Question 1204100
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I'll convert the system of equations into a matrix as shown below.
Then I'll get that matrix into row echelon form (REF)


Each matrix is presented as a table. In my opinion, the grid lines help separate things to make the entries look more cleaner. 
Normally however, matrix notation will not have these helpful grid lines.
<table border = "1" cellpadding = "5"><tr><td>1</td><td>2</td><td>-3</td><td>a</td></tr><tr><td>2</td><td>6</td><td>-11</td><td>b</td></tr><tr><td>1</td><td>-2</td><td>7</td><td>c</td></tr></table>

<table border = "1" cellpadding = "5"><tr><td>1</td><td>2</td><td>-3</td><td>a</td><td></td></tr><tr><td>2</td><td>6</td><td>-11</td><td>b</td><td></td></tr><tr><td>0</td><td>-4</td><td>10</td><td>c-a</td><td>R3  -  R1 --> R3</td></tr></table>
Notation like R3 - R1 --> R3 means we subtract rows 3 and 1, and store the results into row 3. 
<table border = "1" cellpadding = "5"><tr><td>1</td><td>2</td><td>-3</td><td>a</td><td></td></tr><tr><td>0</td><td>2</td><td>-5</td><td>b-2a</td><td>R2  -  2*R1 --> R2</td></tr><tr><td>0</td><td>-4</td><td>10</td><td>c-a</td><td></td></tr></table>

<table border = "1" cellpadding = "5"><tr><td>1</td><td>2</td><td>-3</td><td>a</td><td></td></tr><tr><td>0</td><td>1</td><td>-5/2</td><td>(b-2a)/2</td><td>(1/2)*R2 --> R2</td></tr><tr><td>0</td><td>-4</td><td>10</td><td>c-a</td><td></td></tr></table>

<table border = "1" cellpadding = "5"><tr><td>1</td><td>2</td><td>-3</td><td>a</td><td></td></tr><tr><td>0</td><td>1</td><td>-5/2</td><td>(b-2a)/2</td><td></td></tr><tr><td>0</td><td>0</td><td>0</td><td>-5a+2b+c</td><td>R3 + 4*R2 --> R3</td></tr></table>
We have gone from the matrix {{{(matrix(3,4,1,2,-3,a,2,6,-11,b,1,-2,7,c))}}} to the matrix {{{(matrix(3,4,1,2,-3,a,0,1,-5/2,(b-2a)/2,0,0,0,-5a+2b+c))}}}


The last row leads to the equation
0x+0y+0z = -5a+2b+c
or
0 = -5a+2b+c


Solve for c to get
c = 5a-2b


This will mean we have infinitely many solutions if and only if c = 5a-2b
Otherwise, -5a+2b+c is nonzero and it causes a contradiction, and hence no solutions.


The values of 'a' and b can be any two real numbers you want.


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Let's look at an example that has infinitely many solutions.
a = 1, b = 2
c = 5a-2b = 5*1-2*2 = 1
Therefore this system
{{{system(x+2y-3z=1,2x+6y-11z=2,x-2y+7z=1)}}} 
is consistent and dependent.
It has infinitely many solutions.
I'll let the student verify this claim, and the later claims mentioned below.


Another example with infinitely many solutions.
a = 5, b = 3
c = 5a-2b = 5*5-2*3 = 19
This system
{{{system(x+2y-3z=5,2x+6y-11z=3,x-2y+7z=19)}}} 
is consistent and dependent.
It has infinitely many solutions.


Let's look at an example of a system that has no solutions.
a = 1, b = 2, c = 3
These a,b,c values do not satisfy the equation c = 5a-2b
Therefore this system shown below has no solutions (it is inconsistent)
{{{system(x+2y-3z=1,2x+6y-11z=2,x-2y+7z=3)}}}



As you can see, we cannot pick a trio of a,b,c values to have the system produce exactly one unique solution. 
Either we have infinitely many solutions, or none at all.
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