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Finite Mathematics : Finite Mathematics

Study concepts, example questions & explanations for Finite Mathematics

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1 Diagnostic Test 22 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #1 : Solve A System Of Equations In Three Variables Using Augmented Matrices

$A = \begin{bmatrix} 0.7 & -0.4 \\ -0.8 & 0.2 \end{bmatrix}$ and $B= \begin{bmatrix} 0. 3 & 0.4 \\ 0.8 & 0.8 \end{bmatrix}$ .

True or false: A+ B = I , where is the two-by-two identity matrix.

Possible Answers:

False

True

Correct answer:

True

Explanation:

First, it must be established that A+ B is defined. This is the case if and only and have the same number of rows, which is true, and they have the same number of columns, which is also true. A +B is therefore defined.

Addition of two matrices is performed by adding corresponding elements, so

$A +B= \begin{bmatrix} 0.7 & -0.4 \\ -0.8 & 0.2 \end{bmatrix}+ \begin{bmatrix} 0. 3 & 0.4 \\ 0.8 & 0.8 \end{bmatrix}$

$= \begin{bmatrix} 0.7+0.3 & -0.4+0.4 \\ -0.8+0.8 & 0.2+0.8 \end{bmatrix}$

$= \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}$

This is the two-by-two identity . Therefore, A+ B = I .

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Example Question #31 : Finite Mathematics

You are given that the inverse of the matrix

$A = \begin{bmatrix} 1 & -2 & 3 \\ -2 & 3 & 1 \\ -1 & 1 & 3 \end{bmatrix}$

$A^{-1} = \begin{bmatrix} 8 &9 & -11 \\ 5 & 6 & -7 \\ 1 & 1 & -1 \end{bmatrix}$

Using this information, solve the linear system

x-2y + 3z = 24

-2x+ 3y + z = 40

-x+ y+ 3z = -10

From these five choices, select the correct value of .

Possible Answers:

640

160

430

662

Correct answer:

430

Explanation:

The given linear system can be rewritten as the matrix equation

AX = B ,

where

$A = \begin{bmatrix} 1 & -2 & 3 \\ -2 & 3 & 1 \\ -1 & 1 & 3 \end{bmatrix}$ , $X= \begin{bmatrix} x \\ y \\ z \end{bmatrix}$ , and $B = \begin{bmatrix} 24 \\ 40 \\ -10 \end{bmatrix}$ .

This equation can be restated as

$A^{-1}(AX) =A^{-1} B$

$(A^{-1}A)X =A^{-1} B$

$IX =A^{-1} B$

$X =A^{-1} B$

We are already given $A^{-1}$ , so:

$X =A^{-1} B$

$X = \begin{bmatrix} 8 &9 & -11 \\ 5 & 6 & -7 \\ 1 & 1 & -1 \end{bmatrix} \begin{bmatrix} 24 \\ 40 \\ -10 \end{bmatrix}$

Multiply two matrices by multiplying the rows in the first by the column in the second; this is done by adding the products of entries in corresponding positions, as follows:

$X = \begin{bmatrix} 8 (24)+9(40) +( -11)(-10) \\ 5(24) + 6(40) +( -7)(-10) \\ 1 (24)+ 1(40 )+( -1)(-10) \end{bmatrix}$

$X = \begin{bmatrix} 662 \\ 430 \\ 74 \end{bmatrix}$

We are only interested in the value of , which is 430.

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Example Question #32 : Finite Mathematics

It is suggested (but not necessary) that you use a calculator with matrix capability to work this problem.

A village is famous for being the most prolific producer in the world of ambrosia, butterbeer, and silphium. Every year, they try to meet the world's demand for 20,000 kg of ambrosia, 30,000 kg of butterbeer, and 15,000 kg of silphium. However, the production of these items requires the consumption of these items as well:

The production of 1 kg of ambrosia requires the use of 0.05 kg of butterbeer and 0.08 kg of silphium;

The production of 1 kg of butterbeer requires the use of 0.04 kg of ambrosia and 0.09 kg of silphium; and,

The production of 1 kg of silphium requires the use of 0.06 kg of ambrosia and 0.07 kg of butterbeer.

Use a Leontief model to determine how much silphium (nearest hundred kilograms) must be produced to meet the demand.

Possible Answers:

$18,500 \textup{ kg }$

$18,800 \textup{ kg }$

$18,700 \textup{ kg }$

$18,600 \textup{ kg }$

$18,400 \textup{ kg }$

Correct answer:

$18,700 \textup{ kg }$

Explanation:

The Leontief model for an input-output system requires the following steps.

