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Mathematics LibreTexts

8.E: Systems of Equations and Matrices (Exercises)

 

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8.1: Systems of Linear Equations: Gaussian Elimination

\subsection{Exercises}

(Review Exercises) In Exercises \ref{reviewsystemfirst} - \ref{reviewsystemlast}, take a trip down memory lane and solve the given system using substitution and/or elimination. Classify each system as consistent independent, consistent dependent, or inconsistent. Check your answers both algebraically and graphically.

\begin{multicols}{2}

\begin{enumerate}

\item $\left\{ \begin{array}{rcr} x+2y & = & 5 \\ x & = & 6 \end{array} \right.$ \label{reviewsystemfirst}

\item $\left\{ \begin{array}{rcr} 2y-3x & = & 1 \\ y & = & -3 \end{array} \right.$

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\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

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\item $\left\{ \begin{array}{rcr} \frac{x+2y}{4} & = & -5 \\[5pt] \frac{3x-y}{2} & = & 1 \end{array} \right.$

\item $\left\{ \begin{array}{rcr} \frac{2}{3} x-\frac{1}{5}y & = & 3 \\[5pt] \frac{1}{2}x+\frac{3}{4}y& = & 1 \end{array} \right.$

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\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

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\item $\left\{ \begin{array}{rcr} \frac{1}{2}x-\frac{1}{3}y & = & -1 \\ [5pt] 2y-3x & = & 6 \end{array} \right.$

\item $\left\{ \begin{array}{rcr} x+4y & = & 6 \\ [5pt] \frac{1}{12}x+\frac{1}{3}y& = & \frac{1}{2} \end{array} \right.$

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\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

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\item $\left\{ \begin{array}{rcr} 3y-\frac{3}{2}x & = & -\frac{15}{2} \\ [5pt] \frac{1}{2}x-y & = & \frac{3}{2} \end{array} \right.$

\item $\left\{ \begin{array}{rcr} \frac{5}{6}x+\frac{5}{3}y & = & -\frac{7}{3} \\ [5pt] -\frac{10}{3}x-\frac{20}{3}y & = & 10 \end{array} \right.$ \label{reviewsystemlast}

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\end{enumerate}

\end{multicols}

In Exercises \ref{triangfirst} - \ref{trianglast}, put each system of linear equations into triangular form and solve the system if possible. Classify each system as consistent independent, consistent dependent, or inconsistent.

\begin{multicols}{2}

\begin{enumerate}

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\item $\left\{ \begin{array}{rcr} -5x + y & = & 17 \\ x + y & = & 5 \end{array} \right.$ \label{triangfirst}

\item $\left\{ \begin{array}{rcr} x + y + z & = & 3 \\ 2x - y + z & = & 0 \\ -3x + 5y + 7z & = & 7 \end{array} \right.$

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\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

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\item \label{dependentsystemmuliple} $\left\{ \begin{array}{rcr} 4x - y + z & = & 5 \\ 2y + 6z & = & 30 \\ x + z & = & 5 \end{array} \right.$

\item $\left\{ \begin{array}{rcr} 4x - y + z & = & 5 \\ 2y + 6z & = & 30 \\ x + z & = & 6 \end{array} \right.$

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\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

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\item $\left\{ \begin{array}{rcr} x + y + z & = & -17 \\ y - 3z & = & 0 \end{array} \right.$

\item $\left\{ \begin{array}{rcr} x-2y+3z & = & 7 \\ -3x+y+2z & = & -5 \\ 2x+2y+z & = & 3 \end{array} \right.$

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\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

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\item $\left\{ \begin{array}{rcr} 3x-2y+z & = & -5 \\ x+3y-z & = & 12 \\ x+y+2z & = & 0 \end{array} \right.$

\item $\left\{ \begin{array}{rcr} 2x-y+z& = & -1 \\ 4x+3y+5z & = & 1 \\ 5y+3z & = & 4 \end{array} \right.$

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\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

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\item $\left\{ \begin{array}{rcr} x-y+z & = & -4 \\ -3x+2y+4z & = & -5 \\ x-5y+2z & = & -18 \end{array} \right.$

\item $\left\{ \begin{array}{rcr} 2x-4y+z & = & -7 \\ x-2y+2z & = & -2 \\ -x+4y-2z & = & 3 \end{array} \right.$

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\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

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\item $\left\{ \begin{array}{rcr} 2x-y+z & = & 1 \\ 2x+2y-z & = & 1 \\ 3x+6y+4z & = & 9 \end{array} \right.$

\item $\left\{ \begin{array}{rcr} x-3y-4z & = & 3 \\ 3x+4y-z & = & 13 \\ 2x-19y-19z & = & 2 \end{array} \right.$

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\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

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\item $\left\{ \begin{array}{rcr} x+y+z & = & 4 \\ 2x-4y-z& = & -1 \\ x-y & = & 2 \end{array} \right.$

\item $\left\{ \begin{array}{rcr} x-y+z & = & 8 \\ 3x+3y-9z & = & -6 \\ 7x-2y+5z & = & 39 \end{array} \right.$

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\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

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\item $\left\{ \begin{array}{rcr} 2x-3y+z & = & -1 \\ 4x-4y+4z & = & -13 \\ 6x-5y+7z & = & -25 \end{array} \right.$

\item $\left\{ \begin{array}{rcr} 2x_{\mbox{\tiny$1$}} + x_{\mbox{\tiny$2$}} - 12x_{\mbox{\tiny$3$}} - x_{\mbox{\tiny$4$}} & = & 16 \\

-x_{\mbox{\tiny$1$}} + x_{\mbox{\tiny$2$}} + 12x_{\mbox{\tiny$3$}} - 4x_{\mbox{\tiny$4$}} & = & -5 \\

3x_{\mbox{\tiny$1$}} + 2x_{\mbox{\tiny$2$}} - 16x_{\mbox{\tiny$3$}} - 3x_{\mbox{\tiny$4$}} & = & 25 \\

x_{\mbox{\tiny$1$}} + 2x_{\mbox{\tiny$2$}} - 5x_{\mbox{\tiny$4$}} & = & 11 \end{array} \right.$

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\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

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\item $\left\{ \begin{array}{rcr} x_{\mbox{\tiny$1$}} - x_{\mbox{\tiny$3$}} & = & -2 \\

2x_{\mbox{\tiny$2$}} - x_{\mbox{\tiny$4$}} & = & 0 \\

x_{\mbox{\tiny$1$}} - 2x_{\mbox{\tiny$2$}} + x_{\mbox{\tiny$3$}} & = & 0 \\

-x_{\mbox{\tiny$3$}} + x_{\mbox{\tiny$4$}} & = & 1 \end{array} \right.$

\item $\left\{ \begin{array}{rcr} x_{\mbox{\tiny$1$}} - x_{\mbox{\tiny$2$}} - 5x_{\mbox{\tiny$3$}} + 3x_{\mbox{\tiny$4$}} & = & -1 \\

x_{\mbox{\tiny$1$}} + x_{\mbox{\tiny$2$}} + 5x_{\mbox{\tiny$3$}} - 3x_{\mbox{\tiny$4$}} & = & 0 \\

x_{\mbox{\tiny$2$}} + 5x_{\mbox{\tiny$3$}} - 3x_{\mbox{\tiny$4$}} & = & 1 \\

x_{\mbox{\tiny$1$}} - 2x_{\mbox{\tiny$2$}} - 10x_{\mbox{\tiny$3$}} + 6x_{\mbox{\tiny$4$}} & = & -1 \end{array} \right.$ \label{trianglast}

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\end{enumerate}

\end{multicols}

\begin{enumerate}

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\item Find two other forms of the parametric solution to Exercise \ref{dependentsystemmuliple} above by reorganizing the equations so that $x$ or $y$ can be the free variable.

\item A local buffet charges $\$7.50$ per person for the basic buffet and $\$9.25$ for the deluxe buffet (which includes crab legs.) If 27 diners went out to eat and the total bill was $\$227.00$ before taxes, how many chose the basic buffet and how many chose the deluxe buffet?

\item At The Old Home Fill'er Up and Keep on a-Truckin' Cafe, Mavis mixes two different types of coffee beans to produce a house blend. The first type costs \$3 per pound and the second costs \$8 per pound. How much of each type does Mavis use to make 50 pounds of a blend which costs \$6 per pound?

\item Skippy has a total of $\$$10,000 to split between two investments. One account offers $3\%$ simple interest, and the other account offers $8\%$ simple interest. For tax reasons, he can only earn $\$500$ in interest the entire year. How much money should Skippy invest in each account to earn $\$500$ in interest for the year?

\item A $10 \%$ salt solution is to be mixed with pure water to produce 75 gallons of a $3\%$ salt solution. How much of each are needed?

\item \label{herbalteablend} At The Crispy Critter's Head Shop and Patchouli Emporium along with their dried up weeds, sunflower seeds and astrological postcards they sell an herbal tea blend. By weight, Type I herbal tea is 30\% peppermint, 40\% rose hips and 30\% chamomile, Type II has percents 40\%, 20\% and 40\%, respectively, and Type III has percents 35\%, 30\% and 35\%, respectively. How much of each Type of tea is needed to make 2 pounds of a new blend of tea that is equal parts peppermint, rose hips and chamomile?

\item Discuss with your classmates how you would approach Exercise \ref{herbalteablend} above if they needed to use up a pound of Type I tea to make room on the shelf for a new canister.

\item If you were to try to make 100 mL of a $60\%$ acid solution using stock solutions at $20\%$ and $40\%$, respectively, what would the triangular form of the resulting system look like? Explain.

\end{enumerate}

\newpage

\subsection{Answers}

\begin{multicols}{2}

\begin{enumerate}

\item Consistent independent \\

Solution $\left(6, -\frac{1}{2}\right)$

\item Consistent independent \\

Solution $\left(-\frac{7}{3}, -3\right)$

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\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

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\item Consistent independent \\

Solution $\left(-\frac{16}{7}, -\frac{62}{7}\right)$

\item Consistent independent \\

Solution $\left(\frac{49}{12}, -\frac{25}{18}\right)$

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\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

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\item Consistent dependent\\

Solution $\left(t, \frac{3}{2}t+3\right)$ \\

for all real numbers $t$

\item Consistent dependent\\

Solution $\left(6-4t, t\right)$ \\

for all real numbers $t$

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\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

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\item Inconsistent \\

No solution

\item Inconsistent \\

No solution

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\end{enumerate}

\end{multicols}

Because triangular form is not unique, we give only one possible answer to that part of the question. Yours may be different and still be correct.

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\begin{multicols}{2} \raggedcolumns

\item $\left\{ \begin{array}{rcr} x + y & = & 5 \\ y & = & 7 \end{array} \right.$

Consistent independent\\

Solution $(-2, 7)$

\end{multicols}

\begin{multicols}{2} \raggedcolumns

\item $\left\{ \begin{array}{rcr} x - \frac{5}{3}y - \frac{7}{3}z & = & -\frac{7}{3} \\ [3pt] y + \frac{5}{4}z & = & 2 \\ z & = & 0 \end{array} \right.$

Consistent independent\\

Solution $(1, 2, 0)$

\end{multicols}

\begin{multicols}{2} \raggedcolumns

\item $\left\{ \begin{array}{rcr} x - \frac{1}{4}y + \frac{1}{4}z & = & \frac{5}{4} \\ [3pt] y + 3z & = & 15 \\ 0 & = & 0 \end{array} \right.$

Consistent dependent\\

Solution $(-t + 5, -3t + 15, t)$\\

for all real numbers $t$

\end{multicols}

\begin{multicols}{2} \raggedcolumns

\item $\left\{ \begin{array}{rcr} x - \frac{1}{4}y + \frac{1}{4}z & = & \frac{5}{4} \\ [3pt] y + 3z & = & 15 \\ 0 & = & 1 \end{array} \right.$

Inconsistent\\

No solution

\end{multicols}

\begin{multicols}{2} \raggedcolumns

\item $\left\{ \begin{array}{rcr} x + y + z & = & -17 \\ y - 3z & = & 0 \end{array} \right.$

\vspace{.25in}

Consistent dependent\\

Solution $(-4t - 17, 3t, t)$\\

for all real numbers $t$

\end{multicols}

\begin{multicols}{2} \raggedcolumns

\item $\left\{ \begin{array}{rcr} x-2y+3z & = & 7 \\ y - \frac{11}{5}z & = & -\frac{16}{5} \\ z & = & 1 \\ \end{array} \right.$

\vspace{.25in}

Consistent independent\\

Solution $(2,-1,1)$

\end{multicols}

\begin{multicols}{2} \raggedcolumns

\item $\left\{ \begin{array}{rcr} x+y+2z & = & 0 \\ y - \frac{3}{2}z & = & 6 \\ z & = & -2 \\ \end{array} \right.$

\vspace{.25in}

Consistent independent\\

Solution $(1,3,-2)$

\end{multicols}

\begin{multicols}{2} \raggedcolumns

\item $\left\{ \begin{array}{rcr} x - \frac{1}{2} y + \frac{1}{2} z & = & -\frac{1}{2} \\ [3pt] y + \frac{3}{5} z & = & \frac{3}{5} \\ 0 & = & 1 \\ \end{array} \right.$

\vspace{.25in}

Inconsistent\\

no solution

\end{multicols}

\begin{multicols}{2} \raggedcolumns

\item $\left\{ \begin{array}{rcr} x-y+z & = & -4 \\ y - 7z & = & 17 \\ z & = & -2 \\ \end{array} \right.$

