Question 1207192
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{{{system(2x-3y-9z=20,x+3z=-2, -3x+y-4z=-2)}}}
is equivalent to
{{{system(2x-3y-9z=20,1x+0y+3z=-2, -3x+1y-4z=-2)}}}


That system converts to this augmented matrix.
<table border = "1" cellpadding = "5"><tr><td>2</td><td>-3</td><td>-9</td><td>20</td></tr><tr><td>1</td><td>0</td><td>3</td><td>-2</td></tr><tr><td>-3</td><td>1</td><td>-4</td><td>-2</td></tr></table>
Normally the grid lines aren't present to separate each item. But I decided to make it into a table format.


Let's apply Gauss Jordan Elimination to get the matrix into Reduced Row Echelon Form (RREF).
<table border = "1" cellpadding = "5"><tr><td>2</td><td>-3</td><td>-9</td><td>20</td></tr><tr><td>1</td><td>0</td><td>3</td><td>-2</td></tr><tr><td>-3</td><td>1</td><td>-4</td><td>-2</td></tr></table>

<table border = "1" cellpadding = "5"><tr><td>1</td><td>0</td><td>3</td><td>-2</td><td>R1 <--> R2</td></tr><tr><td>2</td><td>-3</td><td>-9</td><td>20</td><td></td></tr><tr><td>-3</td><td>1</td><td>-4</td><td>-2</td><td></td></tr></table>

<table border = "1" cellpadding = "5"><tr><td>1</td><td>0</td><td>3</td><td>-2</td><td></td></tr><tr><td>0</td><td>-3</td><td>-15</td><td>24</td><td>R2 - 2*R1 --> R2</td></tr><tr><td>-3</td><td>1</td><td>-4</td><td>-2</td><td></td></tr></table>

<table border = "1" cellpadding = "5"><tr><td>1</td><td>0</td><td>3</td><td>-2</td><td></td></tr><tr><td>0</td><td>-3</td><td>-15</td><td>24</td><td></td></tr><tr><td>0</td><td>1</td><td>5</td><td>-8</td><td>R3 + 3*R1 --> R3</td></tr></table>

<table border = "1" cellpadding = "5"><tr><td>1</td><td>0</td><td>3</td><td>-2</td><td></td></tr><tr><td>0</td><td>1</td><td>5</td><td>-8</td><td>R2 <--> R3</td></tr><tr><td>0</td><td>-3</td><td>-15</td><td>24</td><td></td></tr></table>

<table border = "1" cellpadding = "5"><tr><td>1</td><td>0</td><td>3</td><td>-2</td><td></td></tr><tr><td>0</td><td>1</td><td>5</td><td>-8</td><td></td></tr><tr><td>0</td><td>0</td><td>0</td><td>0</td><td>R3 + 3*R2 --> R3</td></tr></table>



Here is a step-by-step calculator that is very useful to row reduce matrices
<a href="http://www.math.odu.edu/~bogacki/lat/">http://www.math.odu.edu/~bogacki/lat/</a>
It is called "linear algebra toolkit".
Click the "Enter" link and then go to "Row operation calculator". Let me know if you have any questions about this calculator. 



More practice with gauss-jordan elimination
<a href="https://www.algebra.com/algebra/homework/coordinate/Linear-systems.faq.question.1203611.html">https://www.algebra.com/algebra/homework/coordinate/Linear-systems.faq.question.1203611.html</a>


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To briefly summarize, we have gone from this matrix 
{{{(matrix(3,4,2,-3,-9,20,1,0,3,-2,-3,1,-4,-2))}}}
to this matrix
{{{(matrix(3,4,1,0,3,-2,0,1,5,-8,0,0,0,0))}}}
The row of all zeros tells us that we will have infinitely many solutions. This system is consistent and dependent.


The 2nd matrix converts back to this system
{{{system(x+3z=-2,y+5z=-8)}}}
and this is what results when we get each z term to the other side
{{{system(x=-3z-2,y=-5z-8)}}}


Therefore each of the infinitely many solutions are of the form (x,y,z) = (-3z-2,-5z-8,z) where z is any real number.


Examples:
If z = 0 then (x,y,z) = (-3z-2,-5z-8,z)= (-3*0-2,-5*0-8,0) = (-2,-8,0) is a solution
If z = 1 then (x,y,z) = (-3z-2,-5z-8,z)= (-3*1-2,-5*1-8,1) = (-5,-13,1) is a solution
If z = 2 then (x,y,z) = (-3z-2,-5z-8,z)= (-3*2-2,-5*2-8,2) = (-8,-18,2) is a solution
If z = 3 then (x,y,z) = (-3z-2,-5z-8,z)= (-3*3-2,-5*3-8,3) = (-11,-23,3) is a solution
All of these solution points are located on the same straight line. 
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