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Chapter 2: Problem 3

Refer to the following matrices: $$ \begin{array}{l} A=\left[\begin{array}{rrrr} 2 & -3 & 9 & -4 \\ -11 & 2 & 6 & 7 \\ 6 & 0 & 2 & 9 \\ 5 & 1 & 5 & -8 \end{array}\right] \\ B=\left[\begin{array}{rrr} 3 & -1 & 2 \\ 0 & 1 & 4 \\ 3 & 2 & 1 \\ -1 & 0 & 8 \end{array}\right] \end{array} $$ $$ \begin{array}{l} C=\left[\begin{array}{lllll} 1 & 0 & 3 & 4 & 5 \end{array}\right] \\ D=\left[\begin{array}{r} 1 \\ 3 \\ -2 \\ 0 \end{array}\right] \end{array} $$ Find \(b_{13}, b_{31}\), and \(b_{43}\).

Short Answer

Expert verified
In matrix B, we have \(b_{13} = 2\), \(b_{31} = 3\), and \(b_{43} = 8\).

Step by step solution

01

Locate the element \(b_{13}\)

To find \(b_{13}\), we look at the first row and the third column of matrix B. The element located at this position is 2. So, \(b_{13} = 2\).
02

Locate the element \(b_{31}\)

To find \(b_{31}\), we look at the third row and the first column of matrix B. The element located at this position is 3. So, \(b_{31} = 3\).
03

Locate the element \(b_{43}\)

To find \(b_{43}\), we look at the fourth row and the third column of matrix B. The element located at this position is 8. So, \(b_{43} = 8\). Thus, we have found the required elements: $$b_{13} = 2, b_{31} = 3, \text{ and } b_{43} = 8.$$

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Matrices
A matrix is a mathematical object that can be thought of as an array of numbers arranged in rows and columns. It's a systematic way to organize and handle multiple numeric values. They are highly useful in various fields, including physics, computer graphics, statistics, and more. Matrices are particularly valuable when solving systems of linear equations or representing linear transformations, among many other applications.

For instance, the matrices referenced in our exercise represent various arrangements of numbers, each with a specific number of rows and columns. Matrix A has four rows and four columns, while matrix B is a 4x3 matrix, signifying it has four rows and three columns. By understanding the structure and properties of matrices, students can manipulate them to perform operations such as addition, subtraction, multiplication, and finding inverses, which are fundamental concepts in linear algebra.
Matrix Notation
Matrix notation is a standardized way of representing the elements of a matrix. Elements within a matrix are typically denoted by a lowercase letter with two subscript indices such as \( a_{ij} \), where \( i \) represents the row number and \( j \) represents the column number. This notation simplifies referencing specific values within a matrix, making it easier to perform calculations and communicate mathematical concepts.

Understanding the Indices

The first index, \( i \), corresponds to the row position, while the second index, \( j \), corresponds to the column position. For example, in our exercise, \( b_{13} \) refers to the element in the first row and third column of matrix B. It is essential to always keep the order of indices consistent to avoid confusion when locating elements within a matrix.
Element Location in Matrices
Identifying the location of an element within a matrix is a basic skill in matrix algebra. Based on the notation \( a_{ij} \), you can determine the exact position of any element. To find an element, you first move down to the \( ith \) row, and then across to the \( jth \) column.

In the provided exercise, the element \( b_{13} \) is found by locating the first row and then moving to the third column, resulting in the element 2. Similarly, \( b_{31} \) is found by moving to the third row and the first column, which reveals the element 3. Lastly, \( b_{43} \) is in the fourth row and the third column, and the number located there is 8.

Practical Tip for Students

When you're trying to locate elements in a matrix, it's useful to draw a line (physically or mentally) down from the top of the desired column and across from the side of the desired row. The intersection of these two lines will be the element you're looking for. This visualization can greatly speed up the process of finding elements, especially in larger matrices.

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Most popular questions from this chapter

For the opening night at the Opera House, a total of 1000 tickets were sold. Front orchestra seats cost $$\$ 80$$ apiece, rear orchestra seats cost $$\$ 60$$ apiece, and front balcony seats cost $$\$ 50$$ apiece. The combined number of tickets sold for the front orchestra and rear orchestra exceeded twice the number of front balcony tickets sold by 400. The total receipts for the performance were $$\$ 62,800$$. Determine how many tickets of each type were sold.

