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So I have a problem.. I already found out what kind of matrix it is.. So all main diagonals of this matrix are 0.. the rest is 1.. it's not 4x4 or 3x3 etc.. it's size is nxn.. does anyone of you know a formular to calculate the determinant for this nxn matrix or how it's called? Thanks alot

Here is an example

$$\begin{pmatrix}0&1&1&1\\1&0&1&1\\1&1&0& 1\\1& 1& 1& 0\end{pmatrix}$$ I hope you can help me out and thanks alot

Pp..
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MenMan
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  • I guess my answer was not sufficient for you. $(-1)^n(1-n)$ is correct. Maybe someone will offer an alternative method. http://math.stackexchange.com/questions/1104686/calculate-the-determinants-of-the-following-matrices/1104777#1104777 – David P Jan 15 '15 at 22:40
  • I still don't understand how that's the solution but still thank you for all your efforts – MenMan Jan 15 '15 at 22:44

3 Answers3

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Suppose you know about eigenvalues. The determinant is the product of the eigenvalues (including multiplicities). If $J$ is the $n\times n$ all-ones matrix, and $I$ is the $n\times n$ identity matrix, then your matrix is $J-I$. Use the fact that $J$ has rank 1 with constant row-sums to find the eigenvalues of $J$, then use this to find the eigenvalues of $J-I$.

As a side note, suppose you also know about graph theory. Then your matrix is the adjacency matrix of the complete graph on $n$ vertices.

Randy E
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  • I understand the matrix is the adjacency matrix of the complete graph on $n$ vertices, how do you find the determinant from knowing this? – HBeel Jan 17 '15 at 11:35
  • That was more of a side note than a hint to find the determinant, but there is a combinatorial interpretation of the determinant of the adjacency matrix of a graph $G$. If $H$ is a spanning subgraph of $G$ whose connected components are $s$ cycles and $t$ edges (and no isolated vertices), then $H$ contributes $2^s(-1)^t$ to the determinant of $G$. Take the sum of $2^s(-1)^t$ over all such subgraphs $H$ to get the determinant. For the complete graph, this means you are adding the signs of all permutations of a set of size with no fixed points. I doubt this is the best method, though. – Randy E Jan 17 '15 at 12:59
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Let $A_n$ be this $n\times n$ matrix.

Hints:

  1. Let $B_n$ be $A_n$, but with the first row replaced to $(1,0,0,0,\dots)$, and let $C_n$ be again $A_n$ but with the first $0$ of first row replaced to $1$.
  2. Observe that $\det A_n+\det B_n=\det C_n$.
  3. Express $\det B_n$ using $\det A_{n-1}$.
  4. Subtract first row from each other row in $C_n$, and determine $\det C_n$.
Berci
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Although I suspect your professor/the author wanted to use part of the already answered question (here) and apply it to get the determinant for this one, here is another way using diagonalization:

Verify that $A = PDP^{-1}$ where

$$D = \left(\begin{matrix}-1 & 0 & \cdots & 0 & 0\\ 0 & -1 & \cdots & 0 & 0\\ \vdots & \vdots & \ddots & 0 & 0\\ 0 & 0 & \cdots & -1 & 0\\0 & 0 & \cdots & 0 & n-1\end{matrix}\right)$$

and

$$P = \left(\begin{matrix}-1 & -1 & \cdots & -1 & 1\\ 0 & 0 & \cdots & 1 & 1\\ \vdots & \vdots & \cdot^{\large \cdot^{\large \cdot}} & 0 & 1\\ 0 & 1 & \cdots & 0 & 1\\1 & 0 & \cdots & 0 & 1\end{matrix}\right)$$

Note you can just find the eigenvalues and their multiplicities, and dont actually have to compute it. In doing this you will find $\lambda =-1$ has geometric multiplicity $n-1$, and $\lambda=n-1$ has multiplicity $1$. But as this is a task in of itself, for your consideration here is $P^{-1}$

$$P^{-1} = \dfrac{1}{n}\left(\begin{matrix}-1 & -1 & \cdots & -1 & n-1\\ -1 & -1 & \cdots & n-1 & -1\\ \vdots & \vdots & \cdot^{\large \cdot^{\large \cdot}} & -1 & -1\\ -1 & n-1 & \cdots & -1 & -1\\1 & 1 & \cdots & 1 & 1\end{matrix}\right)$$

Once you are convinced of this, we have

$$\det(A) = \det(PDP^{-1}) = \det(D) = (-1)^{n-1}(n-1) $$

which is the same as $(-1)^n(1-n)$.

Community
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David P
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  • Thanks again for your efforts.. I understand now why it's (-1)^n(1-n) at least I think so.. but I try to prove it with induction atm.. anyway still thank you very much David – MenMan Jan 16 '15 at 00:29
  • Can I beg you for one last favor.. how can you solve this task with induction.. I tried it but I don't really know how to continue after inserting the n+1 for n.. etc.. I always either end up in some nonsense etc – MenMan Jan 16 '15 at 21:16
  • First subtract the second row from the first then perform cofactor expansion along the first row. You will see how to apply the inductive hypothesis this way. – David P Jan 16 '15 at 21:25
  • Thanks I could prove it.. could you answer my questions to the b at the other question I posed.. since I think you did the cofactor expansion wrong unless I misunderstood your calculations – MenMan Jan 17 '15 at 00:08