Calculate the determinant of this magic square (which is from Albrecht   Dürer's Melancholia)      $$\begin{pmatrix} 16  &amp;   3  &amp;   2  &amp;  13 \\   5  &amp;  10  &amp;  11  &amp;   8   \\   9  &amp;   6  &amp;   7  &amp;  12 \\   4  &amp;  15  &amp;  14  &amp;   1   \end{pmatrix}$$   I'm very unsafe about the term "magic square" and I couldn't find much reliable information on the internet either about calculating their determinant.. : /  Wikipedia says: A magic square is a $n \times n$ square grid (where $n$ is the number of cells on each side) filled with distinct positive integers in the range $1,2,.., n^2$ such that each cell contains a different integer and the sum of the integers in each row, column and diagonal is equal [...].  All in all, I don't really see a difference between a normal square matrix and a magic square when it comes to calculate their determinants. Am I right?  If this is the case, I would just calculate the determinant of this magic square by using formula of Laplace choosing the first column of the magic square as pivot column. I did it on paper and I get $$\det =0$$  So basically my question is: When I got this magic square, can I consider it as a normal square matrix when it comes to calculating its determinant? Or do you calculate its determinant completely different?
121715
The determinant of even-magic square matrix is 0.   For example, this 4x4 magic square matrix:  {{16, 2, 3, 13},   {5, 11, 10, 8},  {9, 7, 6, 12},  {4, 14, 15, 1}}   And this 6x6 matrix:     {{35, 1, 6, 26, 19, 24},      {30, 5, 7, 21, 23, 25},      {31, 9, 2, 22, 27, 20},     {8, 28, 33, 17, 10, 15},      {3, 32, 34, 12, 14, 16},      {4, 36, 29, 13, 18, 11}}   So is true for 8x8,10x10...  Is there anything special here? Is it possible to recognize this kind of matrix and conclude that its determinant is zero?  Reference for magic square matrix:
68904
Let $A_1,\ldots,A_n$ be square matrices (not necessarily symmetric). Taking the trace of a matrix product and a cyclic permutation of the product yields: \begin{align*} \text{trace}(A_nA_{n-1}\cdots A_1) = \text{trace}(A_1A_{n}\cdots A_{2}) \end{align*} The above holds for any cyclic permutation (informally, taking an element off the end and appending it onto the other end), but not, in general, for arbitrary permutations of the matrix product.  On the other hand, the determinant of a matrix product satisfies the following property $\text{det}(A_nA_{n-1}\cdots A_1) = \text{det}(A_n)\text{det}(A_{n-1})\cdots \text{det}(A_1)$. Thus, \begin{align*} \text{det}(A_nA_{n-1}\cdots A_1) = \text{det}(A_1A_{2}\cdots A_n) = \text{det}(\text{"any permutation of the product of $A$ matrices"}) \end{align*}  The determinant can be thought of as the volume of its matrix argument whereas one can think of the trace as being the derivative of the determinant (near the identity matrix), that is, the infinitesimal change in volume.  What is the intuition behind the trace property only holding for cyclic permutations where the determinant is the same for any permutation of the matrix product? Can this be explained in a way that is related to the trace being the determinant's derivative?
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!
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?
25085
Let $x_1,...,x_n $ are distinct real numbers.  Is it a formula for the Vandermonde type determinant $V(x_1, \cdots,x_n)$  whose last column is $x_1^k,\ \cdots,\ x_n^k$, where $k \geq n$, instead of $x_1^{n-1},\ \cdots,\ x_n^{n-1}$?  Thanks
115720
Singmaster (1981) writes, on page 32 of his Notes on Rubik’s Magic Cube:     Frank Barnes observes that the group of the cube is generated by two moves:   \begin{align*} \alpha &amp;= L^2 B R D^{-1} L^{-1} &amp;=(RF,RU,RB,UB,LD,LB,LU,BD,DF,FL,RD)&amp; \\ &amp;&amp;\cdot (FUR,UBR,LDB,LBU,DLF,BDR,DFR)\\ \beta &amp;= UFRUR^{-1}U^{-1}F^{-1} &amp;=(UF,UL)_+(UR)_+(UBR,UFL)_-(URF)_+ \end{align*}   Observe that $\alpha^7$ is an $11$-cycle of edges and $\alpha^{11}$ is a $7$-cycle of corners, that $\beta$ affects the edge and corner left fixed by $\alpha$, and that $$\beta^2 = (UF)_+(UL)_+(UBR)_-(UFL)_-(UFR)_-$$ [...] The remaining details are left as an exercise.   I hadn't seen this notation before, so I'll explain it here.     Notation like $(LU, BD, DF)$ means an edge cycle, in which:         The $L$-$U$ edge moves to the $B$-$D$ edge's place, with the $L$ half ending up on the $B$ face, and the $U$ half ending up on the $D$ face.   Similarly $BD \to DF$ and $DF \to LU$.         The notation for corners is similar.      Notation like $(UF, UL)_+$ is a twisted cycle: again, $UF \to UL$, but now $UL \to FU$; the final edge gets flipped when cycling back to the first edge. For corners, the notation is similar, but corners rotate, they don't flip. A subscript $+$ means clockwise rotation, a subscript $-$ means counterclockwise rotation.      $(UR)_+$ means a single edge is flipped. $(UBR)_-$ means a single corner is rotated counterclockwise.   I would like to show that this is indeed true, by writing each element in $\{F,B,L,R,U,D\}$ as a product of elements in $\{\alpha, \beta, \alpha^{-1}, \beta^{-1}\}$ – preferably, having those product be as short as possible. How would I go about finding them? (I'm okay with using software like GAP – if it is at all computationally possible.)
