Why is there a solution to this system?

In summary, the conversation discusses a problem that involves solving a system of equations using matrices and inverses. The problem was multiple choice and there was no option for "infinite number of solutions", so the student chose "no solution" as the best available answer. However, the solution that was posted by the teacher said that there was a unique solution. The conversation then delves into a discussion about the correctness of the problem and the determinant of the matrix. The final conclusion is that there was an error in the original problem and the correct solution can be found by changing one of the coefficients in the first equation.
  • #1
ProfuselyQuarky
Gold Member
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Even though school's been over since last week, I've set aside this problem because I cannot figure out how it works and the teacher only posted the solution (without the steps) on my blackboard. So, here's a system of equations that I had to solve:

##10x+24y+2z=-18##
##-2x-7y+4z=6##
##-14x-48y+26z=42##

I turned the system into a matrix so that I could solve it using inverses:

##\begin{bmatrix}
10&24&2\\
-2&-7&4\\
-14&-48&26\\
\end{bmatrix}\cdot
\begin{bmatrix}
x\\
y\\
z\\
\end{bmatrix}=
\begin{bmatrix}
-18\\
6\\
42\\
\end{bmatrix}##

But before I did so, I tried to find the determinant of the matrix to see whether ##|A|\neq 0## or not.

##|A|=10
\begin{vmatrix}
-7&4\\
-48&26\\
\end{vmatrix}-24
\begin{vmatrix}
-2&4\\
-14&26\\
\end{vmatrix}+2
\begin{vmatrix}
-2&-7\\
-14&-48\\
\end{vmatrix}=0
##

The matrix is singular, so that means that the system doesn't have a unique solution. The problem was multiple choice, and there was no option for "infinite number of solutions", so I picked "no solution" (badly written multiple choice question, imo). The solution that was posted said that there was, in fact a unique solution:

##
\left\{
\begin{array}{ll}
x=22\\
y=-10\\
z=-5
\end{array}
\right.
##

I've tried to understand what's going on, but I can't figure out why there's a solution.
 
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  • #2
Did you check the answer to see if it really is a solution?
 
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  • #3
I would double-check your numbers...perhaps what you copied down and the teacher's version of the problem are different.
##A [22,-10,-5]^T = [-30, 6, 42]^T ##
In general, infinite number of solutions and no solution are not the same. If one of the multiple choice options was one of the infinite solutions, you should have chosen it as the "best available answer".
 
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  • #4
am I being dum?
10*22+24*(-10)+2*-5 = -30 not -18

Other 2 equations work out though
 
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  • #5
I suggest that you check if you copied your teacher's writing correctly by comparing with your friend's note.

If you do a row reduction on the augmented matrix between the coefficient and the RHS, the reduced form has a pivot in the last column which indicates that this system of equations does not have a solution.
 
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  • #6
Simon Bridge said:
Did you check the answer to see if it really is a solution?
Yes, I have. It doesn't work.
RUber said:
In general, infinite number of solutions and no solution are not the same. If one of the multiple choice options was one of the infinite solution
I know. But there wasn't an option for infinite solution.
blue_leaf77 said:
I suggest that you check if you copied your teacher's writing correctly by comparing with your friend's note.
I didn't copy anything. All my work was on a printed sheet that had the problem (typed up and clear to read). Fortunately, he's stopped writing problems by hand because it's chicken scratch.
 
  • #7
ProfuselyQuarky said:
Yes, I have. It doesn't work.

Methinks that the question was written wrong on the sheet then. The proof is in the fact that if you put in the values for x,y,z into equation one, you don't get the correct answer.
 
  • #8
cpscdave said:
Methinks that the question was written wrong on the sheet then. The proof is in the fact that if you put in the values for x,y,z into equation one, you don't get the correct answer.
That's what I thought, but he's hardly ever wrong and I was wondering if I was missing something :frown:
 
  • #9
Given that 2 of the 3 systems work out correctly. I'd think he likely made a error entering the first system.

If you change it to 11x +25y -2z it works out.

If you're just looking for practice you could try solving it using that system instead and see if you come to the correct answer :)
 
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  • #10
cpscdave said:
Given that 2 of the 3 systems work out correctly. I'd think he likely made a error entering the first system.

If you change it to 11x +25y -2z it works out.

If you're just looking for practice you could try solving it using that system instead and see if you come to the correct answer :)
Okay, thanks. There's always room for more practice problems :)
 
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  • #11
The determinant is not 0. I got -8.
 
  • #13
mathman said:
The determinant is not 0. I got -8.
How?
 
  • #14
ProfuselyQuarky said:
How?
I took the original expression and did the arithmetic.
 
  • #15
mathman said:
I took the original expression and did the arithmetic.

I think you did it incorrectly.
 
  • #16
micromass said:
I think you did it incorrectly.
You are right. I stand corrected.
 
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Related to Why is there a solution to this system?

1. Why is there a solution to this system?

A system has a solution if and only if the equations in the system are consistent, meaning they have a common point of intersection. This point of intersection represents the solution to the system.

2. What factors determine if a system has a solution?

The number of equations and variables in the system determine if a solution exists. If there are more equations than variables, the system may have no solution. If there are more variables than equations, the system may have infinitely many solutions.

3. Can a system have more than one solution?

Yes, a system can have infinitely many solutions if there are more variables than equations. This means that there are multiple points of intersection between the equations.

4. How do we find the solution to a system?

The solution to a system can be found through various methods such as substitution, elimination, or graphing. These methods involve manipulating the equations to isolate variables and solve for their values.

5. Why is it important to find the solution to a system?

Finding the solution to a system is important because it helps us understand the relationships between the variables and how they affect each other. It also allows us to make predictions and solve real-world problems by using the equations in the system.

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