```Question 115252
<pre><font size = 5 color = "blue"><b>
Use Gauss-Jordan elimination method to solve

2x-5y+5z=17
x-2y+3z=9
-x+3y=-4

Line up the terms vertically:

2x - 5y + 5z = 17
x - 2y + 3z =  9
-x + 3y      = -4

Fill up the "hole" in the bottom
equation with "+ 0z:

2x - 5y + 5z = 17
x - 2y + 3z =  9
-x + 3y + 0z = -4

Fill in all invisible 1's for
coefficients:

2x - 5y + 5z = 17
1x - 2y + 3z =  9
-1x + 3y + 0z = -4

Erase all the letters and
equal signs:

2  - 5  + 5    17
1  - 2  + 3     9
-1  + 3  + 0    -4

Erase the plus signs
and move the minus
signs close to the
numbers as negative
signs:

2   -5    5    17
1   -2    3     9
-1    3    0    -4

Draw a vertical line
where the equal signs
were and put brackets
around the whole thing.

[ 2   -5    5  | 17]
[ 1   -2    3  |  9]
[-1    3    0  | -4]

This is called the
augmented matrix

The idea is to get 0's
in the lower left three
positions, where the
three red numbers are
below:

[ 2   -5    5  | 17]
[<font color = "red"> 1</font>   -2    3  |  9]
[<font color = "red">-1    3</font>    0  | -4]

To get a 0 where the 1 is,
temporarily multiply the top
row thru by 1 and the middle
row thru by -2.

1[ 2   -5    5  | 17]
-2[<font color = "red"> 1</font>   -2    3  |  9]
[<font color = "red">-1    3</font>    0  | -4]

[ 2   -5    5  | 17]
[<font color = "red">-2</font>    4   -6  |-18]
[<font color = "red">-1    3</font>    0  | -4]

Add the top row to the middle
row and leave the top row as
it is:

[ 2   -5    5  | 17]
[<font color = "red"> 0</font>   -1   -1  | -1]
[<font color = "red">-1    3</font>    0  | -4]

To get a 0 where the -1 at the
bottom left is, multiply the
top row by 1 and the bottom row
by 2:

1[ 2   -5    5  | 17]
[<font color = "red"> 0</font>   -1   -1  | -1]
2[<font color = "red">-1    3</font>    0  | -4]

[ 2   -5    5  | 17]
[<font color = "red"> 0</font>   -1   -1  | -1]
[<font color = "red">-2    6</font>    0  | -8]

Add the top row to the bottom
row, leaving the top row as it
is:

[ 2   -5    5  | 17]
[<font color = "red"> 0</font>   -1   -1  | -1]
[<font color = "red"> 0    1</font>    5  |  9]

To get a 0 where the 1 is,
multiply the middle row by 1
and add to the bottom row:

[ 2   -5    5  | 17]
1[<font color = "red"> 0</font>   -1   -1  | -1]
1[<font color = "red"> 0    1</font>    5  |  9]

[ 2   -5    5  | 17]
[<font color = "red"> 0</font>   -1   -1  | -1]
[<font color = "red"> 0    0</font>    4  |  8]

Now that there are 0's
in those three positions,
we rewrite the augmented
matrix as a system of
equations, by putting the
variables and equal signs
back in:

[ 2x  -5y   5z = 17]
[<font color = "red"> 0x</font>  -1y  -1z = -1]
[<font color = "red"> 0x   0y</font>   4z =  8]

Erase the brackets, the
terms with 0 coefficients,
the 1's, and move the negative signs
left as minus signs:

2x - 5y + 5z = 17
-y -  z = -1
4z =  8

Solve the bottom equation for z:

4z = 8
z = 2

Substitute z = 2 into the middle
equation:

-y -  z = -1
-y -  2 = -1
-y =  1
y = -1

Substitute y = -1 and z = 2 into
the top equation:

2x - 5y + 5z = 17
2x - 5(-1) + 5(2) = 17
2x + 5 + 10 = 17
2x + 15 = 17
2x =  2
x =  1

So the solution is

(x, y, z) = (1, -1, 2)

Edwin</pre></font></b>
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