#LyX 1.4.2 created this file. For more info see http://www.lyx.org/ \lyxformat 245 \begin_document \begin_header \textclass article \begin_preamble %% \usepackage{amsfonts} % \setlength{\topmargin}{-.8in} \setlength{\headheight}{0in} \setlength{\headsep}{0in} \setlength{\footskip}{0in} \setlength{\evensidemargin}{-.5in} \setlength{\oddsidemargin}{-.5in} \setlength{\textheight}{9.3in} \setlength{\textwidth}{7in} \setlength{\parindent}{0in} \newcommand{\rank}{\operatorname{rank}} \newcommand{\spann}{\operatorname{span}} \end_preamble \language english \inputencoding auto \fontscheme default \graphics default \paperfontsize 12 \spacing single \papersize default \use_geometry false \use_amsmath 1 \cite_engine basic \use_bibtopic false \paperorientation portrait \secnumdepth 3 \tocdepth 3 \paragraph_separation indent \defskip medskip \quotes_language english \papercolumns 1 \papersides 1 \paperpagestyle empty \tracking_changes false \output_changes true \end_header \begin_body \begin_layout Standard \series bold Exam\InsetSpace ~ 1 \hfill Math\InsetSpace ~ 314 \hfill Sample \newline Name: \series default \begin_inset ERT status collapsed \begin_layout Standard \backslash rule{2.4in}{.01in} \end_layout \end_inset \hfill \series bold Score: \series default \begin_inset ERT status collapsed \begin_layout Standard \backslash rule{1in}{.01in} \end_layout \end_inset \end_layout \begin_layout Standard \begin_inset ERT status collapsed \begin_layout Standard \backslash hspace{.25in} \end_layout \end_inset \shape italic Instructions: \begin_inset ERT status collapsed \begin_layout Standard \backslash /{} \end_layout \end_inset \shape default Show your work in the spaces provided below for full credit. Use the reverse side for additional space, \shape italic but clearly so indicate. \begin_inset ERT status collapsed \begin_layout Standard \backslash /{} \end_layout \end_inset \shape default You must clearly identify answers and show supporting work to receive any credit. Exact answers (e.g., \begin_inset Formula $\pi$ \end_inset ) are preferred to inexact (e.g., \begin_inset Formula $3.14$ \end_inset ). Point values of problems are given in parentheses. Notes or text in \emph on any \emph default form not allowed. Calculator is allowed. \newline \begin_inset ERT status collapsed \begin_layout Standard \backslash rule[.2in]{7in}{.01in} \end_layout \end_inset (25) \series bold 1. \series default In each of the following problems, clearly identify the augmented matrix of the system and give its rank. \end_layout \begin_layout Standard (i) Use Gauss-Jordan elimination to find the general solution to this system: \begin_inset Formula \begin{eqnarray*} x_{1}+x_{2}+x_{4} & = & 1\\ 2x_{1}+2x_{2}+x_{3}+x_{4} & = & 1\\ 2x_{1}+2x_{2}+2x_{4} & = & 2\end{eqnarray*} \end_inset \end_layout \begin_layout Standard (ii) Use Gaussian elimination with back solving to find the solution to this system. Here, \begin_inset Formula $b_{1},b_{2}$ \end_inset are constants and \begin_inset Formula $x_{1},x_{2}$ \end_inset the unknowns. \begin_inset Formula \begin{eqnarray*} x_{1}-x_{2} & = & b_{1}\\ 2x_{1}+2x_{2} & = & b_{2}\end{eqnarray*} \end_inset \end_layout \begin_layout Standard \begin_inset ERT status collapsed \begin_layout Standard \backslash vspace{.2in} \end_layout \end_inset (14) \series bold 2. \series default Given the matrix \begin_inset Formula $A=\left[\begin{array}{ccc} 1 & -2 & 1\\ 0 & 2 & 0\\ -1 & 0 & 1\end{array}\right]$ \end_inset and vector \begin_inset Formula $\mathbf{b}=\left[\begin{array}{c} 3\\ 0\\ 1\end{array}\right]$ \end_inset , find the inverse of the matrix \begin_inset Formula $A$ \end_inset and use this to solve the system \begin_inset Formula $A\mathbf{x}=\mathbf{b}$ \end_inset . \end_layout \begin_layout Standard \begin_inset ERT status collapsed \begin_layout Standard \backslash vspace{.3in} \end_layout \end_inset (13) \series bold 3. \series default Calculate the following: \end_layout \begin_layout Standard (i) The solutions to the equation \begin_inset Formula $z^{3}=-8.