Chapter 3: Problem 29
Use matrices to solve each system of equations. $$ \left\\{\begin{array}{l} x+y+z=6 \\ x+2 y+z=8 \\ x+y+2 z=7 \end{array}\right. $$
Short Answer
Expert verified
The solution is \( x = 2 \), \( y = 1 \), \( z = 3 \).
Step by step solution
01
Write the System as a Matrix Equation
First, express the system of equations as a matrix equation of the form \( AX = B \), where \( A \) is the coefficient matrix, \( X \) is the column matrix of variables, and \( B \) is the column matrix of constants. The given system is:\[\begin{bmatrix} 1 & 1 & 1 \ 1 & 2 & 1 \ 1 & 1 & 2 \end{bmatrix} \begin{bmatrix} x \ y \ z \end{bmatrix} = \begin{bmatrix} 6 \ 8 \ 7\end{bmatrix}.\]
02
Check Matrix Determinant
Ensure that the coefficient matrix \( A \) has a non-zero determinant to confirm that the system has a unique solution. For matrix \( A \):\[ A = \begin{bmatrix} 1 & 1 & 1 \ 1 & 2 & 1 \ 1 & 1 & 2 \end{bmatrix}.\]Calculate the determinant of \( A \):\[ \text{det}(A) = 1(2 \cdot 2 - 1 \cdot 1) - 1(1 \cdot 2 - 1 \cdot 1) + 1(1 \cdot 1 - 2 \cdot 1) = 4 - 1 + (-1) = 2.\]Since \( \text{det}(A) = 2 eq 0 \), the system has a unique solution.
03
Find the Inverse of Matrix A
Compute the inverse of matrix \( A \) using the formula for the inverse of a 3x3 matrix. The inverse is given by:\[ A^{-1} = \frac{1}{\text{det}(A)} \begin{bmatrix} 2 & -1 & -1 \ -1 & 1 & 0 \ -1 & 0 & 1 \end{bmatrix}.\]Since the determinant is 2, the inverse is:\[ A^{-1} = \frac{1}{2} \begin{bmatrix} 2 & -1 & -1 \ -1 & 1 & 0 \ -1 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & -\frac{1}{2} & -\frac{1}{2} \ -\frac{1}{2} & \frac{1}{2} & 0 \ -\frac{1}{2} & 0 & \frac{1}{2} \end{bmatrix}.\]
04
Multiply Inverse by Constants Matrix
Multiply the inverse of matrix \( A \) by the constants matrix \( B \):\[ X = A^{-1}B = \begin{bmatrix} 1 & -\frac{1}{2} & -\frac{1}{2} \ -\frac{1}{2} & \frac{1}{2} & 0 \ -\frac{1}{2} & 0 & \frac{1}{2} \end{bmatrix} \begin{bmatrix} 6 \ 8 \ 7 \end{bmatrix}.\]Perform the matrix multiplication:\[ X = \begin{bmatrix} 1(6) + (-\frac{1}{2})(8) + (-\frac{1}{2})(7) \ (-\frac{1}{2})(6) + \frac{1}{2}(8) + 0(7) \ (-\frac{1}{2})(6) + 0(8) + \frac{1}{2}(7) \end{bmatrix} = \begin{bmatrix} 2 \ 1 \ 3 \end{bmatrix}.\]
05
Interpret the Results
The resulting matrix \( X \) provides the values for the variables \( x \), \( y \), and \( z \). Thus, the solution to the system of equations is:- \( x = 2 \)- \( y = 1 \)- \( z = 3 \)
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Systems of Equations
A system of equations is a set of equations that share the same variables. In our exercise, we have three equations involving the variables \( x \), \( y \), and \( z \). The goal is to find values for these variables that satisfy all equations simultaneously. Solving such systems allows us to understand relationships among variables and deduce a common solution.
- Consistent systems have at least one solution. In this example, since the determinant is non-zero, our system is consistent and has exactly one solution.
- Inconsistent systems do not have any solutions.
Matrix Equation
A matrix equation rewrites a system of equations in a compact form using matrices. Our original set of equations is expressed as the matrix equation \( AX = B \). Here:
- \( A \) is the coefficient matrix, representing the coefficients of the variables in each equation.
- \( X \) is the column matrix containing the variables \( x \), \( y \), and \( z \).
- \( B \) is the column matrix of constants, representing the other side of the equations.
Determinant of a Matrix
The determinant of a matrix is a special number that provides various insights into the matrix's properties. For our 3x3 coefficient matrix \( A \), the determinant helps us determine if the system of equations has a unique solution.
- If the determinant is zero, the system has no unique solution and is said to be singular.
- If the determinant is non-zero (as it is here, with a value of 2), the system has a unique solution.
Inverse of a Matrix
The inverse of a matrix \( A \), denoted \( A^{-1} \), is the matrix that, when multiplied by \( A \), results in the identity matrix. The identity matrix acts like the number 1 in matrix arithmetic.
- Finding the inverse involves dividing the adjoint of \( A \) by its determinant.
- The matrix must be square (same number of rows as columns) and have a non-zero determinant to have an inverse.