Let be the technology matrix for the system, the entries of which are the amounts of each product consumed in the manufacture of each product. Letting the rows/columns represent, in order, ambrosia, butterbeer, and silphium,

$A = \begin{bmatrix} 0 & 0.05 & 0.08 \\ 0.04 & 0 & 0.09 \\ 0.06 & 0.07 & 0\end{bmatrix}$

Let be the external demand vector, which gives the amount of each product demanded by outside consumers:

$D= \begin{bmatrix} 20,000 \\ 30,000 \\ 15,000 \end{bmatrix}$

The output vector , which gives the amount of each product that must be produced to meet the demand, can be obtained through the matrix equation

$X = (I- A )^{-1}D$

$= \left ( \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} - \begin{bmatrix} 0 & 0.05 & 0.08 \\ 0.04 & 0 & 0.09 \\ 0.06 & 0.07 & 0\end{bmatrix} \right )^{-1} \begin{bmatrix} 20,000 \\ 30,000 \\ 15,000 \end{bmatrix}$

$= \begin{bmatrix} 1 &-0.05 & -0.08 \\ -0.04 & 1 & -0.09 \\ -0.06 & -0.07 & 1 \end{bmatrix} ^{-1} \begin{bmatrix} 20,000 \\ 30,000 \\ 15,000 \end{bmatrix}$

$\approx \begin{bmatrix} 1.007394521 &0.056366243 & 0.085664523 \\0.046025673 & 1.008915193 & 0.094484421 \\ 0.063665468 & 0.074006038 &1.011753781 \end{bmatrix} \begin{bmatrix} 20,000 \\ 30,000 \\ 15,000 \end{bmatrix}$

$\approx \begin{bmatrix} 23,123.85 \\ 32,605.24\\ 18,669.80 \end{bmatrix}$

Silphium is represented by the bottom entry; the amount the village must produce rounds to 18,700 kg.

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Example Question #33 : Finite Mathematics

It is suggested (but not necessary) that you use a calculator with matrix capability to work this problem.

The production of 1 kg of ambrosia requires the use of 0.05 kg of butterbeer and 0.08 kg of silphium;

The production of 1 kg of butterbeer requires the use of 0.04 kg of ambrosia and 0.09 kg of silphium; and,

The production of 1 kg of silphium requires the use of 0.06 kg of ambrosia and 0.07 kg of butterbeer.

Use a Leontief model to determine how much ambrosia (nearest hundred kilograms) must be produced to meet the demand.

Possible Answers:

$23,500 \textup{ kg}$

$23,300 \textup{ kg}$

$23,100 \textup{ kg}$

$23,400 \textup{ kg}$

$23,200 \textup{ kg}$

Correct answer:

$23,100 \textup{ kg}$

Explanation:

The Leontief model for an input-output system requires the following steps.

$A = \begin{bmatrix} 0 & 0.05 & 0.08 \\ 0.04 & 0 & 0.09 \\ 0.06 & 0.07 & 0\end{bmatrix}$

Let be the external demand vector, which gives the amount of each product demanded by outside consumers:

$D= \begin{bmatrix} 20,000 \\ 30,000 \\ 15,000 \end{bmatrix}$

The output vector , which gives the amount of each product that must be produced to meet the demand, can be obtained through the matrix equation

$X = (I- A )^{-1}D$

$= \begin{bmatrix} 1 &-0.05 & -0.08 \\ -0.04 & 1 & -0.09 \\ -0.06 & -0.07 & 1 \end{bmatrix} ^{-1} \begin{bmatrix} 20,000 \\ 30,000 \\ 15,000 \end{bmatrix}$

$\approx \begin{bmatrix} 23,123.85 \\ 32,605.24\\ 18,669.80 \end{bmatrix}$

Ambrosia is represented by the top entry; the amount the village must produce rounds to 23,100 kg.

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Example Question #31 : Finite Mathematics

$A= \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$

True or false: $A^{2} = \begin{bmatrix} 1 & 4 \\ 9 & 16 \end{bmatrix}$

Possible Answers:

False

True

Correct answer:

False

Explanation:

$A^{2} = AA=\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$

Two matrices are multiplied by multiplying the rows of the first by the columns of the second - that is, adding the products of the entries in corresponding positions:

$A^{2} = \begin{bmatrix} 1(1)+2(3) & 1(2)+2(4) \\3(1)+4(3) & 3(2)+4(4) \end{bmatrix}$

$= \begin{bmatrix} 7 & 10 \\ 15 & 22 \end{bmatrix}$

$\ne \begin{bmatrix} 1 & 4 \\ 9 & 16 \end{bmatrix}$

The statement is false.

