\documentclass{amsbook} \usepackage{enumitem} \newtheorem{theorem}{Theorem}[chapter] \newtheorem{lemma}[theorem]{Lemma} \theoremstyle{definition} \newtheorem{definition}[theorem]{Definition} \newtheorem{example}[theorem]{Example} \newtheorem{xca}[theorem]{Exercise} \theoremstyle{remark} \newtheorem{remark}[theorem]{Remark} \numberwithin{section}{chapter} \numberwithin{equation}{chapter} \begin{document} \frontmatter \title{Minqi's Solutions to the Exercises and Problems of Michael Artin (2011), Algebra (2nd Edition)} \author{Minqi Pan} \date{\today} \maketitle \setcounter{page}{4} \tableofcontents \mainmatter \chapter{Matrices} \section{Exercise 1.1: The Row Index and the Column Index} What are the entries $a_{21}$ and $a_{23}$ of the matrix $\begin{bmatrix} 1 & 2 & 5\\ 2 & 7 & 8\\ 0 & 9 & 4 \end{bmatrix}$? \begin{proof}[Answer] $2$ and $8$. \end{proof} \section{Exercise 1.2: Matrix Multiplication is not Commutative} Determine the products $AB$ and $BA$ for the following values of $A$ and $B$: $A=\begin{bmatrix} 1&2&3\\ 3&3&1 \end{bmatrix}, B=\begin{bmatrix} -8 & -4\\ 9 & 5\\ -3 & -2\end{bmatrix}; A=\begin{bmatrix} 1&4\\ 1&2\end{bmatrix}, B=\begin{bmatrix} 6 & -4\\ 3 & 2\end{bmatrix}.$ \begin{proof}[Answer] For the 1st pair, $AB=\begin{bmatrix} 1&0\\ 0&1\end{bmatrix}, BA=\begin{bmatrix} -20 & -28 & -28\\ 24 & 33 & 32\\ -9 & -12 & -11\end{bmatrix}.$ For the 2nd pair, $AB=\begin{bmatrix} 18 & 4\\ 12 & 0 \end{bmatrix}, BA=\begin{bmatrix} 2 & 16 \\ 5 & 16 \end{bmatrix}.$ \end{proof} \section{Exercise 1.3: Row$\times$Column and Column$\times$Row} Let $A=\begin{bmatrix} a_1 & \dots & a_n \end{bmatrix}$ be a row vector, and let $B=\begin{bmatrix} b_1\\ \vdots\\ b_n \end{bmatrix}$ be a column vector. Compute the products $AB$ and $BA$. \begin{proof}[Answer] $AB=a_1b_1+a_2b_2+\dots+a_nb_n.$ $BA=\begin{bmatrix} b_1a_1 & b_1a_2 & \dots & b_1a_n \\ b_2a_1 & b_2a_2 & \dots & b_2a_n \\ \vdots \\ b_na_1 & b_na_2 & \dots & b_na_n \end{bmatrix}.$ \end{proof} \section{Exercise 1.4: Matrix Multiplication is Associative} Verify the associative law for the matrix product $\begin{bmatrix} 1 & 2\\ 0 & 1\end{bmatrix}, \begin{bmatrix} 0 & 1 & 2\\ 1 & 1 & 3 \end{bmatrix}, \begin{bmatrix} 1\\ 4\\ 3\end{bmatrix}$. \begin{proof}[Answer] $\left( \begin{bmatrix} 1 & 2\\ 0 & 1\end{bmatrix} \begin{bmatrix} 0 & 1 & 2\\ 1 & 1 & 3 \end{bmatrix} \right) \begin{bmatrix} 1\\ 4\\ 3\end{bmatrix} =\begin{bmatrix} 2 &3 &8\\ 1 &1 &3 \end{bmatrix}\begin{bmatrix} 1\\ 4\\ 3\end{bmatrix}=\begin{bmatrix} 38\\ 14\end{bmatrix}.$ $\begin{bmatrix} 1 & 2\\ 0 & 1\end{bmatrix} \left(\begin{bmatrix} 0 & 1 & 2\\ 1 & 1 & 3 \end{bmatrix} \begin{bmatrix} 1\\ 4\\ 3\end{bmatrix} \right) =\begin{bmatrix} 1 &2\\ 0 &1\end{bmatrix}\begin{bmatrix} 10\\ 14\end{bmatrix}=\begin{bmatrix} 38\\ 14\end{bmatrix}$ \end{proof} \section{Exercise 1.5: The Number of Multiplications Required to Multiply 3 Matrices} Let $A$, $B$ and $C$ be matrices of size $l\times m$, $m\times n$ and $n\times p$. How many multiplications are required to compute the product $AB$? In which order should the triple product $ABC$ be computed, so as to minimize the number of multiplications required? \begin{proof}[Answer] $AB$ has $l\times n$ entries, each in need of $m$ multiplications. Therefore $l\times m\times n$ multiplications are required to compute the product $AB$. There are $2$ ways to compute $ABC$: \begin{enumerate} \item Compute $(AB)C$ with $l\times m\times n + l\times n\times p$ multiplications; \item Compute $A(BC)$ with $m\times n\times p + l\times m\times p$ multiplications. \end{enumerate} Therefore, \begin{enumerate} \item If $lmn+lnp>mnp+lmp$, the product $ABC$ should be computed in the order of $A(BC)$. \item If \$lmn+lnp