Context (skippable)  I was asked (by a friend who is preparing for an exam) whether there was a special trick to compute the determinant of the following matrix. I didn't see anything beyond using the standard computations (like using "Gauss" to compute the value).  Then I asked another math student who, while quite bright, is a bit rusty in linear algebra and using sagemath we empirically found the below formula. Of course we were both confused as to a) whether it actually always holds and b) why it holds.  Actual question  Let $n\in\mathbb N$ be a positive integer. Let $I_n\in\mathbb R^{n\times n}$ be the identity matrix and let $1_n\in\mathbb R^{n\times n}$ be the all-one matrix, that is, the matrix for which every entry is $1$. Now I am confused as to why the following (empirically found) statement holds (or does not):  $$\forall n\in\mathbb N:\det(1_n-I_n)=(-1)^{n-1}(n-1)$$    For illustration purposes, here is the matrix for $n=4$ (with the determinant being $-3$): \begin{pmatrix} 0&amp;1&amp;1&amp;1\\ 1&amp;0&amp;1&amp;1\\ 1&amp;1&amp;0&amp;1\\ 1&amp;1&amp;1&amp;0 \end{pmatrix}
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 $0$.
21095
I was hoping that someone could point me in the right direction on how I could start answering the following question:     Based on the fact that $\det(AB)=\det(A)\det(B)$ for all square $n \times n$ matrices $A,B$, explain in every detail why:   $$\det(A^{-1}) = 1/\det(A).$$
56996
I'm trying to calculate the following determinant: $$\begin{vmatrix} a_0 &amp; a_1 &amp; a_2 &amp; \dots &amp; a_n \\ a_0 &amp; x &amp; a_2 &amp; \dots &amp; a_n \\ a_0 &amp; a_1 &amp; x &amp; \dots &amp; a_n \\ \dots &amp; \dots &amp; \dots &amp; \dots &amp; \dots \\ a_0 &amp; a_1 &amp; a_2 &amp; \dots &amp; x \end{vmatrix} = $$ $$ = \begin{vmatrix} a_0 &amp; a_1 &amp; a_2 &amp; \dots &amp; a_n \\ a_0 &amp; a_1 &amp; a_2 &amp; \dots &amp; a_n \\ a_0 &amp; a_1 &amp; a_2 &amp; \dots &amp; a_n \\ \dots &amp; \dots &amp; \dots &amp; \dots &amp; \dots \\ a_0 &amp; a_1 &amp; a_2 &amp; \dots &amp; a_n \end{vmatrix} + \begin{vmatrix} 0 &amp; 0 &amp; 0 &amp; \dots &amp; 0 \\ 0 &amp; x - a_1 &amp; 0 &amp; \dots &amp; 0 \\ 0 &amp; 0 &amp; x - a_2 &amp; \dots &amp; 0 \\ \dots &amp; \dots &amp; \dots &amp; \dots &amp; \dots \\ 0 &amp; 0 &amp; 0 &amp; \dots &amp; x-a_n \end{vmatrix} = 0 + 0 = 0 $$  Still, experimental results contradict, since for one example I get a non-zero determinant.  What am I doing wrong?
71132
I have encountered an statement several times while proving determinant of a block matrix.      $$\det\pmatrix{A&amp;0\\0&amp;D}\; = \det(A)\det(D)$$   where $A$ is $k\times k$ and $D$ is $n\times n$ matrix. How to prove this?  Thanks in advance.
62232
I reduced a certain determinant to the following which has the form: \begin{vmatrix} 1&amp;1&amp;\cdots &amp; 1 \\ -a_1&amp;a_2&amp;\cdots &amp;a_n\\ -a_1^{2}&amp;a_2^{2}&amp;\cdots&amp;a_n^{2}\\ \vdots&amp;\vdots&amp;\ddots&amp; \vdots\\ -a_1^{n-1}&amp;a_2^{n-1}&amp;\cdots&amp;a_n^{n-1}\\ \end{vmatrix}  To clarify a bit, it is exactly the Vandermonde determinant except the first column is negative, while the $(1,1)$-entry is still $1$.   I think the key is to apply Vandermonde determinant yet I can't proceed. There may be a quick answer to this, however. Any hints?
8559
Hi guys I am working with this and I am trying to prove to myself that n by n matrices of the type zero on the diagonal and 1 everywhere else are invertible.  I ran some cases and looked at the determinant and came to the conclusion that we can easily find the determinant by using the following $\det(A)=(-1)^{n+1}(n-1)$. To prove this I do induction  n=2 we have the $A=\begin{bmatrix} 0 &amp; 1\\  1 &amp; 0 \end{bmatrix}$ $\det(A)=-1$ and my formula gives me the same thing (-1)(2-1)=-1  Now assume if for $n \times n$ and $\det(A)=(-1)^{n+1}(n-1)$  Now to show for  a matrix B of size $n+1 \times n+1$. I am not sure I was thinking to take the determinant of the $n \times n$ minors but I am maybe someone can help me. Also is there an easier way to see this is invertible other than the determinant? I am curious.