\vspace{.25in}

Consistent independent\\

Solution $(1,3,-2)$

\end{multicols}

\begin{multicols}{2} \raggedcolumns

\item $\left\{ \begin{array}{rcr} x-2y+2z & = & -2 \\ y & = & \frac{1}{2} \\ z & = & 1 \\ \end{array} \right.$

\vspace{.25in}

Consistent independent\\

Solution $\left(-3,\frac{1}{2},1\right)$

\end{multicols}

\begin{multicols}{2} \raggedcolumns

\item $\left\{ \begin{array}{rcr} x-\frac{1}{2} y+\frac{1}{2} z & = & \frac{1}{2} \\ [3pt] y - \frac{2}{3} z & = & 0 \\ z & = & 1 \\ \end{array} \right.$

\vspace{.25in}

Consistent independent\\

Solution $\left(\frac{1}{3},\frac{2}{3},1\right)$

\end{multicols}

\begin{multicols}{2} \raggedcolumns

\item $\left\{ \begin{array}{rcr} x-3y-4z & = & 3 \\ y + \frac{11}{13} z & = & \frac{4}{13} \\ 0 & = & 0 \\ \end{array} \right.$

\vspace{.25in}

Consistent dependent\\

Solution $\left(\frac{19}{13} t + \frac{51}{13},-\frac{11}{13} t+\frac{4}{13},t\right)$\\

for all real numbers $t$

\end{multicols}

\begin{multicols}{2} \raggedcolumns

\item $\left\{ \begin{array}{rcr} x+y+z & = & 4 \\ y + \frac{1}{2} z & = & \frac{3}{2} \\ 0 & = & 1 \\ \end{array} \right.$

\vspace{.25in}

Inconsistent\\

no solution

\end{multicols}

\begin{multicols}{2} \raggedcolumns

\item $\left\{ \begin{array}{rcr} x- y + z & = & 8 \\ y -2z & = & -5 \\ z & = & 1 \\ \end{array} \right.$

\vspace{.25in}

Consistent independent\\

Solution $\left(4,-3,1\right)$

\end{multicols}

\begin{multicols}{2} \raggedcolumns

\item $\left\{ \begin{array}{rcr} x- \frac{3}{2} y + \frac{1}{2} z & = & -\frac{1}{2} \\[3pt] y + z & = & -\frac{11}{2} \\ 0 & = & 0 \\ \end{array} \right.$

\vspace{.25in}

Consistent dependent\\

Solution $\left(-2t - \frac{35}{4},-t - \frac{11}{2},t\right)$\\

for all real numbers $t$

\end{multicols}

\begin{multicols}{2} \raggedcolumns

\item $\left\{ \begin{array}{rcr} x_{\mbox{\tiny$1$}} + \frac{2}{3}x_{\mbox{\tiny$2$}} - \frac{16}{3}x_{\mbox{\tiny$3$}} - x_{\mbox{\tiny$4$}} & = & \frac{25}{3} \\ [3pt]

x_{\mbox{\tiny$2$}} + 4x_{\mbox{\tiny$3$}} - 3x_{\mbox{\tiny$4$}} & = & 2 \\

0 & = & 0 \\

0 & = & 0 \end{array} \right.$

Consistent dependent\\

Solution $(8s - t + 7, -4s + 3t + 2, s, t)$\\

for all real numbers $s$ and $t$

\end{multicols}

\begin{multicols}{2} \raggedcolumns

\item $\left\{ \begin{array}{rcr} x_{\mbox{\tiny$1$}} - x_{\mbox{\tiny$3$}} & = & -2 \\ [3pt]

x_{\mbox{\tiny$2$}} - \frac{1}{2}x_{\mbox{\tiny$4$}} & = & 0 \\ [3pt]

x_{\mbox{\tiny$3$}} - \frac{1}{2} x_{\mbox{\tiny$4$}} & = & 1 \\ [3pt]

x_{\mbox{\tiny$4$}} & = & 4 \end{array} \right.$

Consistent independent\\

Solution $(1, 2, 3, 4)$

\end{multicols}

\begin{multicols}{2} \raggedcolumns

\item $\left\{ \begin{array}{rcr} x_{\mbox{\tiny$1$}} - x_{\mbox{\tiny$2$}} - 5x_{\mbox{\tiny$3$}} + 3x_{\mbox{\tiny$4$}} & = & -1 \\

x_{\mbox{\tiny$2$}} + 5x_{\mbox{\tiny$3$}} - 3x_{\mbox{\tiny$4$}} & = & \frac{1}{2} \\

0 & = & 1 \\

0 & = & 0 \end{array} \right.$

Inconsistent\\

No solution

\end{multicols}

\item If $x$ is the free variable then the solution is $(t, 3t, -t + 5)$ and if $y$ is the free variable then the solution is $\left(\frac{1}{3}t, t, -\frac{1}{3}t + 5\right)$.

\item $13$ chose the basic buffet and $14$ chose the deluxe buffet.

\item Mavis needs 20 pounds of \$3 per pound coffee and 30 pounds of \$8 per pound coffee.

\item Skippy needs to invest $\$$6000 in the $3\%$ account and $\$$4000 in the $8 \%$ account.

\item $22.5$ gallons of the $10 \%$ solution and $52.5$ gallons of pure water.

\item $\frac{4}{3}- \frac{1}{2}t$ pounds of Type I, $\frac{2}{3} - \frac{1}{2}t$ pounds of Type II and $t$ pounds of Type III where $0 \leq t \leq \frac{4}{3}$.

\end{enumerate}

\closegraphsfile

8.2: Systems of Linear Equations: Augmented Matrices

\subsection{Exercises}

In Exercises \ref{rreffirst} - \ref{rreflast}, state whether the given matrix is in reduced row echelon form, row echelon form only or in neither of those forms.

\begin{multicols}{3}

\begin{enumerate}

\item $\left[ \begin{array}{rr|r}

1 & 0 & 3 \\

0 & 1 & 3 \\

\end{array} \right]$ \label{rreffirst}

\item $\left[ \begin{array}{rrr|r}

3 & -1 & \hphantom{-}1 & 3 \\

2 & -4 & 3 & 16 \\

1 & -1 & 1 & 5 \\

\end{array} \right]$

\item $\left[ \begin{array}{rrr|r}

1 & 1 & 4 & 3 \\

0 & 1 & 3 & 6 \\

0 & 0 & 0 & 1 \\

\end{array} \right]$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{3}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\left[ \begin{array}{rrr|r}

1 & 0 & 0 & 0 \\

0 & 1 & 0 & 0 \\

0 & 0 & 0 & 1 \\

\end{array} \right]$

\item $\left[ \begin{array}{rrrr|r}

1 & 0 & 4 & 3 & 0 \\

0 & 1 & 3 & 6 & 0 \\

0 & 0 & 0 & 0 & 0

\end{array} \right]$

\item $\left[ \begin{array}{rrr|r}

1 & 1 & 4 & 3 \\

0 & 1 & 3 & 6 \\

\end{array} \right]$ \label{rreflast}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

In Exercises \ref{decodefirst} - \ref{decodelast}, the following matrices are in reduced row echelon form. Determine the solution of the corresponding system of linear equations or state that the system is inconsistent.

\begin{multicols}{3}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\left[ \begin{array}{rr|r}

1 & 0 & -2 \\

0 & 1 & 7 \\

\end{array} \right]$ \label{decodefirst}

\item $\left[ \begin{array}{rrr|r}

1 & 0 & 0 & -3 \\

0 & 1 & 0 & 20 \\

0 & 0 & 1 & 19

\end{array} \right]$

\item $\left[ \begin{array}{rrrr|r}

1 & 0 & 0 & 3 & 4 \\

0 & 1 & 0 & 6 & -6 \\

0 & 0 & 1 & 0 & 2

\end{array} \right]$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{3}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\left[ \begin{array}{rrrr|r}

1 & 0 & 0 & 3 & 0 \\

0 & 1 & 2 & 6 & 0 \\

0 & 0 & 0 & 0 & 1

\end{array} \right]$

\item $\left[ \begin{array}{rrrr|r}

1 & \hphantom{-}0 & -8 & 1 & 7 \\

0 & 1 & 4 & -3 & 2 \\

0 & 0 & 0 & 0 & 0 \\

0 & 0 & 0 & 0 & 0

\end{array} \right]$

\item $\left[ \begin{array}{rrr|r}

1 & \hphantom{-}0 & 9 & -3 \\

0 & 1 & -4 & 20 \\

0 & 0 & 0 & 0

\end{array} \right]$ \label{decodelast}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

In Exercises \ref{solveaugfirst} - \ref{solveauglast}, solve the following systems of linear equations using the techniques discussed in this section. Compare and contrast these techniques with those you used to solve the systems in the Exercises in Section \ref{LinSystems}.

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\left\{ \begin{array}{rcr} -5x + y & = & 17 \\ x + y & = & 5 \end{array} \right.$ \label{solveaugfirst}

\item $\left\{ \begin{array}{rcr} x + y + z & = & 3 \\ 2x - y + z & = & 0 \\ -3x + 5y + 7z & = & 7 \end{array} \right.$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\left\{ \begin{array}{rcr} 4x - y + z & = & 5 \\ 2y + 6z & = & 30 \\ x + z & = & 5 \end{array} \right.$

\item $\left\{ \begin{array}{rcr} x-2y+3z & = & 7 \\ -3x+y+2z & = & -5 \\ 2x+2y+z & = & 3 \end{array} \right.$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\left\{ \begin{array}{rcr} 3x-2y+z & = & -5 \\ x+3y-z & = & 12 \\ x+y+2z & = & 0 \end{array} \right.$

\item $\left\{ \begin{array}{rcr} 2x-y+z& = & -1 \\ 4x+3y+5z & = & 1 \\ 5y+3z & = & 4 \end{array} \right.$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\left\{ \begin{array}{rcr} x-y+z & = & -4 \\ -3x+2y+4z & = & -5 \\ x-5y+2z & = & -18 \end{array} \right.$

\item $\left\{ \begin{array}{rcr} 2x-4y+z & = & -7 \\ x-2y+2z & = & -2 \\ -x+4y-2z & = & 3 \end{array} \right.$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\left\{ \begin{array}{rcr} 2x-y+z & = & 1 \\ 2x+2y-z & = & 1 \\ 3x+6y+4z & = & 9 \end{array} \right.$

\item $\left\{ \begin{array}{rcr} x-3y-4z & = & 3 \\ 3x+4y-z & = & 13 \\ 2x-19y-19z & = & 2 \end{array} \right.$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\left\{ \begin{array}{rcr} x+y+z & = & 4 \\ 2x-4y-z& = & -1 \\ x-y & = & 2 \end{array} \right.$

\item $\left\{ \begin{array}{rcr} x-y+z & = & 8 \\ 3x+3y-9z & = & -6 \\ 7x-2y+5z & = & 39 \end{array} \right.$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\left\{ \begin{array}{rcr} 2x-3y+z & = & -1 \\ 4x-4y+4z & = & -13 \\ 6x-5y+7z & = & -25 \end{array} \right.$

\item $\left\{ \begin{array}{rcr} x_{\mbox{\tiny$1$}} - x_{\mbox{\tiny$3$}} & = & -2 \\

2x_{\mbox{\tiny$2$}} - x_{\mbox{\tiny$4$}} & = & 0 \\

x_{\mbox{\tiny$1$}} - 2x_{\mbox{\tiny$2$}} + x_{\mbox{\tiny$3$}} & = & 0 \\

-x_{\mbox{\tiny$3$}} + x_{\mbox{\tiny$4$}} & = & 1 \end{array} \right.$ \label{solveauglast}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item It's time for another meal at our local buffet. This time, 22 diners (5 of whom were children) feasted for $\$162.25$, before taxes. If the kids buffet is $\$4.50$, the basic buffet is $\$7.50$, and the deluxe buffet (with crab legs) is $\$9.25$, find out how many diners chose the deluxe buffet.

\item Carl wants to make a party mix consisting of almonds (which cost $\$7$ per pound), cashews (which cost $\$5$ per pound), and peanuts (which cost $\$2$ per pound.) If he wants to make a $10$ pound mix with a budget of $\$35$, what are the possible combinations almonds, cashews, and peanuts? (You may find it helpful to review Example \ref{lucasmixex} in Section \ref{LinSystems}.)

\item Find the quadratic function passing through the points $(-2,1)$, $(1,4)$, $(3,-2)$

\item At 9 PM, the temperature was $60^{\circ}$F; at midnight, the temperature was $50^{\circ}$F; and at 6 AM, the temperature was $70^{\circ}$F . Use the technique in Example \ref{matrixcurvefitting} to fit a quadratic function to these data with the temperature, $T$, measured in degrees Fahrenheit, as the dependent variable, and the number of hours after 9 PM, $t$, measured in hours, as the independent variable. What was the coldest temperature of the night? When did it occur?

\item The price for admission into the Stitz-Zeager Sasquatch Museum and Research Station is \$15 for adults and \$8 for kids 13 years old and younger. When the Zahlenreich family visits the museum their bill is \$38 and when the Nullsatz family visits their bill is \$39. One day both families went together and took an adult babysitter along to watch the kids and the total admission charge was \$92. Later that summer, the adults from both families went without the kids and the bill was \$45. Is that enough information to determine how many adults and children are in each family? If not, state whether the resulting system is inconsistent or consistent dependent. In the latter case, give at least two plausible solutions.

\item Use the technique in Example \ref{matrixcurvefitting} to find the line between the points $(-3, 4)$ and $(6, 1)$. How does your answer compare to the slope-intercept form of the line in Equation \ref{slopeintercept}?

