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\def\picture #1 wide #2 high #3{\vbox to #2{\hbox to #1{\hfil}\vfil
  \special{picture #3}}}
\def\picturetop #1 wide #2 high #3{\vtop to #2{\vfil\hbox to #1{\hfil}
  \special{picture #3}}}

\newcount\problemcount
\def\newproblem{\advance\problemcount by 1
		\leavevmode\llap{\the\problemcount.~}}

\def\endproblem{\vskip\interproblemspace\vfil\goodbreak}

\def\answers#1#2#3#4#5{\smallskip\theanswers{#1}{#2}{#3}{#4}{#5}\par
    \vskip\interproblemspace\vfil\goodbreak}

\def\TF{\smallskip
    {\it (T)~True\qquad (F)~False}\vskip\interproblemspace\vfil\goodbreak}

\def\theanswers#1#2#3#4#5{\leavevmode
  {\it (a)~}#1\hfil{\it (b)~}#2\hfil{\it (c)~}#3\hfil{\it (d)~}#4\hfil{\it (e)~}#5}

\def\endanswerline{\endgraf\hfilneg} % to force next line in \theanswers

\newcount\answercount
\newcount\groupcount
\newcount\columncount
\newcount\columnend
\newcount\columnsize
\def\answersheet#1#2#3#4#5{{
%#1 = start, #2 = stop, #3 = group size, #4 = no. of columns, #5 text for each answer
  \columncount=#4 \advance\columncount by 1 \divide\hsize by\columncount
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  \advance\columnsize by1\advance\columnsize by\columncount
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    \advance\columnend by \columnsize
    \ifnum#2<\columnend \global\columnend=#2\fi 
    \hfill
    \vtop{\loop\ifnum\answercount<\columnend \global\advance\answercount by 1
      \leavevmode\llap{\number\answercount.~}\enspace
      #5\endgraf
				  \ifnum\groupcount<#3 \advance\groupcount by 1\else\groupcount=1\bigskip\fi
				  \repeat}\hfill\groupcount=1
  \repeat}}

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{\bf Math 226: Calculus IV}\hfill Name:\hbox to 2.5truein{\hrulefill}\medskip
{\bf Exam I}\quad{\sl February 13, 1990}\hfill
Instructor:\hbox to 2.5truein{\hrulefill}\medskip
\leavevmode\hfill Section:\hbox to 2.5truein{\hrulefill}

\bigskip
Record your answers to the multiple choice problems by placing an $\times$
through one letter for each problem on this answer sheet. There are 17
questions worth 6 points each.
\bigskip\bigskip

{\baselineskip=20pt
  \answersheet{1}{17}{3}{2}{\theanswers{}{}{}{}{}}
}

\bigskip\bigskip
\leavevmode
\hfill Score \hbox to 1truein{\hrulefill}

\vfil\eject

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%    Questions
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\def\A{{\bf A}}
\def\B{{\bf B}}
\def\C{{\bf C}}
\def\D{{\bf D}}
\def\E{{\bf E}}
\def\F{{\bf F}}
\def\X{{\bf X}}
\def\V{{\bf V}}
\def\0{{\bf 0}}

\pageno=1
\footline={\hfil{\rm \folio}\hfil}
\bigskip\bigskip

\newproblem
Let 
$$A = \pmatrix{2&-3&4\cr7&0&1\cr}\qquad B = \pmatrix{-1&2&2\cr6&-1&1\cr}$$
Find $A - 3B$.
\answers
{$\pmatrix{5&-9&-2\cr-11&3&-2\cr}$}
{$\pmatrix{-1&3&10\cr25&-3&4\cr}$}
{$\pmatrix{1&-1&2\cr1&1&0\cr}$\endanswerline}
{$\pmatrix{3&-6&-6\cr-18&3&-3\cr}$}
{{\sl Not defined}}

\newproblem
Calculate the matrix product
$$\pmatrix{1&-2\cr3&1\cr}\pmatrix{2&1&-3\cr5&7&0\cr}$$

