Determine values of h - augmented matrix

[Boxiom]

Homework Statement

Homework Equations

The Attempt at a Solution

Tried to get it on reduced echelon form, but I haven't done problems like this before so I don't know what I'm supposed to do.



Thanks!

 

[Karnage1993]

What have you tried? Show us the row operations you've done so we can work from there.

 

[Boxiom]

The thing is I don't know where to start. By multiplying with -4 and adding to second row, I have:

1 5 -4
0 20-h 0

And this doesn't really tell me anything, as I have never done matrix with variables before.

 

[SteamKing]

Ignore the augmented -column.

What condition must hold with the matrix of coefficients in order for the system to be consistent? (hint: it involves a determinant)

 

[Karnage1993]

The thing is I don't know where to start. By multiplying with -4 and adding to second row, I have:

1 5 -4
0 20-h 0

And this doesn't really tell me anything, as I have never done matrix with variables before.

You didn't calculate that properly. Try again.

 

[Boxiom]

Woops, I meant -20+h.

 

[Karnage1993]

Yes that was the answer I was looking for, but in this case, it doesn't matter since you could just multiply by -1. So,

Rewriting the last row in terms of variables, we have:

$0*x + (h - 20)*y = 0$
$(h - 20)*y = 0$

Recall that $a*b = 0$ if and only if $a = 0$ or $b = 0$. One case is that $y = 0$ and that would yield a consistent solution. What must be the second one?

 

[Boxiom]

So if h is 20 the solution would also be consistent?

 

[Karnage1993]

Yup, that's correct. Notice that plugging in $h = 20$, and then row reducing, we get a row of only $0s$ which is consistent. If $h$ was anything but 20, we would have the 2nd row look like: $0 \ r \ | \ 0$ which is inconsistent for any $r \in \mathbb{R}, r \not = 0$.

 

[Boxiom]

Alright, that made sense. Thanks for the help :)