Solving for Nontrivial Solutions of a System of Linear Equations

In summary, the system has a nontrivial solution when the determinant of the coefficient matrix is equal to 0. This occurs when a = 3 or a = -1. The reduced row echelon form of the matrix can be used to determine these values.
  • #1
nietzsche
186
0

Homework Statement



For what values of a does the system

(a-1)x + 2y = 0
2x + (a-1)y = 0

have a nontrivial solution?

Homework Equations


The Attempt at a Solution



Argh... I'm really bad at linear algebra, can't seem to grasp the concepts.

I know that if I make this into a matrix

[tex]
\left[
\begin{array}{cc}
a-1 & 2\\
2 & a-1
\end{array}
\right]
[/tex]

that the only way the system will have a nontrivial solution is if the reduced row echelon form is not the 2x2 identity, i.e. the matrix is singular. But I don't know how to use those ideas to solve it. Any help would be great!
 
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  • #2


If the determinant is 0, what does that tell you about the system?
 
  • #3


Pengwuino said:
If the determinant is 0, what does that tell you about the system?

For some reason, my prof hasn't taught us anything about the determinant. I've read about it, and I sort of have an idea of how it works, but I don't think our prof wants us to use it to solve problems.
 
  • #4


Ok, so start reducing it to echelon form. Divide the first row by a-1. Under what conditions can you fail to get ones on the diagonal?
 
  • #5


my intuition is telling me that a = 3 ... i did it like this:

[tex]
\left[
\begin{array}{cc}
2 & a-1\\
a-1 & 2
\end{array}
\right]
\\
\text{then}
\\
\left[
\begin{array}{cc}
1 & \frac{a-1}{2}\\
\frac{a-1}{2} & 1
\end{array}
\right]
[/tex]

so from this we see that [tex]\frac{a-1}{2} = 1[/tex] because that would make row 2 a multiple of row 1, and subtracting it out would leave a row of zeros. so a = 3.

does it make sense? i feel like I'm missing something.
 
  • #6


That makes a lot of sense. Sure a=3 is a problem. But keep going with the row reduction. Multiply the first row by (a-1)/2 and subtract it from the second row. There is another value of a that creates a problem.
 
Last edited:
  • #7


i was thinking that maybe a = 1 is also a problem, but if you substitute a = 1 into the original matrix, you can transform it into the 2x2 identity.

but if i try to put it into rref i end up having to divide by a-1, and if a=1 then it is undefined.

so I'm confused about that...
 
  • #8


nietzsche said:
my intuition is telling me that a = 3 ... i did it like this:

[tex]
\left[
\begin{array}{cc}
2 & a-1\\
a-1 & 2
\end{array}
\right]
\\
\text{then}
\\
\left[
\begin{array}{cc}
1 & \frac{a-1}{2}\\
\frac{a-1}{2} & 1
\end{array}
\right]
[/tex]

so from this we see that [tex]\frac{a-1}{2} = 1[/tex] because that would make row 2 a multiple of row 1, and subtracting it out would leave a row of zeros. so a = 3.

does it make sense? i feel like I'm missing something.

Like I edited my last post to say, just multiply the first row by (a-1)/2 and subtract from the second row. Then ask yourself when you can get a second row of zeros.
 
  • #9


ah, i see

a = -1

thanks again dick. you're really saving my skin.
 
  • #10


nietzsche said:
ah, i see

a = -1

thanks again dick. you're really saving my skin.

Well, you are helping by using the hints and thinking about them. Not everyone does that.
 

1. What are nontrivial solutions in a system of linear equations?

Nontrivial solutions in a system of linear equations refer to solutions that are not equal to zero or do not involve all unknown variables being equal to zero.

2. How do you solve for nontrivial solutions in a system of linear equations?

To solve for nontrivial solutions in a system of linear equations, you can use methods such as substitution, elimination, or Gauss-Jordan elimination. These methods involve manipulating the equations to isolate and solve for the unknown variables.

3. Can a system of linear equations have multiple nontrivial solutions?

Yes, a system of linear equations can have multiple nontrivial solutions if there are more unknown variables than equations, or if the equations are not linearly independent. In these cases, there may be an infinite number of possible solutions.

4. How do you determine if a system of linear equations has no nontrivial solutions?

If a system of linear equations has no nontrivial solutions, it means that the equations are inconsistent and there is no set of values that can satisfy all of the equations simultaneously. This can be determined by using methods such as substitution or elimination and arriving at a statement that is always false.

5. What is the importance of finding nontrivial solutions in a system of linear equations?

Finding nontrivial solutions in a system of linear equations is important in many fields of science and mathematics, such as engineering, physics, and economics. These solutions can represent real-world scenarios and provide valuable information for decision making and problem solving.

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