How do I find solution depending on parameters a,b in RR?

2x+ay-3z=2
x+y-z=1
-x+by+z=3

1 Answer
Jan 2, 2018

If b!=-1
then {(x=1+(4(a-3))/(b+1)),(y=4/(b+1)),(z=(4(a-2))/(b+1)):}

but if b=-1 then there's no solution.

Explanation:

{(2x+ay-3z=2),(x+y-z=1),(-x+by+z=3):}

I'm using Gaussian elimination with smart choice of the first equation and substitution.

Add second equation to third and subtract from first two times
{(2x-color(red)(2x)+ay-color(red)(2y)-3z-(color(red)(-2z))=2-color(red)(2)larr),(x+y-z=1rarr), (color(red)(x)-x+by+color(red)(y)+z+(color(red)(-z))=3+color(red)(1)larr):}

Combine like terms
{(color(red)((a-2)y-z=0)),(x+y-z=1),(color(red)((b+1)y=4)):}

We got y
{((a-2)y-z=0),(x+y-z=1),(color(red)(y=4/(b+1))):}
Assuming b!=-1. If b=-1 then 0=4 we have contradiction and no solutions.

Substitute
{((a-2)color(red)(4/(b+1))-z=0larr),(x+color(red)(4/(b+1))-z=1larr),(y=4/(b+1)rarr):}

Move known terms to the right
{(-z=-color(red)((4(a-2))/(b+1))),(x-z=1-color(red)(4/(b+1))),(y=4/(b+1)):}

We got z
{(color(red)(z=(4(a-2))/(b+1))),(x-z=1-4/(b+1)),(y=4/(b+1)):}

Substitute
{(z=(4(a-2))/(b+1)rarr),(x-color(red)((4(a-2))/(b+1))=1-4/(b+1)larr),(y=4/(b+1)):}

Move known terms to the right
{(z=(4(a-2))/(b+1)),(x=1-4/(b+1)+color(red)((4(a-2))/(b+1))),(y=4/(b+1)):}

Simplify and we finally got it
{(color(red)(x=1+(4(a-2)-4)/(b+1)=1+(4(a-3))/(b+1))),(y=4/(b+1)),(z=(4(a-2))/(b+1)):}