67647
Suppose A is a 4x4 matrix such that $\det(A)=\frac{1}{64}$ then $\det(4A^{-1})^T$   I created a 2x2 matrix $B$ and transposed it both had the same determinant  I then found $\det(B)$ and $\det(B^{-1})$ the results were inverses of each other I then tried found the $\det(2B)$    but I cannot see what is the relationship between the determinant of a matrix and the determinant of that matrix multiplied by a constant?
17498
Find Maximum value of Determinant of $3 \times 3$ Matrix  with  entries $\pm 1$  My try:  I considered a matrix as :  $$A=\begin{bmatrix} 1 &amp;-1  &amp;-1 \\  -1 &amp;1  &amp;-1 \\   -1&amp;-1  &amp;1  \end{bmatrix}$$  we have $$Det(A)=-4$$ and maximum is $4$, but how can we show that these are max and min values?   I also tried as follows:  By definition Determinant of a matrix is dot product of elements of any row with corresponding Cofactors  hence  $$Det(A)=a_{11}C_{11}+a_{12}C_{12}+a_{13}C_{13}$$  By cauchy Scwartz Inequality we have  $$a_{11}C_{11}+a_{12}C_{12}+a_{13}C_{13} \le \left(\sqrt{a_{11}^2+a_{12}^2+a_{13}^2}\right)\left(\sqrt{C_{11}^2+C_{12}^2+C_{13}^2}\right)$$  any way to proceed here?
14391
Is the following statement true or false? There exists $A \in M_{3,3}(\mathbb{Z})$ with determinant $2$ such that $$A\begin{pmatrix}2\\1\\4\end{pmatrix} = \begin{pmatrix}4\\-8\\16\end{pmatrix}$$   I first thought of eigenvalues but it doesn't look like the second vector is a multiple of the first. What are the conditions on $A$ so that it's "true"? Or is it always false?
1470
This question aims to create an &quot;&quot;  of numerous questions that ask about determinants of specific matrices (I may have missed a few):              The general question of this type is  Let $A$ be a square matrix of rank$~1$, let $I$ the identity matrix of the same size, and $\lambda$ a scalar. What is the determinant of $A+\lambda I$?  A clearly very closely related question is  What is the characteristic polynomial of a matrix $A$ of rank$~1$?
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?
60108
I'm seeking for a proof of the following:     Let $A$ be an invertible matrix. Then the determinant of $A^{-1}$ equals:   $$\left|A^{-1}\right|=|A|^{-1} $$   I don't know where to begin the proof. Any suggestions?
100008
Actually I need to show that $\det(AB) = \det(A)\det(B)$ if $A$ is a singular matrix.  The determinant of $A$ is $0$ if $A$ is singular, so $\det(AB)$ has to be $0$ as well, but I have problems showing that $AB$ is singular if $A$ is singular. How can I show that?
70619
I am convinced that the following identity is true: \begin{equation} \det\begin{bmatrix}  1+a_1^2 &amp; a_1 a_2 &amp; a_1 a_3 &amp; \ldots &amp; a_1a_n \\  a_1a_2 &amp; 1+a_2^2 &amp; a_2a_3 &amp; \ldots &amp; a_2a_n \\  \ldots &amp;\ldots &amp;\dots &amp; \ldots &amp; \ldots \\ a_na_1 &amp;a_na_2 &amp; a_na_3 &amp;\ldots &amp;1+a_n^2 \end{bmatrix}= 1+ a_1^2+a_2^2+\dots+a_n^2. \end{equation} Can you help me proving it?    This determinant comes out in the computation of the volume form on a n-dimensional surface on $\mathbb{R}^{n+1}$ described by the equation  \begin{equation} x_{n+1}=f(x_1\ldots x_n). \end{equation} The volume form is given by $\sqrt{g}dx_1\ldots dx_{n}$, where $g$ is the determinant above with $a_j=\partial_{x_j}f$. The result proven in this question &amp; answer shows that the volume form is  $$ \sqrt{1+\sum_{j=1}^n \left(\frac{\partial f}{\partial x_j}\right)^2}\, dx_1\ldots dx_n.$$
32977
I wrote an answer to  question based on determinants, but subsequently deleted it because the OP is interested in non-square matrices, which effectively blocks the use of determinants and thereby undermined the entire answer. However, it can be salvaged if there exists a function $\det$ defined on all real-valued matrices (not just the square ones) having the following properties.   $\det$ is real-valued $\det$ has its usual value for square matrices $\det(AB)$ always equals $\det(A)\det(B)$ whenever the product $AB$ is defined. $\det(A) \neq 0$ iff $\det(A^\top) \neq 0$   Does such a function exist?
