- #1

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*The augmented matrix of a system of linear equations in the unknowns x, y, z, u and v, has the form*

[tex]\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)[/tex]

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)

[tex]\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)[/tex]

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?