# Find a basis for $\Span(S)$, where $S$ is a Set of Four Vectors

## Problem 710

Find a basis for $\Span(S)$ where $S= \left\{ \begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix} , \begin{bmatrix} -1 \\ -2 \\ -1 \end{bmatrix} , \begin{bmatrix} 2 \\ 6 \\ -2 \end{bmatrix} , \begin{bmatrix} 1 \\ 1 \\ 3 \end{bmatrix} \right\}$.

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## Solution.

We will first use the leading $1$ method. Consider the system
$x_{1} \begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix} +x_{2} \begin{bmatrix} -1 \\ -2 \\ -1 \end{bmatrix} +x_{3} \begin{bmatrix} 2 \\ 6 \\ -2 \end{bmatrix} +x_{4} \begin{bmatrix} 1 \\ 1 \\ 3 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} .$ The augmented matrix for this system is
$\left[\begin{array}{cccc|c} 1 & -1 & 2 & 1 & 0 \\ 2 & -2 & 6 & 1 & 0 \\ 1 & -1 & -2 & 3 & 0 \end{array}\right] \xrightarrow{\substack{R_{2}-2R_{1} \\ R_{3}-R_{1}}} \left[\begin{array}{cccc|c} 1 & -1 & 2 & 1 & 0 \\ 0 & 0 & 2 & -1 & 0 \\ 0 & 0 & -4 & 2 & 0 \end{array}\right]$ $\to \left[\begin{array}{cccc|c} 1 & -1 & 0 & 2 & 0 \\ 0 & 0 & 1 & -1/2 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right].$ Since the above matrix has leading $1$’s in the first and third columns, we can conclude that the first and third vectors of $S$ form a basis of $\Span(S)$. Thus
$\left\{ \begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix} , \begin{bmatrix} 2 \\ 6 \\ -2 \end{bmatrix} \right\}$ is a basis for $\Span(S)$.

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