/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 10 Solve the following systems of e... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 10

Solve the following systems of equations using matrix inverses. (The second can be done quickly once the first has been solved. Can you see why?) \(x+y+z=1\) (i) \(x+2 y+3 z=2\) $$ x+y+z=1 $$ $$ \begin{aligned} x+2 y+3 z=2 & \text { (ii) } x+2 y+4 z=2 \\ x+4 y+9 z=3 ; & x+3 y+9 z=3 \end{aligned} $$

Short Answer

Expert verified
Both systems have the solution: \(x = 1\), \(y = 1\), \(z = -1\).

Step by step solution

01

Write the System as Matrices

Write the system of equations in matrix form. For the equations \(x + 2y + 3z = 2\), \(x + 4y + 9z = 3\), and \(x + y + z = 1\), the coefficient matrix \(A\), variable matrix \(X\), and constant matrix \(B\) are: \[A = \begin{bmatrix} 1 & 2 & 3 \ 1 & 4 & 9 \ 1 & 1 & 1 \end{bmatrix}, X = \begin{bmatrix} x \ y \ z \end{bmatrix},B = \begin{bmatrix} 2 \ 3 \ 1 \end{bmatrix}.\] The system is written as: \(AX = B\).
02

Find the Inverse of the Matrix A

To solve the system, find the inverse of matrix \(A\), \(A^{-1}\). The inverse is found by the formula using the adjugate matrix and determinant: \[A^{-1} = \frac{1}{\text{det}(A)} \text{adj}(A).\] Calculate the determinant of \(A\): \[\text{det}(A) = 1(4 \cdot 1 - 9 \cdot 1) - 2(1 \cdot 1 - 9 \cdot 1) + 3(1 \cdot 1 - 4 \cdot 1) = -1.\] Compute the adjugate of \(A\) and then \(A^{-1}\).
03

Compute A^{-1}

The inverse \(A^{-1}\) given \(\text{det}(A) = -1\) and the adjugate is calculated is: \[A^{-1} = \begin{bmatrix} 5 & 6 & -3 \ 0 & 1 & -1 \ -1 & -2 & 1 \end{bmatrix}.\]
04

Calculate X using A^{-1}B

Now calculate \(X\) using \(A^{-1}B\): \[X = A^{-1}B = \begin{bmatrix} 5 & 6 & -3 \ 0 & 1 & -1 \ -1 & -2 & 1 \end{bmatrix}\begin{bmatrix} 2 \ 3 \ 1 \end{bmatrix} = \begin{bmatrix} 1 \ 1 \ -1 \end{bmatrix}.\]Thus, the solution is \(x = 1\), \(y = 1\), \(z = -1\).
05

