Problem 36 - College Algebra

Question

For the following exercises, solve a system using the inverse of a \(3 \times 3\) matrix. $$ \begin{array}{l}{4 x+4 y+4 z=40} \\ {2 x-3 y+4 z=-12} \\ {-x+3 y+4 z=9}\end{array} $$

Step-by-Step Solution

Answer

The solutions are \(x = 1\), \(y = -2\), and \(z = 5\).

Step 1: Write the system in matrix form

The given system of equations can be written in matrix form as \(AX = B\), where \(A\) is the coefficient matrix, \(X\) is the column matrix of variables \(\begin{bmatrix} x \ y \ z \end{bmatrix}\), and \(B\) is the constant matrix.

Step 2: Define the coefficient matrix

The coefficient matrix \(A\) is constructed by placing the coefficients of \(x, y, z\) from each equation in the rows. So, \(A\) is: \[A = \begin{bmatrix} 4 & 4 & 4 \ 2 & -3 & 4 \ -1 & 3 & 4 \end{bmatrix}\]

Step 3: Define the constant matrix

The constants from the right side of the equations form matrix \(B\): \[B = \begin{bmatrix} 40 \ -12 \ 9 \end{bmatrix}\]

Step 4: Find the inverse of matrix A

Calculate the inverse of matrix \(A\), denoted as \(A^{-1}\). Use a calculator or software to find:\[A^{-1} = \begin{bmatrix} 0.208 & 0.333 & -0.5 \ -0.083 & -0.333 & 0.5 \ 0.125 & -0.0 & 0.0 \end{bmatrix}\]

Step 5: Apply the inverse matrix to solve for X

To find the solution matrix \(X\), multiply the inverse matrix \(A^{-1}\) by matrix \(B\):\[X = A^{-1}B = \begin{bmatrix} 0.208 & 0.333 & -0.5 \ -0.083 & -0.333 & 0.5 \ 0.125 & -0.0 & 0.0 \end{bmatrix} \begin{bmatrix} 40 \ -12 \ 9 \end{bmatrix} = \begin{bmatrix} 1 \ -2 \ 5 \end{bmatrix}\]

Step 6: Interpret the solution

The solution matrix \(X\) reveals the values of \(x, y, z\). Hence, \(x = 1\), \(y = -2\), and \(z = 5\).

Key Concepts

System of Equations

A system of equations is a collection of two or more equations with a set of unknowns. In this exercise, we have three equations, each involving three variables: \(x\), \(y\), and \(z\). The key objective here is to find values for these variables that satisfy all equations simultaneously. This is crucial in many real-world applications such as engineering, physics, and economics, where multiple conditions must be met at once. To solve this system using matrix inversion, we first rewrite the equations in matrix form, allowing for more streamlined and efficient computational solutions. This transformation is foundational, especially in computer algorithms and numerical methods where matrix operations are a fundamental tool.

Coefficient Matrix

The coefficient matrix, commonly represented as \(A\), contains all the coefficients from the system of equations. For our example, using the provided equations, \(A\) is constructed as follows:

• First row: Coefficients of the first equation (\(4, 4, 4\)).
• Second row: Coefficients of the second equation (\(2, -3, 4\)).
• Third row: Coefficients of the third equation (\(-1, 3, 4\)).

Hence, \(A\) is formed as:\[A = \begin{bmatrix} 4 & 4 & 4 \ 2 & -3 & 4 \ -1 & 3 & 4 \end{bmatrix}\]The coefficient matrix is a compact representation of the linear relationships described by the equations. Its properties, such as invertibility and determinants, are crucial for solving the system via matrix operations.

Inverse Matrix

The inverse matrix, denoted as \(A^{-1}\), is a special matrix that, when multiplied by the original matrix \(A\), yields the identity matrix. This is somewhat analogous to dividing by a number in regular arithmetic. However, not all matrices have inverses. For \(A\) to be invertible, it must be a square matrix (same number of rows as columns) and its determinant must be non-zero. In this exercise, we find that \(A^{-1}\) is:\[A^{-1} = \begin{bmatrix} 0.208 & 0.333 & -0.5 \ -0.083 & -0.333 & 0.5 \ 0.125 & 0.0 & 0.0 \end{bmatrix}\]Finding the inverse is often done using computational tools due to complexity, particularly for large matrices. Once the inverse is known, it can be used to solve the system of equations efficiently.

Matrix Multiplication

Matrix multiplication is a core operation in linear algebra, involving the combination of rows from one matrix with the columns of another. For our system of equations, after computing \(A^{-1}\), we use matrix multiplication to solve for the variable matrix \(X\). Here’s how it works:To solve \(AX = B\);1. Compute \(X = A^{-1}B\).The multiplication follows this principle:- Multiply each element of the rows from \(A^{-1}\) with the corresponding elements in the columns of \(B\), and sum the results to form a single element of the solution matrix.Performing the multiplication, we find:\[X = A^{-1}B = \begin{bmatrix} 1 \ -2 \ 5 \end{bmatrix}\]This result reveals the values of the variables, providing a powerful method to solve complex systems through algebraic manipulation in matrix form.