Question 1136571
(a) Consider the following system of linear equations:
x+ 5y+ z= 0
x+ 6y- z= 0
2x +ay +bz= c
Find values of a, b, and c such that the above system of linear equations has:
<pre>
{{{system(x+5y+z=0, x+6y-z=0, 2x+ay+bz=c)}}}

Add the first and second equations:

{{{2x+11y = 0}}}
{{{2x=-11y}}}
{{{x=-11y/2}}}

Subtract the first equation from the second:

{{{y-2z=0}}}
{{{-2z=-y}}}
{{{z=y/2}}}

So from the first two equations, if there are any solutions, they can only be

{{{matrix(1,3,(matrix(1,5,x,",",y,",",z)),""="",(matrix(1,5,-11y/2,",",y,",",y/2)))}}}

Substitute in the 3rd given equation: 

{{{2x+ay+bz=c}}}

{{{2(-11y/2)+ay+b(y/2)=c}}}
{{{-11y+ay+by/2=c}}}
{{{-22y+2ay+by=2c}}}
{{{(-22+2a+b)y=2c}}}

Let's answer (ii) first.
</pre>(ii) an infinite number of solution;<pre>
If -22+2a+b=0 and 2c=0 then this equation becomes

{{{0y=0}}}, which has infinitely many solutions for y

That is when 2a+b=22 and c=0, for instance when a=7, b=8, c=0,

there are infinitely many solutions.

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If we solve for y

 {{{y=2c/(-22+2a+b)}}} 

Substituting in

{{{matrix(1,3,(matrix(1,5,x,",",y,",",z)),""="",(matrix(1,5,-11y/2,",",y,",",y/2)))}}}


{{{matrix(1,3,(matrix(1,5,x,",",y,",",z)),""="",(matrix(1,5,(-11c)/(-22+2a+b),",",2c/(-22+2a+b),",",c/(-22+2a+b))))}}}

Let's answer (iii)
</pre>(iii) no solution<pre>
That's when y is undefined.

So that is when the denominator -22+2a+b equals 0 and the numerator is
not equal to 0, for instance when a=7, b=8, c=1, there are no solutions.

Finally we answer (i)
</pre>(i) exactly one solution;<pre> 
There will be only 1 solution when -22+2a+b is not 0
For instance when a=4 and b=5 and c=1.

Edwin</pre>