# Finding the span of 3 vectors in $\mathbb R^4$?

I'm aware that very similar questions have been asked but I could really do with someone walking me through this question.

What is the span of the set

$$\underline a = \begin{pmatrix}1 \\ -1 \\ 2 \\ 1 \end{pmatrix},\ \underline b = \begin{pmatrix}2 \\ -3 \\ 4 \\ 1 \end{pmatrix},\ \underline c = \begin{pmatrix}1 \\ 1 \\ 2 \\ 3 \end{pmatrix}$$

I know I need to find all $$\underline b = \lambda_1\underline a +\lambda_2\underline b + \lambda_3\underline c$$

So find the solution to the equation

$$\begin{pmatrix}1 & 2 & 1 \\ -1 & -3 & 1 \\ 2 & 4 & 2 \\ 1 & 1 & 3\end{pmatrix}\begin{pmatrix}\lambda_1 \\ \lambda_2 \\ \lambda_3 \end{pmatrix}=\begin{pmatrix}b_1 \\ b_2 \\ b_3 \\ b_4\end{pmatrix}$$

But I can't interpret the answer.

Observe that $$\left( \begin{array}{c} 1 \\ 1 \\ 2 \\ 3 \\ \end{array} \right) = 5\left( \begin{array}{c} 1 \\ -1 \\ 2 \\ 1 \\ \end{array} \right)-2\left( \begin{array}{c} 2 \\ -3 \\ 4 \\ 1 \\ \end{array} \right),$$ and $$\underline a$$ is not a scalar multiple of $$\underline b$$. Hence, $$\operatorname{Span}(\{\underline a,\underline b,\underline c\}) = \operatorname{Span}(\{\underline a,\underline b\})$$.
All that you should be looking for is the reduced echelon form (REF) of the matrix: $$\begin{pmatrix}1&2&1\\ \:\:-1&-3&1\\ \:\:2&4&2\\ \:\:1&1&3\end{pmatrix}$$ Which becomes by applying Gaussian elimination: $$\begin{pmatrix}1&2&1\\ 0&1&-2\\ 0&0&0\\ 0&0&0\end{pmatrix}$$ Now the basis of the column space of this matrix is also the basis of the span of the three vectors, and to find it we simply take the vectors in the original matrix which correspond to the columns in the REF that have a leading one: $$\begin{pmatrix}1\\ -1\\ 2\\ 1\end{pmatrix}\:,\:\begin{pmatrix}2\\ \:-3\\ \:4\\ \:1\end{pmatrix}$$