Answers to Study Guide Questions

Unit 1

A linear equation is an equation of the form $a x + b = 0$ , where $a$ and $b$ are any real numbers.
For a linear equation $y = a x + b$ , a linear inequality is of the form $a x + b \leq c$ , for some constant $c$ .
A break-even analysis identifies the point at which two distinct events are equal.
No, the number 4 is in the interval, but it does not satisfy the equation.
If a line passes through the two points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ , then its slope is $\frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ .
A linear equation in two variables is an equation of the form $a x + b y = c$ , where $a$ , $b$ , and $c$ are any real numbers.
An equation of a horizontal line is an equation of the form $y = b$ , for some constant $b$ .
An equation of a vertical line is an equation of the form $x = a$ , for some constant $a$ .
An equilibrium point is the point at which supply and demand are equal.
The slope of a line is zero when the line is horizontal.
The slope of a line with variables $x$ and $y$ is defined as the rate of change of $y$ with respect to $x$ because if the slope of the line is $m$ , then if $x$ increases by one unit, the variable $y$ increases by $m$ units. Thus, the variable $y$ increases $m$ units per unit increase in $x$ .
A set of data is the information obtained through a period of time about the relationship between two or more variables, one of which depends on the other or others.
Regression analysis is a process for finding a function that provides a useful mathematical model for a set of given data.
Linear regression is a technique used to obtain as a mathematical model a line that best fits the given data.

Unit 2

The dependent variables are the person (a person has only one income), the item in a supermarket, the student, the day, the item manufactured, the item sold, the square, and the number.
We use the vertical line test when we want to know whether a curve sketched on the Cartesian plane is a function or not.
The notation $f (x)$ identifies a function where the variable $f$ depends on the variable $x$ .
If $p = m - n x$ is a price-demand function, then $x$ is the number of items that can be sold at price $p$ . If $x = 0$ , then $p = 0$ . Thus, $m$ is the price at which no item can be sold. Finally, $n$ is the rate at which the price is decreasing. For every item sold, the price decreases $n$ dollars.
Profit $P$ is revenue $R$ minus cost $C$ . If $P (a) = 0$ , then $R (a) = C (a)$ . Therefore, $a$ is the number of items that must be sold in order to break even, that is, to experience no gains and no losses.
If $f (x) = \sqrt{x}$ , then, $f (\frac{- 1}{2})$ does not exist, because $f (\frac{- 1}{2}) = \sqrt{\frac{- 1}{2}}$ , and the square root of a negative number is not defined.
The domain of the function $f (x) = \sqrt{|x|}$ is all real numbers because $|x| \geq 0$ for all real numbers. Thus, $\sqrt{|x|}$ is defined for any number $x$ , and the domain is all real numbers.
To shift the graph of $f$ upward by $5$ units, we add the constant $5$ to $f$ ; hence, $f (x) + 5$ . To stretch the graph $2$ units vertically, we multiply it by $2$ . Therefore, the required function is $g (x) = 2 (f (x) + 5) = 2 f (x) + 10$ .
We represent the graph of a function $f (x)$ on the Cartesian plane because for each value of $f (x) = b$ we have a pair ( $a$ , $b$ ) on the Cartesian plane. The collection of all these points forms the graph or visual representation of $f$ .
If $a > 0$ , the parabola opens upwards. If $a < 0$ , the parabola opens downwards.
If $b^{2} - 4 a c > 0$ , the parabola does not have roots, which means that it does not cross the $x$ -axis.
Yes. The polynomial function $g (x) = x^{2} + 1$ is nonzero for any real number $x$ ; hence, the domain of the rational function $f (x) g (x)$ is all real numbers.
Yes. The function $g (x) = 1$ is a polynomial function; hence, any polynomial function $f (x)$ is rational, because $f (x) = \frac{f (x)}{1}$ .
The exponential function $a^{x}$ is decreasing for any $0 < a < 1$ .
The base $b$ of an exponential function $b^{x}$ is positive because if $b$ is negative, then $b^{1 ∕ 2} = \sqrt{b}$ is not defined as a real number. So, for an exponential function $b^{x}$ to be defined for all real numbers $x$ , $b$ must be positive.
Yes, if $e^{x} = e^{y}$ , then $x = y$ . Any horizontal line crosses the graph of an exponential function at one and only one point; hence, $e^{x}$ is unique for each real number $x$ , and $e^{x} = e^{y}$ only if $x = y$ .
Compound growth is an application of exponential functions because the function that corresponds to compound growth is of the form $A (t) = P {(1 + \frac{r}{m})}^{m t}$ , which is an exponential function with base $1 + \frac{r}{m}$ .
The function or mathematical model that best fits a data set of an exponential regression is the function $y = a b^{x}$ , where $a$ and $b$ are constants and $b > 0$ .
The exponential and logarithmic functions are invertible because any horizontal line crosses the graphs of any exponential and logarithmic function at one and only one point. Therefore, these graphs are one-to-one.
$\log_{64} 8 = \frac{1}{2}$ because $\log_{b} x$ is the number to which we must raise $b$ to obtain $x$ . Since $64^{1 / 2} = 8$ , we have $\log_{64} 8 = \frac{1}{2}$ .
It is true that if $\ln x = \ln y$ , then $x = y$ , because $\ln x$ is a one-to-one function.
The properties of the logarithmic functions are “dual” to the properties of the exponential functions because the sum and product are dual properties. The properties of the logarithmic function listed on page 109 and the properties of the exponential function listed on page 96 correspond to each other, if we change plus or minus into product. For instance, $\log_{b} M N = \log_{b} M + \log_{b} N$ corresponds to $e^{M + N} = (e^{M}) (e^{N})$ .
Any exponential function can be represented as an exponential function in base $e$ because if $a^{x}$ is any exponential function, then $a^{x} = e^{k x}$ , where $k = \ln a$ .
Any logarithmic function can be represented as a natural logarithmic function because if $\log_{b} x$ is any logarithmic function, then $\log_{b} x = \frac{\ln x}{k}$ , where $k = \ln b$ .