\item With the help of your classmates, find at least two different row echelon forms for the matrix \[\left[ \begin{array}{rr|r}

1 & 2 & 3 \\

4 & 12 & 8 \\

\end{array} \right]\]

\end{enumerate}

\newpage

\subsection{Answers}

\begin{multicols}{2}

\begin{enumerate}

\item Reduced row echelon form

\item Neither

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item Row echelon form only

\item Reduced row echelon form

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item Reduced row echelon form

\item Row echelon form only

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $(-2, 7)$

\item $(-3, 20, 19)$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $(-3t + 4, -6t - 6, 2, t)$ \\

for all real numbers $t$

\item Inconsistent

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $(8s - t + 7, -4s + 3t + 2, s, t)$ \\ for all real numbers $s$ and $t$

\item $(-9t - 3, 4t + 20, t)$ \\ for all real numbers $t$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $(-2, 7)$

\item $(1, 2, 0)$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $(-t + 5, -3t + 15, t)$\\

for all real numbers $t$

\item $(2,-1,1)$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $(1,3,-2)$

\item Inconsistent

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $(1,3,-2)$

\item $\left(-3,\frac{1}{2},1\right)$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\left(\frac{1}{3},\frac{2}{3},1\right)$

\item $\left(\frac{19}{13} t + \frac{51}{13},-\frac{11}{13} t+\frac{4}{13},t\right)$\\

for all real numbers $t$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item Inconsistent

\item $\left(4,-3,1\right)$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\left(-2t - \frac{35}{4},-t - \frac{11}{2},t\right)$\\

for all real numbers $t$

\item $(1, 2, 3, 4)$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item This time, 7 diners chose the deluxe buffet.

\item If $t$ represents the amount (in pounds) of peanuts, then we need $1.5 t - 7.5$ pounds of almonds and $17.5 - 2.5t$ pounds of cashews. Since we can't have a negative amount of nuts, $5 \leq t \leq 7$.

\item $f(x) = -\frac{4}{5} x^2+\frac{1}{5} x + \frac{23}{5}$

\item $T(t) = \frac{20}{27} t^2 - \frac{50}{9} t + 60$. Lowest temperature of the evening $\frac{595}{12} \approx 49.58^{\circ}$F at 12:45 AM.

\newpage

\item Let $x_{\mbox{\tiny$1$}}$ and $x_{\mbox{\tiny$2$}}$ be the numbers of adults and children, respectively, in the Zahlenreich family and let $x_{\mbox{\tiny$3$}}$ and $x_{\mbox{\tiny$4$}}$ be the numbers of adults and children, respectively, in the Nullsatz family. The system of equations determined by the given information is

$\left\{ \begin{array}{rcr} 15x_{\mbox{\tiny$1$}} + 8x_{\mbox{\tiny$2$}} & = & 38 \\

15x_{\mbox{\tiny$3$}} + 8x_{\mbox{\tiny$4$}} & = & 39 \\

15x_{\mbox{\tiny$1$}} + 8x_{\mbox{\tiny$2$}} + 15x_{\mbox{\tiny$3$}} + 8x_{\mbox{\tiny$4$}} & = & 77 \\

15x_{\mbox{\tiny$1$}} + 15x_{\mbox{\tiny$3$}} & = & 45 \end{array} \right.$

We subtracted the cost of the babysitter in E3 so the constant is 77, not 92. This system is consistent dependent and its solution is $\left(\frac{8}{15}t + \frac{2}{5}, -t + 4, -\frac{8}{15}t + \frac{13}{5}, t \right)$. Our variables represent numbers of adults and children so they must be whole numbers. Running through the values $t = 0, 1, 2, 3, 4$ yields only one solution where all four variables are whole numbers; $t = 3$ gives us $(2, 1, 1, 3)$. Thus there are 2 adults and 1 child in the Zahlenreichs and 1 adult and 3 kids in the Nullsatzs.

\end{enumerate}

\closegraphsfile

8.3: Matrix Arithmetic

\subsection{Exercises}

For each pair of matrices $A$ and $B$ in Exercises \ref{easymatarithfirst} - \ref{easymatarithlast}, find the following, if defined

\begin{multicols}{3}

\begin{itemize}

\item $3A$

\item $-B$

\item $A^2$

\end{itemize}

\end{multicols}

\begin{multicols}{3}

\begin{itemize}

\item $A-2B$

\item $AB$

\item $BA$

\end{itemize}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\item $A = \left[ \begin{array}{rr} 2 & -3 \\ 1 & 4 \end{array} \right]$, $B=\left[ \begin{array}{rr} 5 & -2 \\ 4 & 8 \end{array} \right]$ \label{easymatarithfirst}

\item $A = \left[ \begin{array}{rr} -1 & 5 \\ -3 & 6 \end{array} \right]$, $B=\left[ \begin{array}{rr} 2 & 10 \\ -7 & 1 \end{array} \right]$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $A = \left[ \begin{array}{rr} -1 & 3 \\ 5 & 2 \end{array} \right]$, $B=\left[ \begin{array}{rrr} 7 & 0 & 8 \\ -3 & 1 & 4 \end{array} \right]$

\item $A = \left[ \begin{array}{rr} 2 & 4 \\ 6 & 8 \end{array} \right]$, $B=\left[ \begin{array}{rrr} -1 & 3 & -5 \\ 7 & -9 & 11 \end{array} \right]$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $A = \left[ \begin{array}{r} 7 \\ 8 \\ 9 \end{array} \right]$, $B=\left[ \begin{array}{rrr} 1 & 2 & 3 \end{array} \right]$

\item $A = \left[ \begin{array}{rr} 1 & -2 \\ -3 & 4 \\ 5 & -6 \end{array} \right]$, $B=\left[ \begin{array}{rrr} -5 & 1 & 8 \end{array} \right]$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $ A = \left[ \begin{array}{rrr} 2 & -3 & 5 \\ 3 & 1 &-2 \\ -7 & 1 & -1 \end{array} \right]$, $B= \left[ \begin{array}{rrr} 1 & 2 & 1 \\ 17 & 33 & 19 \\ 10 & 19 & 11 \end{array} \right]$ \label{easymatarithlast}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

In Exercises \ref{matarithfirst} - \ref{matarithlast}, use the matrices \[A = \left[ \begin{array}{rr} 1 & 2 \\ 3 & 4 \end{array} \right] \;\;\; B = \left[ \begin{array}{rr} 0 & -3 \\ -5 & 2 \end{array} \right] \;\;\; C = \left[ \begin{array}{rrr} 10 & -\frac{11}{2} & 0 \\ \frac{3}{5} & 5 & 9 \end{array} \right]\] \[ D = \left[ \begin{array}{rr} 7 & -13 \\ -\frac{4}{3} & 0 \\ 6 & 8 \end{array} \right] \;\;\; E = \left[ \begin{array}{rrr} 1 & \hphantom{-}2 & 3 \\ 0 & 4 & -9 \\ 0 & 0 & -5 \end{array} \right] \] to compute the following or state that the indicated operation is undefined.

\begin{multicols}{3}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $7B - 4A$ \label{matarithfirst}

\item $AB$

\item $BA$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{3}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $E + D$

\item $ED$

\item $CD + 2I_{2}A$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{3}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $A - 4I_{2}$

\item $A^2 - B^2$

\item $(A+B)(A-B)$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{3}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $A^2-5A-2I_{2}$

\item $E^2 + 5E-36I_{3}$

\item $EDC$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{3}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $CDE$

\item $ABCEDI_{2}$ \label{matarithlast}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item Let $A = \left[ \begin{array}{rrr} a & b & c \\ d & e & f \end{array} \right] \;\;\; E_{\mbox{\tiny$1$}} = \left[ \begin{array}{rr} 0 & 1 \\ 1 & 0 \end{array} \right] \;\;\; E_{\mbox{\tiny$2$}} = \left[ \begin{array}{rr} 5 & 0 \\ 0 & 1 \end{array} \right] \;\;\; E_{\mbox{\tiny$3$}} = \left[ \begin{array}{rr} 1 & -2 \\ 0 & 1 \end{array} \right]$

\smallskip

Compute $E_{\mbox{\tiny$1$}}A$, $\; E_{\mbox{\tiny$2$}}A$ and $E_{\mbox{\tiny$3$}}A$. What effect did each of the $E_{i}$ matrices have on the rows of $A$? Create $E_{\mbox{\tiny$4$}}$ so that its effect on $A$ is to multiply the bottom row by $-6$. How would you extend this idea to matrices with more than two rows?

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\phantomsection

\label{Markovchain}

In Exercises \ref{MCfirst} - \ref{MClast}, consider the following scenario. In the small village of Pedimaxus in the country of Sasquatchia, all 150 residents get one of the two local newspapers. Market research has shown that in any given week, 90\% of those who subscribe to the Pedimaxus Tribune want to keep getting it, but 10\% want to switch to the Sasquatchia Picayune. Of those who receive the Picayune, 80\% want to continue with it and 20\% want switch to the Tribune. We can express this situation using matrices. Specifically, let $X$ be the `state matrix' given by \[X = \left[ \begin{array}{r} T \\ P \end{array} \right]\] where $T$ is the number of people who get the Tribune and $P$ is the number of people who get the Picayune in a given week. Let $Q$ be the `transition matrix' given by \[Q = \left[ \begin{array}{rr} 0.90 & 0.20 \\ 0.10 & 0.80 \end{array} \right]\] such that $QX$ will be the state matrix for the next week.

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item \label{MCfirst} Let's assume that when Pedimaxus was founded, all 150 residents got the Tribune. (Let's call this Week 0.) This would mean \[X = \left[ \begin{array}{r} 150 \\ 0 \end{array} \right]\] Since 10\% of that 150 want to switch to the Picayune, we should have that for Week 1, 135 people get the Tribune and 15 people get the Picayune. Show that $QX$ in this situation is indeed \[QX = \left[ \begin{array}{r} 135 \\ 15 \end{array} \right]\]

\item Assuming that the percentages stay the same, we can get to the subscription numbers for Week 2 by computing $Q^{2}X$. How many people get each paper in Week 2?

\item Explain why the transition matrix does what we want it to do.

\item If the conditions do not change from week to week, then $Q$ remains the same and we have what's known as a \index{stochastic process} \index{Markov Chain} {\bf Stochastic Process}\footnote{More specifically, we have a Markov Chain, which is a special type of stochastic process.} because Week $n$'s numbers are found by computing $Q^{n}X$. Choose a few values of $n$ and, with the help of your classmates and calculator, find out how many people get each paper for that week. You should start to see a pattern as $n \rightarrow \infty$.

\item If you didn't see the pattern, we'll help you out. Let \[X_{s} = \left[ \begin{array}{r} 100 \\ 50 \end{array} \right].\] Show that $QX_{s} = X_{s}$ This is called the {\bf steady state} \index{steady state} because the number of people who get each paper didn't change for the next week. Show that $Q^{n}X \rightarrow X_{s}$ as $n \rightarrow \infty$.

\item Now let \[S = \left[ \begin{array}{rr} \frac{2}{3} & \frac{2}{3} \\ [3pt] \frac{1}{3} & \frac{1}{3} \end{array} \right]\] Show that $Q^{n} \rightarrow S$ as $n \rightarrow \infty$.

\item \label{MClast} Show that $SY = X_{s}$ for any matrix $Y$ of the form \[Y = \left[ \begin{array}{r} y \\ 150 - y \end{array} \right]\] This means that no matter how the distribution starts in Pedimaxus, if $Q$ is applied often enough, we always end up with 100 people getting the Tribune and 50 people getting the Picayune.

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item Let $z = a + bi$ and $w = c + di$ be arbitrary complex numbers. Associate $z$ and $w$ with the matrices \[Z = \left[ \begin{array}{rr} a & b \\ -b & a \end{array} \right] \;\; \mbox{and} \;\; W = \left[ \begin{array}{rr} c & d \\ -d & c \end{array} \right]\] Show that complex number addition, subtraction and multiplication are mirrored by the associated \emph{matrix} arithmetic. That is, show that $Z + W$, $Z - W$ and $ZW$ produce matrices which can be associated with the complex numbers $z + w$, $z - w$ and $zw$, respectively.

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item Let \[A = \left[ \begin{array}{rr} 1 & 2 \\ 3 & 4 \end{array} \right] \; \mbox{and} \; B = \left[ \begin{array}{rr} 0 & -3 \\ -5 & 2 \end{array} \right]\] Compare $(A + B)^{2}$ to $A^{2} + 2AB + B^{2}$. Discuss with your classmates what constraints must be placed on two arbitrary matrices $A$ and $B$ so that both $(A + B)^{2}$ and $A^{2} + 2AB + B^{2}$ exist. When will $(A + B)^{2} = A^{2} + 2AB + B^{2}$? In general, what is the correct formula for $(A + B)^{2}$?

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\phantomsection

\label{triangularmatrices}

In Exercises \ref{triangexfirst} - \ref{triangexlast}, consider the following definitions. A square matrix is said to be an \index{upper triangular matrix} \index{matrix ! upper triangular} {\bf upper triangular matrix} if all of its entries below the main diagonal are zero and it is said to be a {\bf lower triangular matrix} \index{lower triangular matrix} \index{matrix ! lower triangular} if all of its entries above the main diagonal are zero. For example, \[E = \left[ \begin{array}{rrr} 1 & \hphantom{-}2 & 3 \\ 0 & 4 & -9 \\ 0 & 0 & -5 \end{array} \right]\] from Exercises \ref{matarithfirst} - \ref{matarithlast} above is an upper triangular matrix whereas \[F = \left[ \begin{array}{rr} 1 & 0 \\ 3 & 0 \end{array} \right]\] is a lower triangular matrix. (Zeros are allowed on the main diagonal.) Discuss the following questions with your classmates.

