
% extra credit ch3 2250, S2015
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{\large\sf Problem #1}. {\bf #2}\par}
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\begin{document}
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 {\bf Name~}\underline{~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~}\hfill
 {Class Time~}\underline{~~~~~~~~}\newline


\begin{center}
\bf Math 2250 Extra Credit Problems \\
\bf Chapter 3 \\
{\small\bf S2015}
\end{center}

{\bf Submitted work}. Please submit one stapled package with this sheet
on top. Kindly check-mark the problems submitted and
label the paper \fbox{\bf Extra Credit}. Label each solved problem with
its corresponding problem number, e.g., \fbox{Xc10.3-20}.

{\bf Definition}. A {\bf frame sequence} is a sequence of displayed
equations, each display having the same number of equations and
unknowns. Each displayed set is required to be the result of the
previous display by applying exactly one fundamental operation: swap,
combination, multiply. The same terminology applies to matrices which
represent a system of linear algebraic equations.

\Prob{Xc3.1-16}{(Elimination)}
Solve the system below by hand using frame sequences and report one of
the three possibilities: no solution, unique solution, infinitely many
solutions.

$$
\begin{array}{rcrcrcl}
x &+& 5y &+& 6z &=& 3, \\
5x &+& 2y &-& 10z &=& 1, \\
8x &+& 17y &+& 8z &=& 5. \\
\end{array}
$$

\Prob{Xc3.1-26}{(Systems of equations)}
Give an example of a $3\times 3$ system of equations which illustrates
three planes, two of which intersect in a line, and that line lies
entirely in the third plane.


\Prob{Xc3.2-14}{(Echelon systems)}
Solve the system below using frame sequences and report one of the three
possibilities: no solution, unique solution, infinitely many solutions.
Use variable list order $x$, $y$, $z$, $w$.

$$
\begin{array}{rcrcrcrcl}
3x &-& 6y &+& z  &+& 13w &=& 15, \\
3x &-& 6y &+& 3z &+& 21w &=& 21, \\
2x &-& 4y &+& 5z &+& 26w &=& 23. \\
\end{array}
$$

\Prob{Xc3.2-24}{(Three possibilities with symbols)}
Solve the system below for all values of $a$, $b$ using frame sequences
and report one of the three possibilities: no solution, unique solution,
infinitely many solutions. If the system has a solution, then report the
general solution.

$$
\begin{array}{rcrcrcrcl}
x &+& ay &=& b, \\
ax &+& (a-b)y &=& a.
\end{array}
$$

\vfill\newpage

\Prob{Xc3.3-10}{(RREF)}
Show the frame sequence steps to $\rref(A)$ and attach a technology answer
check. Alternatively, do the whole problem in a workbench or computer algebra system.

$$
A=\left(\begin{array}{rrrr}
1 & -4 & -2 \\
3 & -12 & 1 \\
2 & -8 & 5
\end{array}
\right)
$$

\Prob{Xc3.3-20}{(RREF)}
Show the frame sequence steps to $\rref(A)$ and attach an answer
check. Alternative: do the whole problem using techology.

$$
A=\left(\begin{array}{rrrrrr}
1 & -4 & -2 & 4 & 0 \\
3 & -12 & 1 & 5 & 0 \\
2 & -8 & 5  & 5 & 1
\end{array}
\right)
$$

\Prob{Xc3.4-20}{(Vector general solution)}
Find the general solution in vector form $\.x$, expressed as a
linear combination of column vectors using symbols $t_1$, $t_2$, $t_3$
\ldots (as many symbols as needed for the free variables). 

$$
\begin{array}{rcrcrcrcrcrcrcrc}
x_1 &-& x_2 & &     &+& 7x_4 &+& 3x_5 &=& 0, \\
    & &     & & x_3 &-& x_4  &-& 2x_5 &=& 0, \\
    & &     & &     & &      & &  0  &=& 0, \\
    & &     & &     & &      & &  0  &=& 0, \\
    & &     & &     & &      & &  0  &=& 0. \\
\end{array}
$$


\Prob{Xc3.4-40}{(Superposition)}
~\newline
(a) Add the two systems below to prove that sums of solutions are again
solutions. You will show that $\.x=\colvector{x_1+x_2\\ y_1+y_2}$ is a
solution, given that 
$\colvector{x_1\\ y_1}$
and
$\colvector{x_2\\ y_2}$
are solutions of the homogeneous equation.