Unit 3

In the formula $A = P (1 + r t)$ , $P$ is the amount to be invested at an annual rate $r$ to yield an amount $A$ after $t$ years.
We use the formula $P = \frac{A}{1 + r t}$ when we want to know the present value we must invest at a rate $r$ for a period of $t$ years in order to get an amount $A$ .
We use the formula $r = \frac{A - P}{P t}$ when we want to know at what rate we must invest an amount $P$ if we want to obtain an amount $A$ after a period of $t$ years.
We use the formula $t = \frac{A - P}{P r}$ when we want to know the number of years we must invest an amount $P$ at a rate $r$ if we want to obtain an amount $A$ .
Simple interest gives the amount earned after $t$ years, without reinvestment of the interest. Compound interest gives the amount earned after $n$ periods, when the interest is reinvested after each period.
Compound interest gives the amount earned after $n$ periods. Continuous interest gives the amount earned when the sum invested is compounded for infinitely many very small periods $n$ (that is, if $n$ tends to infinity).
The investment doubles in value, so we use the continuous compound interest formula $2 P = P e^{r t}$ . Solving for $t$ :
$2 = e^{r t}$
$t = \frac{\ln 2}{r}$ .
The investment doubles in value, so we use the continuous compound interest formula $2 P = P e^{r t}$ . Solving for $r$ :
$2 = e^{r t}$
$r = \frac{\ln 2}{t}$ .
The annual percentage yield is the simple interest rate that will produce the same amount of revenue in one year as a given nominal annual rate compounded a set number of times will produce in one year.
An annuity is any sequence of equal periodic payments.
If the interest is compounded $m$ times a year, then $i = \frac{r}{m}$ , and after the first period the amount is
$\begin{array}{l} P M T + P M T \frac{r}{m} + P M T & = P M T (2 + \frac{r}{m}) \\ P M T + P M T (2 + \frac{r}{m}) + P M T (2 + \frac{r}{m}) (\frac{r}{m}) & = P M T (1 + (2 + \frac{r}{m}) (1 + \frac{r}{m})) \end{array}$
A sinking fund is an account that is established to accumulate funds to meet future obligations.
An amortization is the process by which a debt is reduced to zero in a given length of time by making equal periodic payments that include compounded interest.
Present value is the amount of each periodic payment made until a fixed amount is zero. An amortization payment is the amount of the periodic payments to be made until a present value is zero.
As the periods are monthly, we would use formula (5) on page 161, adjusting it slightly. The loan is $P V$ , $i = \frac{r}{12}$ and $n = 12 m$ . Therefore, $P M T = P V \frac{\frac{r}{12}}{1 - {(1 - \frac{r}{12})}^{12 m}}$ .

