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?

 

[TD]

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?

 

[TD]

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?

 

[TD]

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