$ \end_inset Express them in polar form and graph them in the complex plane. \end_layout \begin_layout Standard \begin_inset ERT status collapsed \begin_layout Standard \backslash vspace{.1in} \end_layout \end_inset (ii) \begin_inset Formula ${\displaystyle \frac{2+i}{2-i}=}$ \end_inset \end_layout \begin_layout Standard \begin_inset ERT status collapsed \begin_layout Standard \backslash vspace{.1in} \end_layout \end_inset (iii) \begin_inset Formula $\left|\begin{array}{cccc} 1 & 1 & 0 & 1\\ 1 & 2 & 1 & 1\\ 0 & 0 & 1 & 3\\ 0 & 0 & 2 & 0\end{array}\right|=$ \end_inset \end_layout \begin_layout Standard \begin_inset ERT status collapsed \begin_layout Standard \backslash vspace{.1in} \end_layout \end_inset (20) \series bold 4. \series default Carry out these calculations or indicate they are impossible. You are given that \begin_inset Formula $\mathbf{a}=\left[\begin{array}{c} 1\\ 2\end{array}\right]$ \end_inset , \begin_inset Formula $\mathbf{b}=\left[\begin{array}{cc} 3 & 2\end{array}\right]$ \end_inset , \begin_inset Formula $C=\left[\begin{array}{cc} 1 & 0\\ 1+\imath & 2\end{array}\right]$ \end_inset , and \begin_inset Formula $D=\left[\begin{array}{cc} 1 & -1\\ 2 & 2\end{array}\right]$ \end_inset . \end_layout \begin_layout Standard (a) \begin_inset Formula $ba$ \end_inset \end_layout \begin_layout Standard \begin_inset ERT status collapsed \begin_layout Standard \backslash vspace{.2in} \end_layout \end_inset (b) \begin_inset Formula $\mathbf{a}\mathbf{b}$ \end_inset \end_layout \begin_layout Standard \begin_inset ERT status collapsed \begin_layout Standard \backslash vspace{.2in} \end_layout \end_inset (c) \begin_inset Formula $C\mathbf{b}$ \end_inset \end_layout \begin_layout Standard \begin_inset ERT status collapsed \begin_layout Standard \backslash vspace{.2in} \end_layout \end_inset (d) \begin_inset Formula $\mathbf{b}D$ \end_inset \end_layout \begin_layout Standard \begin_inset ERT status collapsed \begin_layout Standard \backslash vspace{.2in} \end_layout \end_inset (e) \begin_inset Formula $D+D^{T}$ \end_inset \end_layout \begin_layout Standard \begin_inset ERT status collapsed \begin_layout Standard \backslash vspace{.2in} \end_layout \end_inset (f) \begin_inset Formula $C^{*}C$ \end_inset \end_layout \begin_layout Standard \begin_inset ERT status collapsed \begin_layout Standard \backslash vspace{.2in} \end_layout \end_inset (g) \begin_inset Formula $C^{-1}$ \end_inset \end_layout \begin_layout Standard \begin_inset ERT status collapsed \begin_layout Standard \backslash vspace{.2in} \end_layout \end_inset (18) \series bold 5. \series default Fill in the blanks or answer True/False (T/F): \end_layout \begin_layout Standard (a) If \begin_inset Formula $P$ \end_inset and \begin_inset Formula $Q$ \end_inset are invertible and \begin_inset Formula $A$ \end_inset a matrix such that \begin_inset Formula $PAQ$ \end_inset is defined, then \begin_inset Formula $\mbox{rank}PAQ=$ \end_inset \begin_inset ERT status collapsed \begin_layout Standard \backslash rule{1.3in}{.01in} \end_layout \end_inset \end_layout \begin_layout Standard \begin_inset ERT status collapsed \begin_layout Standard \backslash bigskip \end_layout \begin_layout Standard \end_layout \end_inset (b) If \begin_inset Formula $A$ \end_inset is a \begin_inset Formula $3\times3$ \end_inset matrix, then in terms of \begin_inset Formula $\det(A)$ \end_inset , we can say that \begin_inset Formula $\det(-2A)=$ \end_inset \begin_inset ERT status collapsed \begin_layout Standard \backslash rule{1in}{.01in} \end_layout \end_inset . \end_layout \begin_layout Standard \begin_inset ERT status collapsed \begin_layout Standard \backslash bigskip \end_layout \begin_layout Standard \end_layout \end_inset (c) As an elementary