Unit 4

The graphing method used to solve a system of linear equations with two variables is not very exact. The solution is the intersection of the two lines, and the coordinates may not be easy to establish.
To determine whether a system of linear operators with two variables is inconsistent, we find the slope of each line. If the slopes are equal, the lines are either parallel or the same. If the $y$ -intersects are different, the lines are parallel. Thus, a system is inconsistent if the slopes are equal and the $y$ -intercepts are different.
We know that a system of linear equations with two variables has only one solution if the slopes of the lines are not equal.
Two systems of linear equations are equivalent when their solutions are equal.
We know that the mathematical model needed to solve a problem is a system of linear equations with two variables if the problem requires us to know two distinct quantities.
The augmented matrix that corresponds to a system of four linear equations of two variables has four rows and two columns, hence $4 \times 2$ .
We know that a system of three linear equations and two variables has only one solution if each linear equation has a different slope, or if after we perform the row operations to the augmented matrix the diagonal of the matrix consists of ones.
When we perform row operations to an augmented matrix, the corresponding systems of equations are equivalent. The more ones and zeros an augmented matrix has, the easier is to find the solution of the corresponding system.
From the second row, we conclude that the second variable is 0, and from the first row, we conclude that the first variable is 0; hence, the solution is $(0, 0)$ .
From the second row, we conclude that the second variable is $a$ , and from the first row, we conclude that the first variable is equal to $- a$ . Therefore, the solution is $(- a, a)$ .
We know from the reduced form of an augmented matrix that a system is inconsistent if we have a row that yields a contradiction, for instance $0 0 ... 0 a$ where $a \neq 0$ .
We know from the reduced form of an augmented matrix that a system has only one solution if the matrix has the dimensions $m \times n$ , and if the reduced form has a pivot in each of the first $n - 1$ columns.
We know from the reduced form of an augmented matrix that a system has infinitely many solutions if the matrix has the dimensions $m \times n$ , and if the reduced form has no pivots in the first $n - 1$ columns.
If $A$ and $B$ are matrices of the same dimensions, the statements $A + B = B + A$ and $A + A + A = 3 A$ are both true. The $c_{i j}$ entry of the matrix $A + B$ is equal to the sum of the $a_{i j}$ entry of $A$ and the $b_{i j}$ entry of $B;$ hence, $c_{i j} = a_{i j} + b_{i j}$ . Since $a_{i j} + b_{i j} = b_{i j} + a_{i j}$ , we have that $c_{i j}$ is the corresponding entry of $B + A$ . Thus, $A + B = B + A$ . If $a_{i j}$ is the entry of $A$ , then the $c_{i j}$ entry of $A + A + A$ is $c_{i j} = a_{i j} + a_{i j} + a_{i j} = 3 a_{i j}$ . Since $3 a_{i j}$ is the entry of $3 A$ , we conclude that $A + A + A = 3 A$ .
If $A$ and $B$ are square matrices, it is not necessarily true that $A B = B A$ . Exercise 46 on page 221 gives two square matrices such that $A B \neq B A$ .
If $A$ and $B$ are square matrices, it is not necessarily true that if $A B = 0$ , then $A = 0$ or $B = 0$ . Exercise 45 on page 221 gives two nonzero square matrices such that $A B = 0$ .
If $A$ and $B$ are square matrices, it is not necessarily true that if $A B = 0$ , then $B A = 0$ . Exercise 46 on page 221 gives an example of two square matrices such that $B A = 0$ and $A B \neq 0$ .
A matrix is invertible if its reduced form is the identity matrix.
If $R$ and $S$ are two rows in a square matrix $M$ , and $R = k S$ for some constant $k$ , then the matrix $M$ is not invertible. We established in the previous question that the reduced form of the matrix $M$ must be the identity. However, if $R = k S$ , then the reduced form of the matrix has a row of zeros; hence, it is not the identity.
It is true that if a matrix $M$ is invertible, then it has only one inverse $M^{- 1}$ . If $M$ has two inverses, $M^{- 1}$ and $A^{- 1}$ , then $M^{- 1} M = I$ and $M A^{- 1} = I$ ; hence $M^{- 1} M A^{- 1} = M^{- 1} I = M^{- 1}$ . Therefore, $I A^{- 1} = A^{- 1} = M^{- 1}$ . So, they are equal.
It is true that if a matrix $M$ is invertible, then its inverse matrix $M^{- 1}$ is also invertible. The inverse of $M^{- 1}$ is the matrix $M$ ; that is, ${(M^{- 1})}^{- 1} = M$ , because $M^{- 1} {(M^{- 1})}^{- 1} = I = M^{- 1} M$ . This statement says that ${(M^{- 1})}^{- 1}$ and $M$ is the inverse of $M^{- 1}$ . From the previous question, we know that the inverse is unique; hence, ${(M^{- 1})}^{- 1} = M$ .
We consider only square matrices when we look for the inverse of a matrix because if a matrix $M$ is not square $—$ say it is $m \times n$ $—$ then its inverse $M^{- 1}$ must be $r \times m$ , so that the product $M^{- 1} M$ is well defined. But if $r \neq n$ , then the product $M M^{- 1}$ is not well defined. So, we do not have $M^{- 1} M = M M^{- 1}$ .
We have a left and a right distributive property for matrices because we can have matrices $A$ and $B$ such that $A B \neq B A$ . So, it may be the case that $A (B + C) \neq (B + C) A$ .
It is not true that for any nonzero matrices $A$ , $B$ , $C$ , if $A B = A C$ , then $B = C$ , because $A B = A C$ implies that $A B - A C = A (B - C) = 0$ , and we indicated in question 16 on page 42 that this statement does not imply that $A = 0$ or $B - C = 0$ . Since $A \neq 0$ , we cannot conclude that $B - C = 0$ ; therefore, $B$ may not be equal to $C$ .
For $A$ invertible, if $A B = A C$ , then $A^{- 1} A B = A^{- 1} A C$ ; hence, $I B = I C$ and $B = C$ .