$$
\left\{
\begin{array}{rcrcrcrcl}
ax_1 &+& by_1 &=& 0, \\
cx_1 &+& dy_1 &=& 0.
\end{array}
\right.\quad\quad
\left\{
\begin{array}{rcrcrcrcl}
ax_2 &+& by_2 &=& 0, \\
cx_2 &+& dy_2 &=& 0.
\end{array}
\right.
$$

~\newline
(b) Add the two systems below to prove the superposition principle.
You will show that $\.x=\colvector{x_1+x_3\\ y_1+y_3}$ is a
solution, given that 
$\colvector{x_1\\ y_1}$
solves the homogeneous problem 
and
$\colvector{x_3\\ y_3}$
solves the non-homogeneous problem.

$$
\left\{
\begin{array}{rcrcrcrcl}
ax_1 &+& by_1 &=& 0, \\
cx_1 &+& dy_1 &=& 0.
\end{array}
\right.\quad\quad
\left\{
\begin{array}{rcrcrcrcl}
ax_3 &+& by_3 &=& e, \\
cx_3 &+& dy_3 &=& f.
\end{array}
\right.
$$

\Prob{Xc3.5-16}{(Inverse by frame sequence)}
Calculate the frame sequence from $C=\aug(A,I)$ to $\rref(C)$ and
report $A^{-1}$. Perform a hand answer check for the inverse matrix. No
technology please, all with pencil and paper.

$$A=\threebythree{1 & -2 & 2}{3 & 0 & 1}{1 & -1 & 2}$$


\Prob{Xc3.5-44a}{(Inverses and frame sequences I)}
~\newline
(a) Suppose $A$ is $8\times 8$ and $60$ entries are ones. Explain why $A^{-1}$
does not exist.

(b) Suppose that $A$ is invertible and $3\times 3$. A frame sequence is started
with $A$ and gives final frame (not the $\rref$)
$$\threebythree{1 & 1 & 2}{ 0 & 1 & -1}{0 & 0 & 3}$$
The steps used to arrive at the final frame are 
(1) {\tt combo(1,2,-3)},
(2) {\tt swap(2,3)},
(3) {\tt combo(1,2,-1)},
(4) {\tt combo(2,3,1)},
(5) {\tt mult(2,-1)}.
Find the matrix $A$.


\Prob{Xc3.5-44b}{(Inverses and frame sequences II)}
~\newline
Invent a particular $3\times 3$ invertible matrix $A_1$ and display a
frame sequence $A_1$ to $A_6$ (or slightly longer) involving documented
steps of combo, swap and mult (one of each at least). Then write the
frame sequence in the form

                $$A_6 = E_5 E_4 E_3 E_2 E_1 A_1$$

where $E_1, \ldots, E_5$ are the elementary matrices representing the combo,
swap and mult operations. Finally, check your answer by multiplying
out the right side of the above identity, showing the multiplication gives
$A_6$ (which should be $\rref(A_1)=I$).

{\bf Example}. The same problem but for $2\times 2$ matrix $A_1$.

\begin{small}
 $A_1 = \twobytwo{0&3}{2&4}$

 $A_2 =\twobytwo{0&3}{1&2}$, {\tt mult(2,1/2)}, $E_1 = \twobytwo{1&0}{0&1/2}$
 
 $A_3=\twobytwo{1&2}{0&3}$, {\tt swap(1,2)}, $ E_2 = \twobytwo{0&1}{1&0}$

 $A_4 = \twobytwo{1&2}{0&1}$, {\tt mult(2,1/3)}, $E_3=\twobytwo{1&0}{0&1/3}$

 $A_5 = \twobytwo{1&0}{0&1}$, {\tt combo(2,1,-2)}, $E_4=\twobytwo{1&-2}{0&1}$

Then 
     $$\begin{array}{lcl}
     A_5 &=& E_4 A_4 \\
         &=& E_4 E_3 A_3 \\
         &=& E_4 E_3 E_2 A_2 \\
         &=& E_4 E_3 E_2 E_1 A_1
     \end{array}$$

Multiply out the four elementary matrices by hand to get
%
%                      [-2/3  1/2]
%        E4 E3 E2 E1 = [ 1/3   0 ]
%
      $$ E_4 E_3 E_2 E_1 = \twobytwo{-2/3&1/2}{1/3&0}$$
and then
%
%                         [1 0]
%        E4 E3 E2 E1 A1 = [0 1] = I
%
      $$ E_4 E_3 E_2 E_1 A_1 = \twobytwo{1&0}{0&1} = I$$
 
