Troubleshooting a System of Linear Equations

In summary, the conversation is about a person seeking help with a problem they have been struggling with all day. They provide equations and an augmented matrix, but they make a mistake in reducing the matrix which leads to incorrect solutions. After receiving help, they realize their mistake and feel embarrassed.
  • #1
Exulus
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0
I'm hoping someone can come along and give me a tap on the head for being so silly, but I've been trying this problem *all day* and i just can't seem to get a correct answer, even when checked with other people, we're all stumped! What we have:

[itex]3x - 2y + 5z = 0[/itex]
[itex]x + y + 5z = 5[/itex]
[itex]x - 2y - z = -4[/itex]

I've then put this into an augmented matrix, and converted to reduce row echelon form which gives me these equations:

[itex]x + z = 2[/itex]
[itex]y + z = 3[/itex]

So I try a general solution set as (t, 1+t, 2-t) however it doesn't seem to work with the equations at the beginning. so i looked at what i'd written and went back a step before row echelon form which has these equations:

[itex] x + y + 5z = 5[/itex]
[itex] y + z = 3[/itex]

Giving a solution set of (4t-10, t, 3-t) but still this doesn't seem to be consistent. Any ideas where I am going wrong? Thanks :)
 
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  • #2
You made a mistake reducing your matrix.

You should get x = 0 , y = 5/3 z = 2/3.
 
  • #3
Hmm, are you sure? Because the question asks for the "general solutions" rather than any specific one. According to the way i reduced my matrix, there are infinitely many solutions. I'll have a go at typing out what I've done:

[itex]\left(\begin{array}{ccc|c}3&-2&5&0\\1&1&5&5\\1&-2&-1&-4\end{array}\right)[/itex]

R1 <-> R2

[itex]\left(\begin{array}{ l c c | r } 1 & 1 & 5 & 5 \\ 3 & -2 & 5 & 0 \\ 1 & -2 & -1 & -4 \end{array}\right)[/itex]

R2 -> R2 - 3R1
R3 -> R3 - R1

[itex]\left(\begin{array}{ccc|c}1&1&5&5\\0&-5&-10&-15\\0&-3&-6&-9\end{array}\right)[/itex]

R2 -> -1/5 x R2
R3 -> -1/3 x R3

[itex]\left(\begin{array}{ccc|c}1&1&5&5\\0&1&2&3\\0&1&2&3\end{array}\right)[/itex]

R3 -> R3 - R2
R1 -> R1 - R2

[itex]\left(\begin{array}{ccc|c}1&0&3&2\\0&1&2&3\\0&0&0&0\end{array}\right)[/itex]

Sorry for the dodgy formatting..it doesn't seem to like me..hope that made sense though? If anyone knows how to stop it cutting off the top of the numbers then that would be great :)
 
Last edited:
  • #4
Well there's your problem. In your original post you dropped the coefficients for z.
So you have x+3z =2, y+2z=3, z=t. So x=2-3t, y=3-2t, z=t.
 
  • #5
I can't believe i didnt see that :smile: I knew i must've done something silly! Thanks for pointing that stupid mistake out...i think i'll go crawl in a hole now :biggrin:
 

Related to Troubleshooting a System of Linear Equations

1. What is a system of linear equations?

A system of linear equations is a set of two or more equations that contain two or more variables. The solution to a system of linear equations is the set of values that make all of the equations true simultaneously.

2. How do you solve a system of linear equations?

There are several methods for solving a system of linear equations, including substitution, elimination, and graphing. These methods involve manipulating the equations to isolate one variable and then using that value to solve for the other variables.

3. Can a system of linear equations have more than one solution?

Yes, a system of linear equations can have one, infinite, or no solutions. A system with one solution has a unique set of values for the variables that satisfy all of the equations. A system with infinite solutions has multiple sets of values that satisfy the equations. A system with no solutions has no set of values that satisfy all of the equations.

4. What is the significance of solving a system of linear equations?

Solving a system of linear equations can have several real-world applications, such as finding the intersection point of two lines, determining the optimum solution in linear programming problems, and predicting future values in mathematical models.

5. Can a system of linear equations be solved using matrices?

Yes, a system of linear equations can be solved using matrices. The equations can be written in matrix form and then solved using matrix operations, such as row reduction, to find the solution. This method is particularly useful for systems with a large number of equations and variables.

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