\documentstyle[12pt]{article} \pagestyle{headings} % macros \newcounter{probno} \newcommand{\prob}[1]{\addtocounter{probno}{1} \par\vfill\noindent\theprobno . (#1 point(s)) } \newcommand{\true }{\addtocounter{probno}{1} \par\smallskip\noindent\theprobno .(1 point)(Circle one: True False.) } \newcommand{\by }{\times} \newcommand{\norm}[1]{||#1||} \begin{document} \begin{flushright} A. Miller\\ M340 Exam 1\\ Feb 23, 1990 \end{flushright} {\bf Put your answers on this exam. Put your work on separate sheets of paper to be handed in with the exam. There are a total of 50 points.} \bigskip \noindent Name\underline{\phantom{AAAAAAAAAAAAAAAAAAAAAAAAAAAA}} Score\underline{\phantom{AAAAAAAA}} \bigskip $$ A= \left[ \begin{array}{ccc} 1 & 1+i & 0 \\ -1 & 2i & 3 \end{array} \right] % \;\;B= \left[ \begin{array}{c} 0\\ 1+i \\ 1 \end{array} \right] % \;\;C= \left[ \begin{array}{ccc} i\\1\\ i \end{array} \right] % $$ Compute the following or say that it is undefined. \prob 1 $AB$= \prob 1 $B+C$= \prob 1 $BA$= \prob 1 $i A$= \prob 1 $A^{\prime}$= \prob 1 $\langle B,C\rangle$= \prob 1 $|\langle B,C\rangle|$= \prob 1 $\norm{C}$= \newpage \prob 1 If the norm of a vector $Q$ is 6 what is the norm of ${iQ}$? \prob 2 If $\langle U,V\rangle=2+3i$ what is $\langle V,2U\rangle$? \prob 2 If $B$ is the inverse of the matrix $A$, then what is the inverse of the matrix $6A$? \prob 2 Give the definition of the inverse of an $n\by n$ matrix $A$. (Your answer should consist of a grammatically correct sentence.) \prob 2 Give an example other than the identity of an orthogonal symmetric $2\by 2$ matrix. \prob 1 Give an example of a trivial $3\by 1$ vector. \prob 1 Give an example of a nontrivial $2\by 1$ vector. \prob 2 For what values of $a,b,c$ does the equation $$ \left[ \begin{array}{ccccc} 1 & 1 & 0 & 1 & 0 \\ 0 & 0 & 1 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 \\ \end{array} \right] X= \left[ \begin{array}{c} a \\ b \\ c \\ \end{array} \right] $$ have \begin{enumerate} \begin{enumerate} \item no solutions? \item a unique solution? \item infinitely many solutions? \end{enumerate} \end{enumerate} \prob 2 What are the slack columns of the matrix $A$? $$A= \left[ \begin{array}{ccccc} 0 & 1 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 \\ \end{array} \right] $$ \newpage \prob 4 Find the inverse of $A$ or show that the inverse does not exist by exhibiting a nontrivial vector $X$ such that $AX=[0]$. (Remember: Put your work on a separate sheet of paper.) \begin{enumerate} \begin{enumerate} \item $$A= \left[ \begin{array}{cccc} 0 & -1 & 0 & 2 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & -1 \\ \end{array} \right] $$ \vfill \item $$A= \left[ \begin{array}{ccc} 1 & 2 & 1 \\ 0 & 1 & 1 \\ 1 & 3 & 2 \\ \end{array} \right] $$ \end{enumerate} \end{enumerate} \prob 2 Find an elementary matrix $E$ such that $EA$ is in row echelon form or say that no such $E$ exists. $$A= \left[ \begin{array}{ccccc} 0 & 1 & 0 & 2 & 0 \\ 1 & 0 & 0 & 0 & 2 \\ 0 & 0 & 1 & 0 & -1 \\ \end{array} \right] $$ \newpage \true If $U$ and $V$ are $n\by n$ uppertriangular matrices then so is $UV$. \true If $D$ is a diagonal $n\by n$ matrix, then so is $cD^{n+1}$ where $c$ is a scalar. \true If $A,B,C$ and $D$ are $n\by n$ matrices, then $(AB)(CD)=A(BC)D$. \true If $A$ and $B$ are $n\by n$ matrices and $c$ is a scalar, then $A(cB)=B(cA)$. \true If $A,B$, and $C$ are $n\by n$ matrices and $A+C=C+B$, then $A=B$. \true If $A$ is an $n\by n$ invertible matrix, then the matrix equation $AX=B$ has a unique solution. \true If $A$ and $B$ are symmetric $n\by n$ matrices, then $AB$ is a symmetric matrix. \true If $A$ and $B$ are symmetric $n\by n$ matrices, then $A+B$ is a symmetric matrix. \true If $A,B$, and $C$ are $n\by n$ matrices and $AC=BC$, then either $A=B$ or $C$ is the zero matrix. \true It is possible to find $5\by 1$ vectors $U$ and $V$ such that $\norm{U}$=2, $\norm{V}=3$ and $\norm{U+V}=4$. \true It is possible to find $17\by 1$ vectors $U$ and $V$ such that $\norm{U}$=4, $\norm{V}=2$ and $\norm{U+V}=7$. \true It is possible to find $3\by 1$ vectors $U$ and $V$ such that $\langle U,V\rangle=5$ and $\norm{U}=\norm{V}=2$. \true For any $n$ there exists an $n\by n$ elementary matrix E such that $EA=A^T$ for any $n\by n$ matrix $A$. \true For any $m\by n$ matrix $A$ it is possible to find an invertible $m\by m$ matrix $M$ such that $MA$ is in row echelon form. \true The sum of two $n\by n$ invertible matrices is invertible. \bigskip \prob 6 Suppose $A$ is an $n\by n$ matrix such that $\| AX\|=\|X\|$ for all $n\by 1$ vectors $X$. {\bf On the back of this page} prove that $\langle AX,AY\rangle= \langle X,Y\rangle$ for all $n\by 1$ vectors $X$ and $Y$. \newpage \begin{verbatim} exam 1 solutions by minimat ** Diary file: C:\CLASS\340JAN90\EX1 opened at Date/Time: 90-02-23 10:38:24** (Type DIARY OFF to close the diary.) #> i=sqrt(-1) i = ( 0.000 + 1.000*i) #> a=[1 1+i 0; -1 2*i 3] a = ( 1.000 + 0.000*i) ( 1.000 + 1.000*i) ( 0.000 + 0.000*i) ( -1.000 + 0.000*i) ( 0.000 + 2.000*i) ( 3.000 + 0.000*i) #> b=[0;1+i;1] b = ( 0.000 + 0.000*i) ( 1.000 + 1.000*i) ( 1.000 + 0.000*i) #> c=[i;1;i] c = ( 0.000 + 1.000*i) ( 1.000 + 0.000*i) ( 0.000 + 1.000*i) #> a*b ans = ( 0.000 + 2.000*i) ( 1.000 + 2.000*i) #> b+c ans = ( 0.000 + 1.000*i) ( 2.000 + 1.000*i) ( 1.000 + 1.000*i) #> b*a [71] : Incompatible sizes for multiplication #> i*a ans = ( 0.000 + 1.000*i) ( -1.000 + 1.000*i) ( 0.000 + 0.000*i) ( 0.000 - 1.000*i) ( -2.000 + 0.000*i) ( 0.000 + 3.000*i) #> a' ans = ( 1.000 + 0.000*i) ( -1.000 + 0.000*i) ( 1.000 - 1.000*i) ( 0.000 - 2.000*i) ( 0.000 + 0.000*i) ( 3.000 + 0.000*i) #> b*c [71] : Incompatible sizes for multiplication #> inner(b,c) ans = 1.000 #> norm(c) ans = 1.732 #> sqrt(3 ) ans = 1.732 #> q=6 q = 6.000 #> norm(i*q) ans = 6.000 #> u=2+3*i, v=1 u = ( 2.000 + 3.000*i) v = 1.000 #> inner(u,v) ans = ( 2.000 + 3.000*i) #> inner(v,2*u) ans = ( 4.000 - 6.000*i) #> a=rand(3,3) a = 0.598 0.882 0.295 0.189 0.486 0.161 0.476 0.125 0.469 #> b=inv(a) b = 3.967 -7.197 -0.027 -0.225 2.667 -0.774 -3.966 6.595 2.366 #> inv(6*a) ans = 0.661 -1.200 -0.004 -0.037 0.444 -0.129 -0.661 1.099 0.394 #> a=[0 -1 0 2; 1 0 0 0 ; 0 0 1 0 ; 0 1 0 -1] a = 0.000 -1.000 0.000 2.000 1.000 0.000 0.000 0.000 0.000 0.000 1.000 0.000 0.000 1.000 0.000 -1.000 #> inv(a) ans = 0.000 1.000 0.000 0.000 1.000 0.000 0.000 2.000 0.000 0.000 1.000 0.000 1.000 0.000 0.000 1.000 #> randsoln(a) ans = 0.000 0.000 0.000 0.000 #> a=[ 1 2 1; 0 1 1; 1 3 2 ] a = 1.000 2.000 1.000 0.000 1.000 1.000 1.000 3.000 2.000 #> inv(a) [82] : Vanishing determinant #> randsoln(a) ans = 0.483 -0.483 0.483 #> a=[0 1 0 2 0 ; 1 0 0 0 2; 0 0 1 0 -1] a = 0.000 1.000 0.000 2.000 0.000 1.000 0.000 0.000 0.000 2.000 0.000 0.000 1.000 0.000 -1.000 #> [b m]=refm(a) b = 1.000 0.000 0.000 0.000 2.000 0.000 1.000 0.000 2.000 0.000 0.000 0.000 1.000 0.000 -1.000 m = 0.000 1.000 0.000 1.000 0.000 0.000 0.000 0.000 1.000 #> round(rand(15,1)) ans = 0.000 0.000 0.000 0.000 1.000 0.000 0.000 0.000 1.000 0.000 0.000 1.000 1.000 0.000 1.000 #> quit ** Diary file: C:\CLASS\340JAN90\EX1 closed at Date/Time: 90-02-23 14:41:09** \end{verbatim} \end{document}