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item Give an example of a matrix which is neither upper triangular nor lower triangular. \label{triangexfirst}

\item Is the product of two $n \times n$ upper triangular matrices always upper triangular?

\item Is the product of two $n \times n$ lower triangular matrices always lower triangular?

\item Given the matrix \[A = \left[ \begin{array}{rr} 1 & 2 \\ 3 & 4 \end{array} \right]\] write $A$ as $LU$ where $L$ is a lower triangular matrix and $U$ is an upper triangular matrix?

\item Are there any matrices which are simultaneously upper and lower triangular? \label{triangexlast}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\newpage

\subsection{Answers}

\begin{enumerate}

\item For $A = \left[ \begin{array}{rr} 2 & -3 \\ 1 & 4 \end{array} \right]$ and $B=\left[ \begin{array}{rr} 5 & -2 \\ 4 & 8 \end{array} \right]$

\begin{multicols}{3}

\begin{itemize}

\item $3A = \left[ \begin{array}{rr} 6 & -9 \\ 3 & 12 \end{array} \right]$

\item $-B = \left[ \begin{array}{rr} -5 & 2 \\ -4 & -8 \end{array} \right]$

\item $A^2 = \left[ \begin{array}{rr} 1 & -18 \\ 6 & 13 \end{array} \right]$

\end{itemize}

\end{multicols}

\begin{multicols}{3}

\begin{itemize}

\item $A-2B = \left[ \begin{array}{rr} -8 & 1 \\ -7 & -12 \end{array} \right]$

\item $AB = \left[ \begin{array}{rr} -2 & -28 \\ 21 & 30 \end{array} \right]$

\item $BA = \left[ \begin{array}{rr} 8 & -23 \\ 16 & 20 \end{array} \right]$

\end{itemize}

\end{multicols}

\item For $A = \left[ \begin{array}{rr} -1 & 5 \\ -3 & 6 \end{array} \right]$ and $B=\left[ \begin{array}{rr} 2 & 10 \\ -7 & 1 \end{array} \right]$

\begin{multicols}{3}

\begin{itemize}

\item $3A = \left[ \begin{array}{rr} -3 & 15 \\ -9 & 18 \end{array} \right]$

\item $-B = \left[ \begin{array}{rr} -2 & -10 \\ 7 & -1 \end{array} \right]$

\item $A^2 = \left[ \begin{array}{rr} -14 & 25 \\ -15 & 21 \end{array} \right]$

\end{itemize}

\end{multicols}

\begin{multicols}{3}

\begin{itemize}

\item $A-2B = \left[ \begin{array}{rr} -5 & -15 \\ 11 & 4 \end{array} \right]$

\item $AB = \left[ \begin{array}{rr} -37 & -5 \\ -48 & -24 \end{array} \right]$

\item $BA = \left[ \begin{array}{rr} -32 & 70 \\ 4 & -29 \end{array} \right]$

\end{itemize}

\end{multicols}

\item For $A = \left[ \begin{array}{rr} -1 & 3 \\ 5 & 2 \end{array} \right]$ and

$B=\left[ \begin{array}{rrr} 7 & 0 & 8 \\ -3 & 1 & 4 \end{array} \right]$

\begin{multicols}{3}

\begin{itemize}

\item $3A = \left[ \begin{array}{rr} -3 & 9 \\ 15 & 6\end{array} \right]$

\item $-B = \left[ \begin{array}{rrr} -7 & 0 & -8 \\ 3 & -1 & -4 \end{array} \right]$

\item $A^2 = \left[ \begin{array}{rr} 16 & 3 \\ 5 & 19 \end{array} \right]$

\end{itemize}

\end{multicols}

\begin{multicols}{3}

\begin{itemize}

\item $A-2B$ is not defined

\item $AB = \left[ \begin{array}{rrr} -16 & 3 & 4 \\ 29 & 2 & 48 \end{array} \right]$

\item $BA$ is not defined

\end{itemize}

\end{multicols}

\item For $A = \left[ \begin{array}{rr} 2 & 4 \\ 6 & 8 \end{array} \right]$ and $B=\left[ \begin{array}{rrr} -1 & 3 & -5 \\ 7 & -9 & 11 \end{array} \right]$

\begin{multicols}{3}

\begin{itemize}

\item $3A = \left[ \begin{array}{rr} 6 & 12 \\ 18 & 24 \end{array} \right]$

\item $-B = \left[ \begin{array}{rrr} 1 & -3 & 5 \\ -7 & 9 & -11 \end{array} \right]$

\item $A^2 = \left[ \begin{array}{rr} 28 & 40 \\ 60 & 88 \end{array} \right]$

\end{itemize}

\end{multicols}

\begin{multicols}{3}

\begin{itemize}

\item $A-2B$ is not defined

\item $AB = \left[ \begin{array}{rrr} 26 & -30 & 34 \\ 50 & -54 & 58 \end{array} \right]$

\item $BA$ is not defined

\end{itemize}

\end{multicols}

\pagebreak

\item For $A = \left[ \begin{array}{r} 7 \\ 8 \\ 9 \end{array} \right]$ and $B=\left[ \begin{array}{rrr} 1 & 2 & 3 \end{array} \right]$

\begin{multicols}{2}

\begin{itemize}

\item $3A = \left[ \begin{array}{r} 21 \\ 24 \\ 27 \end{array} \right]$

\item $-B = \left[ \begin{array}{rrr} -1 & -2 & -3 \end{array} \right] \vphantom{\left[ \begin{array}{r} 21 \\ 24 \\ 27 \end{array} \right]}$

\end{itemize}

\end{multicols}

\begin{multicols}{2}

\begin{itemize}

\item $A^2$ is not defined

\item $A-2B$ is not defined

\end{itemize}

\end{multicols}

\begin{multicols}{2}

\begin{itemize}

\item $AB = \left[ \begin{array}{rrr} 7 & 14 & 21 \\ 8 & 16 & 24 \\ 9 & 18 & 27 \end{array} \right]$

\item $BA = [50] \vphantom{\left[ \begin{array}{rrr} 7 & 14 & 21 \\ 8 & 16 & 24 \\ 9 & 18 & 27 \end{array} \right]}$

\end{itemize}

\end{multicols}

\item For $A = \left[ \begin{array}{rr} 1 & -2 \\ -3 & 4 \\ 5 & -6 \end{array} \right]$ and $B=\left[ \begin{array}{rrr} -5 & 1 & 8 \end{array} \right]$

\begin{multicols}{2}

\begin{itemize}

\item $3A = \left[ \begin{array}{rr} 3 & -6 \\ -9 & 12 \\ 15 & -18 \end{array} \right]$

\item $-B = \left[ \begin{array}{rrr} 5 & -1 & -8 \end{array} \right] \vphantom{\left[ \begin{array}{rr} 3 & -6 \\ -9 & 12 \\ 15 & -18 \end{array} \right]}$

\end{itemize}

\end{multicols}

\begin{multicols}{2}

\begin{itemize}

\item $A^2$ is not defined

\item $A-2B$ is not defined

\end{itemize}

\end{multicols}

\begin{multicols}{2}

\begin{itemize}

\item $AB$ is not defined

\item $BA = \left[ \begin{array}{rr} 32 & -34 \end{array} \right]$

\end{itemize}

\end{multicols}

\item For $ A = \left[ \begin{array}{rrr} 2 & -3 & 5 \\ 3 & 1 &-2 \\ -7 & 1 & -1 \end{array} \right]$ and $B= \left[ \begin{array}{rrr} 1 & 2 & 1 \\ 17 & 33 & 19 \\ 10 & 19 & 11 \end{array} \right]$

\begin{multicols}{2}

\begin{itemize}

\item $3A = \left[ \begin{array}{rrr} 6 & -9 & 15 \\ 9 & 3 &-6 \\ -21 & 3 & -3 \end{array} \right]$

\item $-B = \left[ \begin{array}{rrr} -1 & -2 & -1 \\ -17 & -33 & -19 \\ -10 & -19 & -11 \end{array} \right]$

\end{itemize}

\end{multicols}

\begin{multicols}{2}

\begin{itemize}

\item $A^2 = \left[ \begin{array}{rrr} -40 & -4 & 11 \\ 23 & -10 & 15 \\ -4 & 21 & -36 \end{array} \right]$

\item $A-2B = \left[ \begin{array}{rrr} 0 & -7 & 3 \\ -31 & -65 & -40 \\ -27 & -37 & -23 \end{array} \right]$

\end{itemize}

\end{multicols}

\begin{multicols}{2}

\begin{itemize}

\item $AB = \left[ \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right]$

\item $BA = \left[ \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right]$

\end{itemize}

\end{multicols}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $7B - 4A = \left[ \begin{array}{rr} -4 & -29 \\ -47 & -2 \end{array} \right]$

\item $AB = \left[ \begin{array}{rr} -10 & 1 \\ -20 & -1 \end{array} \right]$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $BA = \left[ \begin{array}{rr} -9 & -12 \\ 1 & -2 \end{array} \right]$

\item $E + D \vphantom{\left[ \begin{array}{rr} -9 & -12 \\ 1 & -2 \end{array} \right]}$ is undefined

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $ED = \left[ \begin{array}{rr} \frac{67}{3}& 11 \\[3pt] -\frac{178}{3} & -72 \\ -30 & -40 \end{array} \right]$

\item $CD + 2I_{2}A = \left[ \begin{array}{rr} \frac{238}{3} & -126 \\[3pt] \frac{863}{15} & \frac{361}{5} \end{array} \right] \vphantom{\left[ \begin{array}{rr} \frac{67}{3}& 11 \\[3pt] -\frac{178}{3} & -72 \\ -30 & -40 \end{array} \right]}$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $A - 4I_{2} = \left[ \begin{array}{rr} -3 & 2 \\ 3 & 0 \end{array} \right]$

\item $A^2 - B^2 = \left[ \begin{array}{rr} -8 & 16 \\ 25 & 3 \end{array} \right]$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $(A+B)(A-B) = \left[ \begin{array}{rr} -7 & 3 \\ 46 & 2 \end{array} \right]$

\item $A^2-5A-2I_{2} = \left[ \begin{array}{rr} 0 & 0 \\ 0 & 0 \end{array} \right]$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $E^2 + 5E-36I_{3} = \left[ \begin{array}{rrr} -30 & 20 & -15 \\ 0 & 0 & -36 \\ 0 & 0 & -36 \end{array} \right] \vphantom{\left[ \begin{array}{rrr} \frac{3449}{15} & -\frac{407}{6} & 99 \\[3pt] -\frac{9548}{15} & -\frac{101}{3} & -648 \\ -324 & -35 & -360 \end{array} \right]}$

\item $EDC = \left[ \begin{array}{rrr} \frac{3449}{15} & -\frac{407}{6} & 99 \\[3pt] -\frac{9548}{15} & -\frac{101}{3} & -648 \\ -324 & -35 & -360 \end{array} \right]$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $CDE \vphantom{\left[ \begin{array}{rr} -\frac{90749}{15} & -\frac{28867}{5} \\[3pt] -\frac{156601}{15} & -\frac{47033}{5} \end{array} \right]}$ is undefined

\item $ABCEDI_{2} = \left[ \begin{array}{rr} -\frac{90749}{15} & -\frac{28867}{5} \\[3pt] -\frac{156601}{15} & -\frac{47033}{5} \end{array} \right]$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $E_{\mbox{\tiny$1$}}A = \left[ \begin{array}{rrr} d & e & f \\ a & b & c\end{array} \right]\;\;$ $E_{\mbox{\tiny$1$}}$ interchanged $R1$ and $R2$ of $A$.\\

$E_{\mbox{\tiny$2$}}A = \left[ \begin{array}{rrr} 5a & 5b & 5c \\ d & e & f \end{array} \right]\;\;$ $E_{\mbox{\tiny$2$}}$ multiplied $R1$ of $A$ by 5.\\

$E_{\mbox{\tiny$3$}}A = \left[ \begin{array}{rrr} a - 2d & b - 2e & c - 2f \\ d & e & f \end{array} \right]\;\;$ $E_{\mbox{\tiny$3$}}$ replaced $R1$ in $A$ with $R1 - 2R2$.\\

$E_{\mbox{\tiny$4$}} = \left[ \begin{array}{rr} 1 & 0 \\ 0 & -6 \end{array} \right]\;\;$

\end{enumerate}

\closegraphsfile

8.4: Systems of Linear Equations: Matrix Inverses

\subsection{Exercises}

In Exercises \ref{findmatinversefirst} - \ref{findmatinverselast}, find the inverse of the matrix or state that the matrix is not invertible.

\begin{multicols}{2}

\begin{enumerate}

\item $A = \left[ \begin{array}{rr} 1 & 2 \\ 3 & 4 \end{array} \right]$ \label{findmatinversefirst}

\item $B = \left[ \begin{array}{rr} 12 & -7 \\ -5 & 3 \end{array} \right]$ \label{matrixB}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $C = \left[ \begin{array}{rr} 6 & 15 \\ 14 & 35 \end{array} \right]$