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Example Question #31 : Systems Of Linear Equations: Matrices

$A= \begin{bmatrix}- \frac{4}{7} & \frac{2}{9} \\ \\ -\frac{5}{6}& \frac{3}{11} \end{bmatrix}$ and $B= \begin{bmatrix} \frac{5}{7} & \frac{2}{9} \\ \\ -\frac{5}{6}& -\frac{8}{11} \end{bmatrix}$ .

True or false: A- B = I , where is the two-by-two identity matrix.

Possible Answers:

True

False

Correct answer:

False

Explanation:

First, it must be established that A- B is defined. This is the case if and only and have the same number of rows, which is true, and they have the same number of columns, which is also true. A -B is therefore defined.

Subtraction of two matrices is performed by subtracting corresponding elements, so

$A-B= \begin{bmatrix}- \frac{4}{7} & \frac{2}{9} \\ \\ -\frac{5}{6}& \frac{3}{11} \end{bmatrix}- \begin{bmatrix} \frac{5}{7} & \frac{2}{9} \\ \\ -\frac{5}{6}& -\frac{8}{11} \end{bmatrix}$

$= \begin{bmatrix}- \frac{4}{7}-\frac{5}{7} & \frac{2}{9} - \frac{2}{9} \\ \\ -\frac{5}{6} - \left ( -\frac{5}{6} \right )& \frac{3}{11} -\left ( -\frac{8}{11} \right )\end{bmatrix}$

$= \begin{bmatrix} {\color{Red} -1} & 0 \\ 0 & 1 \end{bmatrix}$

However, $I = \begin{bmatrix} {\color{Red} 1 }& 0 \\ 0 & 1 \end{bmatrix}$ .

Therefore, $A-B \ne I$ .

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Example Question #32 : Finite Mathematics

2x+ 5y = 17

4x-y = 13

Use Cramer's rule to evaluate .

Possible Answers:

$x= \frac{21}{11}$

$x= \frac{41}{11}$

$x= -\frac{69}{22}$

None of the other choices gives the correct response.

$x= -\frac{59}{22}$

Correct answer:

$x= \frac{41}{11}$

Explanation:

By Cramer's rule, the value of is equal to $x= \frac{D_{x}}{D}$ , where is the determinant of the matrix of coefficients

$D =\begin{vmatrix} 2 & 5 \\ 4 & -1 \end{vmatrix}$ ,

and $D_{x}$ is the same matrix with the x-coefficients replaced by the constants:

$D_{x}= \begin{vmatrix} 17 & 5 \\ 13 & -1 \end{vmatrix}$

The determinant of a two-by-two matrix is equal to the product of the entries in the main diagonal minus the product of the other two entries. Therefore,

$D_{x}= \begin{vmatrix} 17 & 5 \\ 13 & -1 \end{vmatrix} = 17(-1)-5(13)= -17 -65 = -82$

$D= \begin{vmatrix} 2 & 5 \\ 4 & -1 \end{vmatrix} = 2(-1) - 5(4) = -2 - 20 = -22$

and

$x= \frac{D_{x}}{D} = \frac{-82}{-22} = \frac{41}{11}$ .

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Example Question #33 : Systems Of Linear Equations: Matrices

The initial simplex tableau for a minimization linear programming problem is given below:

$\begin{bmatrix} 2 & 3 & 2 &1 & 0 & 0 & 1,200 \\ 1 & 2 & 4 & 0 & 1 & 0 & 600\\ 0 & 4 & 3 & 0 & 0 & 1 & 900 \\ -1 & -1 & -2 & 0&0&0 & 0 \end{bmatrix}$

After the first pivot of the simplex method is performed, what will be the entry at bottom right?

Possible Answers:

400

600

300

1,200

Correct answer:

300

Explanation:

First, the pivot element must be identified. Select the pivot column by spotting the "most negative" element in the bottom, or objective, row. This is the in Column 3, making that the pivot column.

$\begin{bmatrix} 2 & 3 & 2 &1 & 0 & 0 & 1,200 \\ 1 & 2 & 4 & 0 & 1 & 0 & 600\\ 0 & 4 & 3 & 0 & 0 & 1 & 900 \\ -1 & -1 & {\color{Red} \textbf{-2}} & 0&0&0 & 0 \end{bmatrix}$

Select the pivot row by dividing the rightmost element in each other row by the pivot-column element in that row. The "least positive" quotient determines the pivot row:

$\begin{bmatrix} 2 & 3 & {\color{Blue} \textbf{2}} &1 & 0 & 0 & {\color{Blue} \textbf{1,200}} \\ 1 & 2 & {\color{Blue} \textbf{4}} & 0 & 1 & 0 & {\color{Blue} \textbf{600}}\\ 0 & 4 & {\color{Blue} \textbf{3}} & 0 & 0 & 1 & {\color{Blue} \textbf{900}} \\ -1 & -1 & {\color{Red} \textbf{-2}} & 0&0&0 & 0 \end{bmatrix}\begin{matrix} 1,200 \div 2 = 600 \\ 600 \div 4 = {\color{Red} \textbf{150}} \\ 900 \div 3 = 300\\\; \end{matrix}$

The "least positive" of the three quotients is 150, in Row 2. The pivot element is the 4 in Row 2, Column 3.

