Question

I am to use the Gauss Jordan Method to solve for the system of equations. If the system has infinitely many solutions I am to give the z arbitrary. I have the following equation I am to solve for which is going to be x=5y+4z=1 and 3x-2y+3z=-2 for these I have arrived at the answers of -7z-1/13, 9z-5/13, z Have I worked these correctly?

Answer 1

x + 5y + 4z = 1
3x + 2y + 3z = -2

Guess a value of z such as z = 1

Then
x + 5y + 4(1) = 1
3x + 2y + 3(1) = -2

x + 5y + 4 = 1
3x + 2y + 3 = -2

x + 5y = -3
3x + 2y = -5

Now you have a system in two equations x and y which you should be able to solve.

Answer 2

Okay when I work it this time I have \[\frac{-7z-12 }{ 13 }, \frac{ 9z-5 }{ 13}, z \]

Answer 3

Still not getting the right answer here. Any help?

Answer 4

If I recall, gauss jordan involves reducing the coefficient matrix right?

Answer 5

So z is the free variable?

Answer 6

Can you write the coefficient matrix? Then can you write it in reduced row echelon?

Answer 7

oh and wio, yea z is free (arbitrary)

Answer 8

\[
\begin{bmatrix}
1&5&4\\
3&-2&3
\end{bmatrix}
\begin{bmatrix}
x\\
y\\
z
\end{bmatrix}
=
\begin{bmatrix}
1\\
-2
\end{bmatrix}
\]

Answer 9

Hmm, @ShortStaticBurst I'll show how my work and we'll see if your answer matches up.

Answer 10

Okay.

Answer 11

I'm doing \(R_2 = R_2-3R_1\)
\[
\begin{bmatrix}
1&5&4\\
0&-17&-9
\end{bmatrix}
\begin{bmatrix}
x\\
y\\
z
\end{bmatrix}
=
\begin{bmatrix}
1\\
-5
\end{bmatrix}
\]

Answer 12

I'm doing \(R_2 =5 R_2+17R_1\)
\[
\begin{bmatrix}
1&5&4\\
0&0&23
\end{bmatrix}
\begin{bmatrix}
x\\
y\\
z
\end{bmatrix}
=
\begin{bmatrix}
1\\
-8
\end{bmatrix}
\]

Answer 13

actually, doing it this way makes \(y\) the free variable.

Answer 14

But it doesn't matter.

Answer 15

Now I am subtracting \(5y\) in the top equation \[
\begin{bmatrix}
1&4\\
0&23
\end{bmatrix}
\begin{bmatrix}
x\\
z
\end{bmatrix}
=
\begin{bmatrix}
1-5y\\
-8
\end{bmatrix}
\]

Answer 16

Actually, let me double check my work for a moment...

Answer 17

@ShortStaticBurst Did you use a different method than what I am doing?

Answer 18

I think my method is a bit... confusing. Let me consider an easier route.

Answer 19

Yes I did. But I don't trust the method the book suggests.

Answer 20

We add a new equation: \(z=z\), and it gives us:\[
\begin{bmatrix}
1&5&4\\
3&-2&3\\
0&0&1
\end{bmatrix}
\begin{bmatrix}
x\\
y\\
z
\end{bmatrix}
=
\begin{bmatrix}
1\\
-2\\
z
\end{bmatrix}
\]Does this make sense to you?

Answer 21

I'm doing \(R_2 = R_2-3R_1\)
\[
\begin{bmatrix}
1&5&4\\
0&-17&-9\\
0&0&1
\end{bmatrix}
\begin{bmatrix}
x\\
y\\
z
\end{bmatrix}
=
\begin{bmatrix}
1\\
-5\\
z
\end{bmatrix}
\]

Answer 22

I used the augmented matrix from the book. two equations expressing x and y in terms of z

Answer 23

There are many ways to approach this problem, so I don't know which method your book did in particular.

Answer 24

I'm doing \(R_2 = -R_2/17\)
\[
\begin{bmatrix}
1&5&4\\
0&1&9/17\\
0&0&1
\end{bmatrix}
\begin{bmatrix}
x\\
y\\
z
\end{bmatrix}
=
\begin{bmatrix}
1\\
5/17\\
z
\end{bmatrix}
\]

Answer 25

At this point, I can back substitute.

Answer 26

I have: \(y + 9z/17 = 5/17\)
So this means \[y = \frac{5-9z}{17}\]

Answer 27

Then we have in the top row: \[
x+\frac{5-9z}{17}+4z=1
\]

Answer 28

Whoops, that should be: \[
x+5\left(\frac{5-9z}{17}\right)+4z=1
\]

Answer 29

The question is confusing because there are infinite solutions as far as I can tell. In fact one of the three choices to answer with has that it could be a solution set where z is any real number

Answer 30

Yeah, the solution will be a line.

Answer 31

There is 1 free variable.

Answer 32

I really find the whole thing horribly confusing. It seems to be you should solve it with three spaces for the solution set or two spaces and a z for the solution set.
This question is not one I have found to be easy to come by.

Answer 33

All you are doing is adding a third equation.. it could be \(z=t\):\[
x=5y+4z=1 \\ 3x-2y+3z=-2 \\ z=t\]or if could be \(y=t\), or even \(x=t\).

Answer 34

In this case \(t\) can be any number. It could be 2, 0, or -122. It's just representing a value that \(z\), \(x\), or \(y\) might be set to...

Answer 35

Consider if you have just the equation \(x+y=2\). It's obvious this is a line.\[
y = -x+2
\]

Answer 36

Since there is one equation and two variables, there is one free variable.

Answer 37

Okay.