\answers
{$\pmatrix{2&-2&-3\cr15&7&0\cr}$}
{$\pmatrix{-8&-13\cr11&13\cr}$}
{$\pmatrix{17&1\cr22&5\cr-3&6\cr}$\endanswerline}
{$\pmatrix{-8&-13&-3\cr11&10&-9\cr}$}
{{\sl Not defined}}

\newproblem
Which of the following matrices are in row-echelon form?
$$
A = \pmatrix{
1&3&-2&0\cr
0&0&1&5\cr
0&0&0&0\cr}
\qquad
B = \pmatrix{
0&1&5\cr
1&-3&2\cr}
\qquad
C = \pmatrix{
0&2&4\cr
0&0&1\cr}
$$
$$
D = \pmatrix{
0&1&0&4\cr
0&0&1&3\cr
0&0&0&1\cr}
\qquad
E = \pmatrix{
1&0&3\cr
0&4&5\cr}
$$
\answers
{$A$ only}
{$A$ and $D$ only}
{$A$ and $E$ only}
{$A$, $C$, $D$, and $E$ only\endanswerline}
{$A$, $B$, $C$, $D$, and $E$}

\newproblem
Translate the following system of equations into an augmented matrix.
$$\eqalign{
x_1 - x_5 &= -1\cr
2x_2 - 3x_4 &=2\cr
x_2 + x_3 + x_5 &= -5\cr
3x_1 - 2x_3 + x_4 &= 1\cr}$$
\answers
{\vtop{\hbox{
$\pmatrix{
1&2&0&-3&-1&:&1\cr
1&1&0&1&0&:&-5\cr
3&0&-2&1&0&:&1\cr
}$}\bigskip}}
{\vtop{\hbox{
$\pmatrix{
1&-1&:&-1\cr
2&-3&:&2\cr
1&1&:&-5\cr
3&-2&:&1\cr
}$}}\endanswerline}
{\vtop{\hbox{
$\pmatrix{
1&0&-1&:&-1\cr
2&-3&0&:&2\cr
1&1&1&:&-5\cr
3&-2&1&:&1\cr
}$}\bigskip}}
{\vtop{\hbox{
$\pmatrix{
1&0&0&0&-1&:&-1\cr
0&2&0&-3&0&:&2\cr
0&1&1&0&1&:&-5\cr
3&0&-2&1&0&:&1\cr
}$}}\endanswerline}
{\vtop{\hbox{
$\pmatrix{
1&0&0&3&:&-1\cr
0&2&1&0&:&2\cr
0&-3&0&1&:&-5\cr
-1&0&1&0&:&1\cr
}$}}}

\newproblem
Which of the following is the general solution of the system of
equations associated to the augmented matrix
$$\pmatrix{
1&-1&1&0&-1&:&1\cr
0&1&-2&1&-1&:&3\cr
0&0&1&-3&4&:&2\cr}$$
after back-substitution?
\answers
{\vtop{
\hbox{$x_1 = 2 + 3x_4 - 4x_5$}
\hbox{$x_2 = 3 + 2x_3 -  x_4 +  x_5$}
\hbox{$x_3 = 1 + x_2 - x_3 + x_5$}
}}
{\vtop{
\hbox{$x_1 = 1 -  x_2 +  x_3 - x_5$}
\hbox{$x_2 = 3 - 2x_3 +  x_4 - x_5$}
\hbox{$x_3 = 2 - 3x_4 + 4x_5$}
}}
{\vtop{
\hbox{$x_1 = 6 + 2x_4 - 2x_5$}
\hbox{$x_2 = 7 + 5x_4 - 7x_5$}
\hbox{$x_3 = 2 + 3x_4 - 4x_5$}
\bigskip
}\endanswerline}
{\vtop{
\hbox{$x_1 = 6$}
\hbox{$x_2 = 7$}
\hbox{$x_3 = 2$}
}}
{\vtop{
\hbox{$x_1 = 1 + 2x_4 - 2x_5$}
\hbox{$x_2 = 3 + 5x_4 - 7x_5$}
\hbox{$x_3 = 2 + 3x_4 - 4x_5$}
}}

