Question 1042313
which condition must be placed on a,b,c, so that the system 
x+2y-3z=a
2x+6y-11z=b
x-2y+7z=c
is consistent ? can the system have unique solution.
<pre><b>
{{{(matrix(3,5,

1,2,-3,"|",a,
2,6,-11,"|",b,
1,-2,7,"|",c))}}}

-2R1+1R2->R2

{{{matrix(3,1,-2,1,"")}}}{{{(matrix(3,5,

1,2,-3,"|",a,
2,6,-11,"|",b,
1,-2,7,"|",c))}}}

{{{(matrix(3,5,

1,2,-3,"|",a,
0,2,-5,"|",-2a+b,
1,-2,7,"|",c))}}}

-1R1+1R3->R3


{{{matrix(3,1,-1,"",1)}}}{{{(matrix(3,5,

1,2,-3,"|",a,
0,2,-5,"|",-2a+b,
1,-2,7,"|",c))}}}

{{{(matrix(3,5,

1,2,-3,"|",a,
0,2,-5,"|",-2a+b,
0,-4,10,"|",c-a))}}}

2R2+1R3->R3

{{{matrix(3,1,"",2,1)}}}{{{(matrix(3,5,

1,2,-3,"|",a,
0,2,-5,"|",-2a+b,
0,-4,10,"|",c-a))}}}

{{{(matrix(3,5,

1,2,-3,"|",a,
0,2,-5,"|",-2a+b,
0,0,0,"|",2(-2a+b)+(c-a)))}}}

To be consistent, since all the elements
but the last one on the bottom row are 0,
the last one must be 0 also:

{{{2(-2a+b)+(c-a) = 0}}}

{{{-4a+2b+c-a = 0}}}

{{{-5a+2b+c=0}}}

Or if you prefer c = 5a-2b

There will only be a unique solution if a=b=c=0

and that unique solution will be (x,y,z) = (0,0,0)

Edwin</pre></b>