102918
I would like to simplify  $\det\begin{pmatrix}1&amp;x_1^2&amp;x_1^3&amp;x_1^4\\1&amp;x_2^2&amp;x_2^3&amp;x_2^4\\1&amp;x_3^2&amp;x_3^3&amp;x_3^4\\1&amp;x_4^2&amp;x_4^3&amp;x_4^4 \end{pmatrix}$  which is not exactly like the Vandemonde matrix by the exponents, but I was wondering if there is a way to compute it in $\mathbb{Q}[x_1,x_2,x_3,x_4]$  I tried computing it like a normal matrix but after a very long computation I don't find an "easy" formula to describe it, I also tried following the general proof of the Vandermone matrix but without success. I think there is more behind it, does anyone know about this?
116266
I'm trying to find a closed-form solution $\forall$ odd integer $n\ge 3$ for the determinant of a matrix with some structure on it. After some manipulation, I've reduced it to the following matrix: $\small\begin{bmatrix}\boldsymbol{t_{1}^{n}-t_{a}^{n}} &amp; \boldsymbol{t_{2}^{n}-t_{a}^{n}} &amp; \boldsymbol{\cdots} &amp; \boldsymbol{t_{a-1}^{n}-t_{a}^{n}} &amp; nt_{1}^{n-1} &amp; \cdots &amp; nt_{a-1}^{n-1} &amp; nt_{a}^{n-1}\\ \boldsymbol{t_{1}^{n-1}-t_{a}^{n-1}} &amp; \boldsymbol{t_{2}^{n-1}-t_{a}^{n-1}} &amp; \boldsymbol{\cdots} &amp; \boldsymbol{t_{a-1}^{n-1}-t_{a}^{n-1}} &amp; (n-1)t_{1}^{n-2} &amp; \cdots &amp; (n-1)t_{a-1}^{n-2} &amp; (n-1)t_{a}^{n-2}\\ \boldsymbol{\vdots} &amp; \boldsymbol{\vdots} &amp; \boldsymbol{\ddots} &amp; \boldsymbol{\vdots} &amp; \vdots &amp; \ddots &amp; \vdots &amp; \vdots\\ \boldsymbol{t_{1}^{2}-t_{a}^{2}} &amp; \boldsymbol{t_{2}^{2}-t_{a}^{2}} &amp; \boldsymbol{\cdots} &amp; \boldsymbol{t_{a-1}^{2}-t_{a}^{2}} &amp; 2t_{1} &amp; \cdots &amp; 2t_{a-1} &amp; 2t_{a}\\ \boldsymbol{t_{1}-t_{a}} &amp; \boldsymbol{t_{2}-t_{a}} &amp; \boldsymbol{\cdots} &amp; \boldsymbol{t_{a-1}-t_{a}} &amp; 1 &amp; \cdots &amp; 1 &amp; 1 \end{bmatrix}_{n\times n}$ where $a:=\frac{n+1}{2}$, the bold block is $n\times(\frac{n+1}{2}-1)$, and the non-bold block is $n \times \frac{n+1}{2}$. Although it has some similarities with the  or , it's not the same. Using some values of n, its determinant looks pretty simple, which leads me to think that there should be a closed-form solution: $n=3$: $$ det\left(  \left[\begin{array}{ccc} {t_{1}}^3-{t_{2}}^3 &amp; 3\,{t_{1}}^2 &amp; 3\,{t_{2}}^2\\ {t_{1}}^2-{t_{2}}^2 &amp; 2\,t_{1} &amp; 2\,t_{2}\\ t_{1}-t_{2} &amp; 1 &amp; 1 \end{array}\right] \right)= -{\left(t_{1}-t_{2}\right)}^4 $$ $n=5$: $$ det\left(      \left[\begin{array}{ccccc} {t_{1}}^5-{t_{3}}^5 &amp; {t_{2}}^5-{t_{3}}^5 &amp; 5\,{t_{1}}^4 &amp; 5\,{t_{2}}^4 &amp; 5\,{t_{3}}^4\\ {t_{1}}^4-{t_{3}}^4 &amp; {t_{2}}^4-{t_{3}}^4 &amp; 4\,{t_{1}}^3 &amp; 4\,{t_{2}}^3 &amp; 4\,{t_{3}}^3\\ {t_{1}}^3-{t_{3}}^3 &amp; {t_{2}}^3-{t_{3}}^3 &amp; 3\,{t_{1}}^2 &amp; 3\,{t_{2}}^2 &amp; 3\,{t_{3}}^2\\ {t_{1}}^2-{t_{3}}^2 &amp; {t_{2}}^2-{t_{3}}^2 &amp; 2\,t_{1} &amp; 2\,t_{2} &amp; 2\,t_{3}\\ t_{1}-t_{3} &amp; t_{2}-t_{3} &amp; 1 &amp; 1 &amp; 1 \end{array}\right] \right)= -{\left(t_{1}-t_{2}\right)}^4\,{\left(t_{1}-t_{3}\right)}^4\,{\left(t_{2}-t_{3}\right)}^4 $$ I was wondering if there is a known closed-form solution for this determinant, or if it could be found using the determinant of a generalized Vandermonde matrix Thanks!