how to go about computing following determinant? I tried using Gaussian elimination on some special cases and figured there might be some pattern, maybe a recurrence relation involved, but I just can't see it.  $\begin{vmatrix}   1 &amp; 2 &amp; 3 &amp; \cdots &amp; n-1 &amp; n \\   2 &amp; 3 &amp; 4 &amp; \cdots &amp; n &amp; 1 \\   3 &amp; 4 &amp; 5 &amp; \cdots &amp; 1 &amp; 2 \\   \vdots  &amp; \vdots  &amp; \ddots &amp; \vdots  \\   n &amp; 1 &amp; 2 &amp; \cdots &amp; n-2 &amp; n-1   \end{vmatrix}$
99163
I'm stuck in this question: How calculate this determinant ? $$\Delta=\left|\begin{array}{cccccc} 1&amp;2&amp;3&amp;\cdots&amp;\cdots&amp;n\\ n&amp;1&amp;2&amp;\cdots&amp;\cdots&amp; n-1\\ n-1&amp;n&amp;1&amp;\cdots&amp;\cdots&amp;n-2\\ \vdots &amp;\ddots &amp; \ddots&amp;\ddots&amp;&amp;\vdots\\ \vdots &amp;\ddots &amp; \ddots&amp;\ddots&amp;\ddots&amp;\vdots\\ 2&amp;3&amp;4&amp;\cdots&amp;\cdots&amp;1 \end{array}\right|$$ Thanks a lot.
19918
I'm having troubles solving the following determinant:  $$\left|   \begin{array}{cccc}     a_1 &amp; x &amp; \ldots &amp; x \\    x &amp; a_2 &amp; \ldots &amp; \vdots \\     \vdots &amp; \ldots &amp; \ddots &amp; x \\     x &amp; \ldots &amp; x &amp; a_n \\   \end{array} \right|$$  I have tried coming up with recurrent formula, but there's always one lingering row and column. I also tried finding values of x for which determinant is 0, hoping I could get a nice polynomial solution, with no luck.  By subtracting first row from every other I get somewhat nicer result, but there's still that one row/column which I can't solve. If all other columns are added to the first one and then I subtract the first row, then the column doesn't depend on x anymore, but I haven't found 1..n much more useful.
53005
$$\begin{bmatrix} 1235 &amp;2344 &amp;1234 &amp;1990\\ 2124 &amp; 4123&amp; 1990&amp; 3026 \\ 1230 &amp;1234 &amp;9095 &amp;1230\\ 1262 &amp;2312&amp; 2324 &amp;3907  \end{bmatrix}$$  Clearly, its determinant is not zero and, hence, the matrix is invertible.   Is there a more elegant way to do this?  Is there a pattern among these entries?
116960
I have this determinant:   $A=\begin{vmatrix} 1 &amp; a_{1} &amp; ... &amp; a_{1}^{n-2} &amp; a_{1}^{n} \\  1 &amp; a_{2} &amp; ... &amp; a_{2}^{n-2} &amp; a_{2}^{n}\\   .&amp; . &amp; ... &amp; . &amp; .\\   .&amp; . &amp; ... &amp; . &amp; .\\  1 &amp; a_{n} &amp; ... &amp; a_{n}^{n-2} &amp; a_{n}^{n} \end{vmatrix}$  Prove that: $A=\left ( a_{1}+a_{2}+...+a_{n} \right )\det\left ( B \right ),$ inside $B$ is the Vandermonde matrix.  Actually, I have not yet thought of a solution. Hope everyone helps me with the solution of this problem.
97342
I'm just starting to learn linear algebra and I understand Gaussion elimination but I don't understand this proof:    Is the heart of this the fact that all the while when I've been row-reducing matrices to reduced row-echelon form... I've actually been multiplying a matrix by a series of a elementary matrices?  What does the notation $[A\ \ 0]$ mean?  How is the equation $A = E_1^{-1}...E_k^{-1}$ arrived at?
