Question

help me plz again

Answer 1

@Nnesha

Answer 2

@paki

Answer 3

@dan815

Answer 4

@jim_thompson5910

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

i do not sorry

Answer 9

they are simply the numbers in front of the variables

Answer 10

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

Answer 11

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

Answer 12

so four and five right

Answer 13

along the first equation, what are the 3 coefficients?

Answer 14

4 5 and 1x right

Answer 15

read across the row

Answer 16

not column

Answer 17

do you see what I mean?

Answer 18

ohh ok i got you hold on

Answer 19

so 4 and three for the first one right

Answer 20

don't forget about the -1

Answer 21

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

Answer 22

ohh yeah sorry lol

Answer 23

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

Answer 24

-32 on the right side, not 12

Answer 25

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

Answer 26

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

Answer 27

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

Answer 28

agreed? or no?

Answer 29

yes i agree

Answer 30

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

Answer 31

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
\[
\left[\begin{array}{ccc}
12\\
-32\\
20
\end{array}
\right]
\]

Answer 32

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?

Answer 33

yeah

Answer 34

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

Answer 35

makes since now

Answer 36

this is where gauss jordan elimination begins

Answer 37

ok

Answer 38

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

Answer 39

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

Answer 40

that's ok

Answer 41

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

Answer 42

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

Answer 43

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

Answer 44

OKim ready

Answer 45

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


\[
\left[\begin{array}{ccc|c}
1 & 4 & 6 & -32\\
0 & -17 & -21 & 140\\
5 & 3 & 9 & 20
\end{array}
\right]
\begin{array}{c}
\\
R_2 - 4R_1 \rightarrow R_2\\
\\
\end{array}
\]

Answer 46

OK GOTCHA

Answer 47

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


\[
\left[\begin{array}{ccc|c}
1 & 4 & 6 & -32\\
0 & -17 & -21 & 140\\
0 & -17 & -21 & 180\\
\end{array}
\right]
\begin{array}{c}
\\
\\
R_3 - 5R_1 \rightarrow R_3\\
\end{array}
\]

Answer 48

ok i got that

Answer 49

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

\[
\left[\begin{array}{ccc|c}
1 & 4 & 6 & -32\\
0 & -17 & -21 & 140\\
0 & 0 & 0 & 40\\
\end{array}
\right]
\begin{array}{c}
\\
\\
R_3 - R_2 \rightarrow R_3\\
\end{array}
\]

Answer 50

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`

Answer 51

wow that was alot

Answer 52

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.