I know how to solve for the determinant of a matrix, but I'm struggling to understand what it represents.
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What is the geometric meaning of the determinant of a matrix? I know that "The determinant of a matrix represents the area of ​​a rectangle." Perhaps this phrase is imprecise, but I would like to know something more, please.  Thank you very much.
88846
i have to prove:   $$\begin{vmatrix} -2a &amp;a+b  &amp;a+c \\   b+a&amp;-2b  &amp;b+c \\   c+a&amp;c+b  &amp;-2c  \end{vmatrix} = 4(a+b)(b+c)(c+a)$$   I have tried many calculations between the rows and columns of the determinant to get to the answer i want to solve the exercise however none of them gave me something right. Can someone help?
16411
Suppose that $A \in \mathbb{R}^{n\times n }$, and $x,y\in \mathbb{R}^n$.    Show that there exist numbers $a,b$ so that $\det{(A + sxy^*)}=a+bs$   Show that if $\det{(A)}\neq 0$ then $a = \det{(A)}$ and $b = \det{(A)} \cdot y^*A^{-1}x$.   Is it true that if $\det{(A)}=0$ then $a=0$?    My answer so far: The third question seems trivial and silly, and I don't understand why it's on there. I have solved the first one as follows: take $\beta$ to be any basis whose first element is $x$. Then with respect to basis $\beta$, $A+ sxy^*$ has the form   $$\left(  \begin{matrix} \alpha_{11} + sy_1 &amp; \alpha_{12} + sy_2  &amp; \dotsb &amp; \alpha_{1n}+sy_n \\ \alpha_{21}&amp;\dotsb&amp;&amp;\alpha_{2n} \\ \vdots &amp; \ddots &amp;&amp;\vdots\\ \alpha_{n1} &amp; \dotsb &amp;&amp;\alpha_{nn} \end{matrix} \right),$$  from which it's clear that the determinant is a polynomial in $s$ of degree $\leq 1$.    I am less clear on the second question.   For a limited case, note that if $A$ is diagonalizable and $x$ is an eigenvector of $A$ with eigenvalue $\lambda_x \neq 0$, then adding $sxy^*$ to $A$ has the effect of preserving all the eigenvectors, bumping up $\lambda_x$ to $\lambda_x + sy^*x$, and since the determinant is the product of all the eigenvalues, the product of $\lambda_1 \lambda_2  \dotsb (\lambda_x + sy^*x)\dotsb \lambda_n $ is   $$\det{(A)} + \det{(A)}\frac{sy^*x}{\lambda_x}=\det{(A)}\cdot sy^*(\frac{x}{\lambda_x}) = \det{(A)}\cdot s y^*(A^{-1}x).$$  Note that $y^*A^{-1}x$ is $\alpha_1 y_1 + \dotsb + \alpha_n y_n$, for $\alpha_i$ such that $x = \alpha_1 A_1 + \dotsb + \alpha_n A_n$, where $A_i$ are the columns of $A$.  But this is all I've got. Can someone give me a hint? I'm not looking for you to feed me the answer. Thanks as always
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I was reading over some notes of mine from the Linear Algebra class I took a while ago in order to prepare myself for Computer Graphics course I'll be taking this Fall and I ran across a sentence that I can't quite grasp. It states that     Since we have a non-zero vector $x$ in the nullspace of $A - \lambda I$, then $\det(A - \lambda I) = 0$.   From reading Wikipedia page on determinants I got that     ... the system has a unique solution exactly when the determinant is nonzero; when the determinant is zero there are either no solutions or many solutions.   So if we have vector $x$ in the nullspace of $A-\lambda I$, then we have either 0 solutions or many solutions. But I still don't understand why $\det(A - \lambda I)$ has to be equal zero?
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The cross product $\vec{a} \times \vec{b}$ can be written as the determinant of the matrix:  $$\left| \begin{matrix} \vec{i} &amp; \vec{j} &amp; \vec{k} \\ a_i &amp; a_j &amp; a_k \\ b_i &amp; b_j &amp; b_k \end{matrix}\right|$$  Is there any physical significance to this matrix, or is it just some mathematical trick?