matrix, \begin_inset Formula $\left[\begin{array}{ccc} 1 & 3 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1\end{array}\right]$ \end_inset is \begin_inset ERT status collapsed \begin_layout Standard \backslash rule{1in}{.01in} \end_layout \end_inset . \end_layout \begin_layout Standard \begin_inset ERT status collapsed \begin_layout Standard \backslash bigskip \end_layout \begin_layout Standard \end_layout \end_inset (d) As elementary matrices, \begin_inset Formula $E_{ij}(a)^{-1}=$ \end_inset \begin_inset ERT status collapsed \begin_layout Standard \backslash rule{1in}{.01in} \end_layout \end_inset and \begin_inset Formula $E_{ij}(a)^{T}=$ \end_inset \begin_inset ERT status collapsed \begin_layout Standard \backslash rule{1in}{.01in} \end_layout \end_inset . \end_layout \begin_layout Standard \begin_inset ERT status collapsed \begin_layout Standard \backslash bigskip \end_layout \begin_layout Standard \end_layout \end_inset (e) If two rows of a determinant \begin_inset Formula $|A|$ \end_inset are interchanged then the new determinant is \begin_inset ERT status collapsed \begin_layout Standard \backslash rule{1in}{.01in} \end_layout \end_inset . \end_layout \begin_layout Standard \begin_inset ERT status collapsed \begin_layout Standard \backslash bigskip \end_layout \begin_layout Standard \end_layout \end_inset (f) If \begin_inset Formula $A\mathbf{x}=\mathbf{b}$ \end_inset has a unique solution for some particular \begin_inset Formula $\mathbf{b}$ \end_inset then \begin_inset Formula $A$ \end_inset is invertible (T/F): \end_layout \begin_layout Standard \begin_inset ERT status collapsed \begin_layout Standard \backslash bigskip \end_layout \begin_layout Standard \end_layout \end_inset (g) Every nonzero square matrix has an inverse (T/F): \end_layout \begin_layout Standard \begin_inset ERT status collapsed \begin_layout Standard \backslash bigskip \end_layout \begin_layout Standard \end_layout \end_inset (h) Any homogeneous linear system with more unknowns than equations has a nontrivial solution (T/F): \end_layout \begin_layout Standard \begin_inset ERT status collapsed \begin_layout Standard \backslash bigskip \end_layout \begin_layout Standard \end_layout \end_inset (i) The rank of the matrix \begin_inset Formula $A$ \end_inset is the same as the rank of \begin_inset Formula $A^{T}$ \end_inset (T/F): \end_layout \begin_layout Standard \begin_inset ERT status collapsed \begin_layout Standard \backslash bigskip \end_layout \begin_layout Standard \end_layout \end_inset (j) For \begin_inset Formula $2\times2$ \end_inset matrices \begin_inset Formula $A$ \end_inset and \begin_inset Formula $B$ \end_inset , \begin_inset Formula $AB=BA$ \end_inset (T/F): \end_layout \begin_layout Standard \begin_inset ERT status collapsed \begin_layout Standard \backslash bigskip \end_layout \begin_layout Standard \end_layout \end_inset (k) If \begin_inset Formula $A$ \end_inset and \begin_inset Formula $B$ \end_inset are matrices such that \begin_inset Formula $AB=0$ \end_inset , then \begin_inset Formula $A=0$ \end_inset or \begin_inset Formula $B=0$ \end_inset (T/F): \end_layout \begin_layout Standard \begin_inset ERT status collapsed \begin_layout Standard \backslash bigskip \end_layout \begin_layout Standard \end_layout \end_inset (l) If \begin_inset Formula $A$ \end_inset is both upper triangular and lower triangular, then \begin_inset Formula $A$ \end_inset is diagonal (T/F): \end_layout \begin_layout Standard \begin_inset ERT status collapsed \begin_layout Standard \backslash vspace{.2in} \end_layout \end_inset (10) \series bold 6. \series default Give the definition of an invertible matrix. Then use the definition and basic facts to prove that if \begin_inset Formula $A$ \end_inset is invertible and \begin_inset Formula $A$ \end_inset is symmetric, then so is \begin_inset Formula $A^{-1}$ \end_inset . \end_layout \end_body \end_document