help me plz again

- anonymous

help me plz again

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- anonymous

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## More answers

- anonymous

@dan815

- anonymous

@jim_thompson5910

- jim_thompson5910

how far did you get? were you able to set up the matrix?

- anonymous

i was not i was so confused on this like i dont know where to start

- jim_thompson5910

well you first pull out the coefficients of each equation. Do you see what they are?

- anonymous

i do not sorry

- jim_thompson5910

they are simply the numbers in front of the variables

- jim_thompson5910

example: `4x` has the coefficient of `4`

- jim_thompson5910

the `-y` is really `-1y`

- anonymous

so four and five right

- jim_thompson5910

along the first equation, what are the 3 coefficients?

- anonymous

4 5 and 1x right

- jim_thompson5910

read across the row

- jim_thompson5910

not column

- jim_thompson5910

do you see what I mean?

- anonymous

ohh ok i got you hold on

- anonymous

so 4 and three for the first one right

- jim_thompson5910

don't forget about the -1

- jim_thompson5910

the first equation is really `4x - 1y + 3z = 12`

- anonymous

ohh yeah sorry lol

- anonymous

so the seciond one is 1x+4y+6z=12

- jim_thompson5910

-32 on the right side, not 12

- anonymous

opps my bad i looked at it wrong haha i acciedently looked up at the end sorry

- anonymous

5x+3y+9z=20 for the last

- jim_thompson5910

So if we just focus on the coefficients on the left side, we have this 3x3 matrix
\[
\left[\begin{array}{ccc}
4 & -1 & 3\\
1 & 4 & 6\\
5 & 3 & 9
\end{array}
\right]
\]

- jim_thompson5910

agreed? or no?

- anonymous

yes i agree

- jim_thompson5910

the next matrix is simply the matrix of variables written as a column like this
\[
\left[\begin{array}{ccc}
x\\
y\\
z
\end{array}
\right]
\]

- jim_thompson5910

and then finally we have a third matrix of the values on the right side of the equation
the next matrix is simply the matrix of variables written as a column like this
\[
\left[\begin{array}{ccc}
12\\
-32\\
20
\end{array}
\right]
\]

- jim_thompson5910

putting this all together, we form this matrix equation
\[
\left[\begin{array}{ccc}
4 & -1 & 3\\
1 & 4 & 6\\
5 & 3 & 9
\end{array}
\right]
*
\left[\begin{array}{ccc}
x\\
y\\
z
\end{array}
\right]
=
\left[\begin{array}{ccc}
12\\
-32\\
20
\end{array}
\right]
\]
making sense so far?

- anonymous

yeah

- jim_thompson5910

next we form an augmented matrix by ignoring the variables temporarily. We just combine the first and third matrix like this
\[
\left[\begin{array}{ccc|c}
4 & -1 & 3 & 12\\
1 & 4 & 6 & -32\\
5 & 3 & 9 & 20
\end{array}
\right]
\]
the vertical bar separates the two matrices

- anonymous

makes since now

- jim_thompson5910

this is where gauss jordan elimination begins

- anonymous

ok

- jim_thompson5910

do you know how to do gauss jordan elimination? or no?

- anonymous

i really dont i have treid and i keep faling at it this is whay im here now hahaha

- jim_thompson5910

that's ok

- jim_thompson5910

The first thing I would do is swap rows 1 and 2. We denote this with notation R1 <--> R2. I'm doing this to make the element in row1,column1 be a `1`
\[
\left[\begin{array}{ccc|c}
4 & -1 & 3 & 12\\
1 & 4 & 6 & -32\\
5 & 3 & 9 & 20
\end{array}
\right]
\]
\[
\left[\begin{array}{ccc|c}
1 & 4 & 6 & -32\\
4 & -1 & 3 & 12\\
5 & 3 & 9 & 20
\end{array}
\right]
\begin{array}{c}
\phantom\\
R_1 \leftrightarrow R_2\\
\phantom\\
\end{array}
\]
I couldn't get the thing to line up properly, but the `R1 <--> R2` should either be in row 1 or row 2

- jim_thompson5910

let me know when you're ready for the next step

- anonymous

OK GIVE ME A SECOND TO GET THIS DOWN sorry for the caps i for got i had them on

- anonymous

OKim ready

- jim_thompson5910

Now we want to make that `4` in row2,column1 to be a `0`. We do this by computing `R2-4*R1` and replacing all of R2 with that result
\[
\left[\begin{array}{ccc|c}
1 & 4 & 6 & -32\\
0 & -17 & -21 & 140\\
5 & 3 & 9 & 20
\end{array}
\right]
\begin{array}{c}
\\
R_2 - 4R_1 \rightarrow R_2\\
\\
\end{array}
\]

- anonymous

OK GOTCHA

- jim_thompson5910

Next we want to make that `5` in row3,column1 to be a `0`. We do this by computing `R3-5*R1` and replacing all of R3 with that result
\[
\left[\begin{array}{ccc|c}
1 & 4 & 6 & -32\\
0 & -17 & -21 & 140\\
0 & -17 & -21 & 180\\
\end{array}
\right]
\begin{array}{c}
\\
\\
R_3 - 5R_1 \rightarrow R_3\\
\end{array}
\]

- anonymous

ok i got that

- jim_thompson5910

We could divide the second row by -17 to make that first '-17' in that row equal to '1', but I'm going to skip that step. Instead I'm going to subtract R2 and R3 and replace R3
R3-R2 --> R3
\[
\left[\begin{array}{ccc|c}
1 & 4 & 6 & -32\\
0 & -17 & -21 & 140\\
0 & 0 & 0 & 40\\
\end{array}
\right]
\begin{array}{c}
\\
\\
R_3 - R_2 \rightarrow R_3\\
\end{array}
\]

- jim_thompson5910

The last row of this matrix
\[
\left[\begin{array}{ccc|c}
1 & 4 & 6 & -32\\
0 & -17 & -21 & 140\\
0 & 0 & 0 & 40\\
\end{array}
\right]
\]
tells us that `0x + 0y + 0z = 40` which essentially turns into `0+0+0 = 40` and that turns into `0 = 40`

- anonymous

wow that was alot

- jim_thompson5910

but `0 = 40` is NEVER true regardless of what x,y,z are. Since we have a contradiction, we have an inconsistent system that has NO solutions at all.

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