# B Why is there a solution to this system?

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1. Jun 6, 2016

### ProfuselyQuarky

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.

2. Jun 6, 2016

### Simon Bridge

Did you check the answer to see if it really is a solution?

3. Jun 6, 2016

### RUber

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".

4. Jun 6, 2016

### cpscdave

am I being dum?
10*22+24*(-10)+2*-5 = -30 not -18

Other 2 equations work out though

5. Jun 6, 2016

### blue_leaf77

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.

6. Jun 6, 2016

### ProfuselyQuarky

Yes, I have. It doesn't work.
I know. But there wasn't an option for infinite solution.
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. Jun 6, 2016

### cpscdave

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. Jun 6, 2016

### ProfuselyQuarky

That's what I thought, but he's hardly ever wrong and I was wondering if I was missing something

9. Jun 6, 2016

### cpscdave

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 :)

10. Jun 6, 2016

### ProfuselyQuarky

Okay, thanks. There's always room for more practice problems :)

11. Jun 6, 2016

### mathman

The determinant is not 0. I got -8.

12. Jun 6, 2016

### Staff: Mentor

I got 0.

13. Jun 6, 2016

### ProfuselyQuarky

How?

14. Jun 7, 2016

### mathman

I took the original expression and did the arithmetic.

15. Jun 7, 2016

### micromass

I think you did it incorrectly.

16. Jun 8, 2016

### mathman

You are right. I stand corrected.