12150
How do I calculate the determinant of the following $n\times n$ matrices  $$\begin {bmatrix} 0 &amp; 1 &amp; \ldots &amp; 1 \\ 1 &amp; 0 &amp; \ldots &amp; 1 \\ \vdots &amp; \vdots &amp; \ddots &amp; \vdots \\ 1 &amp; 1 &amp; ... &amp; 0 \end {bmatrix}$$  and the same matrix but one of columns replaced only with $1$s?  In the  above matrix all off-diagonal elements are $1$ and diagonal elements are $0$.
33605
\begin{array}{rrrrr|r}     b &amp; a &amp; a &amp; \cdot \cdot \cdot &amp; a \\     a &amp; b &amp; a &amp; \cdot \cdot \cdot &amp; a \\     a &amp; a &amp; b &amp; \cdot \cdot \cdot &amp; a \\     \cdot &amp; \cdot &amp; \cdot &amp; \space &amp; \cdot\\ \cdot &amp; \cdot &amp; \cdot &amp; \space &amp; \cdot\\ a &amp; a &amp; a &amp; \cdot \cdot  \cdot &amp; b   \end{array}  I have the above matrix $A\in M_{n\times n}(F)$ where $F$ is a field and $n\geq1$, $a,b\in F$.  I'm trying to find out how to use row operations to make it into an upper triangular matrix in order to figure out the determinant. But I'm not sure how I would approach it.
87321
A friend of mine asked me to calculate the value of a perticular determinant, but after spending quite some time on it I'm unable to find a solution to it. The given determinant is:  $$         \begin{vmatrix}         0 &amp; 1 &amp; \cdots &amp;n-2&amp;n-1 \\         n-1 &amp; 0 &amp; \cdots &amp;n-3&amp;n-2 \\         \vdots &amp; \vdots &amp; \ddots &amp; \vdots &amp;\vdots\\         1 &amp; 2 &amp; \cdots &amp; n-1 &amp;0 \\         \end{vmatrix} $$  Also my friend is supposed to solve this determinant by more or less elementary methods, like elementary row operations, transpose... and no eigenvalues, nor adjoints.    First of all it seems like the value of the determinant is:  $$         \frac{1}{2n}\begin{vmatrix}         0 &amp; n &amp; \cdots &amp;n&amp;n \\         n &amp; 0 &amp; \cdots &amp;n&amp;n \\         \vdots &amp; \vdots &amp; \ddots &amp; \vdots &amp;\vdots\\         n &amp; n &amp; \cdots &amp; n &amp;0 \\         \end{vmatrix} $$  Anyway I tried adding all the rows to the last one and by taking $\frac{n(n-1)}{2}$ in front to have a row of 1s, but it didn't helped me much.  Then I tried to use the fact that:  $$         \begin{vmatrix}         A &amp; C \\         0 &amp; B \\         \end{vmatrix} = \det(A) \cdot \det(B) $$, where $A,B$ are square matrices. By subtracting the $(\frac{n}{2} + i)$-th column from the $i$-th ($i=0,1,2...$) and then adding the $i$-th row to the $(\frac{n}{2} + i)$-th row, one can obtain the wanted form, but again nothing more. Also this method pretty much fails when $n$ is odd.  My best attepmt was to notice that $A_{ij} + A_{ji} = n$ and using $\det(A^{T}) = \det(A)$, but unfortunatelly  $$\begin{vmatrix}         0 &amp; n &amp; \cdots &amp;n&amp;n \\         n &amp; 0 &amp; \cdots &amp;n&amp;n \\         \vdots &amp; \vdots &amp; \ddots &amp; \vdots &amp;\vdots\\         n &amp; n &amp; \cdots &amp; n &amp;0 \\         \end{vmatrix} \not = \begin{vmatrix}         0 &amp; 1 &amp; \cdots &amp;n-2&amp;n-1 \\         n-1 &amp; 0 &amp; \cdots &amp;n-3&amp;n-2 \\         \vdots &amp; \vdots &amp; \ddots &amp; \vdots &amp;\vdots\\         1 &amp; 2 &amp; \cdots &amp; n-1 &amp;0 \\         \end{vmatrix} + \left( \begin{vmatrix}         0 &amp; 1 &amp; \cdots &amp;n-2&amp;n-1 \\         n-1 &amp; 0 &amp; \cdots &amp;n-3&amp;n-2 \\         \vdots &amp; \vdots &amp; \ddots &amp; \vdots &amp;\vdots\\         1 &amp; 2 &amp; \cdots &amp; n-1 &amp;0 \\         \end{vmatrix}\right)^T$$ I tried to decompose the LHS by using the $n-$linearity of the determinants, by then one gets $2^n$ determinants on the RHS similar to the first one, but things are messy.