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Given is an $n \times n$ matrix $M_n$ with the following structure — $$M_n = \left( \begin{array}{ccccc} a_1 &amp; 0 &amp; \dots &amp; 0 &amp; b_1 \\ b_2 &amp; a_2 &amp;  &amp;  &amp; 0\\  &amp; \ddots &amp; \ddots &amp;  &amp; \vdots \\  &amp;  &amp; b_{n-1} &amp; a_{n-1} &amp; 0\\  &amp;  &amp;  &amp; b_n &amp; a_n\end{array} \right),$$ with $a_i$ and $b_i \in \mathbb{R}$ where $i \in \{1,\dots,n\}$.  Is there a name for this specific type of matrix? The non-zero entries in each row are shifted one position to the right compared to the previous row, which reminds me of a circulant matrix (though in that case, the rows share the same values, which is generally not the case for $M_n$).  Is there a short/fast way (i.e. not just using ) to compute the determinant of this type of matrix?   It might help to consider an example, so here's $M_5$: $$M_5 = \left( \begin{array}{ccccc} a_1 &amp; 0 &amp; 0 &amp; 0 &amp; b_1 \\ b_2 &amp; a_2 &amp; 0 &amp; 0 &amp; 0\\ 0 &amp; b_3 &amp; a_3 &amp; 0 &amp; 0\\ 0 &amp; 0 &amp; b_4 &amp; a_4 &amp; 0\\ 0 &amp; 0 &amp; 0 &amp; b_5 &amp; a_5\end{array} \right) = \left( \begin{array}{ccccc} a_1 &amp;  &amp;  &amp;  &amp; b_1 \\ b_2 &amp; a_2 &amp;  &amp;  &amp; \\  &amp; b_3 &amp; a_3 &amp;  &amp; \\  &amp;  &amp; b_4 &amp; a_4 &amp; \\  &amp;  &amp;  &amp; b_5 &amp; a_5\end{array} \right)$$
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Given a real valued matrix $A$ that satisfies $AA^{\top} = I$ and $\det(A)&lt;0$, calculate $\det(A + I)$.  My start : Since $A$ satisfies  $AA^{\top} = I$, $A$ is a unitary matrix. The determinant of a unitary matrix with real entries is either $+1$ or $-1$. Since we know that $\det(A)&lt;0$, it follows that $\det(A)=-1$.
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Prove that $\det(kA) = k^n \det(A)$ for and ($n \times n$) matrix.   I have tried looking at this a couple of ways but can't figure out where to start. It's confusing to me since the equation for a determinant is such a weird summation.
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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?
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$$\det(A^T) = \det(A)$$ Using the geometric definition of the determinant as the area spanned by the columns, could someone give a geometric interpretation of the property?
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Whenever I needed to find the inverse of a matrix, I was told to check if its determinant is not zero. However, once I directly applied the Gauss-Jordan's method for finding the inverse of matrix whose determinant was zero. The inverse matrix that I got looked pretty normal like any other (if there wasn't a mistake).  I want to know how does the determinant of the matrix is related to inverse of matrix or why is that if determinant is zero then inverse doesn't exist? What exactly is inverse?
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Why is the determinant of an elementary permutation matrix equal to $-1$?  I am brand new to determinants and I've tried expanding it and using cofactor expansion, but it's messy and complicated. I would prefer if someone could show me using expansion, but alternative methods are welcome. Thanks.
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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?
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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!
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I need to calculate the following determinant in order to prove the following equality:  $$\det\begin{pmatrix} a &amp; b &amp; c &amp; d \\ -b &amp; a &amp; -d &amp; c \\ -c &amp; d &amp; a &amp; -b \\ -d &amp; -c &amp; b &amp; a \end{pmatrix} = (a^2+b^2+c^2+d^2)^2.$$  I tried using Gauss-algorithm to get an easier matrix, but I'm not sure if I did it correctly.  Calling the $4$ lines $I$, $II$, $III$ and $IV$, I did:  (1) $II \cdot a$  (2) $III \cdot a$  (3) $IV \cdot a$  After this I did:  (4) $II' + I \cdot b$  (5) $III' + I \cdot c$  (6) $IV' + I \cdot d$  So finally I got the following matrix:  $$\begin{pmatrix} a &amp; b &amp; c &amp; d \\ 0 &amp; a^2+b^2 &amp; bc-ad &amp; ac+bd \\ 0 &amp; ad+bc &amp; a^2+c^2 &amp; cd-ab \\ 0 &amp; bd-ac &amp; ab-cd &amp; a^2+d^2 \end{pmatrix}.$$  I thought this would make the determinant a bit easier, unfortunately I must have done something wrong. Is multiplication with single lines allowed as I have done it?  Thank you very much.
122719
Consider matrices of the following form: \begin{bmatrix} a &amp; x &amp; \cdots \\  x &amp; b &amp; x &amp;\cdots \\ \vdots &amp; x &amp; c &amp;\cdots \\   &amp; \vdots &amp; x &amp; \ddots &amp; \\ \end{bmatrix} where all $x$ are the same scalar. Is there anything that can be done to calculate the determinant easily? Obviously if $$x=a=b=c=\cdots $$ then the matrix is singular and the determinant is zero. Assume this is not the case.   Do they have any other interesting properties?
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Show that the determinant of a matrix $A$ is equal to the product of its eigenvalues $\lambda_i$.   So I'm having a tough time figuring this one out. I know that I have to work with the characteristic polynomial of the matrix $\det(A-\lambda I)$. But, when considering an $n \times n$ matrix, I do not know how to work out the proof. Should I just use the determinant formula for any $n \times n$ matrix? I'm guessing not, because that is quite complicated. Any insights would be great.