Matlab, Maple and Mathematica can do the answer check.
Maple uses this plan:
\begin{tabular}{l}
\verb|  A1:=Matrix([[0,3],[2,4]]);|\\ 
\verb|  E1:=Matrix([[1,0],[0,1/2]]);|\\ 
\vdots \\
\verb|  E4.E3.E2.E1.A1;|
\end{tabular}
The last line gives the identity $A_5 = E_4 E_3 E_2 E_1 A_1$, which
 completes the answer check. 

\end{small}



\Prob{Xc3.6-6}{(Determinants and the four rules)}
Calculate $\det(A)$ using only the four rules {\em triang, swap, combo,
mult}. Check the answer using technology.

$$
A=\left(\begin{array}{rrrrrr}
1 & -4 & -2 & 4 & 0 \\
3 & -12 & 1 & 5 & 0 \\
2 & -8 & 5  & 5 & 1 \\
0 & -8 & 5  & 5 & 1 \\
0 & 0 & 5  & 5 & 1
\end{array}
\right)
$$

\Prob{Xc3.6-20}{(Determinants, hybrid rules)}
Calculate $\det(A)$ using the four rules {\em triang, swap, combo,
mult} plus the cofactor rule. Check the answer using technology.

$$
A=\left(\begin{array}{rrrrrr}
1 & -4 & -2 & 4 & 0 \\
3 & -12 & 1 & 5 & 0 \\
0 & -12 & 0 & 5 & 0 \\
0 & -12 & 1 & 0 & 0 \\
2 & -8 & 5  & 5 & 1
\end{array}
\right)
$$




\Prob{Xc3.6-32}{(Cramer's Rule)}
Calculate $x$, $y$ and $z$ using Cramer's rule. Check the answer with technology.

$$\threebythree{1 & -2 & 2}{3 & 0 & 1}{1 & -1 & 2}\colvector{x\\ y\\ z} =
\colvector{1\\ 2\\ 3}$$


\Prob{Xc3.6-40}{(Adjugate formula)}
Find the inverse of the matrix $A$ using the formula $A^{-1} = \frac{\mbox{adjugate}}{\mbox{determinant}}$.

$$
A=\left(\begin{array}{rrrrrr}
1 & -4 & -2 & 4 \\
3 & -1 & 1 & 5 \\
0 & -1 & 0 & 1 \\
2 & 0 & -1  & 0
\end{array}
\right)
$$

% A:=matrix([[1 , -4 , -2 , 4],
% [3 , -1 , 1 , 5],
% [0 , -1 , 0 , 1],
% [2 , 0 , -1  , 0]]);
% 
% matrix([[-1/3, 0, 4/3, 2/3], 
%         [5/12, 1/4, -35/12, -7/12], 
%         [-2/3, 0, 8/3,1/3], 
%         [5/12, 1/4, -23/12, -7/12]])


\Prob{Xc3.6-40}{(Adjugate formula)}
Find the entry in row 4 and column 2 of the adjugate matrix for $A$,
using only determinants.



$$
A=\left(\begin{array}{rrrrrr}
1 & -4 & -2 & 4 \\
3 & -1 & -1 & 3 \\
0 & -1 & 0 & 1 \\
2 & 0 & -1  & 0
\end{array}
\right)
$$


\Prob{Xc3.6-60}{(Induction)}
Assume that $B_1=1$ and $B_2=2$. 
Assume $B_{k+2}=2B_k+B_{k+1}$
for each integer $k=1,2,3,\dots$.

Let ${\cal Q}_n$ denote the statement that $B_{k}=2^{k-1}$
for $1 \le k \le n$.  Prove by mathematical induction that all statements 
${\cal Q}_n$ are true.

Problem note: You must prove that ${\cal Q}_1$ and ${\cal Q}_2$ are true, individually.
Mathematical induction then applies to the sequence of statements ${\cal
Q}_3$, ${\cal Q}_4$, \ldots,
in short, to statements ${\cal P}_j={\cal Q}_{j+2}$, $j=1,2,3, \ldots$.

\bigskip

{\bf End of extra credit problems chapter 3.}

\end{document}
