Linear Equations Solutions: Augmented Matrix and Unknown Variables

[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?

 

[courtrigrad]

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:

$$\left(\begin{array}{c} -4+3s+2t\\ s\\ 3-t\\ t\\ -1\\ \end{array} \right) = \left(\begin{array}{c} -4\\ 0\\ 3\\ 0\\ -1\\ \end{array} \right) + s\left(\begin{array}{c} 3\\ 1\\ 0\\ 0\\ 0\\ \end{array} \right) + t \left(\begin{array}{c} 2\\ 0\\ -1\\ 1\\ 0\\ \end{array} \right)$$

 

[Architeuthis Dux]

Yes, it is perfectly fine to make u and y arbitrary in this case. This is because the system of linear equations has infinitely many solutions, so we can choose any value for u and y and still have a valid solution. This is known as the "free variable" method, where we choose certain variables to be arbitrary and then solve for the remaining variables in terms of those arbitrary ones. In this case, we can choose u = t and y = s, and then solve for the remaining variables in terms of s and t. Your understanding is correct in that we do not know what z and x are at this point, but by choosing u and y to be arbitrary, we can still find a solution for those variables in terms of s and t.