# How to express a vector as a linear combination of others?

I have 3 vectors, $(0,3,1,-1), (6,0,5,1), (4,-7,1,3)$, and using Gaussian elimination I found that they are linearly dependent. The next question is to express each vector as a linear combination of the other two. Different resources say just to use Gaussian elimination, but I just end up with a matrix in RREF. How can I find different vectors as a linear combination of others?

Let's look at Gaussian elimination: \begin{align} \begin{bmatrix} 0 & 6 & 4 \\ 3 & 0 & -7 \\ 1 & 5 & 1 \\ -1 & 1 & 3 \end{bmatrix} \xrightarrow{\text{swap row 1 and 3}}{}& \begin{bmatrix} 1 & 5 & 1 \\ 3 & 0 & -7 \\ 0 & 6 & 4 \\ -1 & 1 & 3 \end{bmatrix}\\ \xrightarrow{R_2-3R_1}{}& \begin{bmatrix} 1 & 5 & 1 \\ 0 & -15 & -10 \\ 0 & 6 & 4 \\ -1 & 1 & 3 \end{bmatrix}\\ \xrightarrow{R_4+R_1}{}& \begin{bmatrix} 1 & 5 & 1 \\ 0 & -15 & -10 \\ 0 & 6 & 4 \\ 0 & 6 & 4 \end{bmatrix}\\ \xrightarrow{-\frac{1}{15}R_2}{}& \begin{bmatrix} 1 & 5 & 1 \\ 0 & 1 & 2/3 \\ 0 & 6 & 4 \\ 0 & 6 & 4 \end{bmatrix}\\ \xrightarrow{R_3-6R_2}{}& \begin{bmatrix} 1 & 5 & 1 \\ 0 & 1 & 2/3 \\ 0 & 0 & 0 \\ 0 & 6 & 4 \end{bmatrix}\\ \xrightarrow{R_4-6R_2}{}& \begin{bmatrix} 1 & 5 & 1 \\ 0 & 1 & 2/3 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}\\ \xrightarrow{R_1-5R_2}{}& \begin{bmatrix} 1 & 0 & -7/3 \\ 0 & 1 & 2/3 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}\\ \end{align} If $v_1$, $v_2$ and $v_3$ are your vectors, this says that $$v_3=-\frac{7}{3}v_1+\frac{2}{3}v_2$$ because elementary row operations don't change linear relations between the columns.
Since (0,3,1,-1) is a linear combination of (6,0,5,1) and (4,-7,1,3), we can write $(0,3,1,-1) = a(6,0,5,1) + b(4,-7,1,3)$. This gives us 4 relations to solve for a and b. We can proceed similarly for the other two vectors.
• You should solve the linear equations for $a$ and $b$, namely: $$0=6a+4b,$$ $$3=-7b$$$$\vdots$$ However, I think a better way would be just to solve $$\vec{0} = a(6,0,5,1)+b(4,−7,1,3)+c(4,-7,1,3)$$ – Amihai Zivan Oct 2 '13 at 4:51
• please replace in my comment above $c(4,−7,1,3)$ with $c(0,3,1,−1)$ – Amihai Zivan Oct 2 '13 at 7:02