Question

@e.mccormick Use a matrix to solve the linear system of equations: {5x-2y=27, -3x+5y=18}

Answer 1

K. Know how to put them into matrix form?

Answer 2

No. I was told to tag you =) I don't know anything about matrices. My lesson doesn't explain it very well. I was hoping you could sort of teach it to me? =/ Hope I'm not asking too much.

Answer 3

The matrix form for this uses what is called an augmented matrix. Basically it uses a | in place of the = signs and gets rid of the variables:

\(\left[\begin{array}{cc|c}
5 & -2 & 27\\
-3 & 5 & 18
\end{array}\right]\)

Answer 4

Oh, okay.

Answer 5

Now, the thing about solving or reducing matrices is what they call elementary row operations. You can multiply a row by a constant. You can multiply a row by a constant and add it to another row. That is basically it.

Answer 6

Once you get things to the form of:
\(\left[\begin{array}{cc|c}
1 & 0 & x\\
0 & 1 & y
\end{array}\right]\)

Whatever is in the x and y would be the answer.

Answer 7

Okay. So how do I go about reducing it to that form?

Answer 8

I'm doing up an example for the first operation.

Answer 9

Let me take the first row: 5 -2 | 27

I need to multiply it by something to add it to the -3 5 | 18 row in such a way that it would get rid of the 3. Well, to turn 5 into 3 you multiply by \(\dfrac{3}{5}\). That multiplies through the entire row:

\(5\cdot\dfrac{3}{5} -2\cdot\dfrac{3}{5} | 27\cdot\dfrac{3}{5}\implies \)

\(3 -\dfrac{6}{5} |\dfrac{81}{5} \)

So I am going to add that and the last row to replace the last row. Does that make sense?

Answer 10

Ahh, so you have to use multiplication or addition to completely get rid of one of the rows?

Answer 11

Yah, zero it.

Answer 12

Okay. Give me a second to think on that =) I need to write it down and actually go through it myself to see how it works.

Answer 13

\(\begin{array}{cc|c}
3 & -\dfrac{6}{5} & \dfrac{81}{5}\\
-3 & \dfrac{25}{5} & \dfrac{90}{5} \\ \hline
0 & \dfrac{18}{5} & \dfrac{171}{5}
\end{array}\)

Now, that would be my new row 2, but it is ugly.

Answer 14

So I would multiply it by \(\dfrac{5}{18}\) before I put it in place.

Answer 15

That got me to:
\(\left[\begin{array}{cc|c}
5 & -2 & 27\\
0 & 1 & \dfrac{19}{2}
\end{array}\right]\)

Now, it should be super easy to see how to eliminate the -2 in row 1 with that. \(\ddot \smile\) And that is basically how you solve systems in matrix form.

Answer 16

Okay, I'm back. Give me a second to read all this again, please?

Answer 17

Alright. I have a question. Why is it that in the second row, you multiply everything by 3/5 except 3? Shouldn't you do the same thing to everything?

Answer 18

In the second row I multiplied two things by 1. My version of 1 eas 1 for the first one but \(\frac{5}{5}\) for the second and third.

Answer 19

I see.

Answer 20

1 has many faces. \(\frac{e}{e}\) is one. 0 also hides a lot. \(1-1\) is 0. So you can add 0 to anything and multiply anything by 1 and not change it. The adding 0 is useful for completing the square.

Answer 21

Hmmm. This just keeps getting more complicated =)

Answer 22

So I'm trying to take what you told me from the first problem and apply it to the second one I have to do.

Answer 23

Well, this one still needs a little work.

Answer 24

Oh, okay. *facepalm* I thought you were done, haha.

Answer 25

\(\left[\begin{array}{cc|c}
5 & -2 & 27\\
0 & 1 & \dfrac{19}{2}
\end{array}\right]\) vs \(\left[\begin{array}{cc|c}
1 & 0 & x\\
0 & 1 & y
\end{array}\right]\)

So you have the y, but not the x.

Answer 26

Okay. So I'm assuming the top row on the left needs to match the top row on the right? How do I make that happen?

Answer 27

Well, first you need to eliminate the -2. So what multiple of row 2 could you add to row 1 to do that?

Answer 28

Seems you got bumped. Here are some notes on elementary row operations:
http://www.purplemath.com/modules/mtrxrows.htm