Problem 46 - A Graphical Approach to Precalculus with Limits

Question

Solve each system by using the matrix inverse method. $$\begin{aligned} 3 x+2 y-w &=0 \\ 2 x+z+2 w &=5 \\ x+2 y-z &=-2 \\ 2 x-y+z+w &=2 \end{aligned}$$

Step-by-Step Solution

Answer

The solution is \( x = -4 \), \( y = -21 \), \( z = -13 \), \( w = 10 \).

Step 1: Write the System in Matrix Form

To solve the system using the matrix inverse method, first express it in matrix form as \( \mathbf{A} \mathbf{x} = \mathbf{b} \). The coefficient matrix \( \mathbf{A} \) is: \[\mathbf{A} = \begin{bmatrix} 3 & 2 & 0 & -1 \ 2 & 0 & 1 & 2 \ 1 & 2 & -1 & 0 \ 2 & -1 & 1 & 1 \end{bmatrix}\]The variables matrix \( \mathbf{x} \) is: \[\mathbf{x} = \begin{bmatrix} x \ y \ z \ w \end{bmatrix}\]The constants matrix \( \mathbf{b} \) is: \[\mathbf{b} = \begin{bmatrix} 0 \ 5 \ -2 \ 2 \end{bmatrix}\]

Step 2: Find the Inverse of Matrix A

The next step is to find the inverse of the coefficient matrix \( \mathbf{A} \), denoted as \( \mathbf{A}^{-1} \). Use a calculator or software to compute the inverse, since manual calculations for a 4x4 matrix can be lengthy. The inverse \( \mathbf{A}^{-1} \) is: \[\mathbf{A}^{-1} = \begin{bmatrix} 1 & 0 & 1 & -1 \ 4 & 1 & 5 & -3 \ 3 & 1 & 4 & -2 \ -2 & -1 & -2 & 1 \end{bmatrix}\]

Step 3: Multiply Inverse with Constants Matrix

Now, multiply the inverse of the coefficient matrix \( \mathbf{A}^{-1} \) with the constants matrix \( \mathbf{b} \) to find the solution matrix \( \mathbf{x} \):\[\mathbf{x} = \mathbf{A}^{-1} \mathbf{b} = \begin{bmatrix} 1 & 0 & 1 & -1 \ 4 & 1 & 5 & -3 \ 3 & 1 & 4 & -2 \ -2 & -1 & -2 & 1 \end{bmatrix} \begin{bmatrix} 0 \ 5 \ -2 \ 2 \end{bmatrix} = \begin{bmatrix} -4 \ -21 \ -13 \ 10 \end{bmatrix}\]

Step 4: Interpret the Solution Matrix

The solution matrix \( \mathbf{x} = \begin{bmatrix} x \ y \ z \ w \end{bmatrix} = \begin{bmatrix} -4 \ -21 \ -13 \ 10 \end{bmatrix} \) indicates that the solution to the system of equations is \( x = -4 \), \( y = -21 \), \( z = -13 \), and \( w = 10 \).

Key Concepts

Systems of Linear Equations

Systems of linear equations are a collection of linear equations involving the same set of variables. In our example, the system consists of four equations with four variables: \(x, y, z,\) and \(w\). Each equation represents a line in a higher-dimensional space, and the solution to the system is the point where all these lines intersect, if such a point exists.

There are different ways to solve a system of linear equations. One intuitive method involves graphically interpreting the intersection of lines or planes represented by each equation. Yet, in higher dimensions and complex systems, algebraic and matrix methods become more practical.

Understanding these systems is crucial in fields where modeling with equations represents complex relationships, such as engineering, physics, and economics.

Matrix Operations

Matrix operations involve various calculations with matrices, which are rectangular arrays of numbers. Matrices can represent coefficients in systems of equations, making them vital for solving such systems using matrix methods. Key operations include addition, subtraction, multiplication, and finding inverses of matrices.

• Addition and Subtraction: These operations are performed element-wise and require matrices to have the same dimensions.
• Multiplication: More complex than addition, it involves the dot product of rows and columns from two matrices. This operation is crucial when solving systems with matrices.
• Identity Matrix: An identity matrix is a special square matrix with ones on the diagonal and zeros elsewhere. It is the multiplicative identity in matrix algebra, similar to how 1 functions in regular arithmetic.

The computation of these operations requires familiarity with matrix dimensions and properties, which is essential for using the matrix inverse method to resolve equations.

Matrix Inversion

Matrix inversion is a fundamental concept in linear algebra, particularly in solving systems of linear equations. The inverse of a matrix \( \mathbf{A} \) is denoted \( \mathbf{A}^{-1} \), and it has a unique property:\[ \mathbf{A} \cdot \mathbf{A}^{-1} = \mathbf{I} \]where \( \mathbf{I} \) is the identity matrix.

Not all matrices have inverses. A matrix must be square (same number of rows and columns) and its determinant must be non-zero. To find the inverse, especially for larger matrices like 4x4, it is often more practical to use computation tools or calculators.

• If the inverse exists, it can be used to transform the equation \( \mathbf{A} \mathbf{x} = \mathbf{b} \) into \( \mathbf{x} = \mathbf{A}^{-1} \mathbf{b} \), effectively solving for the vector \( \mathbf{x} \).
• Studying matrix inversion deepens understanding of linear transformations and their reversals.

Grasping this concept is essential for efficiently using mathematical tools and approaches in algebra and beyond.

Solution of Equations

The solution of a system of equations is the set of values for the variables that satisfies all equations simultaneously. Using the matrix inverse method, as shown in our example, enables a systematic approach through linear algebra techniques.

After expressing the system as \( \mathbf{A} \mathbf{x} = \mathbf{b} \), we compute the inverse of \( \mathbf{A} \), then multiply it by \( \mathbf{b} \) to isolate \( \mathbf{x} \). The resulting values, \( x = -4 \), \( y = -21 \), \( z = -13 \), and \( w = 10 \), are the solutions that make each original equation true.

• This method is particularly advantageous when dealing with large numbers of equations or when alternative methods like substitution become unwieldy.
• Solutions provide insights into systems behavior, model predictions, and help improve decision-making processes in real-world applications.

Understanding this method encapsulates both the concept of linear solutions and practical application, making it a versatile tool in mathematics.