Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 0.1 Exercise 0.1.5 Determine whether the following functions are well defined. $f : \mathbb{Q} \rightarrow \mathbb{Z}$ given by $f(a/b)…
Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 0.1 Exercise 0.1.4 Denote by $\mathcal{A}$ the set of all $2\times 2$ matrices with real number entries. Let $M…
Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 0.1 Exercise 0.1.3 Denote by $\mathcal{A}$ the set of all $2\times 2$ matrices with real number entries. Let $M…
Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 0.1 Exercise 0.1.2 Denote by $\mathcal{A}$ the set of all $2\times 2$ matrices with real number entries. Let $M…
Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 0.1 Exercise 0.1.1 Denote by $\mathcal{A}$ the set of all $2\times 2$ matrices with real number entries. Let $M…
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Vandermonde matrices have the following form:$$A_n=\begin{bmatrix} 1&x_1&x_1^2&\cdots&x_1^{n-1}\\ 1&x_2&x_2^2&\cdots&x_2^{n-1}\\ \vdots&\vdots&\vdots&\ddots&\vdots\\ 1&x_n&x_n^2&\cdots&x_n^{n-1} \end{bmatrix},$$ where $A_n=[a_{ij}]_{i,j=1}^n$ is an $n\times n$ matrices and the element $a_{ij}=x_i^{j-1}$. The transpose $A_n^T$ is also called a Vandermonde…
Let $A$ be an $n\times n$ matrix. If there exists a positive integer $q$ such that \begin{equation}\label{eq:1}A^{q}=0,\end{equation} then we call $A$ a nilpotent matrix, meaning that one of its powers…
Basic Forms $$\int x^{n} d x=\frac{1}{n+1} x^{n+1}, \quad n \neq-1$$ $$\int \frac{1}{x} d x=\ln |x|$$ $$\int u d v=u v-\int v d u$$ $$\int \frac{1}{a x+b} d x=\frac{1}{a} \ln |a…
Exercise 7.1 Solution: (a) Note that\[\sqrt{x+3}-\sqrt{c+3}=\frac{(x+3)-(c+3)}{\sqrt{x+3}+\sqrt{c+3}}=\frac{x-c}{\sqrt{x+3}+\sqrt{c+3}}.\]We have\begin{align*}f'(c)&=\lim_{x\to c}\frac{f(x)-f(c)}{x-c}\\ &=\lim_{x\to c}\frac{\sqrt{x+3}-\sqrt{c+3}}{x-c}\\ &=\lim_{x\to c}\frac{\sqrt{x+3}-\sqrt{c+3}}{x-c}\\&=\lim_{x\to c}\frac{1}{x-c}\frac{x-c}{\sqrt{x+3}+\sqrt{c+3}} \\ &=\lim_{x\to c}\frac{1}{\sqrt{x+3}+\sqrt{c+3}}\\&=\frac{1}{2\sqrt{c+3}}. \end{align*}(b) Note that\[\frac{1}{x}-\frac{1}{c}=\frac{c-x}{xc}.\]We have\begin{align*}g'(c)&=\lim_{x\to c}\frac{g(x)-g(c)}{x-c}\\ &=\lim_{x\to c}\frac{\frac{1}{x}-\frac{1}{c}}{x-c}\\ &= \lim_{x\to c}\frac{\frac{c-x}{xc}}{x-c}\\ &=\lim_{x\to c}\frac{-1}{xc}=-\frac{1}{c^2}.\end{align*}(c) Note that…