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Chapter 8: Problem 67

apply matrix algebra to solve the system of linear equations. $$\begin{aligned} x+y+z &=1 \\ x-y-z &=-1 \\\\-x+y-z &=-1 \end{aligned}$$

Short Answer

Expert verified
The solution is \( x = -1 \), \( y = 1 \), \( z = 1 \).

Step by step solution

01

Write the System in Matrix Form

The system of equations can be represented in the matrix form as \( A\mathbf{x} = \mathbf{b} \), where \( A \) is the coefficient matrix, \( \mathbf{x} \) is the vector of variables, and \( \mathbf{b} \) is the constant vector. For the given system, \( A = \begin{bmatrix} 1 & 1 & 1 \ 1 & -1 & -1 \ -1 & 1 & -1 \end{bmatrix} \), \( \mathbf{x} = \begin{bmatrix} x \ y \ z \end{bmatrix} \), and \( \mathbf{b} = \begin{bmatrix} 1 \ -1 \ -1 \end{bmatrix} \).
02

Find the Inverse of Matrix A

To solve \( A\mathbf{x} = \mathbf{b} \), we need to compute the inverse of \( A \), denoted as \( A^{-1} \). The inverse \( A^{-1} \) exists if the determinant of \( A \) is non-zero. Calculate \( \det(A) = 1(-1 \cdot -1 - -1 \cdot 1) - 1(1 \cdot -1 - -1 \cdot -1) + 1(1 \cdot 1 - (-1) \cdot 1) = 2 \). Since \( \det(A) eq 0 \), \( A \) is invertible.
03

Calculate the Inverse of Matrix A

To find \( A^{-1} \), use the formula for the inverse: \( A^{-1} = \frac{1}{\det(A)} \text{adj}(A) \), where \( \text{adj}(A) \) is the adjugate of \( A \). Calculate adjugate using the minors and cofactor matrix, resulting in: \( \text{adj}(A) = \begin{bmatrix} 0 & 0 & 2 \ 0 & -2 & 0 \ 2 & 0 & 0 \end{bmatrix} \). Therefore, \( A^{-1} = \frac{1}{2} \begin{bmatrix} 0 & 0 & 2 \ 0 & -2 & 0 \ 2 & 0 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 1 \ 0 & -1 & 0 \ 1 & 0 & 0 \end{bmatrix} \).
04

Solve for the Variable Vector \(\mathbf{x}\)

Using the inverse, calculate \( \mathbf{x} = A^{-1} \mathbf{b} \). Substitute the values into the equation: \( \mathbf{x} = \begin{bmatrix} 0 & 0 & 1 \ 0 & -1 & 0 \ 1 & 0 & 0 \end{bmatrix} \begin{bmatrix} 1 \ -1 \ -1 \end{bmatrix} = \begin{bmatrix} 0 \times 1 + 0 \times (-1) + 1 \times (-1) \ 0 \times 1 + (-1) \times (-1) + 0 \times (-1) \ 1 \times 1 + 0 \times (-1) + 0 \times (-1) \end{bmatrix} = \begin{bmatrix} -1 \ 1 \ 1 \end{bmatrix} \).
05

Write the Solution

The solution to the system of equations, given by \( \mathbf{x} = \begin{bmatrix} x \ y \ z \end{bmatrix} = \begin{bmatrix} -1 \ 1 \ 1 \end{bmatrix} \), is \( x = -1 \), \( y = 1 \), and \( z = 1 \).

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Key Concepts

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

System of Linear Equations
A system of linear equations is a collection of two or more linear equations involving the same set of variables. In this exercise, the system of equations involves three variables: \( x, y, \) and \( z \). Each equation represents a plane in three-dimensional space, and the solution to the system is the point where these planes intersect. Systems of equations can have a single solution, infinitely many solutions, or no solution. To express this system in matrix form, we use
  • the coefficient matrix \( A \)
  • the variable vector \( \mathbf{x} \)
  • the constant vector \( \mathbf{b} \)
This matrix representation allows us to apply matrix algebra techniques to solve the system.
Inverse of a Matrix
The inverse of a matrix \( A \), denoted as \( A^{-1} \), plays a critical role in solving matrix equations. It allows us to find the solution to the matrix equation \( A\mathbf{x} = \mathbf{b} \) by transforming it to \( \mathbf{x} = A^{-1}\mathbf{b} \). However, not all matrices have inverses.
The inverse exists only if the determinant of the matrix is nonzero. In this exercise, the determinant of the matrix \( A \) is calculated to be 2, which is nonzero, confirming the existence of the inverse. Once we know the inverse exists, we can compute it using the formula
\( A^{-1} = \frac{1}{\det(A)}\text{adj}(A) \).
Determinant of a Matrix
The determinant of a matrix is a special number that can be calculated from its elements. It provides key insights into the properties of the matrix, such as invertibility. For a 3x3 matrix like \( A \), the determinant is calculated using a specific formula, which involves sums and products of its elements and their minors.
The determinant is important because:
  • If the determinant is zero, the matrix does not have an inverse.
  • If the determinant is non-zero, the matrix is invertible, and the inverse can be used to find solutions to matrix equations.
In this exercise, the determinant was determined to be 2, indicating that the inverse can indeed be calculated.
Adjugate Matrix
The adjugate matrix, denoted \( \text{adj}(A) \), is the transpose of the cofactor matrix. It is essential in computing the inverse of a matrix. Each element of the adjugate is the determinant of a submatrix, obtained by deleting one row and one column, and it is multiplied by \((-1)^{i+j}\) to account for position.
In this solution, the adjugate of matrix \( A \) was calculated and used to find the inverse. It highlights how the adjugate matrix assists in transforming \( A \) into \( A^{-1} \). The final inverse matrix is obtained by dividing each element of the adjugate by the determinant of \( A \). Understanding the adjugate matrix helps deepen the comprehension of matrix operations and their outcomes in solving systems of equations.

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Most popular questions from this chapter

In calculus, determinants are used when evaluating double and triple integrals through a change of variables. In these cases, the elements of the determinant are functions. Find each determinant. $$\left|\begin{array}{ccc} \sin \phi \cos \theta & -\rho \sin \phi \sin \theta & \rho \cos \phi \cos \theta \\ \sin \phi \sin \theta & \rho \sin \phi \cos \theta & \rho \cos \phi \sin \theta \\ \cos \phi & 0 & -\rho \sin \phi \end{array}\right|$$

In Exercises \(79-84,\) determine whether each statement is true or false. A system of linear inequalities always has a solution.

Evaluate the determinant: $$ \left|\begin{array}{rrrr} 1 & -2 & -1 & 3 \\ 4 & 0 & 1 & 2 \\ 0 & 3 & 2 & 4 \\ 1 & -3 & 5 & -4 \end{array}\right| $$

In calculus, the first steps when solving the problem of finding the area enclosed by a set of curves are similar to those for finding the feasible region in a linear programming problem. In Exercises \(95-98,\) graph the system of inequalities and identify the vertices, that is, the points of intersection of the given curves. $$\begin{aligned} &\begin{array}{l} y \leq x^{2} \\ y \geq x^{3} \end{array}\\\ &x \geq 0 \end{aligned}$$

Determine whether the statements are true or false. $$\text { If } A=\left[\begin{array}{ll}a_{11} & a_{12} \\ a_{21} & a_{22}\end{array}\right], \text { then } A^{-1}=\left[\begin{array}{cc}\frac{1}{a_{11}} & \frac{1}{a_{12}} \\\\\frac{1}{a_{21}} & \frac{1}{a_{22}}\end{array}\right]$$
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