## anonymous one year ago Need help with Applied Linear Algebra

1. anonymous

2. anonymous

I found the reduced echelon form to be : |dw:1443287325608:dw|

3. anonymous

and now I do not know what to do, So I know that there will never be a "No Solution" to this system. and I know there will be a unique solution when a=0 (Correct me if I am wrong) but when will there be infinite solution?

4. anonymous

@ganeshie8

5. anonymous
6. surry99

Assuming your reduced matrix is correct, the only solution would be a=0

7. anonymous

so it will just be a trivial solution

8. surry99

yes

9. anonymous

but what if I just have as a row echelon form like this: |dw:1443288786467:dw| the answer would change then, then "a" will have to equal -1 for their to be a trivial solution

10. anonymous

should i use row echelon form or reduced form?

11. anonymous

$\left|\begin{matrix}1 & -1 &4 \\ 3 & -2 &10\\a&3&-8\end{matrix}\right|=1[\left( -2 \right)\left( -8 \right)-10*3]-(-1)[3*(-8)-10a]+4[3*3+2a]$ =16-30-24-10a+36+8a =-2-2a it has infinite solutions if-2-2a=0 or a=-1 no solution if $a \neq-1$ only trivial solution x1=x2=x3=0

12. anonymous

but homogenous systems never have "no solutions". so when it asks when It has no solution I would have to write "never". wait that doesnt make sense, if a=-1 I dont think there will be a inifinite solutions because then (-2-2a)x3=0 so x3 must equal 0. making all the other variables(x1, x2, x3=0) so there will be only 1 solution not infinite right?

13. anonymous

so ultimately when it asks when does the system have unique solution I would have to write when a=-1. and when it asks when does it have infinite solution, I would have to write "never". is that correct?

14. anonymous

you are correct if ax=0 then either a=0 or x=0