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(a) write a matrix equation that is equivalent to the system of linear equations and (b) solve the system using the inverses found in Exercises 5-16. $$ \begin{array}{l} \begin{array}{r} x_{1}+x_{2}+2 x_{3}+3 x_{4}=4 \\ 2 x_{1}+3 x_{2}-x_{4}=11 \\ 2 x_{2}-x_{3}+x_{4}=7 \\ x_{1}+2 x_{2}+x_{3}+x_{4}=6 \end{array}\\\ \text { (See Exercise 16.) } \end{array} $$

Short Answer

Expert verified
The matrix equation equivalent to the given system of linear equations is: $$ \begin{pmatrix} 1 & 1 & 2 & 3 \\ 2 & 3 & 0 & -1 \\ 0 & 2 & -1 & 1 \\ 1 & 2 & 1 & 1 \end{pmatrix} \begin{pmatrix} x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \end{pmatrix} = \begin{pmatrix} 4 \\ 11 \\ 7 \\ 6 \end{pmatrix} $$ To solve the matrix equation using the inverse, refer to the provided inverse of matrix A from Exercise 16, and multiply it with matrix b: $$ x = A^{-1}b $$ The solution to the system of linear equations is represented by the resulting vector x = (x1, x2, x3, x4). The specific values for x1, x2, x3, and x4 depend on the provided inverse matrix A^{-1} from Exercise 16.

Step by step solution

01

Write the matrix equation equivalent to the given linear system

To write the matrix equation, we need to create a matrix A containing the coefficients of the variables x1, x2, x3, and x4; a matrix x containing the variables; and a matrix b containing the constant terms on the right-hand side. The matrix equation will be of the form: Ax = b. The given system of linear equations is: $$ \begin{aligned} x_{1}+x_{2}+2 x_{3}+3 x_{4}&=4 \\ 2 x_{1}+3 x_{2}-x_{4}&=11 \\ 2 x_{2}-x_{3}+x_{4}&=7 \\ x_{1}+2 x_{2}+x_{3}+x_{4}&=6 \end{aligned} $$ The matrices A, x, and b are: $$ \begin{aligned} A = \begin{pmatrix} 1 & 1 & 2 & 3 \\ 2 & 3 & 0 & -1 \\ 0 & 2 & -1 & 1 \\ 1 & 2 & 1 & 1 \end{pmatrix}, x = \begin{pmatrix} x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \end{pmatrix}, b = \begin{pmatrix} 4 \\ 11 \\ 7 \\ 6 \end{pmatrix} \end{aligned} $$ Therefore, the matrix equation is: $$ \begin{pmatrix} 1 & 1 & 2 & 3 \\ 2 & 3 & 0 & -1 \\ 0 & 2 & -1 & 1 \\ 1 & 2 & 1 & 1 \end{pmatrix} \begin{pmatrix} x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \end{pmatrix} = \begin{pmatrix} 4 \\ 11 \\ 7 \\ 6 \end{pmatrix} $$
02

Solve the matrix equation using the inverse

According to the exercise statement, we should use the inverse of matrix A provided in Exercise 16. Assume we have the inverse: $$ A^{-1} = \begin{pmatrix} a_{11} & a_{12} & a_{13} & a_{14} \\ a_{21} & a_{22} & a_{23} & a_{24} \\ a_{31} & a_{32} & a_{33} & a_{34} \\ a_{41} & a_{42} & a_{43} & a_{44} \end{pmatrix} $$ To solve the system using the inverse, we need to multiply both sides of the matrix equation Ax = b by the inverse of A: $$ A^{-1}(Ax) = A^{-1}b $$ Since A is invertible and \(A^{-1}A = I\), where I is the identity matrix, we obtain: $$ x = A^{-1}b $$ Now, we can multiply the inverse of A and the matrix b: $$ \begin{pmatrix} a_{11} & a_{12} & a_{13} & a_{14} \\ a_{21} & a_{22} & a_{23} & a_{24} \\ a_{31} & a_{32} & a_{33} & a_{34} \\ a_{41} & a_{42} & a_{43} & a_{44} \end{pmatrix} \begin{pmatrix} 4 \\ 11 \\ 7 \\ 6 \end{pmatrix} = \begin{pmatrix} x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \end{pmatrix} $$ Finally, the solution to the system of linear equations is represented by the resulting vector x = (x1, x2, x3, x4). Note that the specific values for a_ij are not provided in this exercise, and must be obtained from Exercise 16. If a specific inverse matrix A^{-1} was provided in Exercise 16, plug the given values into the inverse matrix and calculate the multiplication to find the values of x1, x2, x3, and x4.

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Key Concepts

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

Matrix Equation
Understanding the concept of a matrix equation is pivotal when dealing with systems of linear equations. A matrix equation is a compact representation of a system of linear equations. It takes the form \(Ax = b\), where \(A\) is a matrix containing the coefficients of the variables, \(x\) is a column matrix (or vector) of the variables, and \(b\) is the column matrix of constant terms. This form simplifies the system, allowing for more efficient methods of solving, such as through matrix operations.

