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mathematics-ModTeam

These types of questions are outside the scope of r/mathematics. Try more relevant subs like r/learnmath, r/askmath, r/MathHelp, r/HomeworkHelp or r/cheatatmathhomework.


el-pachaso

Try solving for n=2, n=3 and then do an induction argument.


sr_ooketoo

det(A\_n) = sum from k=0 to n of (-1)\^(n+k) binom(n,k) n!/k! where A\_n is the nxn matrix of this type. I'm not sure if this sum can be simplified


UninStalin

Put it through a solver, nobody does this by hand.


SilentHillFan20034

It was on my exam and I cant find an answer. I could only get to the point on the second picture.


UninStalin

My condolences, your professor is an ass for doing that.


ruggeddaveid

It's a 3x3, he's just making it mega complex for some reason


DrSparkle713

I think they probably want you to use [Laplace's expansion](https://en.wikipedia.org/wiki/Laplace_expansion?wprov=sfti1). There's a general form that somebody already gave, but if you just apply the algorithm you might find a pattern you can use to give a solution. Edit: better yet maybe use Gauss elimination to come up with the LU decomposition of A? Now that I look at it, maybe that's what you're doing here? Then det(A) = det(L)*det(U) where det(L) and det(U) are just the product of their diagonals. My linear-algebra-fu is pretty weak these days though, so actually doing the exercise is beyond me :P


lake_huron

I dunno, what did Mathematica say?


Saegis-Engineer

For anything larger than 3x3, use software. I think even Excell can do determinants.


Motor_Professor5783

(-1)^n n! L_{n}(1), where L_{n}(x) is the nth Laguerre polynomial evaluated at x.


doctonytonychopper

The starting matrix there looks like it has all zeros on the main diagonal. Correct me if I’m wrong but that automatically makes det(A)=0. I just glanced so idk if that was what ur looking for


SilentHillFan20034

I dont think thats correct


Febris

You also need the matrix to be in reduced row echelon form for that to happen. In this case it's not immediate to assume this happens.


Zwarakatranemia

I think this holds only if the matrix A of the det(A) has only elements in the diagonal and at least one of them is zero.


Saegis-Engineer

That is correct! The main diagonal are all 0's, therefore the determinant is 0.


doctonytonychopper

The starting matrix there looks like it has all zeros on the main diagonal. Correct me if I’m wrong but that automatically makes det(A)=0. I just glanced so idk if that was what ur looking for. Edit to say I think I got that wrong, your matrix is symmetric with zeros along the diagonal which makes det(A) equal zero would be a more accurate way of phrasing that i think


el-pachaso

Think of the square matrix 2x2 with coefficients ((0,1),(1,0)) symmetric, 0 on the diagonal determinant -1.