The problems in exercise correspond to those in exercises 15-27, Section 2.1. Use the results of your previous work to help you solve these problems. Lawnco produces three grades of commercial fertilizers. A 100 -lb bag of grade-A fertilizer contains \(18 \mathrm{lb}\) of nitrogen, \(4 \mathrm{lb}\) of phosphate, and \(5 \mathrm{lb}\) of potassium. A \(100-\mathrm{lb}\) bag of grade-B fertilizer contains \(20 \mathrm{lb}\) of nitrogen and \(4 \mathrm{lb}\) each of phosphate and potassium. A 100-lb bag of grade-C fertilizer contains \(24 \mathrm{lb}\) of nitrogen, \(3 \mathrm{lb}\) of phosphate, and \(6 \mathrm{lb}\) of potassium. How many 100 -lb bags of each of the three grades of fertilizers should Lawnco produce if \(26,400 \mathrm{lb}\) of nitrogen, \(4900 \mathrm{lb}\) of phosphate, and \(6200 \mathrm{lb}\) of potassium are available and all the nutrients are used?

Let $$ A=\left[\begin{array}{ll} 3 & 1 \\ 0 & 2 \end{array}\right] \text { and } B=\left[\begin{array}{rr} 4 & -2 \\ 2 & 1 \end{array}\right] $$ a. Compute \((A+B)^{2}\). b. Compute \(A^{2}+2 A B+B^{2}\). c. From the results of parts (a) and (b), show that in general \((A+B)^{2} \neq A^{2}+2 A B+B^{2}\).

Fill in the missing entries by performing the indicated row operations to obtain the rowreduced matrices. $$ \begin{array}{l} \text { }\left[\begin{array}{rrr|r} 0 & 1 & 3 & -4 \\ 1 & 2 & 1 & 7 \\ 1 & -2 & 0 & 1 \end{array}\right] \stackrel{R_{1} \leftrightarrow R_{2}}{\longrightarrow}\left[\begin{array}{rrr|r} \cdot & \cdot & \cdot & \cdot \\ \cdot & \cdot & \cdot & \cdot \\ 1 & -2 & 0 & 1 \end{array}\right]\\\ \stackrel{R_{3}-R_{1}}{\longrightarrow}\left[\begin{array}{ccc|r} 1 & 2 & 1 & 7 \\ 0 & 1 & 3 & -4 \\ \cdot & \cdot & \cdot & \cdot \end{array}\right] \frac{R_{1}+\frac{1}{2} R_{3}}{R_{3}+4 R_{2}}\left[\begin{array}{ccc|c} \cdot & \cdot & \cdot & \cdot \\ 0 & 1 & 3 & -4 \\ \cdot & \cdot & \cdot & \cdot \end{array}\right]\\\ \stackrel{\frac{1}{11} R_{3}}{\longrightarrow}\left[\begin{array}{ccc|c} 1 & 0 & \frac{1}{2} & 4 \\ 0 & 1 & 3 & -4 \\ . & \cdot & . & . \end{array}\right] \frac{R_{1}-\frac{1}{2} R_{3}}{R_{2}-3 R_{3}}\left[\begin{array}{ccc|r} 1 & 0 & 0 & 5 \\ 0 & 1 & 0 & 2 \\ 0 & 0 & 1 & -2 \end{array}\right] \end{array} $$

(a) write each system of equations as a matrix equation and (b) solve the system of equations by using the inverse of the coefficient matrix. $$ \begin{array}{l} \begin{aligned} x_{1}+x_{2}+2 x_{3}+x_{4} &=b_{1} \\ 4 x_{1}+5 x_{2}+9 x_{3}+x_{4} &=b_{2} \\ 3 x_{1}+4 x_{2}+7 x_{3}+x_{4} &=b_{3} \\ 2 x_{1}+3 x_{2}+4 x_{3}+2 x_{4} &=b_{4} \end{aligned}\\\ \text { where } \quad \text { (i) } b_{1}=3, b_{2}=6, b_{3}=5, b_{4}=7\\\ \text { and (ii) } b_{1}=1, b_{2}=-1, b_{3}=0, b_{4}=-4 \end{array} $$
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