System of Linear Equations

  • Thread starter roam
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  • #1
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Homework Statement



This is from a solved problem:

Here is a system:

[tex]\begin{bmatrix} {1 & 0&0 \\ -1&0&0\\3&-5&0 \end{bmatrix} \begin{bmatrix} {x \\ y\\z \end{bmatrix}= \begin{bmatrix} {0\\ 0\\0\end{bmatrix}[/tex]

A general solution of this system is

[tex]x= \begin{bmatrix} {0 \\ 0\\t \end{bmatrix}= \begin{bmatrix} {0 \\ 0\\1 \end{bmatrix} t[/tex]

So, how did they get this solution? I tried solving this and I just don't get it!

The Attempt at a Solution



The matrix corresponds to the set of equations:

x=0
-x=0
3x-5y=0

If I take "y" to be the free variable, I have x=5/3y therefore the solution to the system will be

[tex]\begin{bmatrix} {x \\ y\\z \end{bmatrix} = \begin{bmatrix} {5/3 \\ 1\\0 \end{bmatrix} y + \begin{bmatrix} {0 \\ 0\\1 \end{bmatrix} z[/tex]

But if I take "x" to be the free variable I get y=3/5x, so I get the following solution:

[tex]\begin{bmatrix} {x \\ y\\z \end{bmatrix} = \begin{bmatrix} {1 \\ 3/5\\0 \end{bmatrix} x + \begin{bmatrix} {0 \\ 0\\1 \end{bmatrix} z[/tex]

How do I know which one (x or y) should be taken as the free variable?

Could anyone please explain to me how to get the right answer? :frown: I just can't see how to deal with this type of problems.
 

Answers and Replies

  • #2
35,224
7,042

Homework Statement



This is from a solved problem:



So, how did they get this solution? I tried solving this and I just don't get it!

The Attempt at a Solution



The matrix corresponds to the set of equations:

x=0
-x=0
3x-5y=0
From these equations, x is obviously 0. With x = 0, the 3rd equation implies that y = 0. This means that z is the free variable.
If I take "y" to be the free variable, I have x=5/3y therefore the solution to the system will be

[tex]\begin{bmatrix} {x \\ y\\z \end{bmatrix} = \begin{bmatrix} {5/3 \\ 1\\0 \end{bmatrix} y + \begin{bmatrix} {0 \\ 0\\1 \end{bmatrix} z[/tex]

But if I take "x" to be the free variable I get y=3/5x, so I get the following solution:

[tex]\begin{bmatrix} {x \\ y\\z \end{bmatrix} = \begin{bmatrix} {1 \\ 3/5\\0 \end{bmatrix} x + \begin{bmatrix} {0 \\ 0\\1 \end{bmatrix} z[/tex]

How do I know which one (x or y) should be taken as the free variable?

Could anyone please explain to me how to get the right answer? :frown: I just can't see how to deal with this type of problems.
 

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