Apply Solution to Second System

Since the second system \(x + 2y + 4z = 2\), \(x + 3y + 9z = 3\), \(x + y + z = 1\) only slightly modifies the first system with a net zero effect by using the same inverse method, you can utilize the same \((x = 1, y = 1, z = -1)\) since the coefficients cancel out additional terms.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Systems of Equations
A system of equations consists of multiple equations that share variables. The main goal is to find values for the variables that satisfy all equations simultaneously. In a practical context, systems of equations can model a variety of real-world scenarios, such as calculating intersection points, identifying optimal solutions, or balancing chemical reactions. In mathematical terms, systems of equations can be expressed in different forms:
  • Linear equations, where each term is either a constant or the product of a constant and a single variable.
  • Nonlinear equations, involving terms that are not linear, such as squares or other powers of variables.
For linear equations, matrices provide a structured method to solve these equations, especially when dealing with more than two variables.
Matrix Algebra
Matrix algebra simplifies the process of solving multiple linear equations using concise formats and operations. A matrix is a rectangular array of numbers arranged in rows and columns. Essential operations in matrix algebra include addition, subtraction, multiplication, and finding the inverse. The basic components involve:
  • **Addition and Subtraction:** Two matrices of the same dimensions can be added or subtracted by operating on each corresponding element.
  • **Multiplication:** Involves "dot products" of rows from the first matrix and columns from the second matrix.
  • **Inverse of a Matrix:** Like inverse operations in arithmetic, the inverse of a matrix, denoted as \(A^{-1}\), when multiplied with the original matrix \(A\), yields an identity matrix (equivalent of 1 in matrix terms).
Using these operations, complex systems can be compartmentalized and solved efficiently. Matrix algebra provides the tools necessary to turn theoretical systems into solveable entities.
Determinants
A determinant is a special number calculated from a square matrix, representing certain properties of the matrix. It provides crucial insights into the characteristics of the matrix, such as whether the matrix has an inverse, whether the system of equations is consistent, and whether it includes unique solutions. The determinant can be calculated using:
  • **Expansion:** For a 2x2 matrix, the determinant is calculated using the general formula \(ad-bc\). For higher dimensions, calculate minors and cofactors recursively.
  • **Significance:** A non-zero determinant implies that the matrix has an inverse and the system has a unique solution. In contrast, a zero determinant means the matrix is singular, lacking an inverse, often leading to infinite or no solutions.
Knowing the determinant is fundamental for validating the steps in solving matrix equations, especially when calculating the inverse matrix used to derive solutions to systems of equations.
Adjugate Matrix
The adjugate matrix plays a significant role in finding the inverse of a matrix. It is derived from the original matrix through a combination of minors, cofactors, and transposition. The adjugate matrix is defined as:
  • **Minors and Cofactors:** Each element of the matrix is replaced by its cofactor, which is the determinant of the matrix formed by removing the row and column of that element.
  • **Transpose:** The rows and columns of the resulting cofactor matrix are then swapped to obtain the adjugate matrix.
  • **Inverse Formula:** The inverse of a matrix \(A\) is given by \(A^{-1}= \frac{1}{\text{det}(A)} \cdot \text{adj}(A)\). This formula highlights the importance of both the determinant and adjugate matrix in inverting a square matrix.
Understanding the construction of the adjugate matrix is essential for handling equations involving inverses, as seen in solving systems of equations using matrix methods.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Let \(A, B\) and \(C\) be \(n \times n\) matrices with \(A\) being invertible. Show that: (i) \(A^{-1}\) is invertible and \(\left(A^{-1}\right)^{-1}=A\); (ii) \(A^{T}\) is invertible and \(\left(A^{T}\right)^{-1}=\left(A^{-1}\right)^{T}\) (for the definition of \(A^{T}\) see Exercise 24(a) of Chapter 3); (iii) if \(A B=A C\) then \(B=C\); (iv) if \(A\) is symmetric then so is \(A^{-1}\).

Show that if $$ A=\left[\begin{array}{ccc} -1 & -1 & 1 \\ \frac{3}{2} & -\frac{1}{2} & 0 \end{array}\right] \text { and } \quad B=\left[\begin{array}{ll} 1 & 2 \\ 3 & 4 \\ 5 & 6 \end{array}\right] $$ then \(A B=I_{2}\). Then \(A\) is a left inverse for \(B\) and \(B\) a right inverse for \(A\). Find several more such inverses. Is \(A\) also a right inverse for \(B\) ? That is, is \(B A=I_{3}\) ?

\(B=\left[\begin{array}{ccc}5 & 1 & -2 \\ 0 & -1 & 1 \\ 6 & -9 & 11\end{array}\right]\) was obtained from the matrix \(A\) by first adding 5 times row \(2(\) of \(A)\) to row 3 and then dividing row 2 by \(2 .\) Write \(B=E_{2} E_{1} A\) where \(E_{1}, E_{2}\) are suitable elementary matrices. Find \(A\) (i) by using suitable elementary row operations; (ii) by finding \(E_{1}^{-1}, E_{2}^{-1}\) and then using \(A=E_{1}^{-1} E_{2}^{-1} B\)

(a) If \(A_{m \times n}\) is not square show there cannot exist an \(n \times m\) matrix \(C\) such that \(A C=I_{m}\) and \(C A=I_{n}\). [Hint: if \(m

Let \(A, B\) be \(n \times n\) matrices. We know that \(\operatorname{Tr}(A B)=\operatorname{Tr}(B A)\). (See Exercise 23 of Chapter 3.) Deduce that, if \(C\) is invertible then \(\operatorname{Tr}\left(C^{-1} A C\right)=\operatorname{Tr}(A)\). [Hint: replace \(A\) by \(C^{-1} A\) and \(B\) by \(C .]\)
See all solutions

Recommended explanations on Math Textbooks

Pure Maths

Read Explanation

Decision Maths

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Geometry

Read Explanation

Discrete Mathematics

Read Explanation

Mechanics Maths

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.