Unit 5

The region in the inequality $a x + b y < c$ does not include the line $a x + b y = c$ , while the region given by the inequality $a x + b y \leq c$ does do so.
The intersection of the regions $x < 1$ and $y \geq 1$ is all pairs on the plane $(a, b)$ such that $a < 1$ and $b \geq 1$ . This intersection gives a infinite rectangle with a vertex $(1, 1)$ ; the line $x = 1$ is not included, but the line $y = 1$ is included.
No, we cannot graph lines using the $x$ - and $y$ -intercepts in all cases. If the line is of the form $a x + b y = 0$ , then the $x$ - and $y$ -intercepts are $(0, 0)$ . In this case, we use another point by giving $x$ any value, such as $x = 1$ . Then $y = - a / b$ , and the other point on the line is $(1, - a / b)$ .
To determine whether a point is in the region defined by an inequality, we place the given values into the inequality. Thus, $3 (3) + 2 = 11 > 6$ . Since this inequality holds, the point $(3, 2)$ is on the region.
To find a point not in the region given by an inequality, we look for a point $(a, b)$ such that the inequality does not hold. In this case, let us choose . Note that $3 (- 1) + 1 = - 2 \leq 6$ , so the point is not in the given region.
You know that the mathematical model of a problem is a linear inequality if you are looking for two unknowns, and if the conditions on these variables ask for an “at least” or “at most” value.
The graphing method is accurate in finding the region, but not in finding the intersection of the lines.
In order to have a bounded region, we need at least three linear inequalities whose boundaries are three non-parallel lines.
In order to have an unbounded region, we need a maximum of two linear equations.
Yes. Two linear inequalities will give an empty region if the lines are parallel. The inequalities $x + y < 1$ and $x + y > 3$ give an empty region. [Check it for yourself!]
You would set up the objective function of a problem to be solved using linear programming by establishing the quantity to be maximized or minimized.
The constraints in a problem to be solved using linear programming are the limitations imposed by the problem.
It is important to find the corner points of a feasible region in a problem to be solved using linear programming because the solution of the problem, if any, will be in the intersections.
The linear programming method gives a unique solution when the feasible region is bounded. If the feasible region is unbounded, there may or may not be a a unique solution.

