When studying linear algebra, you are constantly dealing with determinants. I read their definition on Wikipedia, saw all kinds of formulas, but found that if someone will ask me WHAT EXACTLY is a matrix determinant, I could not answer. All people who know math answered this question to me like this: "Well, this is such a thing, here is the formula to calculate it, it is very important in linear algebra." I understand that a determinant is the volume (or area) of a figure formed by vectors. I can’t understand why someone came up with the idea to use them. I realize that the question sounds like a request to explain the multiplication, and yet I ask it, since I am very curious about what a determinant really is.  If possible, answer as detailed and exhaustive as possible.
34167
In my linear algebra class, we just talked about determinants. So far I’ve been understanding the material okay, but now I’m very confused. I get that when the determinant is zero, the matrix doesn’t have an inverse. I can find the determinant of a $2\times 2$ matrix by the formula. Our teacher showed us how to compute the determinant of an $n \times n$ matrix by breaking it up into the determinants of smaller matrices. Apparently there is a way by summing over a bunch of permutations. But the notation is really hard for me and I don’t really know what’s going on with them anymore. Can someone help me figure out what a determinant is, intuitively, and how all those definitions of it are related?
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  $$ \sqrt{1+\sum_{j=1}^n \left(\frac{\partial f}{\partial x_j}\right)^2}\, dx_1\ldots dx_n.$$
12482
Most of the time I see matrix multiplication presented and defined, as a seemingly arbitrary sequence of operations. For example, the  I'm currently reading for a linear algebra course defines the product $AB$ as the $(i, j)$ entry in the $1 \times 1$ matrix that is the product of the ith row of $A$ and the jth column of $B$. Properties of matrix multiplication are subsequently proven based upon this definition. The definition is clear, but why the matrix product is useful is not clear to me as a student. A  I'm referencing defines the product $AB$ in terms of linear combinations.   The problem I have is doing matrix multiplication quickly by hand, particularly when the $A$ is $p \times 1$ and $B$ is $1 \times q$. I would like to know of how to look at or define matrix multiplication, in a manner which makes it easy (for the average student) to compute by hand, while being intuitive and consistent for use for later proofs.   wikipedia has a great .
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?
8744
Let $A$ be a $m \times n$ matrix with entries from some field $F$. Define the determinant rank of $A$ to be the largest possible size of a nonzero minor, i.e. the size of the largest invertible square submatrix of $A$. It is true that the determinant rank is equal to the rank of a matrix, which we define to be the dimension of the row/column space.  It's not difficult to see that $\text{rank} \geq \text{determinant rank}$. If some submatrix of $A$ is invertible, then its columns/rows are linearly indepedent, which implies that the corresponding rows/columns of $A$ are also linearly indepedent.  Is there a nice proof for the converse?
116251
Wikipedia says "In the mathematical discipline of linear algebra, a matrix decomposition or matrix factorization is a factorization of a matrix into a product of matrices. There are many different matrix decompositions; each finds use among a particular class of problems."  But in my opinion decomposition term should be used to represent breaking a matrix in different sub-matrices or some new matrices created after some operation on original matrix which if used together and passed through some algorithm(not necessarily product), shall reproduce the original matrix.  Is there some different terminology to represent what I am expecting to say?
19899
I don't understand why the determinant of the 0x0 matrix equals 1.
117934
Simple calculation show that:  $$ \begin{align} \det(A_2)=\begin{vmatrix}     0&amp; 1  \\    -1&amp; 0   \end{vmatrix}&amp;=1\\ \det(A_4)=\begin{vmatrix}     0&amp; 1 &amp;1 &amp;1 \\    -1&amp; 0 &amp;1&amp;1\\    -1&amp; -1&amp; 0&amp;1\\    -1&amp; -1&amp; -1&amp;0  \end{vmatrix}&amp;=1 \end{align} $$  Here is my question:     Is it true that $\det(A_{2n})=1$ for all $n\in{\mathbb Z_+}$?   With MAPLE, I tried some large $n$. And I guess it is true. But temporarily I have no idea how to show it.
