Question 1203611
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I'll show how to solve this system using matrix row reduction.
The goal is to get the matrix into reduced row echelon form (RREF).


We have this given system
x+4y-6z=-1
2x-y+2z=-7
-x+2y-4z=5


which is the same as
1x+4y-6z=-1
2x-1y+2z=-7
-1x+2y-4z=5


It forms this matrix
<table border = "1" cellpadding = "5"><tr><td>1</td><td>4</td><td>-6</td><td>-1</td></tr><tr><td>2</td><td>-1</td><td>2</td><td>-7</td></tr><tr><td>-1</td><td>2</td><td>-4</td><td>5</td></tr></table>
The last column represents the right hand values -1, -7 and 5. 
The rest of the matrix represents the coefficients.


Normally a matrix does not have separating grid lines, which is unfortunate. But I'll use grid lines to help separate the values. It should hopefully make things look a bit cleaner.


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Here are the steps for RREF.

<table border = "1" cellpadding = "5"><tr><td>1</td><td>4</td><td>-6</td><td>-1</td></tr><tr><td>2</td><td>-1</td><td>2</td><td>-7</td></tr><tr><td>-1</td><td>2</td><td>-4</td><td>5</td></tr></table>

<table border = "1" cellpadding = "5"><tr><td>1</td><td>4</td><td>-6</td><td>-1</td><td></td></tr><tr><td>0</td><td>-9</td><td>14</td><td>-5</td><td>R2  -  2*R1 --> R2</td></tr><tr><td>-1</td><td>2</td><td>-4</td><td>5</td><td></td></tr></table>
Notation like R2 - 2*R1 --> R2 means "subtract off twice of row 1 from row 2. Then store the results in row 2 (we overwrite row 2)".



<table border = "1" cellpadding = "5"><tr><td>1</td><td>4</td><td>-6</td><td>-1</td><td></td></tr><tr><td>0</td><td>-9</td><td>14</td><td>-5</td><td></td></tr><tr><td>0</td><td>6</td><td>-10</td><td>4</td><td>R3+R1 --> R3</td></tr></table>

<table border = "1" cellpadding = "5"><tr><td>1</td><td>4</td><td>-6</td><td>-1</td><td></td></tr><tr><td>0</td><td>1</td><td>-14/9</td><td>5/9</td><td>(-1/9)*R2 -> R2</td></tr><tr><td>0</td><td>6</td><td>-10</td><td>4</td><td></td></tr></table>

<table border = "1" cellpadding = "5"><tr><td>1</td><td>4</td><td>-6</td><td>-1</td><td></td></tr><tr><td>0</td><td>1</td><td>-14/9</td><td>5/9</td><td></td></tr><tr><td>0</td><td>0</td><td>-2/3</td><td>2/3</td><td>R3  -  6*R2 -> R3</td></tr></table>

<table border = "1" cellpadding = "5"><tr><td>1</td><td>4</td><td>-6</td><td>-1</td><td></td></tr><tr><td>0</td><td>1</td><td>-14/9</td><td>5/9</td><td></td></tr><tr><td>0</td><td>0</td><td>1</td><td>-1</td><td>(-3/2)*R3 -> R3</td></tr></table>
The matrix is in row echelon form (REF) but not RREF. This is because we have a lower triangular region of zeros below the main diagonal pivot entries. 


<table border = "1" cellpadding = "5"><tr><td>1</td><td>4</td><td>-6</td><td>-1</td><td></td></tr><tr><td>0</td><td>1</td><td>0</td><td>-1</td><td>R2 + (14/9)*R3 -> R2</td></tr><tr><td>0</td><td>0</td><td>1</td><td>-1</td><td></td></tr></table>

<table border = "1" cellpadding = "5"><tr><td>1</td><td>0</td><td>-6</td><td>3</td><td>R1  -  4*R2 -> R1</td></tr><tr><td>0</td><td>1</td><td>0</td><td>-1</td><td></td></tr><tr><td>0</td><td>0</td><td>1</td><td>-1</td><td></td></tr></table>

<table border = "1" cellpadding = "5"><tr><td>1</td><td>0</td><td>0</td><td>-3</td><td>R1 + 6*R3 -> R1</td></tr><tr><td>0</td><td>1</td><td>0</td><td>-1</td><td></td></tr><tr><td>0</td><td>0</td><td>1</td><td>-1</td><td></td></tr></table>


For more practice with RREF, here is a very useful tool
<a href="http://www.math.odu.edu/~bogacki/lat/">http://www.math.odu.edu/~bogacki/lat/</a>
It is called "linear algebra toolkit". It is a collection of matrix solvers that show step by step solutions. 


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To summarize:
We started with this 3x4 matrix
<table border = "1" cellpadding = "5"><tr><td>1</td><td>4</td><td>-6</td><td>-1</td></tr><tr><td>2</td><td>-1</td><td>2</td><td>-7</td></tr><tr><td>-1</td><td>2</td><td>-4</td><td>5</td></tr></table>
and ended up with this reduced row echelon form (RREF)
<table border = "1" cellpadding = "5"><tr><td>1</td><td>0</td><td>0</td><td><font color=red>-3</font></td></tr><tr><td>0</td><td>1</td><td>0</td><td><font color=red>-1</font></td></tr><tr><td>0</td><td>0</td><td>1</td><td><font color=red>-1</font></td></tr></table>


The original 3x3 sub-block has morphed into the 3x3 identity matrix which has the main diagonal of all "1"s, and the rest are "0"s.
The right hand side of this RREF matrix are the solution values.


Answer:
<font color=red>x = -3
y = -1
z = -1</font>
I'll let the student check each equation. 
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