\item $D = \left[ \begin{array}{rr} 2 & -1 \\ 16 & -9 \end{array} \right]$ \label{matrixD}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $E = \left[ \begin{array}{rrr} 3 & 0 & 4 \\ 2 & -1 & 3 \\ -3 & 2 & -5 \end{array} \right]$ \label{matrixE}

\item $F = \left[ \begin{array}{rrr} 4 & \hphantom{-}6 & -3 \\ 3 & 4 & -3 \\ 1 & 2 & 6 \end{array} \right]$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $G = \left[ \begin{array}{rrr} 1 & \hphantom{1}2 & 3 \\ 2 & 3 & 11 \\ 3 & 4 & 19 \end{array} \right]$

\item $H = \left[ \begin{array}{rrrr} 1 & 0 & -3 & 0 \\ 2 & -2 & 8 & 7 \\ -5 & 0 & 16 & 0 \\ 1 & 0 & 4 & 1 \end{array} \right]$ \label{findmatinverselast}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

In Exercises \ref{2by2inversefirst} - \ref{2by2inverselast}, use one matrix inverse to solve the following systems of linear equations.

\begin{multicols}{3}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\left\{ \begin{array}{rcr} 3x + 7y & = & 26 \\ 5x + 12y & = & 39 \end{array} \right.$ \label{2by2inversefirst}

\item $\left\{ \begin{array}{rcr} 3x + 7y & = & 0 \\ 5x + 12y & = & -1 \end{array} \right.$

\item $\left\{ \begin{array}{rcr} 3x + 7y & = & -7 \\ 5x + 12y & = & 5 \end{array} \right.$ \label{2by2inverselast}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

In Exercises \ref{3by3inversefirst} - \ref{3by3inverselast}, use the inverse of $E$ from Exercise \ref{matrixE} above to solve the following systems of linear equations.

\begin{multicols}{3}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\left\{ \begin{array}{rcr} 3x + 4z & = & 1 \\ 2x - y + 3z & = & 0 \\ \!-3x + 2y - 5z & = & 0 \end{array} \right.$ \label{3by3inversefirst}

\item $\left\{ \begin{array}{rcr} 3x + 4z & = & 0 \\ 2x - y + 3z & = & 1 \\ \!-3x + 2y - 5z & = & 0 \end{array} \right.$

\item $\left\{ \begin{array}{rcr} 3x + 4z & = & 0 \\ 2x - y + 3z & = & 0 \\ \!-3x + 2y - 5z & = & 1 \end{array} \right.$ \label{3by3inverselast}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item This exercise is a continuation of Example \ref{rotationmatrixex} in Section \ref{MatArithmetic} and gives another application of matrix inverses. Recall that given the position matrix $P$ for a point in the plane, the matrix $RP$ corresponds to a point rotated $45^{\circ}$ counterclockwise from $P$ where

\[R = \left[ \begin{array}{rr} \frac{\sqrt{2}}{2} & -\frac{\sqrt{2}}{2} \\[3pt] \frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2} \\ \end{array} \right]\]

\begin{enumerate}

\item Find $R^{-1}$.

\item If $RP$ rotates a point counterclockwise $45^{\circ}$, what should $R^{-1}P$ do? Check your answer by finding $R^{-1}P$ for various points on the coordinate axes and the lines $y=\pm x$.

\item Find $R^{-1}P$ where $P$ corresponds to a generic point $P(x,y)$. Verify that this takes points on the curve $y=\frac{2}{x}$ to points on the curve $x^2-y^2=4$.

\end{enumerate}

\item \label{SasquatchDiet} A Sasquatch's diet consists of three primary foods: Ippizuti Fish, Misty Mushrooms, and Sun Berries. Each serving of Ippizuti Fish is 500 calories, contains 40 grams of protein, and has no Vitamin X. Each serving of Misty Mushrooms is 50 calories, contains 1 gram of protein, and 5 milligrams of Vitamin X. Finally, each serving of Sun Berries is 80 calories, contains no protein, but has 15 milligrams of Vitamin X.\footnote{Misty Mushrooms and Sun Berries are the only known fictional sources of Vitamin X.}

\begin{enumerate}

\item If an adult male Sasquatch requires 3200 calories, 130 grams of protein, and 275 milligrams of Vitamin X daily, use a matrix inverse to find how many servings each of Ippizuti Fish, Misty Mushrooms, and Sun Berries he needs to eat each day.

\item An adult female Sasquatch requires 3100 calories, 120 grams of protein, and 300 milligrams of Vitamin X daily. Use the matrix inverse you found in part (a) to find how many servings each of Ippizuti Fish, Misty Mushrooms, and Sun Berries she needs to eat each day.

\item An adolescent Sasquatch requires 5000 calories, 400 grams of protein daily, but no Vitamin X daily.\footnote{Vitamin X is needed to sustain Sasquatch longevity only.} Use the matrix inverse you found in part (a) to find how many servings each of Ippizuti Fish, Misty Mushrooms, and Sun Berries she needs to eat each day.

\end{enumerate}

\item Matrices can be used in cryptography. Suppose we wish to encode the message `BIGFOOT LIVES'. We start by assigning a number to each letter of the alphabet, say $A=1$, $B=2$ and so on. We reserve $0$ to act as a space. Hence, our message `BIGFOOT LIVES' corresponds to the string of numbers `2, 9, 7, 6, 15, 15, 20, 0, 12, 9, 22, 5, 19.' To encode this message, we use an invertible matrix. Any invertible matrix will do, but for this exercise, we choose

\[ A = \left[ \begin{array}{rrr} 2 & -3 & 5 \\ 3 & 1 &-2 \\ -7 & 1 & -1 \end{array} \right] \]

Since $A$ is $3 \times 3$ matrix, we encode our message string into a matrix $M$ with $3$ rows. To do this, we take the first three numbers, 2 9 7, and make them our first column, the next three numbers, 6 15 15, and make them our second column, and so on. We put $0$'s to round out the matrix.

\[ M = \left[ \begin{array}{rrrrr} 2 & 6 & 20 & 9 & 19 \\ 9 & 15 & 0 & 22 & 0 \\ 7 & 15 & 12 & 5 & 0 \end{array} \right] \]

To encode the message, we find the product $AM$

\[AM = \left[ \begin{array}{rrr} 2 & -3 & 5 \\ 3 & 1 &-2 \\ -7 & 1 & -1 \end{array} \right]\left[ \begin{array}{rrrrr} 2 & 6 & 20 & 9 & 19 \\ 9 & 15 & 0 & 22 & 0 \\ 7 & 15 & 12 & 5 & 0 \end{array} \right] = \left[ \begin{array}{rrrrr} 12 & 42 & 100 & -23 & 38 \\ 1 & 3 & 36 & 39 & 57 \\ -12 & -42 & -152 & -46 & -133 \end{array} \right]\]

So our coded message is `12, 1, $-12$, 42, 3, $-42$, 100, 36, $-152$, $-23$, 39, $-46$, 38, 57, $-133$.' To decode this message, we start with this string of numbers, construct a message matrix as we did earlier (we should get the matrix $AM$ again) and then multiply by $A^{-1}$.

\begin{enumerate}

\item Find $A^{-1}$.

\item Use $A^{-1}$ to decode the message and check this method actually works.

\item Decode the message `14, 37, $-76$, 128, 21, $-151$, 31, 65, $-140$'

\item Choose another invertible matrix and encode and decode your own messages.

\end{enumerate}

\item Using the matrices $A$ from Exercise \ref{findmatinversefirst}, $B$ from Exercise \ref{matrixB} and $D$ from Exercise \ref{matrixD}, show $AB = D$ and $D^{-1} = B^{-1}A^{-1}$. That is, show that $(AB)^{-1} = B^{-1}A^{-1}$.

\item Let $M$ and $N$ be invertible $n \times n$ matrices. Show that $(MN)^{-1} = N^{-1}M^{-1}$ and compare your work to Exercise \ref{fcircginverse} in Section \ref{InverseFunctions}.

\end{enumerate}

\newpage

\subsection{Answers}

\begin{multicols}{2}

\begin{enumerate}

\item $A^{-1} = \left[ \begin{array}{rr} -2 & 1 \\[3pt] \frac{3}{2} & -\frac{1}{2} \end{array} \right]$

\item $B^{-1} = \left[ \begin{array}{rr} 3 & 7 \\ 5 & 12 \end{array} \right]$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $C \vphantom{\left[ \begin{array}{rr} \frac{9}{2} & -\frac{1}{2} \\ 8 & -1 \end{array} \right]}$ is not invertible

\item $D^{-1} = \left[ \begin{array}{rr} \frac{9}{2} & -\frac{1}{2} \\ 8 & -1 \end{array} \right]$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $E^{-1} = \left[ \begin{array}{rrr} -1 & 8 & 4 \\ 1 & -3 & -1 \\ 1 & -6 & -3 \end{array} \right] \vphantom{\left[ \begin{array}{rrr} -\frac{5}{2} & \frac{7}{2} & \frac{1}{2} \\[3pt] \frac{7}{4} & -\frac{9}{4} & -\frac{1}{4} \\[3pt] -\frac{1}{6} & \frac{1}{6} & \frac{1}{6} \end{array} \right]}$

\item $F^{-1} = \left[ \begin{array}{rrr} -\frac{5}{2} & \frac{7}{2} & \frac{1}{2} \\[3pt] \frac{7}{4} & -\frac{9}{4} & -\frac{1}{4} \\[3pt] -\frac{1}{6} & \frac{1}{6} & \frac{1}{6} \end{array} \right]$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $G \vphantom{\left[ \begin{array}{rrrr} 16 & 0 & 3 & 0 \\[3pt] -90 & -\frac{1}{2} & -\frac{35}{2} & \frac{7}{2} \\[3pt] 5 & 0 & 1 & 0 \\[3pt] -36 & 0 & -7 & \hphantom{-}1 \end{array} \right]}$ is not invertible

\item $H^{-1} = \left[ \begin{array}{rrrr} 16 & 0 & 3 & 0 \\[3pt] -90 & -\frac{1}{2} & -\frac{35}{2} & \frac{7}{2} \\[3pt] 5 & 0 & 1 & 0 \\[3pt] -36 & 0 & -7 & \hphantom{-}1 \end{array} \right]$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

The coefficient matrix is $B^{-1}$ from Exercise \ref{matrixB} above so the inverse we need is $(B^{-1})^{-1} = B$.

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\left[ \begin{array}{rr} 12 & -7 \\ -5 & 3 \end{array} \right] \left[ \begin{array}{r} 26 \\ 39 \end{array} \right] = \left[ \begin{array}{r} 39 \\ -13 \end{array} \right] \;$ So $x = 39$ and $y = -13$.

\item $\left[ \begin{array}{rr} 12 & -7 \\ -5 & 3 \end{array} \right] \left[ \begin{array}{r} 0 \\ -1 \end{array} \right] = \left[ \begin{array}{r} 7 \\ -3 \end{array} \right] \;$ So $x = 7$ and $y = -3$.

\item $\left[ \begin{array}{rr} 12 & -7 \\ -5 & 3 \end{array} \right] \left[ \begin{array}{r} -7 \\ 5 \end{array} \right] = \left[ \begin{array}{r} -119 \\ 50 \end{array} \right] \;$ So $x = -119$ and $y = 50$.

\setcounter{HW}{\value{enumi}}

\end{enumerate}

The coefficient matrix is $E = \left[ \begin{array}{rrr} 3 & 0 & 4 \\ 2 & -1 & 3 \\ -3 & 2 & -5 \end{array} \right]$ from Exercise \ref{matrixE}, so $E^{-1} = \left[ \begin{array}{rrr} -1 & 8 & 4 \\ 1 & -3 & -1 \\ 1 & -6 & -3 \end{array} \right]$

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\left[ \begin{array}{rrr} -1 & 8 & 4 \\ 1 & -3 & -1 \\ 1 & -6 & -3 \end{array} \right] \left[ \begin{array}{r} 1 \\ 0 \\ 0 \end{array} \right] = \left[ \begin{array}{r} -1 \\ 1 \\ 1 \end{array} \right] \;$ So $x = -1$, $y = 1$ and $z = 1$.

\item $\left[ \begin{array}{rrr} -1 & 8 & 4 \\ 1 & -3 & -1 \\ 1 & -6 & -3 \end{array} \right] \left[ \begin{array}{r} 0 \\ 1 \\ 0 \end{array} \right] = \left[ \begin{array}{r} 8 \\ -3 \\ -6 \end{array} \right] \;$ So $x = 8$, $y = -3$ and $z = -6$.

\item $\left[ \begin{array}{rrr} -1 & 8 & 4 \\ 1 & -3 & -1 \\ 1 & -6 & -3 \end{array} \right] \left[ \begin{array}{r} 0 \\ 0 \\ 1 \end{array} \right] = \left[ \begin{array}{r} 4 \\ -1 \\ -3 \end{array} \right] \;$ So $x = 4$, $y = -1$ and $z = -3$.

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\addtocounter{enumi}{1}

\item \begin{enumerate} \item The adult male Sasquatch needs: 3 servings of Ippizuti Fish, 10 servings of Misty Mushrooms, and 15 servings of Sun Berries daily.

\item The adult female Sasquatch needs: 3 servings of Ippizuti Fish and 20 servings of Sun Berries daily. (No Misty Mushrooms are needed!)

\item The adolescent Sasquatch requires 10 servings of Ippizuti Fish daily. (No Misty Mushrooms or Sun Berries are needed!)