87095
I have the following $N \times N$ matrix. \begin{vmatrix} 0 &amp; 1 &amp; 1 &amp; \ldots &amp; 1 \\  1 &amp; a_1 &amp; 0 &amp; \ldots &amp; 0 \\  1 &amp; 0 &amp; a_2 &amp; \ldots &amp; 0 \\  \vdots &amp; \vdots&amp; &amp;\ddots&amp; \vdots\\ 1 &amp; 0 &amp; 0 &amp; \ldots &amp; a_n \\  \end{vmatrix}  There seems to be a pattern going on for  of this matrix, but I'm not sure how I would find the determinant for the $N \times N$ one.
40975
I was wondering, given an $n \times n$ square matrix, let function $\det : \left(a_1,a_2,\ldots,a_{n^2}\right) \to \textbf{R}$ give the determinant, where $a_{k}$'s are the entries of the $n \times n$ matrix.  Is this function (determinant) a differentiable kind?  If so, is the derivative continuous? That is, is $d\left(\det\right)$ a continuous function?  Furthermore, if so, to what differentiability class does this $\det$ function belong?   Thanks in advance.
29219
Suppose you have two vectors a and b that you want to take the dot product of, now this is done quite simply by taking each corresponding coordinate of each vector, multiplying them and then adding the result together. At the end of performing our operation we are left with a constant number.  My question therefore is what can we do with this number,why do we calculate it so to speak? I mean it seems almost useless to me compared with the cross product of two vectors (where you end up with an actual vector).
5875
I get the physical significance of vector addition &amp; subtraction. But I don't understand what do dot &amp; cross products mean?   More specifically,    Why is it that dot product of vectors $\vec{A}$ and $\vec{B}$ is defined as $AB\cos\theta$? Why is it that cross product of vectors $\vec{A}$ and $\vec{B}$ is defined as $AB\sin\theta$, times a unit vector determined from the right-hand rule?   To me, both these formulae seem to be arbitrarily defined (although, I know that it definitely wouldn't be the case).   If the cross product could be defined arbitrarily, why can't we define division of vectors? What's wrong with that? Why can't vectors be divided?
70255
I'm a student in an elementary linear algebra course. Without bashing on my professor, I must say that s/he is very poor at answering questions, often not addressing the question itself. Throughout the course, there have been multiple questions that have gone unanswered, but I must have the answer to this one question.  "Why are some vectors written as row vectors and others as column vectors?"  I understand that if I transpose one, it becomes the other. However, I'd like to understand the purpose behind writing a vector in a certain format.  Taking examples from my lectures, I see that when I'm trying to prove linear independence of a group of vectors, the vectors are written as column vectors in a matrix, and the row reduced form is found.  Other times, like trying to find the cross product or just solving a matrix, I see the vectors written as row vectors.  My professor is very vague on the notations and explanations, and it bugs me as a person who needs to know the reason behind every small thing, why this variation occurs in the format. Any input is greatly appreciated.
79397
Can  of finding the determinant be used for finding the determinant of matrices greater than $3\times3$, as I am unable to find any example of a matrix greater than $3\times3$ whose determinant is found by Sarrus' method?  I have tried many questions and Sarrus' method only works for $2\times2$ and $3\times3$ matrices. I have not read about Sarrus' method in any book but only from the internet, so I'm not sure about it.
111876
An entry of an $n \times n$ matrix with nonzero determinant is defined as interesting if by changing this entry (and only this entry) the determinant of the matrix can be made $0$.   Is it true that each entry of every matrix with nonzero determinant is interesting? Is it true that there is an interesting entry in each row of a matrix with nonzero determinant?
64000
Why is the determinant of a rotation matrix equal to 1? I would like a geometric interpretation of this. Just curious.
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?