I have been trying to solve a problem and have given my answer in the form of a determinant, namely: \begin{align*} \det\begin{pmatrix} \binom{q}{2} &amp; q &amp; 1 &amp; 0 &amp; \dots &amp; 0 \\ \binom{q}{3} &amp; \binom{q}{2} &amp; q &amp; 1 &amp; \dots &amp; 0 \\ \binom{q}{4} &amp; \binom{q}{3} &amp; \binom{q}{2} &amp; q &amp; \dots &amp; 0 \\ \binom{q}{5} &amp; \binom{q}{4} &amp; \binom{q}{3} &amp; \binom{q}{2} &amp; \dots &amp; 0 \\ \vdots &amp; \vdots &amp; \vdots &amp; \dots &amp; \ddots &amp; q \\  \binom{q}{n+1} &amp; \binom{q}{n} &amp; \binom{q}{n-1} &amp; \binom{q}{n-2} &amp; \dots &amp; \binom{q}{2}  \end{pmatrix} \end{align*} This is probably enough, but I thought I'd challange myself and dust off my old linear algebra skills. Unfortunately, it is not going well. I have attempted brute forcing it by multiplying out coefficients, but I don't think that is the way to go, because I am getting nowhere. Everytime I touch it I seem to be increasing the complexity rather than decreasing it...   Is there a good way to tackle this problem, or is this as good as it gets?  Any help would be much appreciated!
124232
I trying to prove a statement, which boils down to showing that the determinant of a specific matrix is nonzero. I use the convention that $\binom{n}{k} = 0$ if $k &gt; n$ or $k &lt; 0$. Let $k,l$ be natural numbers such that $k \le l$. Then the $n\times n$-Matrix $A$ is defined to have the entries   $a_{ij} = \binom{l}{k +i - j}$. So it looks like  $A = \left( \begin{array}{cccccc} \binom{l}{k} &amp;  \cdots &amp; \binom{l}{0} &amp; 0 &amp; \cdots &amp; 0 \\  \vdots &amp; \ddots &amp;  &amp; \ddots &amp; \ddots &amp;  \vdots \\ \binom{l}{l} &amp; &amp; \ddots &amp; &amp; \ddots   &amp; 0  \\ 0&amp; \ddots &amp;&amp; \ddots &amp; &amp; \binom{l}{0}\\ \vdots&amp;\ddots&amp;\ddots&amp;&amp;\ddots&amp;\vdots\\ 0&amp;\cdots&amp;0&amp; \binom{l}{l} &amp; \cdots &amp; \binom{l}{k}  \end{array}\right)$.  Clearly the cases $k = l$ and $k = 0$ are trivial, since $A$ is then triangular. My first idea was to use the formula  $\binom{r}{s} = \binom{r-1}{s-1} + \binom{r}{s}$ and add columns/rows to each other. But that does not work out that well...   So if anyone has any ideas, or this matrix is known to be invertible, I would be very thankful.
104885
Let $A = (a_{ij}) \in R^{n\times n}$. Find the determinant if:  $$a_{ij}= |i-j|$$  So we have the symmetric matrix \begin{bmatrix} 0 &amp; 1 &amp; 2 &amp; 3 &amp; 4 &amp; \dots &amp; n-1 \\ 1 &amp; 0 &amp; 1 &amp; 2 &amp; 3 &amp; \dots &amp; n-2 \\ 2 &amp; 1 &amp; 0 &amp; 1 &amp; 2 &amp; \dots &amp; n-3 \\ \vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \ddots &amp; \vdots \\ n-1 &amp; n-2 &amp; n-3 &amp; \dots &amp; \dots &amp; \dots &amp; 0 \end{bmatrix}  But i can't find a way to diagonalize the matrix nor find the determinant by Laplace expansion.  Any ideas ??
