Question

Guass Jordan elimination.

x1 + 6x2 - x3 - 4x4 = 0
-2x1 - 12x2 + 5x3 + 17x4 = 0
3x1 + 18x2 - x3 - 6x4 = 0

Answer 1

first set it up as a matrix
can you do that?

Answer 2

Yes

Answer 3

1 6 -1 -4 0
-2 -12 5 17 0
3 18 -1 -6 0

Answer 4

i got 480

Answer 5

jokes

Answer 6

okay so this is a 3x4 matrix so you need to put it in ref
so start eliminating the bottom first

Answer 7

I would do these operations first
\[R_2 \rightarrow 2R_1 + R_2\]
\[R_3 \rightarrow -3R_1 + R_3\]

Answer 8

ok let me show that step..

Answer 9

1 6 -1 -4 0
0 0 3 9 0
0 0 2 6 0

Then 1/3R2 = R2
and 1/2 R3 = R3

Answer 10

yup thats right

Answer 11

ok I have the next step I think...

Answer 12

1 6 -1 -4 0
0 0 1 3 0
0 0 1 3 0

now R3 + (-1)R2 = R3

Answer 13

oops I left something out sorry...let me fix that.

Answer 14

yea i was about to point out an error in that matrix you have

Answer 15

and yea that operation you metioned is correct

Answer 16

Now Im confused..where is the error?

Answer 17

wait nvm ignore that my mistake keep going

Answer 18

First R1 + R3 = R1

Answer 19

to get...

Answer 20

1 6 0 -1 0
0 0 1 3 0
0 0 1 3 0

Answer 21

then R3 + (-1)R2 = R3

Answer 22

good so far

Answer 23

to get..

Answer 24

1 6 0 -1 0
0 0 1 3 0
0 0 0 0 0

Answer 25

and this is the part I really need help with......past this.

Answer 26

okay technically your done reducing alrready now all you need to do is put it in terms of x1, x2 etc

Answer 27

you get what im saying i do the first row for you
x1 + 6x2 - x3 = 0

Answer 28

Im solving just a sec...

Answer 29

-x4 sorry

Answer 30

alright

Answer 31

so how does....
x1 = -6R + S
x2 = R
x3 = -3S
x4 = S
......look?

Answer 32

Im not sure if Im doing this part correctly...

Answer 33

how you get x3 = -3s?

Answer 34

lets do this step by step

Answer 35

ok

Answer 36

lol nvm your right sorry

Answer 37

did the question also asked to put it in parameterized vector form?

Answer 38

no. What does that mean?

Answer 39

so all you do to finish it up is put it like this:

\[\left(\begin{matrix}x_1 \\ x_2 \\ x_3 \\ x_4\end{matrix}\right) = \left(\begin{matrix}-6R + S\\ R \\ -3S \\ S\end{matrix}\right) = \left(\begin{matrix}-6R \\ R \\ 0 \\ 0\end{matrix}\right) + \left(\begin{matrix}0 \\ 0 \\ -3S \\ S\end{matrix}\right)\]

Answer 40

= \[\left(\begin{matrix}-6 \\ 1 \\ 0 \\ 0\end{matrix}\right)R + \left(\begin{matrix}0 \\ 0 \\ -3 \\ 1\end{matrix}\right)S \]
and your done

Answer 41

aw dam i made a mistake its supposed to be a 1 instead of a 0 on the first row
for the vector with the S

Answer 42

I think I still follow it though. Thank you so much for walking me through this one, this is a new concept for me, seeing lineear algebra for the first time.

Answer 43

no prob glad to help
its all about practice and you'll get used to it