\end{enumerate}

\item \begin{enumerate}

\item $A^{-1} = \left[ \begin{array}{rrr} 1 & 2 & 1 \\ 17 & 33 & 19 \\ 10 & 19 & 11 \end{array} \right] $

\item $ \left[ \begin{array}{rrr} 1 & 2 & 1 \\ 17 & 33 & 19 \\ 10 & 19 & 11 \end{array} \right] \left[ \begin{array}{rrrrr} 12 & 42 & 100 & -23 & 38 \\ 1 & 3 & 36 & 39 & 57 \\ -12 & -42 & -152 & -46 & -133 \end{array} \right] = \left[ \begin{array}{rrrrr} 2 & 6 & 20 & 9 & 19 \\ 9 & 15 & 0 & 22 & 0 \\ 7 & 15 & 12 & 5 & 0 \end{array} \right] \quad \checkmark$

\item `LOGS RULE'

\end{enumerate}

\end{enumerate}

\closegraphsfile

8.5: Determinants and Cramer’s Rule

\subsection{Exercises}

In Exercises \ref{finddetfirst} - \ref{finddetlast}, compute the determinant of the given matrix. (Some of these matrices appeared in Exercises \ref{findmatinversefirst} - \ref{findmatinverselast} in Section \ref{MatMethods}.)

\begin{multicols}{2}

\begin{enumerate}

\item $B = \left[ \begin{array}{rr} 12 & -7 \\ -5 & 3 \end{array} \right]$ \label{finddetfirst}

\item $C = \left[ \begin{array}{rr} 6 & 15 \\ 14 & 35 \end{array} \right]$ \label{matrixC}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $Q = \left[ \begin{array}{rr} x & x^{2} \\ 1 & 2x \end{array} \right] \vphantom{ \left[ \begin{array}{rr} \dfrac{1}{x^{3}} & \dfrac{\ln(x)}{x^{3}} \\[10pt] -\dfrac{3}{x^{4}} & \dfrac{1 - 3\ln(x)}{x^{4}} \end{array} \right]}$

\item $L = \left[ \begin{array}{rr} \dfrac{1}{x^{3}} & \dfrac{\ln(x)}{x^{3}} \\[10pt] -\dfrac{3}{x^{4}} & \dfrac{1 - 3\ln(x)}{x^{4}} \end{array} \right]$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $F = \left[ \begin{array}{rrr} 4 & \hphantom{-}6 & -3 \\ 3 & 4 & -3 \\ 1 & 2 & 6 \end{array} \right]$

\item $G = \left[ \begin{array}{rrr} 1 & \hphantom{1}2 & 3 \\ 2 & 3 & 11 \\ 3 & 4 & 19 \end{array} \right]$ \label{matrixG}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $V = \left[ \begin{array}{rrr} i & j & k \\ -1 & 0 & 5 \\ 9 & -4 & -2 \end{array} \right] \vphantom{\left[ \begin{array}{rrrr} 1 & 0 & -3 & 0 \\ 2 & -2 & 8 & 7 \\ -5 & 0 & 16 & 0 \\ 1 & 0 & 4 & 1 \end{array} \right]}$

\item $H = \left[ \begin{array}{rrrr} 1 & 0 & -3 & 0 \\ 2 & -2 & 8 & 7 \\ -5 & 0 & 16 & 0 \\ 1 & 0 & 4 & 1 \end{array} \right]$ \label{finddetlast}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

In Exercises \ref{solvecramerfirst} - \ref{solvecramerlast}, use Cramer's Rule to solve the system of linear equations.

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\left\{ \begin{array}{rcr} 3x + 7y & = & 26 \\ 5x + 12y & = & 39 \end{array} \right.$ \label{solvecramerfirst}

\item $\left\{ \begin{array}{rcr} 2x-4y & = & 5 \\ 10x + 13y & = & -6 \end{array} \right.$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\left\{ \begin{array}{rcr} x + y & = & 8000 \\ 0.03x + 0.05y & = & 250 \end{array} \right.$

\item $\left\{ \begin{array}{rcr} \frac{1}{2}x - \frac{1}{5}y & = & 1 \\ 6x +7y & = & 3 \end{array} \right.$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\left\{ \begin{array}{rcr} x + y + z & = & 3 \\ 2x - y + z & = & 0 \\ -3x + 5y + 7z & = & 7 \end{array} \right.$

\item $\left\{ \begin{array}{rcr} 3x + y - 2z & = & 10 \\ 4x - y + z & = & 5 \\ x -3y - 4z & = & -1 \end{array} \right.$ \label{solvecramerlast}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

In Exercises \ref{cramersinglefirst} - \ref{cramersinglelast}, use Cramer's Rule to solve for $x_{\mbox{\tiny$4$}}$.

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\left\{ \begin{array}{rcr} x_{\mbox{\tiny$1$}} - x_{\mbox{\tiny$3$}} & = & -2 \\

2x_{\mbox{\tiny$2$}} - x_{\mbox{\tiny$4$}} & = & 0 \\

x_{\mbox{\tiny$1$}} - 2x_{\mbox{\tiny$2$}} + x_{\mbox{\tiny$3$}} & = & 0 \\

-x_{\mbox{\tiny$3$}} + x_{\mbox{\tiny$4$}} & = & 1 \end{array} \right.$ \label{cramersinglefirst}

\item $\left\{ \begin{array}{rcr} 4x_{\mbox{\tiny$1$}} + x_{\mbox{\tiny$2$}} & = & 4 \\

x_{\mbox{\tiny$2$}} - 3x_{\mbox{\tiny$3$}} & = & 1 \\

10x_{\mbox{\tiny$1$}} +x_{\mbox{\tiny$3$}} + x_{\mbox{\tiny$4$}} & = & 0 \\

-x_{\mbox{\tiny$2$}} + x_{\mbox{\tiny$3$}} & = & -3 \end{array} \right.$ \label{cramersinglelast}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\pagebreak

In Exercises \ref{invadjfirst} - \ref{invadjlast}, find the inverse of the given matrix using their determinants and adjoints.

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $B = \left[ \begin{array}{rr} 12 & -7 \\ -5 & 3 \end{array} \right] \vphantom{\left[ \begin{array}{rrr} 4 & \hphantom{-}6 & -3 \\ 3 & 4 & -3 \\ 1 & 2 & 6 \end{array} \right]}$ \label{invadjfirst}

\item $F = \left[ \begin{array}{rrr} 4 & \hphantom{-}6 & -3 \\ 3 & 4 & -3 \\ 1 & 2 & 6 \end{array} \right]$ \label{invadjlast}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item Carl's Sasquatch Attack! Game Card Collection is a mixture of common and rare cards. Each common card is worth $\$0.25$ while each rare card is worth $\$0.75$. If his entire 117 card collection is worth $\$48.75$, how many of each kind of card does he own?

\item How much of a 5 gallon $40\%$ salt solution should be replaced with pure water to obtain 5 gallons of a $15 \%$ solution?

\item How much of a 10 liter $30\%$ acid solution must be replaced with pure acid to obtain 10 liters of a $50\%$ solution?

\item Daniel's Exotic Animal Rescue houses snakes, tarantulas and scorpions. When asked how many animals of each kind he boards, Daniel answered: `We board 49 total animals, and I am responsible for each of their 272 legs and 28 tails.' How many of each animal does the Rescue board? (Recall: tarantulas have 8 legs and no tails, scorpions have 8 legs and one tail, and snakes have no legs and one tail.)

\item This exercise is a continuation of Exercise \ref{SasquatchDiet} in Section \ref{MatMethods}. Just because a system is consistent independent doesn't mean it will admit a solution that makes sense in an applied setting. Using the nutrient values given for Ippizuti Fish, Misty Mushrooms, and Sun Berries, use Cramer's Rule to determine the number of servings of Ippizuti Fish needed to meet the needs of a daily diet which requires 2500 calories, 1000 grams of protein, and 400 milligrams of Vitamin X. Now use Cramer's Rule to find the number of servings of Misty Mushrooms required. Does a solution to this diet problem exist?

\item Let $R = \left[ \begin{array}{rr} -7 & 3 \\ 11 & \hphantom{-} 2 \end{array} \right], \;\;\; S = \left[ \begin{array}{rr} 1 & -5 \\ 6 & 9 \end{array} \right] \;\;\; T = \left[ \begin{array}{rr} 11 & \hphantom{-} 2 \\ -7 & 3 \end{array} \right], \mbox{ and } U = \left[ \begin{array}{rr} -3 & 15 \\ 6 & 9 \end{array} \right]$

\begin{enumerate}

\item Show that $\det(RS) = \det(R)\det(S)$

\item Show that $\det(T) = -\det(R)$

\item Show that $\det(U) = -3\det(S)$

\end{enumerate}

\item For $M$, $N$, and $P$ below, show that $\det(M) = 0$, $\det(N) = 0$ and $\det(P) = 0$. \[M = \left[ \begin{array}{rrr} 1 & 2 & 3 \\ 0 & 0 & 0 \\ 7 & 8 & 9 \end{array} \right], \quad N = \left[ \begin{array}{rrr} 1 & 2 & 3 \\ 1 & 2 & 3 \\ 4 & 5 & 6 \end{array} \right] , \quad P = \left[ \begin{array}{rrr} 1 & 2 & 3 \\ -2 & -4 & -6 \\ 7 & 8 & 9 \end{array} \right] \]

\pagebreak

\item Let $A$ be an arbitrary invertible $3 \times 3$ matrix.

\begin{enumerate}

\item Show that $\det(I_{\mbox{\tiny$3$}}) = 1$.\footnote{If you think about it for just a moment, you'll see that $\det(I_{n}) = 1$ for any natural number $n$. The formal proof of this fact requires the Principle of Mathematical Induction (Section \ref{Induction}) so we'll stick with $n = 3$ for the time being.}

\item Using the facts that $AA^{-1} = I_{3}$ and $\det(AA^{-1}) = \det(A)\det(A^{-1})$, show that \[\det(A^{-1}) = \dfrac{1}{\det(A)}\]

\end{enumerate}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

The purpose of Exercises \ref{eigenfirst} - \ref{eigenlast} is to introduce you to the eigenvalues and eigenvectors of a matrix.\footnote{This material is usually given its own chapter in a Linear Algebra book so clearly we're not able to tell you everything you need to know about eigenvalues and eigenvectors. They are a nice application of determinants, though, so we're going to give you enough background so that you can start playing around with them.} We begin with an example using a $2 \times 2$ matrix and then guide you through some exercises using a $3 \times 3$ matrix. Consider the matrix \[C = \left[ \begin{array}{rr} 6 & 15 \\ 14 & 35 \end{array} \right]\] from Exercise \ref{matrixC}. We know that $\det(C) = 0$ which means that $CX = 0_{\mbox{\tiny$2$} \times \mbox{\tiny$2$}}$ does not have a unique solution. So there is a nonzero matrix $Y$ with $CY = 0_{\mbox{\tiny$2$} \times \mbox{\tiny$2$}}$. In fact, every matrix of the form \[Y = \left[ \begin{array}{r} -\frac{5}{2}t \\[3pt] t \end{array} \right]\] is a solution to $CX = 0_{\mbox{\tiny$2$} \times \mbox{\tiny$2$}}$, so there are infinitely many matrices such that $CX = 0_{\mbox{\tiny$2$} \times \mbox{\tiny$2$}}$. But consider the matrix \[X_{\mbox{\tiny$41$}} = \left[ \begin{array}{r} 3 \\ 7 \end{array} \right]\] It is NOT a solution to $CX = 0_{\mbox{\tiny$2$} \times \mbox{\tiny$2$}}$, but rather, \[CX_{\mbox{\tiny$41$}}= \left[ \begin{array}{rr} 6 & 15 \\ 14 & 35 \end{array} \right] \left[ \begin{array}{r} 3 \\ 7 \end{array} \right] = \left[ \begin{array}{r} 123 \\ 287 \end{array} \right] = 41\left[ \begin{array}{r} 3 \\ 7 \end{array} \right]\] In fact, if $Z$ is of the form \[Z = \left[ \begin{array}{r} \frac{3}{7}t \\[3pt] t \end{array} \right]\] then \[CZ = \left[ \begin{array}{rr} 6 & 15 \\ 14 & 35 \end{array} \right] \left[ \begin{array}{r} \frac{3}{7}t \\[3pt] t \end{array} \right] = \left[ \begin{array}{r} \frac{123}{7}t \\[3pt] 41t \end{array} \right] = 41\left[ \begin{array}{r} \frac{3}{7}t \\[3pt] t \end{array} \right] = 41Z\] for all $t$. The big question is ``How did we know to use $41$?''