71308
How to show that the determinant of the following $(2n+1)×(2n+1)$ matrix $A$?    \begin{equation}     \det A = \begin{array}{|cccccccccc|cc}       1       &amp; -1 &amp; 0 &amp; \dots &amp; 0 &amp; 0 &amp; 0 &amp; \dots &amp; 0 &amp; 0 &amp; &amp; {\color{blue}{\text{row }1}}\\       -1       &amp; 2 &amp; -1 &amp; \dots &amp; 0 &amp; 0 &amp; 0 &amp; \dots &amp; 0 &amp; 0&amp; &amp; {\color{blue}{\text{row }2}} \\       0       &amp; -1 &amp; 2 &amp; \dots &amp; 0 &amp; 0 &amp; 0 &amp; \dots &amp; 0 &amp; 0&amp; &amp; {\color{blue}{\text{row }3}} \\       \vdots  &amp; \vdots&amp;  \vdots   &amp; \ddots      &amp; \vdots    &amp; \vdots &amp; \vdots &amp; \ddots &amp; \vdots&amp; \vdots  &amp; &amp;{\color{blue}{\vdots}}\\       0 &amp; 0 &amp; 0 &amp; \dots &amp; 2 &amp; -1 &amp; 0 &amp; \dots &amp; 0 &amp; 0  &amp;&amp; {\color{blue}{\text{row }j-1}}\\       0 &amp; 0 &amp; 0 &amp; \dots &amp; -1 &amp; 3 &amp; -1 &amp; \dots &amp; 0 &amp; 0 &amp;{\color{blue}{\rightarrow}}&amp;  {\color{blue}{\text{row }j}}\\       0 &amp; 0 &amp; 0 &amp; \dots &amp; 0 &amp; -1 &amp; 2 &amp; \dots &amp; 0 &amp; 0 &amp;&amp; {\color{blue}{\text{row }j+1}} \\       \vdots  &amp; \vdots&amp;  \vdots   &amp; \ddots      &amp; \vdots    &amp; \vdots &amp; \vdots &amp; \ddots &amp; \vdots&amp; \vdots    &amp; &amp;{\color{blue}{\vdots}}    \\     0       &amp; 0 &amp; 0 &amp; \dots &amp; 0 &amp; 0 &amp; 0 &amp; \dots &amp; 2 &amp; -1 &amp;&amp; {\color{blue}{\text{row }2n}} \\         0       &amp; 0 &amp; 0 &amp; \dots &amp; 0 &amp; 0 &amp; 0 &amp; \dots &amp; -1 &amp; 1 &amp;&amp; {\color{blue}{\text{row }2n+1}}\\     \end{array}   \end{equation}  By a direct calculation of  determinant. here $3$ is at the $j$-th row for some $2\leq j\leq 2n$. Therefore, $\det (L_{A_{11}}) = (-1)^n 2^{n-1}.$  if $n+1\leq i\leq 3n+1$ we denote the resulting matrix by $A_{12}$ after deleting the $i$-th row and column of $L_A$. Then, by a similar computation, we have $\det (L_{A_{12}}) = (-1)^n 2^{n}.$  $ \det A = (-1)^n 2^{n-1}n + (-1)^n 2^{n}(2n+1)=(-1)^n 2^{n-1}(5n+2)$  equal  to $(-1)^n 2^{n-1}(5n+2)$ ?
15023
I'd like to show the case $n=4$ for the Vandermonde-determinant. It should look like this:  $V_4 := \det \begin{pmatrix} 1 &amp; 1 &amp; 1 &amp; 1 \\ x_1 &amp; x_2 &amp; x_3 &amp; x_4 \\ x_1^2 &amp; x_2^2 &amp; x_3^2 &amp; x_4^2 \\ x_1^3 &amp; x_2^3 &amp; x_3^3 &amp; x_4^3 \end{pmatrix} = (x_4 - x_3)(x_4 - x_2)(x_4 - x_1)(x_3 - x_2)(x_3 - x_2)(x_2 - x_1)$.  I tried calculating the $\det(\dots)$ via the first row. However, it ends up in a very large calculation.. I'm sure there's a better way to show especially this case with $n=4$.  Thank you for any help ;)
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?
95368
Find the number of $3 \times 3$ matrices whose determinant is $(a)$ positive $(b)$ negative $(c)$ zero and whose elements are taken from $\{-1,1\}$.  This is what I tried: The total number of $3\times 3$ matrices with entries in $\{ 1 , -1\}$ is $2^9$. Now counting each rows or column of matrices as a vectors with entries in $\{-1,1\}$, there are $8$ such vectors and if any two vectors are same then its determinant is $0$. There are such $24+24=48$ ways.
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?
