# Find value h so the matrix has infinitely many solutions?

mr_coffee
Hello everyone I'm confused on what I'm suppose to do to solve this matrices.
|7 -7|5 |
|21 h|15|

What am I trying to make h be? I row reduced to get
|7 -7 |5|
|0 21+h |0|
So if i let h = -21 will this make it have infin. solutions? because then i would have 0 0 0?

Homework Helper
Indeed, by doing so you create 2 lineair dependant equations so when reduving, one cancels out as a 0 row. You then have 1 equation in 2 unknowns, giving an infinite set of solutions.

mr_coffee
Thanks! this brings up another question..
I'm supppose to find a value of k that will make it have no solutions.
I have:
|1 1 4 2|
|1 2 -4 3|
|6 13 k 20|

-6R2 + R3 -> R3
|1 1 4 2|
|1 2 -4 3|
|0 1 24+k 2|

now if i let k = -24 will this make it have no solutions? because 0 cannot equal 2 right?

Homework Helper
Then it would say in the last row: 0 1 0 2, which is still possible since then you have y = 2. You have to reduce it further to see it, I think
You could also compute the determinant of the coëfficiënt matrix and let it equal 0, solve for k.

mr_coffee
When i keep trying to reduce it gets bad...
I ended up with
|1 1 4 2|
|2 3 0 5|
|0 0 -20-k 0|
and now I'm stuck, anyu ideas?

Homework Helper
That's strange, perhaps you made some mistakes because that doesn't seem right to me.
Check your work again or use the other method I gave.

fahd
reduce the matrix in row echelon form,and follow the rules to solve it..its simple