67658
Show that  $$det \begin{bmatrix}         1 &amp; 1 &amp; \cdots &amp;1 \\         \lambda_1 &amp; \lambda_2 &amp; \cdots &amp;\lambda_n \\         \lambda^2_1 &amp; \lambda^2_2 &amp; \cdots &amp;\lambda^2_n \\         \vdots &amp; \vdots &amp; \vdots &amp; \vdots \\         \lambda^{n-1}_1 &amp; \lambda^{n-1}_2 &amp; \cdots &amp;\lambda^{n-1}_n \\ \end{bmatrix} =\prod_{n\ge i&gt;j\ge1}(λ_i−λ_j)$$  My work so far,  I am trying to use induction on $n$.  For the base case n=2, the claim is true, since we have $(\lambda_2 - \lambda_1)$.  I'm not sure how to carry out the inductive step.  Any hints or suggestions are welcome.    Thanks,
16411
Suppose that $A \in \mathbb{R}^{n\times n }$, and $x,y\in \mathbb{R}^n$.    Show that there exist numbers $a,b$ so that $\det{(A + sxy^*)}=a+bs$   Show that if $\det{(A)}\neq 0$ then $a = \det{(A)}$ and $b = \det{(A)} \cdot y^*A^{-1}x$.   Is it true that if $\det{(A)}=0$ then $a=0$?    My answer so far: The third question seems trivial and silly, and I don't understand why it's on there. I have solved the first one as follows: take $\beta$ to be any basis whose first element is $x$. Then with respect to basis $\beta$, $A+ sxy^*$ has the form   $$\left(  \begin{matrix} \alpha_{11} + sy_1 &amp; \alpha_{12} + sy_2  &amp; \dotsb &amp; \alpha_{1n}+sy_n \\ \alpha_{21}&amp;\dotsb&amp;&amp;\alpha_{2n} \\ \vdots &amp; \ddots &amp;&amp;\vdots\\ \alpha_{n1} &amp; \dotsb &amp;&amp;\alpha_{nn} \end{matrix} \right),$$  from which it's clear that the determinant is a polynomial in $s$ of degree $\leq 1$.    I am less clear on the second question.   For a limited case, note that if $A$ is diagonalizable and $x$ is an eigenvector of $A$ with eigenvalue $\lambda_x \neq 0$, then adding $sxy^*$ to $A$ has the effect of preserving all the eigenvectors, bumping up $\lambda_x$ to $\lambda_x + sy^*x$, and since the determinant is the product of all the eigenvalues, the product of $\lambda_1 \lambda_2  \dotsb (\lambda_x + sy^*x)\dotsb \lambda_n $ is   $$\det{(A)} + \det{(A)}\frac{sy^*x}{\lambda_x}=\det{(A)}\cdot sy^*(\frac{x}{\lambda_x}) = \det{(A)}\cdot s y^*(A^{-1}x).$$  Note that $y^*A^{-1}x$ is $\alpha_1 y_1 + \dotsb + \alpha_n y_n$, for $\alpha_i$ such that $x = \alpha_1 A_1 + \dotsb + \alpha_n A_n$, where $A_i$ are the columns of $A$.  But this is all I've got. Can someone give me a hint? I'm not looking for you to feed me the answer. Thanks as always
40601
Assuming that given a $2\times2$ matrix $A$ with the property $A^3=0$ show that $A^2=0$.  Okay so this question is messing with me. When it says $A^3=0$ do you think it means the determinant? How would one approach this? Any help would be much appreciated
40843
Let  $$f_n := \begin{vmatrix} a_1 &amp; b_1 \\ c_1 &amp; a_2 &amp; b_2 \\ &amp; c_2 &amp; \ddots &amp; \ddots \\ &amp; &amp; \ddots &amp; \ddots &amp; b_{n-1} \\ &amp; &amp; &amp; c_{n-1} &amp; a_n \end{vmatrix}$$  Apparently, the determinant of the tridiagional matrix above is given by the recurrence relation   $$f_n = a_n f_{n-1} - c_{n-1} b_{n-1}f_{n-2}$$  with initial values $f_0 = 1$ and $f_{-1} = 0$ (according to Wikipedia). Can anyone please explain to me how they came to this recurrence relation?  I don't really understand how to derive it.
