Question

Solve the system of equations using matrices. Use Gaussian elimination with back-substitution.
x + y + z = -5
x - y + 3z = -1
4x + y + z = -2


A. {( 1, -4, -2)}
B. {( -2, 1, -4)}
C. {( 1, -2, -4)}
D. {( -2, -4, 1)}

Answer 1

first, make a matrix with the coefficients and constants ^_^

Answer 2

i'll show you what the first row looks like to get you started

1 1 1 -5

Answer 3

I don't really know how to do matrices. @DemolisionWolf

Answer 4

so we are given these equations.
x + y + z = -5
x - y + 3z = -1
4x + y + z = -2

lets look at the first one only
x + y + z = -5

this can be written as this, with the coefficients of each term written in, (becuase when the coefficient is '1' we don't write it)
x + y + z = -5
becomes
1x + 1y + 1z = -5

now, i make the first row of the matrix by taking the coefficients from each term and then the constant on the other side of the equal sign, which is 5
1 1 1 -5

the first '1' comes from the x, the next '1' comes from the y, and the third '1' comes from the z



I'll show you quicker on the second equation given:
x - y + 3z = -1
becomes
1x - 1y + 3z = -1
becomes
1 -1 3 -1


make sense?

Answer 5

yes. so it should be:
1 1 1 -5
1 -1 3 -1
4 1 1 -2?

@DemolisionWolf

Answer 6

Yes, it would start with:
\(\left[\begin{array}{ccc|c}
1 & 1 & 1 & -5 \\
1 & -1 & 3 & -1\\
4 & 1 & 1 & -2
\end{array}\right]\)

I use
```
\(\left[\begin{array}{ccc|c}
1 & 1 & 1 & -5 \\
1 & -1 & 3 & -1\\
4 & 1 & 1 & -2
\end{array}\right]\)
```
to make it come out properly.

Answer 7

Okay. So what's the next step?

Answer 8

@e.mccormick

Answer 9

Know what upper echelon form or, row echelon form are? That is what you are going to want in the end.

That is where the triangle has numbers and the lower part has zeros before the numbers. Let me just show it with letters because it is easier than trying to describe it.

\(\left[\begin{array}{ccc|c}
a & b & c & d \\
0 & e & f & g\\
0 & 0 & h & i
\end{array}\right]\)

To get it into that form, you do elementary row operations. Has your book or class covered those?

Answer 10

Nope. They don't look familiar... @e.mccormick

Answer 11

In short:
Elementary Row Operations.

1. Interchange two rows.
2. Multiply a row with a nonzero number.
3. Add a row to another one.
4. Do 2 and 3 at the same time, so multiply a row by a number then add it to another one.

So if you look at this:

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

You want to get the 1 in the fist column of the second row to 0. What would happen if you multiplied the first row by -1, then added it to the second row to replace the second row?

Answer 12

Multiply the first row by -1 meaning by 1,1,1, and -5?

Answer 13

Yes. Each element in the row is multiplied by -1.

Answer 14

So, I would get -1 -1 -1 and 5

Answer 15

Good. And add that to the second row.

Answer 16

Then, adding it to the 2nd row I'd get
0 -2 2 and 4

Answer 17

=)

Here I was working out the steps in case you did not get it, but you did!

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

OK, now, row 2 is done. Nothing else needs to be fixed for Gaussian Elimination. Row 3 needs both he first and 2nd columns changed to 0s. So, think you could do something similar?

Answer 18

Multiply the 3rd row by a number.... but I'm not sure what number

Answer 19

Actually, I would multiply the first row by a number and add it to the 3rd to get rid of the first column. Now, the first column is a 0. What would add to 4 to make 0?

Answer 20

-4

Answer 21

=) And there is what you need to multiply by.

Answer 22

So multiply the first row by 4, then add to the 3rd row?

Answer 23

-4*

Answer 24

Yep!

Answer 25

okay, so
-4 -4 -4 and 20

when added to the third row,
0 3 3 and 18

Answer 26

Watch your signs. =)

Answer 27

Oh! 0 -3 -3 and 18

Answer 28

OK:
\(\left[\begin{array}{ccc|c}
1 & 1 & 1 & -5 \\
0 & -2 & 2 & 4\\
0 & -3 & -3 & 18
\end{array}\right]\)

Now, what could you multiply row 2 by to make the first non-zero number into a 1?

