# Why is it that the non pivot columns of a matrix allow you to freely pick the variable which scales that column.

I was watching this lecture on linear algebra, how to a linear system of equations where AX=0, where A is some matrix multiplying the vector X. In the lecture it was stated that if upon doing row eliminations a non pivot column is found, we can freely assign any variable for it. Why is this so?

For example, consider the reduced row echelon form system $$\left[\begin{array}{rrrrr|r} 1 & 0 & -5 & 0 & -2 & 6 \\ 0 & 1 & 2 & 0 & 11 & 7 \\ 0 & 0 & 0 & 1 & -16 & -9 \end{array}\right]$$ The "pivot" variables are $$\{x_1, x_2, x_4\}$$ and the "nonpivot" variables are $$\{x_3, x_5\}$$. The first row is the equation $$x_1-5\,x_3-2\,x_5=6$$ which allows us to solve for $$x_1$$ in terms of $$\{x_3, x_5\}$$ giving $$x_1=6+5\,x_3+2\,x_5$$.
The second two equations give $$x_2=7-2\,x_3-11\,x_5$$ and $$x_4=-9+16\,x_5$$.
Putting these relations together gives the general solution to the system as $$\left[\begin{array}{r} x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \\ x_{5} \end{array}\right] = \left[\begin{array}{r} 5 \, x_{3} + 2 \, x_{5} + 6 \\ -2 \, x_{3} - 11 \, x_{5} + 7 \\ x_{3} \\ 16 \, x_{5} - 9 \\ x_{5} \end{array}\right] =\left[\begin{array}{r} 6 \\ 7 \\ 0 \\ -9 \\ 0 \end{array}\right]+x_3\left[\begin{array}{r} 11 \\ 5 \\ 1 \\ -9 \\ 0 \end{array}\right]+x_5\left[\begin{array}{r} 8 \\ -4 \\ 0 \\ 7 \\ 1 \end{array}\right]$$ So, we may "freely" choose values of $$x_3$$ and $$x_5$$. Once these values are set, our formula automatically determines values of $$x_1$$, $$x_2$$, and $$x_4$$.