Solutions in matrices

  • Thread starter ultima9999
  • Start date
  • #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)


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?
 

Answers and Replies

  • #2
1,235
1
Yeah you can do that . The basic variables are [tex] x, z , v [/tex]. The free variables are [tex] y, u [/tex]. If we let [tex] y = s [/tex] and [tex] u = t [/tex] then the solution is:

[tex]\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)[/tex]
 
Last edited:

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