## anonymous 3 years ago For which values of the constant a does the equationsystem have infinty many solutions and then doesn't any solutions exist. $ax _{1} +(a-1)x _{2}+(a-2)x _{3}=0$ $(a-3)x _{2}+(a-4)x _{3}=1$ $(a-5)x _{3}=0$ Well I know that the system has infinity many solutions when a=5 but what should I look for then searching for the values there no solutions exist?

1. anonymous

$\left[\begin{matrix}a & a-1&a-2 \\ 0 & a-3&a-4 \\ 0&0&a-5\end{matrix}\right]$

2. anonymous

@satellite73

3. anonymous

i am not sure, but since you have the matrix there, maybe you could take the determinate, set it equal to zero and solve for $$a$$

4. anonymous

If i rember correct then infinity is when the matrix get a full row of zeros and no solution then matrix gets a row of zeros but the answer isn't zero, do you think that can be right?

5. anonymous

the determinant is $$a(a-3)(a-5)$$ so maybe no solution if $$a=0,a=3,a=5$$

6. anonymous

The key says a=0 and a=3 so that seems to be right

7. anonymous

whew

8. anonymous

Hehe so now I know the way to solve these are by using the determinant, thank you satellite73, helpful as always :)

9. anonymous

yw