Answer 29

1?

Answer 30

Well, there is a -2 there...

Answer 31

I thought by the first non-zero number you meant the 1

Answer 32

In row 2.

Answer 33

my mistake! I was looking at the rows.

Answer 34

I forgot to mention something about the upper echelon form you use: It has 1 in the leading spots, so more like:

\(\left[\begin{array}{ccc|c}
1 & b & c & d \\
0 & 1 & f & g\\
0 & 0 & 1 & i
\end{array}\right]\)

This is to get the 1 in the second row. you already have a 1 in the first row.

Answer 35

Okay. So, I have to multiply the second row by a number that will make the -2 a 1

Answer 36

Exactly!

Answer 37

Can I multiply by a fraction?

Answer 38

Yes. =) \(-\frac{1}{2}\) is valid, which I bet is what you wanted to use.

Answer 39

yes! so that would then be
0 1 -1 and -2

Answer 40

\(\left[\begin{array}{ccc|c}
1 & 1 & 1 & -5 \\
0 & 1 & -1 & -2\\
0 & -3 & -3 & 18
\end{array}\right]\)

OK, so know what is next?

Answer 41

To be honest, there is a choice of 2 things to do next. Either is fine.

Answer 42

Could I add that to the third row?

Answer 43

Add what to the 3rd row?

Answer 44

the second row

Answer 45

Well, would just the 2nd row get rid of it, or would you also need to multiply?

Answer 46

I would alsp need to multiply

Answer 47

also*

Answer 48

OK! So get to it! Let me know what you get.

Answer 49

okay. I would get
0 1 1 and -6

Answer 50

Hmmm...

Answer 51

\(3\left[\begin{array}{ccc|c}
0 & 1 & -1 & -2\\
\end{array}\right] + \left[\begin{array}{ccc|c}
0 & -3 & -3 & 18
\end{array}\right]\) becomes

\(\left[\begin{array}{ccc|c}
0 & 3 & -3 & -6\\
\end{array}\right] + \left[\begin{array}{ccc|c}
0 & -3 & -3 & 18
\end{array}\right]\)

When you add those, what do you get?

Answer 52

0 0 -6 and 12

Answer 53

=)

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

OK, that is in reduced upper echelon form.

Now for back substitution. First, the last row is solved.

\(\left[\begin{array}{ccc|c}
0 & 0 & 1 & -2
\end{array}\right]\) means \(y=-2\)

So, this is where Gaussian Elimination gets a little odd. I'll show you with the second row, then you can try it on the first. OK?

Answer 54

okay! =)

Answer 55

Oops, I meant \(z=-2\) The 3rd column is z, the second y, and the first x.

I put what is known as a multiplier into the matrix:

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

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

Then I move it from the left to right using subtraction:

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

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

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

OK. So \(y=-4\).

Can you now use both \(z=-2\) and \(y-4\) to do the same thing to x?

Answer 56

yes, so that means I'd plug the z into the 2nd row and the y into the first row?

Answer 57

You use the solved y and z to plug into the first row to start solving x.

Answer 58

Okay. And by doing that would I get x=1?

Answer 59

Yes!

Answer 60

THANK YOUUU!!!!!!!!

Answer 61

Want to see the whole thing as one post?

Answer 62

No need! I've copied it all down =)

Answer 63

OK. I put in some notation like:
\(-1R_1+R_2 \rightarrow R_2\)

Answer 64

It keeps it clear what was done.

Answer 65

Thank you so much!!! Could I message you if I need help on anything else?

Answer 66

Well, I am going to go home and get food. but you can tag me if you see me. If I don't respond, it means I am helping someone else or otherwise busy. Mods get tagged a lot. Hehe.

Answer 67

haha okay no problem, thank you! Have a good night.

Answer 68

You too. And if you see anyone with a purple A in front of their name, feel free to tag them too. They are my minion army.

Answer 69

hahaha Thank you! Will do.