Problem 44 - A Graphical Approach to College Algebra

Question

Solve each system by using the matrix inverse method. $$\begin{aligned} x+y-z &=6 \\ 2 x-y+z &=-9 \\ x-2 y+3 z &=1 \end{aligned}$$

Step-by-Step Solution

Answer

The solution is \( x = 1 \), \( y = -2 \), \( z = 5 \).

Step 1: Write the System as a Matrix Equation

First, express the given system of equations in the form of a matrix equation. The system can be represented as \( A \mathbf{x} = \mathbf{b} \), where \( A \) is the coefficient matrix, \( \mathbf{x} \) is the variable matrix, and \( \mathbf{b} \) is the constants matrix.\[ A = \begin{bmatrix} 1 & 1 & -1 \ 2 & -1 & 1 \ 1 & -2 & 3 \end{bmatrix}, \quad \mathbf{x} = \begin{bmatrix} x \ y \ z \end{bmatrix}, \quad \mathbf{b} = \begin{bmatrix} 6 \ -9 \ 1 \end{bmatrix} \]

Step 2: Find the Inverse of Matrix A

Next, find the inverse of matrix \( A \). The inverse of a matrix exists only if the determinant is non-zero. We'll check that first and then proceed with finding the inverse.Compute the determinant of \( A \):\[ \text{det}(A) = 1(-1)(3) + 1(1)(1) + (-1)(2)(-2) - (-1)(-1)(1) - 1(2)(3) - 1(-2)(-1) = -19 \]Since \( \text{det}(A) eq 0 \), the inverse \( A^{-1} \) exists, which can be found using the formula:\[ A^{-1} = \frac{1}{\text{det}(A)} \text{Adj}(A) \]Calculate \( A^{-1} \):\[ A^{-1} = \frac{1}{-19} \begin{bmatrix} -1 & 9 & 5 \ 5 & -15 & -1 \ -3 & 5 & -3 \end{bmatrix} = \begin{bmatrix} \frac{1}{19} & -\frac{9}{19} & -\frac{5}{19} \ -\frac{5}{19} & \frac{15}{19} & \frac{1}{19} \ \frac{3}{19} & -\frac{5}{19} & \frac{3}{19} \end{bmatrix} \]

Step 3: Compute \( \mathbf{x} = A^{-1} \mathbf{b} \)

Multiply the inverse of matrix \( A \) by matrix \( \mathbf{b} \) to find \( \mathbf{x} \).\[A^{-1} \mathbf{b} = \begin{bmatrix} \frac{1}{19} & -\frac{9}{19} & -\frac{5}{19} \ -\frac{5}{19} & \frac{15}{19} & \frac{1}{19} \ \frac{3}{19} & -\frac{5}{19} & \frac{3}{19} \end{bmatrix} \begin{bmatrix} 6 \ -9 \ 1 \end{bmatrix}\]To multiply, compute each element of the resulting vector:\[ x = \frac{1}{19}(6) + -\frac{9}{19}(-9) + -\frac{5}{19}(1) = 1 \]\[ y = -\frac{5}{19}(6) + \frac{15}{19}(-9) + \frac{1}{19}(1) = -2 \]\[ z = \frac{3}{19}(6) + -\frac{5}{19}(-9) + \frac{3}{19}(1) = 5 \]

Step 4: State the Solution

The solution \( \mathbf{x} = \begin{bmatrix} x \ y \ z \end{bmatrix} \) is obtained as follows:\[ \begin{bmatrix} x \ y \ z \end{bmatrix} = \begin{bmatrix} 1 \ -2 \ 5 \end{bmatrix} \]Hence, the values of \( x \), \( y \), and \( z \) that satisfy the given system of equations are \( x = 1 \), \( y = -2 \), and \( z = 5 \).

Key Concepts

System of Equations

A system of equations is a collection of two or more equations with the same set of unknowns. In the context of linear algebra, these systems involve equations that are all linear in nature.
To solve these systems, we aim to find values for the unknowns that satisfy all the equations simultaneously.
For example, in the original exercise, the system of equations involves three equations with three unknowns:

• Equation 1: \( x + y - z = 6 \)
• Equation 2: \( 2x - y + z = -9 \)
• Equation 3: \( x - 2y + 3z = 1 \)

Understanding these equations as a system helps us apply methods like matrix inverse to find solutions efficiently.
This approach requires us to write the system in a matrix form which introduces matrices as a powerful tool for handling multiple equations.

Determinant of a Matrix

The determinant of a matrix is a special number that can be calculated from its elements.
It provides important insights about the matrix, such as whether it is invertible.
For a matrix to have an inverse, its determinant must be non-zero. This is crucial when solving a system of equations using the matrix inverse method.
The determinant tells us if the matrix transformations result in a unique solution, no solution, or infinitely many solutions.
In the original exercise, the determinant of matrix \( A \) was calculated to be \(-19\). This indicates that the matrix is invertible, meaning the system of equations has a unique solution.
In summary, determinants play a key role in matrix calculations.

Inverse of a Matrix

The inverse of a matrix is a matrix that, when multiplied with the original matrix, yields the identity matrix.
Not every matrix has an inverse. It exists only if the matrix is square (same number of rows and columns) and its determinant is not zero.
The inverse of a matrix \( A \), denoted as \( A^{-1} \), satisfies the following equation:

• \( A \, A^{-1} = A^{-1} \, A = I \), where \( I \) is the identity matrix.

To find the inverse, use the formula:

• \( A^{-1} = \frac{1}{\text{det}(A)} \text{Adj}(A) \)

Here, \( \text{Adj}(A) \) is the adjugate of \( A \).
In solving the system in the exercise, after confirming that \( ext{det}(A) = -19 \), we computed \( A^{-1} \) to be used in further calculations.
Without an inverse, solving the system using the matrix inverse method would not be possible.

Matrix Multiplication

Matrix multiplication is an operation that takes two matrices and produces a new matrix.
This operation is essential in finding the solutions to a system of equations using the matrix inverse method.
Matrix multiplication is not element-wise; it involves the systematic combination of rows and columns.
Given two matrices \( A \) and \( B \), the product \( AB \) is computed by taking the dot product of rows in \( A \) with columns in \( B \).
In the context of solving equations, matrix multiplication works as follows: once the inverse of matrix \( A \) is found, it is multiplied by the constants matrix \( \mathbf{b} \) to find the variable matrix \( \mathbf{x} \).
The outcome of this multiplication provided the solution of the system, yielding the values of \( x \), \( y \), and \( z \) in the original exercise. Understanding this operation is crucial for determining solutions using matrices.