Given $n\in \mathbb N$, how can I compute the determinant of $(a_{ij})_{1\leq i,j \leq n} \in \mathcal M_{n\times n}(\mathbb R)$ where, for each $1\leq i,j\leq n:$ $$a_{ij}= \begin{cases} \lambda &amp;i=j \\ \mu &amp; i\neq j \end{cases}$$ The only (or the only easy) way to do this I think involves using Laplace's formula, expanding along any row. But doing this everything gets too messy, I can't express it in a good way. I tried to get a recursive relation and prove it via induction but didn't get anywhere. Is there an easier way? Any hints appreciated.
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.
109668
Let $A\in\mathcal{M}_{n}(K)$, where $K$ is a field. Then, we can obtain the characteristic polynomial of $A$ by simply taking $p(\lambda)=\det(A-\lambda I_n)$, which give us something like $$p(\lambda) = (-1)^n\lambda^n + (-1)^{n-1}(\text{tr } A)\lambda^{n-1} +  \cdots + \det A$$ Now, how can we obtain the matrix $A$ knowing the characteristic polynomial?
50917
Given this (very) tricky determinant, how can we calculate it easily?  $$\begin{pmatrix} \alpha + \beta &amp; \alpha \beta &amp; 0 &amp; ... &amp; ... &amp; 0 \\ 1 &amp; \alpha + \beta &amp;  \alpha \beta &amp; 0 &amp; ... &amp; 0 \\ 0 &amp; 1 &amp; \alpha + \beta &amp; \alpha \beta &amp; ... &amp; ... \\ ... &amp; ... &amp; ... &amp; ... &amp; ... &amp; 0 \\ ... &amp; ... &amp; .... &amp; ... &amp; ... &amp; \alpha \beta \\ 0 &amp; 0 &amp; 0 &amp; ... &amp; 1 &amp; \alpha + \beta   \\ \end{pmatrix} \in M_{n\times n}$$  EDIT:     I have to prove it is equal to $\frac{{\alpha}^{n+1} - {\beta}^{n+1}}{\alpha - \beta}$   Any help is appreciated, I just could not find a trick to ease it up!
92383
I find this question very interesting, but I am having trouble figuring out how to approach the problem.   I know that the $\det(A^{-1}) = 1/\det(A)$, but I'm unsure of where to go from here. If someone can send me in the right direction, it would be very much appreciated.   Thanks in advance,  Tommy.
6835
Let be $u,v\in\mathbb{R}^n$, then $\det(I+uv^\intercal)=1+v^\intercal u $ where $I$ denotes the identity matrix of order $n$. How to prove this?   what I did:   let be $A=\{n\in\mathbb{N}: \forall u,v \in \mathbb{R^n}, \det(I+uv^\intercal)\neq1+v^\intercal u  \} $, and suppose $A\neq \varnothing $, then for the well-orderer-principle there exists $n_0\in A$ such $n_0\leq n,\;\forall n\in A$. since $1\notin A,\; n_0\neq 1 \rightarrow n_0-1\in\mathbb{N}\setminus A  $ so: $$\forall u,v\in\mathbb{R^{n_0-1}}: \det(I+uv^\intercal)=1+v^\intercal u $$ Then, let be $u,v\in\mathbb{R}^{n_0}$, so $$\det(I+uv^\intercal)= \displaystyle\sum_{j=1}^{n_0}(-)^{1+j}a_{1j}\det(A_{1j})$$ where $a_{ij}=\begin{cases} u_iv_j+1 &amp; i=j \\ u_iv_j &amp; i\neq j \end{cases}$, and $A_{1j}$ is the  submatrix of $I+uv^\intercal$ that results deleting the i-file and j-column. Then I try to open the term for $j=1$ and try to use the relation for the new matrix of $n_0-1$ order with the determinant because it holds the same form but I have other terms which difficult me the work. Do you know other method?.  Ps. , in page 2 I've found something similar but I don't understand what it means PS. I'm taking a course of numerical analisys
108442
Let A be an invertible $n \times n$ matrix whose inverse is itself. Prove that $\det(A)$ is either $1$ or $-1$.    I'm really lost in class. I don't even know where to start. Please help.