10950
I have the following $n\times n$ matrix:  $$A=\begin{bmatrix} a &amp; b &amp; \ldots &amp; b\\ b &amp; a &amp; \ldots &amp; b\\ \vdots &amp; \vdots &amp; \ddots &amp; \vdots\\ b &amp; b &amp; \ldots &amp; a\end{bmatrix}$$  where $0 &lt; b &lt; a$.  I am interested in the expression for the determinant $\det[A]$ in terms of $a$, $b$ and $n$.  This seems like a trivial problem, as the matrix $A$ has such a nice structure, but my linear algebra skills are pretty rusty and I can't figure it out.  Any help would be appreciated.
87576
Today, at my linear algebra exam, there was this question that I couldn't solve.   Prove that $$\det \begin{bmatrix}  n^{2} &amp; (n+1)^{2} &amp;(n+2)^{2} \\  (n+1)^{2} &amp;(n+2)^{2}  &amp; (n+3)^{2}\\  (n+2)^{2} &amp; (n+3)^{2} &amp; (n+4)^{2} \end{bmatrix} = -8$$   Clearly, calculating the determinant, with the matrix as it is, wasn't the right way. The calculations went on and on. But I couldn't think of any other way to solve it. Is there any way to simplify $A$, so as to calculate the determinant?
100703
Given a positive integer $n$, express$$ f_n(x) = \left|\begin{array}{c c c c c}  1 &amp; x &amp; \cdots &amp; x^{n - 1} &amp; x^n\\ x &amp; 1 &amp; x &amp; \cdots &amp; x^{n - 1} \\ \vdots &amp; x &amp; \ddots &amp; \ddots &amp; \vdots\\ x^{n - 1} &amp; \vdots &amp; \ddots &amp; 1 &amp; x\\ x^n &amp; x^{n - 1} &amp; \cdots &amp; x &amp; 1 \end{array}\right| $$   as a polynomial of $x$.   I tried to find a recurrence relation of $\{f_n\}_{n \geqslant 1}$ using Laplace expansion, but there seems to be no patterns in the minors in the expansion. Is there a somewhat simple recurrence relation of $\{f_n\}_{n \geqslant 1}$ or these determinants can be computed with other methods?
30188
On , I saw the following formula  $$\det\begin{bmatrix}A &amp; B\\ C &amp; D\end{bmatrix} = \det(AD-BC)$$  if $C$ and $D$ commute. Is this always true?   Or is there a good counter example for each $2 \times 2$ block matrices?
15194
I am trying to understand (intuitive explanation will be fine) why determinant is a multilinear function and therefore to learn how elementary row operation affect the determinant.  I understand that it has something to do with the definition of determinant by permutations, due to permutation being a bijection, in each product of the determinant there is just one entry from each row, but what's next?