\2\\-3\end{bmatrix},  \mathbf{v}_2=\begin{bmatrix}7\\10\\-4\\-1\end{bmatrix}, \mathbf{v}_3=\begin{bmatrix}-2\\1\\5\\-4\end{bmatrix}.</math>
are linearly dependent, form the matrix equation,

:<math>\begin{bmatrix}1&7&-2\\4& 10& 1\\2&-4&5\\-3&-1&-4\end{bmatrix}\begin{bmatrix} a_1\\ a_2 \\ a_3 \end{bmatrix} =  \begin{bmatrix}0\\0\\0\\0\end{bmatrix}.</math>

Row reduce this equation to obtain,
:<math>\begin{bmatrix} 1& 7 & -2 \\ 0& -18& 9\\ 0 & 0 & 0\\ 0& 0& 0\end{bmatrix} \begin{bmatrix} a_1\\ a_2 \\ a_3 \end{bmatrix} =  \begin{bmatrix}0\\0\\0\\0\end{bmatrix}.</math>
Rearrange to solve for v<sub>3</sub> and obtain,
:<math>\begin{bmatrix} 1& 7 \\ 0& -18 \end{bmatrix} \begin{bmatrix} a_1\\ a_2  \end{bmatrix} =  -a_3\begin{bmatrix}-2\\9\end{bmatrix}.</math>
This equation is easily solved to define non-zero ''a''<sub>i</sub>,
:<math>a_1 = -3 a_3 /2,  a_2 = a_3/2,</math>
where <math>a_3</math> can be chosen arbitrarily.  Thus, the vectors <math>\mathbf{v}_1, \mathbf{v}_2,</math> and <math>\mathbf{v}_3</math> are linearly dependent.