Problem 49 - A Graphical Approach to College Algebra

Question

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

Step-by-Step Solution

Answer

Solve by finding the inverse of matrix \( A \) and multiplying it by the constants vector.

Step 1: Write as a Matrix Equation

First, express the system of equations in matrix form: \[A \cdot \begin{bmatrix} x \ y \ z \end{bmatrix} = \begin{bmatrix} 1 \ 2 \ 3 \end{bmatrix}\]where the matrix \( A \) and the constants are given by:\[A = \begin{bmatrix} \pi & e & \sqrt{2} \ e & \pi & \sqrt{2} \ \sqrt{2} & e & \pi\end{bmatrix}, \ b = \begin{bmatrix} 1 \ 2 \ 3 \end{bmatrix}\]

Step 2: Calculate the Inverse of the Matrix

Find the inverse of matrix \( A \), denoted as \( A^{-1} \). This step involves calculating the determinant of \( A \) and the adjugate of \( A \), followed by using the formula \[A^{-1} = \frac{1}{\text{det}(A)} \cdot \text{adj}(A)\]Ensure that \( \text{det}(A) eq 0 \) for the matrix to be invertible.

Step 3: Verify the Matrix is Invertible

Compute the determinant of the matrix \( A \). If it is non-zero, the matrix is invertible and we can proceed to find its inverse. Otherwise, the system does not have a unique solution.

Step 4: Solve for the Variables Using Inverse

Once \( A^{-1} \) is computed, solve for the variables \( x, y, z \) using the matrix equation: \[\begin{bmatrix} x \ y \ z \end{bmatrix} = A^{-1} \cdot \begin{bmatrix} 1 \ 2 \ 3 \end{bmatrix}\]Multiply the inverse matrix by the constants vector to find the solution for \( x, y, z \).

Step 5: Interpretation of the Solution

Write the solution vector as \( \begin{bmatrix} x \ y \ z \end{bmatrix} = \begin{bmatrix} a \ b \ c \end{bmatrix} \), where \( a, b, c \) are real numbers obtained in the previous step after solving the matrix equation.

Key Concepts

Systems of Equations

When solving systems of equations, you're essentially finding values for the variables that will satisfy all equations in the set simultaneously. In a system of linear equations, you could have two, three, or more equations with the same number of variables. For instance, in the exercise provided:

• \( \pi x + e y + \sqrt{2} z = 1 \)
• \( e x + \pi y + \sqrt{2} z = 2 \)
• \( \sqrt{2} x + e y + \pi z = 3 \)

This system has three equations and involves the variables \( x, y, \) and \( z \). The goal is to discover values for these variables such that all equations hold true. Solving a system can be achieved in multiple ways, such as substitution, elimination, or matrix methods like the matrix inverse method used here. This particular method is beneficial when working with large systems or when leveraging computer algorithms for solutions.

Matrix Algebra

Matrix algebra is a powerful tool in mathematics used to solve systems of equations, and more broadly, to handle transformations in various dimensions. Matrices are rectangular arrays of numbers arranged in rows and columns. For example, the matrix involved in our exercise is:
\[ A = \begin{bmatrix} \pi & e & \sqrt{2} \ e & \pi & \sqrt{2} \ \sqrt{2} & e & \pi \end{bmatrix} \]
This form allows us to represent and manipulate systems concisely and computationally. In matrix algebra:

• An important operation is matrix multiplication, which is used to express matrix equations such as \( A \cdot X = B \).
• Matrix inversion is used when solving systems, where you find the inverse \( A^{-1} \) to isolate the variable matrix \( X \).

These operations simplify handling equations simultaneously, especially in systems with numerous variables.

Determinant

The determinant of a matrix is a scalar value that provides important information about the matrix and the system of equations. It's calculated using the matrix's elements and helps determine if the matrix is invertible (non-singular). The determinant is vital because:

• If \( \text{det}(A) eq 0 \), the matrix is invertible, meaning there is a unique solution to the corresponding system of equations.
• If \( \text{det}(A) = 0 \), the matrix is not invertible, and the system may have no solutions or infinitely many, making a unique solution impossible.

For example, in this exercise, calculating \( \text{det}(A) \) is crucial. Once confirmed as non-zero, it enables the computation of the inverse using the formula:
\[ A^{-1} = \frac{1}{\text{det}(A)} \cdot \text{adj}(A) \]
This equation shows how the determinant influences the inverse, underscoring its role in solving systems via the matrix inverse method.