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!
87268
Our teacher didn't explain us how to find it so I've had to look up a bit by myself.  I have this matrix :  $$A = \begin{pmatrix} 9 &amp; 4 &amp; 5 \\ -4 &amp; 0 &amp; -3 \\ -6 &amp; -4 &amp; -2 \end{pmatrix}$$  I've found its characteristic polynomial $p_A(\lambda) = -(\lambda -3)(\lambda - 2)^2$.  I've found its eigenspaces : $E_2 = span(\begin{pmatrix}-2 &amp; 1 &amp; 2\end{pmatrix}^T)$, $E_3 = span(\begin{pmatrix}-3 &amp; 2 &amp; 2\end{pmatrix}^T)$.  And I've found the generalized eigenspace of $\lambda = 2$ :  $\ker((A-2I_3)^2) = span(\begin{pmatrix}0 &amp; 1 &amp; 0\end{pmatrix}^T, \begin{pmatrix}1 &amp; 0 &amp; -1\end{pmatrix}^T)$  But I don't know what to do now. I've tried to see what result I would get with $$M = \begin{pmatrix} -3 &amp; -2 &amp; 0 \\ 2 &amp; 1 &amp; 1 \\ 2 &amp; 2 &amp; 0 \end{pmatrix}.$$  But $$M^{-1}AM = \begin{pmatrix} 3 &amp; 0 &amp; 0 \\ 0 &amp; 2 &amp; -2 \\ 0 &amp; 0 &amp; 2 \end{pmatrix}.$$  So there it is. Any help ?
100594
I got this information regarding Riemann matrix: RIEMANN is a N-by-N matrix associated with the Riemann hypothesis, which is true if and only if:  DET(A) = O( N! N^(-1/2+epsilon) ) for every epsilon&gt;0 ('!' denotes factorial).  where, A= Riemann matrix  with,  A = B(2:N+1, 2:N+1),   where  B(i,j) = i-1 if i divides j and           -1 otherwise.  Properties include, with M = N+1:  Each eigenvalue E(i) satisfies ABS(E(i)) &lt;= M - 1/M. i &lt;= E(i) &lt;= i+1 with at most M-SQRT(M) exceptions. All integers in the interval (M/3, M/2] are eigenvalues.  Reference:  F. Roesler, Riemann's hypothesis as an eigenvalue problem, Linear Algebra and Appl., 81 (1986), pp. 153-198.  Now, I have a Matlab Function which creates a 3-by-3 Riemann matrix as:   R=riemann(3)   R =  1    -1     1 -1     2    -1 -1    -1     3  eig(R) ans = 0.5858 2.0000 3.4142  Questions:  What is &quot;O&quot; factor in the meaning of determinant ? How do the 3-by-3 Riemann matrix &quot;R&quot; created by my function concur with properties stated above ? (1st property holds good but other two ??)  P.S.: If u need to take a look at my function,please tell me.
110912
\begin{pmatrix}3&amp;-1&amp;3\\ -2&amp;3&amp;2\\ 1&amp;-3&amp;1\end{pmatrix}  Any help would be greatly appreciated!  I have done this before with other matrices, yet this one is bringing me trouble. I have spent too long on this as I believe it should be easy and yet my answers are always wrong. What Do you do after you get the identity matrix?  What I am doing: So I first row reduce it to the identity matrix:   I first switch $R_1$ and $R_3$ Then $R_3-3R_1$ then $R_2+2R_1$ then $-1/3R_2$ then $R_3-8R_2$ then $R_2-2R_1$ then $3/32R_3$ then $R_1+3R_3$ and finally $R_2+4/3R_3$   I know there is many possible ways to get this too the identity matrix but this worked for me. I then apply all of the above steps to identity matrices and multiply them together to check my work but it never works out to be the original matrix so I am forced to believe something is wrong.