Perform the pivot by first multiplying each entry in Row 2 by $\frac{1}{4}$ in order to make the pivot element a 1:

$\frac{1}{4}R2\rightarrow R2$

$\begin{bmatrix} 2 & 3 & 2 &1 & 0 & 0 & 1,200 \\ \frac{1}{4} & \frac{1}{2} & 1 & 0 & \frac{1}{4} & 0 & 150\\ 0 & 4 & 3 & 0 & 0 & 1 & 900 \\ -1 & -1 & -2 & 0&0&0 & 0 \end{bmatrix}$

Make the other elements in that column zeroes by adding multiples of the pivot row to the other rows, as follows:

$-2R2 + R 1 \rightarrow R1$

$-3R2 + R3 \rightarrow R3$

$2R2 + R4 \rightarrow R4$

$\begin{bmatrix} \frac{3}{2} & 2 & 0 &1 & -\frac{1}{2} & 0 & 900 \\ \frac{1}{4} & \frac{1}{2} & 1 & 0 & \frac{1}{4} & 0 & 150\\ -\frac{3}{4} & \frac{5}{2} & 0 & 0 & -\frac{3}{4} & 1 & 450 \\ -\frac{1}{2} & 0& 0 & 0&\frac{1}{2}&0 & {\color{Red} \textbf{300}} \end{bmatrix}$

The entry at bottom right is 300.

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Example Question #35 : Finite Mathematics

The graph depicts which inequality?

Possible Answers:

x=-4

x>-4

x<-4

$x\leq-4$

$x\geq-4$

Correct answer:

x>-4

Explanation:

To identify the inequality the graph represents first identify range.

Since the red line goes from negative four to infinity, we can right this in mathematical terms as follows,

x>-4

Now, the inequality will remain a greater than sign because there is an open circle at four.

This means that does not equal four, it only equals values greater than four.

Thus, the correct inequality is

x>-4

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Example Question #32 : Finite Mathematics

The initial simplex tableau for a maximization linear programming problem is

$\begin{bmatrix} 1 & 3 & 1 & 0 & 540 \\ 2 & 1 & 0 & 1 & 200 \\ -4 & -7 & 0 & 0 & 0 \end{bmatrix}$

Locate the first pivot element.

Possible Answers:

Row 2, Column 2

Row 1, Column 1

Row 2, Column 1

Row 1, Column 2

Correct answer:

Row 1, Column 2

Explanation:

To select the pivot element while working a maximization problem, first, identify the pivot column by selecting the "most negative" element in the bottom, or objective, row. In the initial tableau, it is the in Column 2:

$\begin{bmatrix} 1 & 3 & 1 & 0 & 540 \\ 2 & 1 & 0 & 1 & 200 \\ -4 & \textbf{{\color{Red} -7}} & 0 & 0 & 0 \end{bmatrix}$

In the other rows, divide the rightmost element by the element in the pivot column. The pivot row is identified by selecting the "least positive" quotient.

$\begin{bmatrix} 1 & 3 & 1 & 0 & 540 \\ 2 & 1 & 0 & 1 & 200 \\ -4 & \textbf{{\color{Red} -7}} & 0 & 0 & 0 \end{bmatrix}\begin{matrix} 540 \div 3 = \textbf{180 }\\ 200 \div 1 = 200 \\ \; \end{matrix}$

The least positive quotient appears in Row 1. This puts the pivot element in Row 1, Column 2.

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Finite Mathematics : Finite Mathematics

Study concepts, example questions & explanations for Finite Mathematics

All Finite Mathematics Resources

Example Questions

Example Question #1 : Solve A System Of Equations In Three Variables Using Augmented Matrices

Example Question #31 : Finite Mathematics

Example Question #32 : Finite Mathematics

Example Question #33 : Finite Mathematics

Example Question #31 : Finite Mathematics

Example Question #31 : Systems Of Linear Equations: Matrices

Example Question #32 : Finite Mathematics

Example Question #33 : Systems Of Linear Equations: Matrices

Example Question #35 : Finite Mathematics

Example Question #32 : Finite Mathematics

All Finite Mathematics Resources

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