103841
What is the determinant of an $n \times n$ matrix $B=A+I$, where $I$ is the $n\times n$ identity matrix and $A$ is the $n\times n$ matrix  $$ A=\begin{pmatrix} a_1 &amp; a_2 &amp; \dots &amp; a_n \\ a_1 &amp; a_2 &amp; \dots &amp; a_n \\ \vdots &amp; \vdots &amp; \ddots &amp; \vdots \\ a_1 &amp; a_2 &amp; \dots &amp; a_n \\ \end{pmatrix} $$
45003
$\newcommand{\adj}{\operatorname{adj}}$Let $A\in \mathbb{M}_n$ ($n \geq\ 2$) be a regular matrix and $\adj(A)$ its adjoint. Prove that if A is regular then $\adj(\adj(A)) = (\det A)^{n-2} A$ (where $adj(adj(A))$ is the adjoint of $\adj(A)$).  For now, I know, or think I do, that $A^{-1} = \adj(A)/\det(A)$  from where we can see that $\adj(A) = A^{-1} \det(A)$  And that's all I know
37738
Determine the trace and determinant of the linear operator (on the space $\mathbb{F^{n\times n}}$) that sends the matrix $M\to AMB$ where $A$ and $B$ are $n\times n$ matricies
7461
Let $A$ be an $n \times n$ matrix and $k$ be a scalar. Prove that $\det(kA)=k^n\det A$.   I really don't know where to start. Can someone give me a hint for this proof?
94384
How can we find the determinant of the following matrix?  $$A = \begin{pmatrix} x_1y_1 &amp; x_1y_2 &amp; x_1y_3 &amp; \cdots &amp; x_1y_{n-1} &amp; x_1y_n \\  x_1y_2 &amp; x_2y_2 &amp; x_2y_3 &amp; \cdots &amp; x_2y_{n-1} &amp; x_2y_n \\  x_1y_3 &amp; x_2y_3 &amp; x_3y_3 &amp; \cdots &amp; x_3y_{n-1} &amp; x_3y_n \\  \vdots &amp;  \vdots &amp;  \vdots &amp;  \ddots &amp;  \vdots &amp;  \vdots \\ x_1y_{n-1} &amp; x_2y_{n-1} &amp; x_3y_{n-1} &amp; \cdots &amp; x_{n-1}y_{n-1} &amp; x_{n-1}y_n \\  x_1y_n &amp; x_2y_n &amp; x_3y_n &amp; \cdots &amp; x_{n-1}y_n &amp; x_ny_n \end{pmatrix}$$  That is, the entry of $A$ is $a_{ij}=x_iy_j$ for $i\leq j$; and $a_{ij}=a_{ji}$ for $i&gt;j$.  I do not have anything new. And I find also it is difficult to find its eigenvalues.
72544
Let $A$ be an $n \times n$ matrix with integer coefficients and nonzero determinant. Can we say something about $ \mathbb{Z}^n /\ \text{im}( \phi )$ (here $\phi : v \mapsto Av$ )?    This problem has arised as I was solving some problem in homology theory.
109652
Let $SL_n(\mathbb{R})$ be the set of $n\times n$ matrix with determinant one. Where can I find a proof that $[SL_n(\mathbb{R}),SL_n(\mathbb{R})]=SL_n(\mathbb{R})$? I've seen this result mentioned in several places, but couldn't find a proof.  For reference, the commutator $[SL_n(\mathbb{R}),SL_n(\mathbb{R})]$ is defined as the group generated by the set of all elements of the form $a^{-1}b^{-1}ab$, with $a,b\in SL_n(\mathbb{R})$.
79378
Let $A=(a_{ij})$ be a $n \times n$ matrix with $a_{ij}=\gcd(i,j) , \forall i,j=1,2, \cdots, n$ , then how do we prove $\det A=\prod_{i=1}^n \phi(i)$ ? , where $\phi$ is the Euler's phi function
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?
94994
Let $M$ be a (real) $n \times n$ matrix. For $1 \leq i, j \leq n$ we denote by $M_{ij}$ the $(n-1) \times (n-1)$ matrix that we get when the $i$th row and $j$th column of $M$ are removed. Now, consider fixed $i$ and $j$ with $i\neq j$. Let $N$ be the $(n-2) \times (n-2)$ matrix that we get when removing both the $i$th and $j$th row and the $i$th and $j$th column from $M$. Then the following identity holds: $$ \det M \det N = \det M_{ii}\det M_{jj} - \det M_{ij} \det M_{ji}. $$  We were able to prove this looking at all different terms that can occur on both sides when evaluating the determinant as a sum of $n!$ terms (for instance, terms containing $a_{ji}a_{ij}$ are counted both $1$ time in the LHS and $1$ time in the RHS).  We are looking for a slick proof that does not involve writing out the determinant.  Any suggestions or approaches to this problem are appreciated!
87100
Let A = $(a_{ij})_{n\times n}$ such that $AB = BA$ for every square matrix $B$ of order $n$.  $(I)$ Prove that $A$ is a diagonal matrix.  [Hint: Let $B_i$ denote the matrix whose $(i,i)$ entry is $1$, and $0$ elsewhere. Show that $a_{ij} = 0$ whenever $i ≠ j$.]  $(II)$ Prove that $A$ is a scalar matrix.  [Hint: For $i ≠ j$, let $B_{ij}$ denote the matrix whose $(i,j)$-entry is $1$ and $0$ elsewhere.Show that, $a_{ii} = a_{jj}$ for all $i ≠ j$.]  Can someone help me with this question using basic linear algebra concepts? Thanks.