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?
86315
I would like to know the proof for:  The determinant of the block matrix $$\begin{pmatrix} A &amp; B\\ C&amp; D\end{pmatrix}$$ equals $$(D-1) \det(A) + \det(A-BC) = (D+1) \det(A) - \det(A+BC)$$ when $A$ is a square matrix, $D$ is a scalar, $C$ is a row vector and $B$ is a column vector.  I appreciate your help.
44409
I know that given a $2N \times 2N$ block matrix with $N \times N$ blocks like   $$\mathbf{S} = \begin{pmatrix} A &amp; B\\  C &amp; D \end{pmatrix}$$  we can calculate $$\det(\mathbf{S})=\det(AD-BD^{-1}CD)$$ and so clearly if $D$ and $C$ commute this reduces to $\det(AD-BC)$, which is a very nice property.   My question is, for a general $nN\times nN$ matrix with $N\times N$ blocks where all of the blocks commute with each other, can we find the determinant in a similar way? That is, by first finding the determinant treating the blocks like scalars and then taking the determinant of the resulting $N\times N$ matrix.  Thanks in advance!
12388
Find the determinant of the following matrix in the terms of $a_1,a_2,\cdots,a_n$ explicitly, $$ \begin{bmatrix} a_1 &amp; a_2 &amp; a_3 &amp; \cdots &amp; a_n\\ a_2 &amp; a_3 &amp; a_4 &amp; \cdots &amp; a_1\\ a_3 &amp; a_4 &amp; a_5 &amp; \cdots &amp; a_2\\ \vdots &amp; \vdots &amp; \vdots &amp; \ddots &amp; \vdots\\ a_n &amp; a_1 &amp; a_2 &amp; \cdots &amp; a_{n-1}\\ \end{bmatrix} $$ When the determinant is zero?
119950
Problem: Let $A$ and $B$ be real orthogonal matrices, $n$x$n$, where $n$ is an odd number. Prove that, at least one of the matrices $A+B$ and $A-B$ has to be singular.  What have I done so far:  -Since  matrices $A$ and $B$ are real orthogonal, it means that their determinants are $-1$ or $+1$  First I observed matrix $A+B$: $$A+B=AI+BI=ABB^T+BAA^T=AB(B^T+A^T)=AB(A+B)^T\Rightarrow$$  $$det(A+B)=det(AB)det(A+B)^T=det(A)det(B)det(A+B)$$  So, if $detA=detB$ we have $det(A+B)=det(A+B)$ which tells me nothing.  But, if  $detA=-detB$ we have $det(A+B)=-det(A+B)$ which can only hapen if $det(A+B)=0$ making $A+B$ singular.  Then, I tried the same for $A-B$: $$A-B=AI-BI=ABB^T-BAA^T=AB(B^T-A^T)=AB(A-B)^T\Rightarrow$$  $$det(A-B)=det(AB)det(A-B)^T=det(A)det(B)det(A-B)$$  So, if $detA=-detB$ we have $det(A-B)=-det(A-B)$ which can happen if $det(A-B)=0$ making $A-B$ singular.  Is this correct or should I do matrix $A-B$ differently? I am also a little confused why does it have to be said that $n$ is an odd number? What should that tell me?  Any help is greatly appreciated.
70275
So, $A$ is a $n \times n$ matrix with integer entries.  The question is to prove that $A^{-1}$ has all integer entries if and only if ${\rm det}\ (A) =\pm 1$ .  I know that $A^{-1}= {\rm adj}(A)/{\rm det}(A)$ but I have no idea where to go from there for the forward direction.  Any help here would be greatly appreciated.  For the backwards direction, I think I am okay.  I plugged the 1 and the -1 options into $ {\rm adj}(A)/{\rm det}(A)$ So, I know that $A^{-1}= {\rm adj}(A)$ or $-{\rm adj}(A)$.  Then I said that because the ${\rm adj}(A)$ is simply the matrix of co-factors of $A$, and because $A$ has all integer entries then ${\rm adj}(A)$ will have all integer entries, which will mean that $A^{-1}$ will have all integer entries.  Is that okay? Or am I missing something else?   Thank you!!``
6722
Given two square matrices $A$ and $B$, how do you show that $$\det(AB) = \det(A)\det(B)$$ where $\det(\cdot)$ is the determinant of the matrix?