How was the solution to this system of linear equations reached?

Question

I have the following system of linear equations:

$−1 + 2 − 4 = 0$

$−2 + 3 + 4 = 0$

$1 − 3 = 0$

$−4 − 5 + 6 = 0$

$4 − 6 + 7 = 0$

$5 − 7 = 0$

Or as a matrix:

$\begin{bmatrix} -1&1&0&-1&0&0&0 \\ 0&-1&1&1&0&0&0 \\ 1&0&-1&0&0&0&0 \\ 0&0&0&-1&-1&1&0 \\ 0&0&0&1&0&-1&1 \\ 0&0&0&0&1&0&-1 \end{bmatrix}$ = $0$

I already know that the solution to that system is a vector space spanned by these 3 vectors:

$L_1 = \begin{bmatrix} 1&1&1&0&0&0&0 \end{bmatrix}$

$L_2 = \begin{bmatrix} 0&1&0&1&0&1&0 \end{bmatrix}$

$L_3 = \begin{bmatrix} 0&0&0&0&1&1&1 \end{bmatrix}$

I was able to row-reduce the matrix to the following form, but don't really know where to go from here.

$\begin{bmatrix} -1&1&0&-1&0&0&0 \\ 0&-1&1&1&0&0&0 \\ 0&0&0&-1&-1&1&0 \\ 0&0&0&0&-1&0&1 \\ 0&0&0&0&0&0&0 \\ 0&0&0&0&0&0&0 \end{bmatrix}$

Answer 1

The end of the story consists in continuing upwards, to obtain a reduced row echelon form. I'll write only the $4$ first (non-zero) rows, and colour the pivots: \begin{align} &\begin{bmatrix} -1&1&0&-1&0&0&0\\0&-1&1&1&0&0&0\\0&0&0&-1&-1&1&0\\0&0&0&0&-1&0&1 \end{bmatrix}\rightsquigarrow \begin{bmatrix} 1&-1&0&1&0&0&0\\0&1&-1&-1&0&0&0\\0&0&0&1&1&-1&0\\0&0&0&0&1&0&-1 \end{bmatrix}\rightsquigarrow\\[2ex] \rightsquigarrow&\begin{bmatrix} 1&0&-1&0&0&0&0\\0&1&-1&-1&0&0&0\\0&0&0&1&0&-1&1\\0&0&0&0&1&0&-1 \end{bmatrix}\rightsquigarrow \begin{bmatrix} \color{red}1&0&-1&0&0&0&0\\0&\color{red}1&-1&0&0&-1&1\\0&0&0&\color{red}1&0&-1&1\\0&0&0&0&\color{red}1&0&-1 \end{bmatrix}\rightsquigarrow \end{align} So you can deduce that $$x_1=x_3,\quad x_2=x_3+x_6-x_7,\quad x_4=x_6-x_7,\quad x_5=x_7.$$ Setting successively $\;(x_3,x_6,x_7)=(1,0,0),\; (0,1,0),\; (0,0,1)$, you get $$\begin{bmatrix}x_1\\x_2\\x_3\\x_4\\x_5\\x_6\\x_7\end{bmatrix}=x_3\begin{bmatrix}1\\1\\1\\0\\0\\0\\0\end{bmatrix}+x_6\begin{bmatrix}0\\1\\0\\1\\0\\1\\0\end{bmatrix}+x_7\begin{bmatrix}0\\-1\\0\\-1\\1\\0\\1\end{bmatrix}$$ Your last vector is just the sum of the last two vectors in this formula.