\newproblem
Transform the following matrix into row-echelon form.
$$\pmatrix{
1&1&4&0&-1&10\cr
-1&2&0&-1&0&0\cr
0&1&1&0&1&3\cr
1&2&5&0&0&13\cr}$$
\answers
{\vtop{\hbox{
$\pmatrix{
1&1&4&0&-1&10\cr
0&1&1&0&1&3\cr
0&0&1&-1&-4&1\cr
0&0&0&0&0&0\cr}$}\bigskip}
}
{\vtop{\hbox{
$\pmatrix{
1&1&4&0&-1&10\cr
0&3&4&-1&-1&10\cr
0&1&1&0&1&3\cr
0&1&1&0&1&3\cr}$}}
\endanswerline}
{\vtop{\hbox{
$\pmatrix{
1&1&4&0&-1&10\cr
0&1&-4&1&1&-10\cr
0&1&1&0&1&3\cr
0&0&0&0&0&0\cr}$}\bigskip}
}
{\vtop{\hbox{
$\pmatrix{
1&1&4&0&-1&10\cr
0&1&4&-1&-1&10\cr
0&0&1&0&1&3\cr}$}}
\endanswerline}
{\vtop{\hbox{
$\pmatrix{
1&1&4&0&0&10\cr
0&1&1&0&0&3\cr
0&0&1&0&0&1\cr}$}}
}

\newproblem
Find the inverse of the following matrix, if it exists.
$$\pmatrix{
1&1&0\cr
0&2&1\cr
1&1&1\cr}$$
\answers
{$\pmatrix{
1&1&0\cr
0&1\over2&1\cr
1&1&1\cr}$}
{$\pmatrix{
1&-1&0\cr
0&-{1\over2}&1\cr
1\over2&-{1\over2}&1\over2\cr}$\endanswerline}
{$\pmatrix{
1&-1&1\cr
1&1&-1\cr
-{1\over2}&0&1\over2\cr}$}
{$\pmatrix{
1\over2&-{1\over2}&1\over2\cr
1\over2&1\over2&-{1\over2}\cr
-1&0&1\cr}$\endanswerline}
{{\sl Does not exist}}

\newproblem
Calculate the determinant of the matrix
$$\pmatrix{
1&2&2\cr
2&1&2\cr
1&2&2}$$
\answers
{0}
{2}
{4}
{8}
{16}
\newproblem
Use expansion by cofactors to find the determinant of the matrix
$$\pmatrix{
-1&5&3&-1\cr
2&1&4&1\cr
6&0&-1&0\cr
0&0&2&0\cr}$$
\answers
{$0$}
{$-8$}
{$-24$}
{$60$}
{$-72$}

\newproblem
Calculate the (1,2)-minor, $M_{(1,2)}$, of the matrix
$$\pmatrix{
3&1&2\cr
5&2&7\cr
1&4&2\cr}$$
\answers
{$-2$}
{3}
{4}
{5}
{$-6$}

\newproblem
Express $B = (1,2,3)$ as a linear combination of the
vectors $V_1 = (1,3,4)$, $V_2 = (1,0,-1)$, and $V_3 =
(0,-2,-4)$, if possible, by finding constants $d_1$, $d_2$,
$d_3$ such that $B = d_1V_1 + d_2V_2 + d_3V_3$.
\answers
{$d_1 = 0$, $d_2 = 1$, $d_3 = -1$}
{$d_1 = 2$, $d_2 = -2$, $d_3 = 0$\endanswerline}
{$d_1 = 1$, $d_2 = -5$, $d_3 = 4$}
{$d_1 = -1$, $d_2 = 0$, $d_3 = 1$\endanswerline}
{{\sl No solution exists}}

\newproblem
Find the rank of the matrix
$$\pmatrix{
1&2&7&4\cr
0&1&3&7\cr
0&0&0&0\cr
1&0&0&0\cr}$$
\answers
{0}
{1}
{2}
{3}
{4}

\newproblem
Which of the following statements are always true:

{\leavevmode\parindent = 20pt
\item{(A)}\enspace If $V_1$, $V_2$, $V_3$, $V_4$, $V_5$
are 4-dimensional vectors, then they are linearly
dependent.