\smallskip

We need a number $\lambda$ such that $CX = \lambda X$ has nonzero solutions. We have demonstrated that $\lambda = 0$ and $\lambda = 41$ both worked. Are there others? If we look at the matrix equation more closely, what we \emph{really} wanted was a nonzero solution to $(C - \lambda I_{\mbox{\tiny$2$}})X = 0_{\mbox{\tiny$2$} \times \mbox{\tiny$2$}}$ which we know exists if and only if the determinant of $C - \lambda I_{\mbox{\tiny$2$}}$ is zero.\footnote{Think about this.} So we computed \[\det(C - \lambda I_{\mbox{\tiny$2$}}) = \det\left(\left[ \begin{array}{rr} 6 - \lambda & 15 \\ 14 & 35 - \lambda \end{array} \right] \right) = (6 - \lambda)(35 - \lambda) - 14 \cdot 15 = \lambda^{2} - 41 \lambda\] This is called the {\bf characteristic polynomial} \index{characteristic polynomial} \index{matrix ! characteristic polynomial} of the matrix $C$ and it has two zeros: $\lambda = 0$ and $\lambda = 41$. That's how we knew to use $41$ in our work above. The fact that $\lambda = 0$ showed up as one of the zeros of the characteristic polynomial just means that $C$ itself had determinant zero which we already knew. Those two numbers are called the {\bf eigenvalues} of $C$. The corresponding matrix solutions to $CX = \lambda X$ are called the {\bf eigenvectors} of $C$ and the `vector' portion of the name will make more sense after you've studied vectors. \index{eigenvalue} \index{eigenvector}

\smallskip

Now it's your turn. In the following exercises, you'll be using the matrix $G$ from Exercise \ref{matrixG}.\[G = \left[ \begin{array}{rrr} 1 & \hphantom{1}2 & 3 \\ 2 & 3 & 11 \\ 3 & 4 & 19 \end{array} \right]\]

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item Show that the characteristic polynomial of $G$ is $p(\lambda) = -\lambda(\lambda - 1)(\lambda - 22)$. That is, compute $\text{det}\left(G - \lambda I_{\mbox{\tiny$3$}}\right)$. \label{eigenfirst}

\item Let $G_{\mbox{\tiny$0$}} = G$. Find the parametric description of the solution to the system of linear equations given by $GX = 0_{\mbox{\tiny$3$} \times \mbox{\tiny$3$}}$.

\item Let $G_{\mbox{\tiny$1$}} = G - I_{\mbox{\tiny$3$}}$. Find the parametric description of the solution to the system of linear equations given by $G_{\mbox{\tiny$1$}}X = 0_{\mbox{\tiny$3$} \times \mbox{\tiny$3$}}$. Show that any solution to $G_{\mbox{\tiny$1$}}X = 0_{\mbox{\tiny$3$} \times \mbox{\tiny$3$}}$ also has the property that $GX = 1X$.

\item Let $G_{\mbox{\tiny$22$}} = G - 22 I_{\mbox{\tiny$3$}}$. Find the parametric description of the solution to the system of linear equations given by $G_{\mbox{\tiny$22$}}X = 0_{\mbox{\tiny$3$} \times \mbox{\tiny$3$}}$. Show that any solution to $G_{\mbox{\tiny$22$}}X = 0_{\mbox{\tiny$3$} \times \mbox{\tiny$3$}}$ also has the property that $GX = 22X$. \label{eigenlast}

\end{enumerate}

\newpage

\subsection{Answers}

\begin{multicols}{2}

\begin{enumerate}

\item $\det(B) = 1$

\item $\det(C) = 0$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\det(Q) = x^{2} \phantom{\dfrac{1}{x^{7}}}$

\item $\det(L) = \dfrac{1}{x^{7}}$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\det(F) = -12$

\item $\det(G) = 0$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\det(V) = 20i + 43j + 4k$

\item $\det(H) = -2$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $x = 39, \; y = -13$

\item $x = \frac{41}{66}, \; y=-\frac{31}{33}$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $x=7500, \; y=500$

\item $x = \frac{76}{47}, \; y=-\frac{45}{47}$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $x = 1, \; y = 2, \; z = 0$

\item $x = \frac{121}{60}, \; y = \frac{131}{60}, \; z = -\frac{53}{60}$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $x_{\mbox{\tiny$4$}} = 4$

\item $x_{\mbox{\tiny$4$}} = -1$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $B^{-1} = \left[ \begin{array}{rr} 3 & 7 \\ 5 & 12 \end{array} \right]$

\item $F^{-1} = \left[ \begin{array}{rrr} -\frac{5}{2} & \frac{7}{2} & \frac{1}{2} \\[3pt] \frac{7}{4} & -\frac{9}{4} & -\frac{1}{4} \\[3pt] -\frac{1}{6} & \frac{1}{6} & \frac{1}{6} \end{array} \right]$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item Carl owns 78 common cards and 39 rare cards.

\item $3.125$ gallons.

\item $\frac{20}{7} \approx 2.85$ liters.

\item The rescue houses 15 snakes, 21 tarantulas and 13 scorpions.

\item Using Cramer's Rule, we find we need 53 servings of Ippizuti Fish to satisfy the dietary requirements. The number of servings of Misty Mushrooms required, however, is $-1120$. Since it's impossible to have a negative number of servings, there is no solution to the applied problem, despite there being a solution to the mathematical problem. A cautionary tale about using Cramer's Rule: just because you are guaranteed a mathematical answer for each variable doesn't mean the solution will make sense in the `real' world.

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\closegraphsfile

8.6: Partial Fraction Decomposition

\subsection{Exercises}

In Exercises \ref{parfracformfirst} - \ref{parfracformlast}, find only the \emph{form} needed to begin the process of partial fraction decomposition. Do not create the system of linear equations or attempt to find the actual decomposition.

\begin{multicols}{2}

\begin{enumerate}

\item $\dfrac{7}{(x - 3)(x + 5)}$ \label{parfracformfirst}

\item $\dfrac{5x + 4}{x(x - 2)(2 - x)}$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\dfrac{m}{(7x - 6)(x^{2} + 9)}$

\item $\dfrac{ax^{2} + bx + c}{x^3(5x + 9)(3x^{2} + 7x + 9)}$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\dfrac{\text{A polynomial of degree } < 9}{(x + 4)^{5}(x^{2} + 1)^{2}}$

\item $\dfrac{\text{A polynomial of degree } < 7}{x(4x - 1)^{2}(x^{2} + 5)(9x^{2} + 16)}$ \label{parfracformlast}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

In Exercises \ref{findparfracfirst} - \ref{findparfraclast}, find the partial fraction decomposition of the following rational expressions.

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\dfrac{2x}{x^{2} - 1}$ \label{findparfracfirst}

\item $\dfrac{-7x + 43}{3x^{2} + 19x - 14}$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\dfrac{11x^{2} - 5x - 10}{5x^{3} - 5x^{2}}$

\item $\dfrac{-2x^{2} + 20x - 68}{x^{3} + 4x^{2} + 4x + 16}$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\dfrac{-x^{2} + 15}{4x^{4} + 40x^{2} + 36}$

\item $\dfrac{-21x^{2} + x - 16}{3x^{3} + 4x^{2} - 3x + 2}$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\dfrac{5x^{4} - 34x^{3} + 70x^{2} - 33x - 19}{(x - 3)^{2}}$

\item $\dfrac{x^{6} + 5x^{5} + 16x^{4} + 80x^{3} - 2x^{2} + 6x - 43}{x^{3} + 5x^{2} + 16x + 80}$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\dfrac{-7x^{2} - 76x - 208}{x^{3} + 18x^{2} + 108x + 216}$

\item $\dfrac{-10x^{4} + x^{3} - 19x^{2} + x - 10}{x^{5} + 2x^{3} + x}$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\dfrac{4x^{3} - 9x^{2} + 12x + 12}{x^{4} - 4x^{3} + 8x^{2} - 16x + 16}$

\item $\dfrac{2x^{2} + 3x + 14}{(x^{2} + 2x + 9)(x^{2} + x + 5)}$ \label{findparfraclast}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item As we stated at the beginning of this section, the technique of resolving a rational function into partial fractions is a skill needed for Calculus. However, we hope to have shown you that it is worth doing if, for no other reason, it reinforces a hefty amount of algebra. One of the common algebraic errors the authors find students make is something along the lines of

\[ \dfrac{8}{x^2 - 9} \neq \dfrac{8}{x^2} - \dfrac{8}{9}\]

Think about why if the above were true, this section would have no need to exist.

\end{enumerate}

\newpage

\subsection{Answers}

\begin{multicols}{2}

\begin{enumerate}

\item $\dfrac{A}{x - 3} + \dfrac{B}{x + 5}$

\item $\dfrac{A}{x} + \dfrac{B}{x - 2} + \dfrac{C}{(x - 2)^{2}}$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\dfrac{A}{7x - 6} + \dfrac{Bx + C}{x^{2} + 9}$

\item $\dfrac{A}{x} + \dfrac{B}{x^{2}} + \dfrac{C}{x^{3}} + \dfrac{D}{5x + 9} + \dfrac{Ex + F}{3x^{2} + 7x + 9}$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\dfrac{A}{x + 4} + \dfrac{B}{(x + 4)^{2}} + \dfrac{C}{(x + 4)^{3}} + \dfrac{D}{(x + 4)^{4}} + \dfrac{E}{(x + 4)^{5}} + \dfrac{Fx + G}{x^{2} + 1} + \dfrac{Hx + I}{(x^{2} + 1)^{2}}$

\item $\dfrac{A}{x} + \dfrac{B}{4x - 1} + \dfrac{C}{(4x - 1)^{2}} + \dfrac{Dx + E}{x^{2} + 5} + \dfrac{Fx + G}{9x^{2} + 16}$

\item $\dfrac{2x}{x^{2} - 1} = \dfrac{1}{x + 1} + \dfrac{1}{x - 1}$

\item $\dfrac{-7x + 43}{3x^{2} + 19x - 14}= \dfrac{5}{3x - 2} - \dfrac{4}{x + 7}$

\item $\dfrac{11x^{2} - 5x - 10}{5x^{3} - 5x^{2}} = \dfrac{3}{x} + \dfrac{2}{x^{2}} - \dfrac{4}{5(x - 1)}$

\item $\dfrac{-2x^{2} + 20x - 68}{x^{3} + 4x^{2} + 4x + 16} = -\dfrac{9}{x + 4} + \dfrac{7x - 8}{x^{2} + 4}$

\item $\dfrac{-x^{2} + 15}{4x^{4} + 40x^{2} + 36} = \dfrac{1}{2(x^{2} + 1)} - \dfrac{3}{4(x^{2} + 9)}$

\item $\dfrac{-21x^{2} + x - 16}{3x^{3} + 4x^{2} - 3x + 2} = -\dfrac{6}{x + 2} - \dfrac{3x + 5}{3x^{2} - 2x + 1}$

\item $\dfrac{5x^{4} - 34x^{3} + 70x^{2} - 33x - 19}{(x - 3)^{2}} = 5x^{2} - 4x + 1 + \dfrac{9}{x - 3} - \dfrac{1}{(x - 3)^{2}}$

\item $\dfrac{x^{6} + 5x^{5} + 16x^{4} + 80x^{3} - 2x^{2} + 6x - 43}{x^{3} + 5x^{2} + 16x + 80} = x^{3} + \dfrac{x + 1}{x^{2} + 16} - \dfrac{3}{x + 5}$

\item $\dfrac{-7x^{2} - 76x - 208}{x^{3} + 18x^{2} + 108x + 216} = -\dfrac{7}{x + 6} + \dfrac{8}{(x + 6)^{2}} - \dfrac{4}{(x + 6)^{3}}$

\item $\dfrac{-10x^{4} + x^{3} - 19x^{2} + x - 10}{x^{5} + 2x^{3} + x} = -\dfrac{10}{x} + \dfrac{1}{x^{2} + 1} + \dfrac{x}{(x^{2} + 1)^{2}}$

\item $\dfrac{4x^{3} - 9x^{2} + 12x + 12}{x^{4} - 4x^{3} + 8x^{2} - 16x + 16}= \dfrac{1}{x - 2} + \dfrac{4}{(x - 2)^{2}} + \dfrac{3x + 1}{x^{2} + 4}$

\item $\dfrac{2x^{2} + 3x + 14}{(x^{2} + 2x + 9)(x^{2} + x + 5)} = \dfrac{1}{x^{2} + 2x + 9} + \dfrac{1}{x^{2} + x + 5}$

\end{enumerate}

\closegraphsfile

8.7: Systems of Non-Linear Equations and Inequalities

\subsection{Exercises}

In Exercises \ref{solvenonlin1first} - \ref{solvenonlin1last}, solve the given system of nonlinear equations. Sketch the graph of both equations on the same set of axes to verify the solution set.

\begin{multicols}{3}

\begin{enumerate}

\item $\left\{\begin{array}{rcr} x^2 - y & = & 4 \\ x^{2} + y^{2} & = & 4 \\ \end{array} \right.$ \label{solvenonlin1first}

\item $\left\{\begin{array}{rcr} x^{2} + y^{2} & = & 4 \\ x^2 - y & = & 5 \\ \end{array} \right.$

\item $\left\{\begin{array}{rcr} x^2+y^2 & = & 16 \\ 16x^{2} + 4y^{2} & = & 64 \\ \end{array} \right.$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{3}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\left\{\begin{array}{rcr} x^2+y^2 & = & 16 \\ 9x^{2} - 16y^{2} & = & 144 \\ \end{array} \right.$

\item $\left\{\begin{array}{rcr} x^2+y^2 & = & 16 \\ \frac{1}{9} y^2 - \frac{1}{16} x^2& = & 1 \\ \end{array} \right.$

\item $\left\{\begin{array}{rcr} x^{2} + y^{2} & = & 16 \\ x-y & = & 2 \\ \end{array} \right.$ \label{solvenonlin1last}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

In Exercises \ref{solvenonlin2first} - \ref{solveninlin2last}, solve the given system of nonlinear equations. Use a graph to help you avoid any potential extraneous solutions.

\begin{multicols}{3}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\left\{\begin{array}{rcr} x^{2} - y^{2} & = & 1 \\ x^{2} + 4y^{2} & = & 4 \\ \end{array} \right.$

\item $\left\{\begin{array}{rcr} \sqrt{x + 1} - y & = & 0 \\ x^{2} + 4y^{2} & = & 4 \\ \end{array} \right.$

\item $\left\{\begin{array}{rcr} x + 2y^{2} & = & 2 \\ x^{2} + 4y^{2} & = & 4 \\ \end{array} \right.$ \label{solvenonlin2first}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{3}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\left\{\begin{array}{rcr} (x - 2)^{2} + y^{2} & = & 1 \\ x^{2} + 4y^{2} & = & 4 \\ \end{array} \right.$

\item $\left\{\begin{array}{rcr} x^{2} + y^{2} & = & 25 \\ y - x & = & 1 \\ \end{array} \right.$

\item $\left\{\begin{array}{rcr} x^{2} + y^{2} & = & 25 \\ x^{2} + (y - 3)^{2} & = & 10 \\ \end{array} \right.$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{3}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\left\{\begin{array}{rcr} y & = & x^{3} + 8 \\ y & = & 10x - x^{2} \\ \end{array} \right. \vphantom{\left\{\begin{array}{rcr} x^{2} + y^{2} & = & 25 \\ 4x^{2} - 9y & = & 0 \\ 3y^{2} - 16x & = & 0\end{array} \right.}$

\item $\left\{\begin{array}{rcr} x^2 - xy & = & 8 \\ y^2 - xy & = & 8 \\ \end{array} \right. \vphantom{\left\{\begin{array}{rcr} x^{2} + y^{2} & = & 25 \\ 4x^{2} - 9y & = & 0 \\ 3y^{2} - 16x & = & 0\end{array} \right.}$

\item $\left\{\begin{array}{rcr} x^{2} + y^{2} & = & 25 \\ 4x^{2} - 9y & = & 0 \\ 3y^{2} - 16x & = & 0\end{array} \right.$ \label{solveninlin2last}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item A certain bacteria culture follows the Law of Uninbited Growth, Equation \ref{lawofuninhibitedgrowth}. After 10 minutes, there are 10,000 bacteria. Five minutes later, there are 14,000 bacteria. How many bacteria were present initially? How long before there are 50,000 bacteria?

