Question

Solve the system of equations using matrices. Use Gauss-Jordan elimination.

3x - 7 - 7z = 7
6x + 4y - 3z = 67
-6x - 3y + z = -62

Answer 1

Know how to put it in matrix form?

Answer 2

No. I've completely forgotten. I'm also in online school and my book won't load for me to figure out how.

Answer 3

OK. Matrix form is just the coefficients with the signs. So:

a + b = c
d - e = f

becomes:

\(\left[\begin{array}{cc|c}
a & b & c\\
d & -e & f
\end{array}\right]\)

The | takes the place of the =.

Here is a really well done example of using Elementary Row Operations to do Gauss/Jordan:

http://www.missouriwestern.edu/cas/Math/ElementaryRowOperations.pdf

Once you have it in matrix form, that is what you need to use.

Answer 4

So in the matrix the first column would be 3,6,6 . Then the second row would be 0,4,3...etc?

Answer 5

second column*

Answer 6

Is 3x - 7 - 7z = 7 supposed to be like that, or 3x - 7x - 7z = 7 ?

Answer 7

oops. I meant 3x - 7y - 7z = 7

Answer 8

On the worksheet it has it like that. I'm not sure if it's a typo or not though.

Answer 9

Well, if it has the y it is an ugly answer. Without the y is is nice integers.

First get the -7 to the other side... but you forgot the signs when you did your columns.

3 6 -6, and 0 4 -3

Answer 10

I need to get some sleep, but that PDF really goes over this very nicely.

Personally, I would start by adding row three and row two to make a new row two, then adding two of row one to row three to make a new row three. From there it should be a lot easier to see how to get things down to:

\(\left[\begin{array}{ccc|c}
1 & 0 & 0 & x\\
0 & 1 & 0 & y\\
0 & 0 & 1 & z
\end{array}\right]\)

Answer 11

we have the problem in the form Ax = b

which is like for a usual equation y=5x, or
5x = y
5x = b
Ax = b

Answer 12

we want to solve for x, so we would divide by A usually:
y=5x
5x=y
x = y/5

in that case:
1.) the x it there just by itself
2.) y was divided by the coefficient A

Answer 13

we want the same result though as if both sides were divided by A
cause then, we would have

Ax = b
x = b/A <-it would be solved for x

Answer 14

so IF this was achieved, in that case the left hand side would be just
an x, right ?

and the right hand side would have been adjusted accordingly
so that the equation balance is maintained

Answer 15

we effectively make adjustments (GAUSS JORDAN) until the result
is the same as if we had divided by A

Answer 16

so,that's why we wanna get
1 0 0
0 1 0
0 0 1 <on the side where the x's were

Answer 17

because that is the identity matrix,multiplying it with an x matrix
will give you the x matrix

just like x=y/5
x*1 = x

Answer 18

SO

the first step is to transcribe the matrix

3x - 7y - 7z = 7 <- constant1 is 7
6x + 4y - 3z = 67
-6x - 3y + z = -62

\[\left[\begin{array}{ccc|c}
3 & -7 & -7 & 7\\
x & y & z & c_2\\
x & y & z & c_3
\end{array}\right]\]

Answer 19

fully transcribed:\[\left[\begin{array}{ccc|c}
3 & -7 & -7 & 7\\
6 & 4 & -3 & 67\\
-6 & -3 & 1 & -62
\end{array}\right]\]

Answer 20

ok so its possible to move lines no issue
\[\left[\begin{array}{ccc|c}

6 & 4 & -3 & 67\\
-6 & -3 & 1 & -62\\
3 & -7 & -7 & 7
\end{array}\right]\]also possible to add lines together and write result into any line that was part of the addition.
like explained in the PDF that is linked above

add the 1. ,2. line and write into 1.:
\[\left[\begin{array}{ccc|c}

0 & 1 & -2 & 5\\
-6 & -3 & 1 & -62\\
3 & -7 & -7 & 7
\end{array}\right]\]

Answer 21

ok that wasnt such a good step actually.

Answer 22

\[\left[\begin{array}{ccc|c}

6 & 4 & -3 & 67\\
-6 & -3 & 1 & -62\\
3 & -7 & -7 & 7
\end{array}\right]\]they want us to have a line that starts with "1"

Answer 23

3 6 9 12 15 18 21 24
6 12 18 24 30 36 42

there's never a difference of just 1

21, and 24 have a difference of 3 though
also 21 = 7x 3
and 24 = 4x 3

Answer 24

\[\left[\begin{array}{ccc|c}

6 & 4 & -3 & 67<times4\\
-6 & -3 & 1 & -62\\
3 & -7 & -7 & 7<times7
\end{array}\right]\]
\[\left[\begin{array}{ccc|c}

24 & 16 & -12 & 268\\
-6 & -3 & 1 & -62\\
21 & -49 & -49 & 49
\end{array}\right]\]

Answer 25

change sign in 3. row and add it to first row
\[\left[\begin{array}{ccc|c}

24 & 16 & -12 & 268\\
-6 & -3 & 1 & -62\\
-21 & +49 & +49 & -49
\end{array}\right]\]

\[\left[\begin{array}{ccc|c}

3 & 65 & 37 & 219\\
-6 & -3 & 1 & -62\\
-21 & +49 & +49 & -49
\end{array}\right]\]

Answer 26

divide first row by 3

\[\left[\begin{array}{ccc|c}

1 & 65/3 & 37/3 & 73\\
-6 & -3 & 1 & -62\\
-21 & +49 & +49 & -49
\end{array}\right]\]

Answer 27

great. now the idea is since whave a pivot, then we can alsonull every other enttry inth first column by just multiplying the one by whatever it is, adding it up to that row, and ding, that row has a zero entry in the first position

Answer 28

the numbers 65/3 and 37/3 of course aren't that great to work with.
maybe we could've gotten a line beginning with 1
that has nicer numbers in the following entries...

Answer 29

@mathessentials You got the matrix wrong. I confirmed with liebealexxx that R1 is 3x - 7 - 7z = 7 which simplifies to 3x +0y - 7z = 14. That also makes the solutions into nice numbers.