87954
Let  $$A_2 = \left[ \begin{array}{cc} 0 &amp; 1\\ 1 &amp; 0 \end{array}\right]$$  $$A_3 = \left[ \begin{array}{ccc} 0 &amp; 1 &amp; 1\\ 1 &amp; 0 &amp; 1 \\ 1 &amp; 1 &amp; 0 \end{array}\right]$$  $$A_4 = \left[ \begin{array}{cccc} 0 &amp; 1 &amp; 1 &amp; 1\\ 1 &amp; 0 &amp; 1 &amp; 1 \\ 1 &amp; 1 &amp; 0 &amp; 1 \\ 1 &amp; 1 &amp; 1 &amp; 0\end{array}\right]$$  and so on for $A_n$.  I was asked to calculate the determinant for $A_1, A_2, A_3, A_4$ and then guess about the determinant for $A_n$ in general. Of course the pattern is clear that  $$ \det A_n = (n-1)(-1)^{n-1} $$  but I was wondering as to what the proof of this is. I tried to be clever with cofactor expansions but I couldn't get anywhere.  Could someone explain it to me please?
78573
Well considering two $n \times n$ matrices does the following hold true: $$\det(A+B) = \det(A) + \det(B)$$ Can there be said anything about $\det(A+B)$?  If $A/B$ are symmetric (or maybe even of the form $\lambda I$) - can then things be said?
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?
71172
Find the determinant of the following matrix: $$A = \begin{bmatrix} 1+x_1^2 &amp;x_1x_2  &amp; ... &amp; x_1x_n \\   x_2x_1&amp;1+x_2^2  &amp;...  &amp; x_2x_n\\   ...&amp; ... &amp; ... &amp;... \\   x_nx_1&amp; x_nx_2  &amp;...  &amp; 1+x_n^2 \end{bmatrix}$$  I computed for the case $n=2$, and $n=3$ and guessed that $\det(A)$ should be $ 1+\sum_{i=1}^n x_i^2 $  but not sure how to proceed for any $n$.
85783
Find the remainder when the determinant $\begin{vmatrix} { 2014 }^{ 2014 } &amp; { 2015 }^{ 2015 } &amp; { 2016 }^{ 2016 } \\ { 2017 }^{ 2017 } &amp; { 2018 }^{ 2018 } &amp; { 2019 }^{ 2019 } \\ { 2020 }^{ 2020 } &amp; { 2021 }^{ 2021 } &amp; { 2022 }^{ 2022 } \end{vmatrix}$ is divided by $5$.   I'm aware that this problem has a number-theoretic solution involving congruence relations. But considering that this was asked as a multiple-choice question in a test, what should be the best way to approach problems like this?  The options were $(a)\quad1\quad (b)\quad2\quad (c)\quad3\quad (d)\quad 4$
109630
We have been given matrix: $$A =          \begin{pmatrix}         a&amp; b&amp; c &amp;d \\         b &amp;−a&amp; d&amp; −c\\         c&amp; −d &amp;−a&amp; b \\         d &amp;c&amp; −b&amp; −a\\         \end{pmatrix}     $$ ...and have been asked to calculate $\det(A)$ using $AA^T$.  We see that:  $$AA^T=\begin{pmatrix}         a^2+ b^2+ c^2+ d^2&amp; 0&amp; 0&amp;0\\         0 &amp;a^2+ b^2+ c^2+ d^2&amp; 0&amp; 0\\         0&amp; 0 &amp;a^2+ b^2+ c^2+ d^2&amp; b \\        0 &amp;0&amp; 0&amp; a^2+ b^2+ c^2+ d^2\\         \end{pmatrix}     $$   So, $\det(AA^T)= (a^2+ b^2+ c^2+ d^2)^4$  Now which should I choose: $(a^2+ b^2+ c^2+ d^2)^2 $ or $ -(a^2+ b^2+ c^2+ d^2)^2$ ?  Please explain me which one and why.
21165
How to simplify this best  $(a_1 + a_2 +a_3+... +a_n)^m$ for  $m=n, m&lt;n, m&gt;n$  I could only get  $\sum_{i=0}^{m}\binom{m}{i}a_i^i\sum_{j=0}^{m-i}\binom{m-i}{j}a_j ... $
121890
I'd like to know if the following problem of elementary linear algebra is already solved / solvable.  For two (singular) $n\times n$ matrices $P$ and $Q$, if $\det(\lambda P+\mu Q)=0$ for any $\lambda,\mu\in\mathbb{R}$, what are conditions on $P$ and $Q$?
20232
How to evaluate this determinant by just using row and column operations ? I'm stuck.Help please!  \begin{vmatrix} -2a &amp; a+b &amp; a+c \\ b+a &amp; -2b &amp; b+c \\ c+a &amp; c+b &amp; -2c \end{vmatrix}  BTW is there any general method to simplify symmetric determinants like this?
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:   $$\det(A^{-1}) = 1/\det(A).$$