Question

Determinants problem

Answer 1

in this kind of way it spreads to infinity,we have to find the value of this determinant

Answer 2

ok

Answer 3

multiply the diagonals

Answer 4

i think u gotta get the bottom half 0 first though

Answer 5

lemme check..

Answer 6

its a N x N matrix

Answer 7

i mean determinant

Answer 8

determinant x^n?

Answer 9

\[\det=\lim_{n \to \infty}x^n\]

Answer 10

http://en.wikipedia.org/wiki/Determinant#Infinite_matrices

Answer 11

it seems like it has nothing to do with limit

Answer 12

If you let\[A_n\]be the nxn matrix with x's on the diagonal and 1's everywhere else, it looks like the determinant is:\[\det(A_n)=(x-1)^{n-1}(x+(n-1))\]

Answer 13

$$\det\begin{bmatrix}x\end{bmatrix}=x\\\det\begin{bmatrix}x&1\\1&x\end{bmatrix}=x^2-1=(x+1)(x-1)\\\det\begin{bmatrix}x&1&1\\1&x&1\\1&1&x\end{bmatrix}=x^3-3x+2=(x-2)(x-1)\\\dots$$@joemath314159 are you sure that latter factor is always \(+\)? i think it needs a \((-1)^n\) factor

Answer 14

oh oops nvm misread!

Answer 15

i'd agree with @joemath314159; there is a pattern in the eigenvalues, namely that \(x-1\) is of multiplicity \(n-1\) and that we have an additional eigenvalue \(x+(n-1)\)

Answer 16

ohh thats cool :)

Answer 17

I'll show a specific example using a 5x5 matrix, but this can be done with any size. \[\left|\begin{array}{ccccc} x &1&1&1&1 \\ 1&x&1&1&1 \\ 1&1&x&1&1 \\ 1&1&1&x&1\\ 1&1&1&1&x \end{array}\right|= -\left|\begin{array}{ccccc} 1 &x&1&1&1 \\ x&1&1&1&1 \\ 1&1&x&1&1 \\ 1&1&1&x&1\\ 1&1&1&1&x \end{array}\right|\]\[=-\left|\begin{array}{ccccc} 1 &x&1&1&1 \\ 0&1-x^2&1-x&1-x&1-x \\ 0&1-x&x-1&0&0 \\ 0&1-x&0&x-1&0\\ 0&1-x&0&0&x-1 \end{array}\right|\]\[=-1\cdot \left|\begin{array}{cccc} 1-x^2&1-x&1-x&1-x \\ 1-x&x-1&0&0 \\ 1-x&0&x-1&0\\ 1-x&0&0&x-1 \end{array}\right|\]\[=-(1-x)^4\left|\begin{array}{cccc} 1+x&1&1&1 \\ 1&-1&0&0 \\ 1&0&-1&0\\ 1&0&0&-1 \end{array}\right|\]\[=-(1-x)^4\left|\begin{array}{cccc} 4+x&0&0&0 \\ 1&-1&0&0 \\ 1&0&-1&0\\ 1&0&0&-1 \end{array}\right|\]\[=-(1-x)^4(4+x)(-1)^3=(x-1)^4(x+4)\]