\setcounter{HW}{\value{enumi}}

\end{enumerate}

Consider the system of nonlinear equations below \[\left\{\begin{array}{rcr} \dfrac{4}{x} + \dfrac{3}{y} & = & 1 \\[10pt] \dfrac{3}{x} + \dfrac{2}{y} & = & -1 \\ \end{array} \right.\] If we let $u = \frac{1}{x}$ and $v = \frac{1}{y}$ then the system becomes \[\left\{\begin{array}{rcr} 4u + 3v & = & 1 \\ 3u + 2v & = & -1 \\ \end{array} \right.\] This associated system of linear equations can then be solved using any of the techniques presented earlier in the chapter to find that $u = -5$ and $v = 7$. Thus $x = \frac{1}{u} = -\frac{1}{5}$ and $y = \frac{1}{v} = \frac{1}{7}$.

\smallskip

We say that the original system is {\bf linear in form} \index{system of equations ! linear in form} because its equations are not linear but a few substitutions reveal a structure that we can treat like a system of linear equations. Each system in Exercises \ref{linearformfirst} - \ref{linearformlast} is linear in form. Make the appropriate substitutions and solve for $x$ and $y$.

\begin{multicols}{3}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\left\{\begin{array}{rcr} 4x^{3} + 3\sqrt{y} & = & 1 \\ 3x^{3} + 2\sqrt{y} & = & -1 \end{array} \right.$ \label{linearformfirst}

\item $\left\{\begin{array}{rcr} 4e^{x} + 3e^{-y} & = & 1 \\ 3e^{x} + 2e^{-y} & = & -1 \end{array} \right.$

\item $\left\{\begin{array}{rcr} 4\ln(x) + 3y^{2} & = & 1 \\ 3\ln(x) + 2y^{2} & = & -1 \end{array} \right.$ \label{linearformlast}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item Solve the following system \[\left\{\begin{array}{rcr} x^{2} + \sqrt{y} + \log_{2}(z) & = & 6 \\[3pt] 3x^{2} - 2\sqrt{y} + 2\log_{2}(z) & = & 5 \\[3pt] -5x^{2} + 3\sqrt{y} + 4\log_{2}(z) & = & 13 \end{array} \right.\]

\setcounter{HW}{\value{enumi}}

\end{enumerate}

In Exercises \ref{nonlinearsysineqfirst} - \ref{nonlinearsysineqlast}, sketch the solution to each system of nonlinear inequalities in the plane.

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\left\{\begin{array}{rcr} x^{2} - y^{2} & \leq & 1 \\ x^{2} + 4y^{2} & \geq & 4 \end{array} \right.$ \label{nonlinearsysineqfirst}

\item $\left\{\begin{array}{rcr} x^{2} + y^{2} & < & 25 \\ x^{2} + (y - 3)^{2} & \geq & 10 \end{array} \right.$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\left\{\begin{array}{rcr} (x - 2)^{2} + y^{2} & < & 1 \\ x^{2} + 4y^{2} & < & 4 \end{array} \right.$

\item $\left\{\begin{array}{rcr} y & > & 10x - x^{2} \\ y & < & x^{3} + 8 \end{array} \right.$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\left\{\begin{array}{rcr} x + 2y^{2} & > & 2 \\ x^{2} + 4y^{2} & \leq & 4 \end{array} \right.$

\item $\left\{\begin{array}{rcr} x^{2} + y^{2} & \geq & 25 \\ y - x & \leq & 1 \end{array} \right.$ \label{nonlinearsysineqlast}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item Systems of nonlinear equations show up in third semester Calculus in the midst of some really cool problems. The system below came from a problem in which we were asked to find the dimensions of a rectangular box with a volume of 1000 cubic inches that has minimal surface area. The variables $x$, $y$ and $z$ are the dimensions of the box and $\lambda$ is called a Lagrange multiplier. With the help of your classmates, solve the system.\footnote{If using $\lambda$ bothers you, change it to $w$ when you solve the system.} \[\left\{\begin{array}{rcr} 2y + 2z & = & \lambda yz \\ 2x + 2z & = & \lambda xz \\ 2y + 2x & = & \lambda xy \\ xyz & = & 1000 \end{array} \right.\]

\item According to Theorem \ref{realfactorization} in Section \ref{ComplexZeros}, the polynomial $p(x) = x^{4} + 4$ can be factored into the product linear and irreducible quadratic factors. In this exercise, we present a method for obtaining that factorization.

\label{factorpolywithnonlinear}

\begin{enumerate}

\item Show that $p$ has no real zeros.

\item Because $p$ has no real zeros, its factorization must be of the form $(x^{2} + ax + b)(x^{2} + cx + d)$ where each factor is an irreducible quadratic. Expand this quantity and gather like terms together.

\item Create and solve the system of nonlinear equations which results from equating the coefficients of the expansion found above with those of $x^{4} + 4$. You should get four equations in the four unknowns $a$, $b$, $c$ and $d$. Write $p(x)$ in factored form.

\end{enumerate}

\item Factor $q(x) = x^{4} + 6x^{2} - 5x + 6$.

\end{enumerate}

\newpage

\subsection{Answers}

\begin{multicols}{2}

\begin{enumerate}

\item $(\pm 2, 0)$, $\left(\pm \sqrt{3}, -1\right)$ \\

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\item No solution \\

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\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $(0, \pm 4)$ \\

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\columnbreak

\item $(\pm 4,0)$ \\

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\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\left(\pm \frac{4 \sqrt{7}}{5}, \pm \frac{12 \sqrt{2}}{5} \right)$ \\

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\columnbreak

\item $\left( 1 + \sqrt{7}, -1 + \sqrt{7} \right)$, $\left( 1 - \sqrt{7}, -1 - \sqrt{7} \right)$ \\

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\begin{multicols}{3}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\left( \pm \frac{2 \sqrt{10}}{5}, \pm \frac{\sqrt{15}}{5} \right)$

\item $(0, 1)$

\item $(0, \pm 1), \; (2, 0)$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{3}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\left( \frac{4}{3}, \pm \frac{\sqrt{5}}{3} \right)$

\item $(3, 4), \; (-4, -3)$

\item $(\pm 3, 4)$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{3}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $(-4, -56), \; (1, 9), \; (2, 16)$

\item $(-2,2)$, $(2,-2)$

\item $(3, 4)$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item Initially, there are $\frac{250000}{49} \approx 5102$ bacteria. It will take $\frac{5\ln(49/5)}{\ln(7/5)} \approx 33.92$ minutes for the colony to grow to 50,000 bacteria.

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\begin{multicols}{3}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\left( -\sqrt[3]{5}, 49 \right)$

\item No solution

\item $\left( e^{-5}, \pm \sqrt{7} \right)$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $(1, 4, 8), \; (-1, 4, 8)$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\left\{\begin{array}{rcr} x^{2} - y^{2} & \leq & 1 \\ x^{2} + 4y^{2} & \geq & 4 \end{array} \right.$ \\

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\item $\left\{\begin{array}{rcr} x^{2} + y^{2} & < & 25 \\ x^{2} + (y - 3)^{2} & \geq & 10 \end{array} \right.$\\

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\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\left\{\begin{array}{rcr} (x - 2)^{2} + y^{2} & < & 1 \\ x^{2} + 4y^{2} & < & 4 \end{array} \right.$ \\

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\columnbreak

\item $\left\{\begin{array}{rcr} y & > & 10x - x^{2} \\ y & < & x^{3} + 8 \end{array} \right.$ \\

\begin{mfpic}[10][1]{-5}{3.1}{-56}{42}

\fillcolor[gray]{.7}

\gfill \btwnfcn{-4,1,0.1}{10*x - (x**2)}{(x**3) + 8}

\gfill \btwnfcn{2,3.2,0.1}{10*x - (x**2)}{(x**3) + 8}

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\arrow \dashed \function{2,3.2,0.1}{10*x - (x**2)}

\arrow \dashed \function{2,3.2,0.1}{(x**3) + 8}

\axes

\tlabel[cc](3,-4){\scriptsize $x$}

\tlabel[cc](0.5,42){\scriptsize $y$}

\xmarks{-4 step 1 until 2}

\ymarks{-56,9,16}

\tlpointsep{4pt}

\tiny

\axislabels {x}{{$-4 \hspace{7pt}$} -4, {$-3 \hspace{7pt}$} -3, {$-2 \hspace{7pt}$} -2, {$-1 \hspace{7pt}$} -1, {$1$} 1, {$2$} 2}

\axislabels {y}{{$-56$} -56, {$9$} 9, {$16$} 16}

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\end{mfpic}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\left\{\begin{array}{rcr} x + 2y^{2} & > & 2 \\ x^{2} + 4y^{2} & \leq & 4 \end{array} \right.$ \\

\begin{mfpic}[30]{-0.5}{2.5}{-1.5}{1.5}

\fillcolor[gray]{.7}

\gfill \btwnfcn{0,2,0.1}{sqrt(1 - (x/2))}{sqrt(1 - ((x**2)/4))}

\gfill \btwnfcn{0,2,0.1}{-sqrt(1 - (x/2))}{-sqrt(1 - ((x**2)/4))}

\parafcn{-1.5708, 1.5708,0.1}{(2*cos(t),sin(t))}

\dashed \parafcn{-1,1,0.1}{(2 - (2*(t**2)), t)}

\axes

\gclear \circle{(2,0), 0.05}

\circle{(2,0), 0.05}

\gclear \circle{(0,1), 0.05}

\circle{(0,1), 0.05}

\gclear \circle{(0,-1), 0.05}

\circle{(0,-1), 0.05}

\tlabel[cc](2.5,-0.25){\scriptsize $x$}

\tlabel[cc](0.25,1.5){\scriptsize $y$}

\xmarks{1}

\tlpointsep{4pt}

\tiny

\axislabels {x}{{$1$} 1, {$2$} 2.1}

\axislabels {y}{{$-1$} -1, {$1$} 1}

\normalsize

\end{mfpic}

\vfill

\columnbreak

\item $\left\{\begin{array}{rcr} x^{2} + y^{2} & \geq & 25 \\ y - x & \leq & 1 \end{array} \right.$ \\

\begin{mfpic}[6]{-7}{7}{-7}{8}

\fillcolor[gray]{.7}

\gfill \btwnfcn{-7,7,0.1}{-7}{1+x}

\gclear \circle{(0,0), 5}

\arc[t]{(-4,-3), (5,0), (3,4)}

\arrow \reverse \function{-7,-4,0.1}{1+x}

\arrow \function{3,7,0.1}{1+x}

\axes

\tlabel[cc](7,-0.5){\scriptsize $x$}

\tlabel[cc](0.5,8){\scriptsize $y$}

\xmarks{-6 step 1 until 6}

\ymarks{-6 step 1 until 7}

\tlpointsep{4pt}

\tiny

\axislabels {x}{{$-5 \hspace{7pt}$} -5, {$-3 \hspace{7pt}$} -3, {$-1 \hspace{7pt}$} -1, {$1$} 1, {$3$} 3, {$5$} 5}

\axislabels {y}{ {$-5$} -5, {$-3$} -3, {$1$} 1, {$3$} 3, {$5$} 5, {$7$} 7}

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\end{mfpic}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $x = 10, \; y = 10, \; z = 10, \lambda = \frac{2}{5}$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item \begin{enumerate}

\addtocounter{enumii}{2}

\item $x^{4} + 4 = (x^{2} - 2x + 2)(x^{2} + 2x + 2)$

\end{enumerate}

\item $x^{4} + 6x^{2} - 5x + 6 = (x^{2} - x + 1)(x^{2} + x + 6)$

\end{enumerate}

\closegraphsfile