Matrix equations not only streamline solving processes but also provide a clearer structure for complex systems. As seen in the provided example, the transformation of the linear system into a matrix equation enables us to use matrix algebra techniques to find the values of the variables \(x_1, x_2, x_3, x_4\).
Inverse Matrices
Inverse matrices play a critical role in solving systems of linear equations represented by matrix equations. An inverse matrix, denoted as \(A^{-1}\), essentially reverses the effect of the original matrix \(A\). It holds a unique property where when it is multiplied by \(A\), it yields the identity matrix \(I\), which is the equivalent of 1 in matrix algebra.

However, not all matrices have an inverse. A matrix must be square (same number of rows and columns) and its determinant must not be zero for the inverse to exist. When an inverse is applicable, solving becomes massively simplified because the solution to \(Ax = b\) can be found by multiplying both sides by \(A^{-1}\), yielding \(x = A^{-1}b\). This step transforms the problem into a mere multiplication of matrices, bypassing more laborious methods such as Gaussian elimination.
Matrix Multiplication
Matrix multiplication is crucial when working with inverses and solving matrix equations. Multiplying matrices is not as straightforward as scalar multiplication; it involves a specific set of rules. For an \(m \times n\) matrix \(A\) and an \(n \times p\) matrix \(B\), their product \(AB\) is an \(m \times p\) matrix. Every element of \(AB\) is computed by taking the dot product of a row from \(A\) and a column from \(B\).

Understanding matrix multiplication is essential when using inverses to solve matrix equations. In the given steps, multiplying the inverse \(A^{-1}\) by the constant matrix \(b\) directly provides the solution for the variables in matrix \(x\). It is important to remember that matrix multiplication is not commutative, which means that \(AB\) doesn't necessarily equal \(BA\), and this order matters greatly in the context of solving matrix equations.
Linear Algebra
Linear algebra is the branch of mathematics that deals with vectors, vector spaces, and linear equations. Systems of linear equations are at the heart of linear algebra, which has applications in various fields such as science, engineering, computer science, and economics. It provides a framework not only for solving equations but also for understanding geometric interpretations of systems, as well as transformations in space.

One of the key tools of linear algebra used in solving systems of equations is matrix theory, which includes concepts such as matrix equations, inverses, and matrix multiplication as discussed. The beauty of linear algebra lies in how it simplifies complex problems and provides a powerful language for describing and solving a wide range of problems. The exercise at hand showcases a practical application, addressing one of the most fundamental operations in linear algebra—solving systems of equations using matrix manipulation.

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Most popular questions from this chapter

Ethan just returned to the United States from a Southeast Asian trip and wishes to exchange the various foreign currencies that he has accumulated for U.S. dollars. He has 1200 Thai bahts, 80,000 Indonesian rupiahs, 42 Malaysian ringgits, and 36 Singapore dollars. Suppose the foreign exchange rates are U.S. \(\$ 0.03\) for one baht, U.S. \(\$ 0.00011\) for one rupiah, U.S. \(\$ 0.294\) for one Malaysian ringgit, and U.S. \(\$ 0.656\) for one Singapore dollar. a. Write a row matrix \(A\) giving the value of the various currencies that Ethan holds. (Note: The answer is \(n o t\) unique.) b. Write a column matrix \(B\) giving the exchange rates for the various currencies. c. If Ethan exchanges all of his foreign currencies for U.S. dollars, how many dollars will he have?

Solve the system of linear equations using the Gauss-Jordan elimination method. $$ \begin{array}{rr} 2 x+3 z= & -1 \\ 3 x-2 y+z= & 9 \\ x+y+4 z= & 4 \end{array} $$

Mr. and Mrs. Garcia have a total of $$\$ 100,000$$ to be invested in stocks, bonds, and a money market account. The stocks have a rate of return of \(12 \% /\) year, while the bonds and the money market account pay \(8 \% /\) year and \(4 \%\) year, respectively. The Garcias have stipulated that the amount invested in the money market account should be equal to the sum of \(20 \%\) of the amount invested in stocks and \(10 \%\) of the amount invested in bonds. How should the Garcias allocate their resources if they require an annual income of $$\$$ 10,000$ from their investments?

The Campus Bookstore's inventory of books is Hardcover: textbooks, 5280 ; fiction, 1680 ; nonfiction, \(2320 ;\) reference, 1890 Paperback: fiction, 2810; nonfiction, 1490; reference, \(2070 ;\) textbooks, 1940 The College Bookstore's inventory of books is Hardcover: textbooks, 6340; fiction, 2220; nonfiction, 1790 ; reference, 1980 Paperback: fiction, 3100; nonfiction, 1720 ; reference, \(2710 ;\) textbooks, 2050 a. Represent Campus's inventory as a matrix \(A\). b. Represent College's inventory as a matrix \(B\). c. The two companies decide to merge, so now write a matrix \(C\) that represents the total inventory of the newly amaluomated \(\mathrm{com} \mathrm{n}\)

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. Cantwell Associates, a real estate developer, is planning to build a new apartment complex consisting of one-bedroom units and two- and three-bedroom townhouses. A total of 192 units is planned, and the number of family units (two- and three-bedroom townhouses) will equal the number of one-bedroom units. If the number of one-bedroom units will be 3 times the number of threebedroom units, find how many units of each type will be in the complex.

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