\item{(B)}\enspace If $V_1$, $V_2$, $V_3$, $V_4$
are linearly dependent, then $V_1$, $V_2$, $V_3$ are
linearly dependent.

\item{(C)}\enspace If $V_1$, $V_2$, $V_3$, $V_4$
are linearly independent, then $V_1$, $V_2$, $V_3$ are
linearly independent.

\item{(D)}\enspace If $V_1$, $V_2$, $V_3$
are linearly dependent, then $V_1$, $V_2$, $V_3$, $V_4$ are
linearly dependent.
\medskip}
\answers
{A, B, and C only}
{A, C, and D only}
{B, C, and D only\endanswerline}
{C, and D only}
{A, B, C, and D}

\newproblem
Find all solutions to the equation $AX = 0$ where
$$A = \pmatrix{1&-1\cr2&-2\cr},\qquad
X = \pmatrix{x_1\cr x_2\cr}$$
\answers
{$\pmatrix{0\cr0\cr}$}
{$\pmatrix{1\cr1\cr}$}
{$t\pmatrix{1\cr1\cr}$,\quad$t$ arbitrary\endanswerline}
{$t\pmatrix{1\cr0\cr} + s\pmatrix{0\cr1\cr}$,\quad$t$, $s$
arbitrary}
{{\sl no solutions exist}}

\newproblem
Determine a maximal subset of linearly
independent vectors from the vectors
$$V_1 = (0,0,0,0)\quad V_2 =(1,0,0,0)\quad V_3 =
(3,2,0,0)\quad V_4 = (5,4,3,0)$$
\answers
{$V_2$}
{$V_2$, $V_3$}
{$V_2$, $V_4$\endanswerline}
{$V_1$, $V_2$, $V_3$}
{$V_2$, $V_3$, $V_4$}

\newproblem
Use the Gram-Schmidt process to produce an
orthonormal set of vectors from the following vectors:
$X_1 =(1,0,0,1)$, $X_2 = (0,2,0,1)$, $X_3 = (0,1,1,0)$.
\answers
{\vtop{
\hbox{$Q_1 =(1,0,0,1)$}
\hbox{$Q_2 ={\sqrt{6}\over6}(-1,2,0,1)$}
\hbox{$Q_3={\sqrt{89}\over89}(2,0,9,-2)$}
\bigskip
}}
{\vtop{
\hbox{$Q_1 ={\sqrt{2}\over2}(1,0,0,1)$}
\hbox{$Q_2 ={\sqrt{2}\over6}(-1,4,0,1)$}
\hbox{$Q_3={\sqrt{10}\over30}(2,1,9,-2)$}
}}
{\vtop{
\hbox{$Q_1 ={\sqrt{2}\over2}(1,0,0,1)$}
\hbox{$Q_2 ={\sqrt{5}\over5}(0,2,0,1)$}
\hbox{$Q_3={\sqrt{2}\over2}(0,1,1,0)$}
}\endanswerline}
{\vtop{
\hbox{$Q_1 ={\sqrt{2}\over2}(1,0,0,1)$}
\hbox{$Q_2 ={\sqrt{6}\over6}(-1,2,0,1)$}
\hbox{$Q_3={1\over2}(1,1,1,-1)$}
}}
{\vtop{
\hbox{$Q_1 ={\sqrt{2}\over2}(1,0,0,1)$}
\hbox{$Q_2 ={\sqrt{6}\over6}(-1,4,0,1)$}
\hbox{$Q_3 ={\sqrt{10}\over10}(-2,-1,1,2)$}
}}

\newproblem
Suppose $y = y(x)$ is a function which satisfies the differential equation $y' +
e^x y = e^x$ and $y(0) = 0$. Solve the equation to find $y(1)$.
\answers
{$1$}
{$1 - e^e$}
{$e^{1\over e}$}
{$1 - e^{-e}$}
{$1 - e^{1-e}$}

\bye