Can someone actually solve this determinant?

Can someone actually solve this determinant?
Thanks

Not doing your homework.

Yeah, but it doesn't take a genius to realize this is just something your professor gave you to see if you understand determinant properties, so you can do your own homework.

wats a determinant lol

Use induction or something

yeah I know but I'm sitting here for 2 hours doing everything and still cant figure this out

Damn really? I am going to help you out of pity. You did not ask to be born retarded.

Dude, it is really obvious. Here is a hint: Add up all the other columns to the first column and see what remains.

The determinant reduces to two smaller n-1 determinants, which reduces to two more smaller determinants etc...

>Being this low IQ
Do what is instructed in and then it reduces to just one smaller determinant.

I've never worked with determinants that are bigger than 3x3 before and I really need to see how the fuck i work with these.

If you really need to just do the special case when n=3. Then the pattern of what happens in the general case will become obvious.

After that just remember that determinants in all dimensions work the same. I mean, you proved all the determinant properties right? You must understand them.

our professor isn't that good desu. He left plenty of things unexplained during his lectures and
I dont have any books yet so i'm trying to work through the net

Just get any book on it. The answer is [math] (a_1 + a_2 + ... + a_n) x^{n-1} [/math] by the way. But it is too long to latex the entire procedure.

Just add up every other column to the first column. You will see then that from the second entry to the last, the first column will be all 0s. So now you reduce the determinant by that column and that will yield a (n-1) times (n-1) determinant that is defined like this:

Every entry in the main diagonal is an x.
Every entry directly below the main diagonal is an -x.
Everything above the main diagonal is a 0.
Everything below the -x's is a 0.

Then you have to work with that determinant separately and prove it evaluates to [math] x^{n-1} [/math]. This you can do by induction very easily, you'll see why.

Is it [spoiler](a1 + ... + an)*(x**(n-1))[/spoiler]?

Why in the ever-loving FUCK would I want to do that?

its clearly a polynomial.

a1x^n + a2x^n-2 + ... + anx

>autism speaks

You can do that, but you still have to repeat that process n times. It ends up being the same as what I said but you have to carry that summation for the rest of the calculation if you do it your way.

Wait, you can reduce a determinant by columns? I thought this only applies with rows.

Yeah, it is actually the same because remember that the determinant of a matrix is the same as the determinant of the transpose of a matrix. And then playing around with rows in the transpose is like playing around with columns in the original matrix.

You gotta put 2 and 2 together man.

>solve this determinant
L0Lno fgt pls

The way I'd do it would be to cycle all the columns so the last one is first. Then it would be almost upper triangular except for the x in the bottom left corner.

Looking at the new first column you'd see the determinant would be
[math]D_n=(-1)^{n-1}(a_n(-x)^{n-1}+(-1)^{n-1}xD_{n-1})=a_nx^{n-1}+xD_{n-1}[/math]


The result follows almost immediately by induction.

Swapping columns for rows and vice versa is an automorphism, my dude. Column reduction, independence, etc all work too.

thanks, that helps a lot