Unit 6

We have three equations, so we introduce three slack variables, one for each equation. We have a total of 5 variables.
We have three equations, so we have three basic variables and two nonbasic variables.
We set the nonbasic variables to zero; hence, we have a system of three equations with three variables, and the augmented matrix is $3 \times 4$ .
A basic solution is a solution of the system as described in question 3. A basic feasible solution is a positive basic solution.
The objective function introduces one variable, and each problem constraint introduces one slack variable, for a total of $n$ slack variables, plus $m$ variables, for a total of $m + n$ variables. Hence, the initial system has $m + n + 1$ variables. The initial simplex tableau is $n + 1$ rows and $m + n + 2$ columns.
If there are no negative indicators in the bottom row, the objective function is as follows:
$P = - a_{1} x_{1} - a_{2} x_{2} - ... - a_{n} x_{n}$ ,

and since the variables $x_{1}, x_{2}, ..., x_{n}$ are positive, the optimal solution is

$x_{1} = x_{2} = ... = x_{n} = 0$ .
The bottom row in the initial simplex tableau is $P + a_{1} x_{1} - a_{2} x_{2} - a_{3} x_{3} = 0$ ; hence, we apply Step 2 a maximum of two times.
If a matrix $A$ has the dimensions $m \times n$ , its transpose has the dimensions $n \times m$ .
The coefficients of the objective function become the last column in the simplex tableau, together with a zero in the last entry of this column.
To say that a problem has no optimal solution means that the feasible region given by the constraint inequalities is empty.
We use the big $M$ method to solve a linear programming problem when the constraint conditions give inequalities of both types $\leq$ and $\geq$ .
No, there is no limit to the number of constraint inequalities or equalities we can have if we wish to apply the big $M$ method.
If the problem has only constraint inequalities of type $\geq$ we cannot use the big $M$ method, because it does not give a feasible tableau.

Unit 7

In a transition matrix, the entry $a_{11}$ is the probability that the system will remain in State 1. Similarly, $a_{22}$ is the probability that the system will remain in State 2.
Once a system enters an absorbing state, it never leaves.
If the entry $a_{i i} = 1$ , the state $i$ is absorbing.
A transition matrix is always square because it has one column and one row for each transition state.
A stationary matrix indicates the long trend of the problem.
In a Markov chain, the existence of a stationary matrix results from a regular transition matrix.
Yes, some non-regular transition matrices may have two different stationary matrices.
We recognize a possible absorbing Markov chain by the fact that the transition matrix has at least one 1 in the main diagonal.
Yes, if the Markov chain has two states, then the transition matrix is $2 \times 2$ , and since one state is absorbing, the matrix has the form
$[\begin{matrix} 1 & 0 \\ a & b \end{matrix}]$

with $a + b = 1$ . If we do the transition diagram, we see that this is an absorbing Markov chain.
The components of a transition matrix in an absorbing Markov chain are the identity matrix $I$ , a zero matrix, and two nonzero matrices $R$ and $Q$ of the nonabsorbing states.
The limiting transition matrix of a Markov chain indicates the long trend of the states.

Unit 8

A strictly determined matrix game is a matrix game with a saddle value.
A nonstrictly determined matrix game is a matrix game with no saddle values.
The entry of each matrix has two possibilities, and there are four entries: $2 (2) (2) (2) = 16$ .
A game is fair when the saddle value is zero.

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