Problem 29 - Intermediate Algebra

Question

Use matrices to solve each system of equations. $$ \left\\{\begin{array}{l} x+y+z=6 \\ x+2 y+z=8 \\ x+y+2 z=7 \end{array}\right. $$

Step-by-Step Solution

Answer

The solution is \( x = 2 \), \( y = 1 \), \( z = 3 \).

Step 1: 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:\[\begin{bmatrix} 1 & 1 & 1 \ 1 & 2 & 1 \ 1 & 1 & 2 \end{bmatrix} \begin{bmatrix} x \ y \ z \end{bmatrix} = \begin{bmatrix} 6 \ 8 \ 7\end{bmatrix}.\]

Step 2: 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.

Step 3: 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:\[ A^{-1} = \frac{1}{2} \begin{bmatrix} 2 & -1 & -1 \ -1 & 1 & 0 \ -1 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & -\frac{1}{2} & -\frac{1}{2} \ -\frac{1}{2} & \frac{1}{2} & 0 \ -\frac{1}{2} & 0 & \frac{1}{2} \end{bmatrix}.\]

Step 4: 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:\[ X = \begin{bmatrix} 1(6) + (-\frac{1}{2})(8) + (-\frac{1}{2})(7) \ (-\frac{1}{2})(6) + \frac{1}{2}(8) + 0(7) \ (-\frac{1}{2})(6) + 0(8) + \frac{1}{2}(7) \end{bmatrix} = \begin{bmatrix} 2 \ 1 \ 3 \end{bmatrix}.\]

Step 5: 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:- \( x = 2 \)- \( y = 1 \)- \( z = 3 \)

Key Concepts

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.

These systems can be solved using various methods like substitution, elimination, or matrix operations. Using matrices often simplifies solving systems, especially when dealing with many variables or equations.

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.

This representation is useful as it allows us to use linear algebra techniques to solve for \( X \). Transforming a system into this form simplifies the computation process, especially when paired with inverse matrices.

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.

Calculating the determinant involves a specific formula for a 3x3 matrix, which ensures the solution's uniqueness and the matrix's invertibility.

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.

Having the inverse of \( A \) allows us to solve the matrix equation \( AX = B \) easily. By multiplying both sides of the equation by \( A^{-1} \), we isolate the variable matrix \( X \) and achieve the solution \( X = A^{-1}B \). This method streamlines the process of solving systems of linear equations.