# Solutions in matrices

1. Nov 13, 2006

### ultima9999

The augmented matrix of a system of linear equations in the unknowns x, y, z, u and v, has the form
$$\left(\begin{array}{ccccc|c} 1 & -3 & 1 & -1 & 0 & -1\\ 0 & 0 & 1 & 1 & 2 & 1\\ 0 & 0 & 0 & 0 & 1 & -1\\ 0 & 0 & 0 & 0 & 0 & 0 \end{array} \right)$$
Write down all solutions of the system.

The answer to this is (x, y, z, u, v) = (-4 + 3s + 2t, s, 3 - t, t, -1)

I have:
v = -1
u = 1 - 2v - z = 3 - z
z = 1 - 2v - u = 3 - u
-3y = -4 + 2t - x
x = -4 + 2t + 3y

Now, I can see that they just went ahead and made u and y arbitrary. It is perfectly fine to do that?

I'm thinking that we can allow u = t because at that point we do not know what z is; and when we solve for y we do not know what x is. Am I right in thinking that?

2. Nov 13, 2006

Yeah you can do that . The basic variables are $$x, z , v$$. The free variables are $$y, u$$. If we let $$y = s$$ and $$u = t$$ then the solution is:
$$\left(\begin{array}{c} -4+3s+2t\\ s\\ 3-t\\ t\\ -1\\ \end{array} \right) = \left(\begin{array}{c} -4\\ 0\\ 3\\ 0\\ -1\\ \end{array} \right) + s\left(\begin{array}{c} 3\\ 1\\ 0\\ 0\\ 0\\ \end{array} \right) + t \left(\begin{array}{c} 2\\ 0\\ -1\\ 1\\ 0\\ \end{array} \right)$$