Problem 40 - College Algebra

Question

For the following exercises, solve a system using the inverse of a \(3 \times 3\) matrix. $$\begin{array}{l}{\frac{1}{10} x-\frac{1}{5} y+4 z=-\frac{41}{2}} \\\ {\frac{1}{5} x-20 y+\frac{2}{5} z=-101} \\ {\frac{3}{10} x+4 y-\frac{3}{10} z=23}\end{array}$$

Step-by-Step Solution

Answer

Use matrix inversion; compute \( A^{-1} \), then multiply by \( \mathbf{b} \). Verify solution by substitution.

Step 1: Write the matrix equation

First, we express the given system of equations in a matrix form, \( A \mathbf{x} = \mathbf{b} \), where \( A \) is the coefficient matrix, \( \mathbf{x} \) is the vector of variables, and \( \mathbf{b} \) is the constant vector. \[A = \begin{pmatrix} \frac{1}{10} & -\frac{1}{5} & 4 \ \frac{1}{5} & -20 & \frac{2}{5} \ \frac{3}{10} & 4 & -\frac{3}{10} \end{pmatrix},\mathbf{x} = \begin{pmatrix} x \ y \ z \end{pmatrix},\mathbf{b} = \begin{pmatrix} -\frac{41}{2} \ -101 \ 23 \end{pmatrix} \]

Step 2: Find the inverse of matrix A

Compute the inverse of the matrix \( A \), denoted as \( A^{-1} \). Use the formula for the inverse of a 3x3 matrix and verify if the determinant is non-zero, ensuring the inverse exists. Calculate the inverse using a calculator or software for accuracy.

Step 3: Multiply the inverse by the constant vector

With \( A^{-1} \) calculated, find \( \mathbf{x} \) by multiplying \( A^{-1} \) with \( \mathbf{b} \). This gives \( \mathbf{x} = A^{-1} \mathbf{b} \), effectively solving for \( x, y, z \).

Step 4: Verify the solution

Substitute the obtained values of \( x, y, z \) back into the original equations to ensure they satisfy all three equations. If the left-hand side equals the right-hand side for each equation, the solution is correct.

Key Concepts

System of Equations

A system of equations consists of multiple equations that are solved together because they share variables. In our exercise, the system consists of three equations involving the variables \( x \), \( y \), and \( z \):

• \( \frac{1}{10} x - \frac{1}{5} y + 4z = -\frac{41}{2} \)
• \( \frac{1}{5} x - 20y + \frac{2}{5}z = -101 \)
• \( \frac{3}{10} x + 4y - \frac{3}{10} z = 23 \)

To solve this system, we treat these equations simultaneously to find a common solution for \( x \), \( y \), and \( z \). It means finding a set of values that satisfy all equations at the same time. One effective method for solving such systems with three variables is by using matrices.

Inverse Matrix

An inverse matrix is crucial for solving a system of equations using matrix algebra. For a matrix \( A \), the inverse, denoted as \( A^{-1} \), is a matrix such that when it is multiplied by \( A \), it results in the identity matrix. The identity matrix is a square matrix with ones on the diagonal and zeros elsewhere.In the context of solving equations, if we have a matrix equation \( A\mathbf{x} = \mathbf{b} \), multiplying both sides by \( A^{-1} \) isolates the vector \( \mathbf{x} \):\[ A^{-1}A\mathbf{x} = A^{-1}\mathbf{b} \]\[ \mathbf{x} = A^{-1}\mathbf{b} \]For this process, it's essential that matrix \( A \) has an inverse. This is only possible if \( A \) is a square matrix (same number of rows and columns) and its determinant is non-zero. The determinant of a matrix is a special number that can be calculated from its elements.

3x3 Matrix

A 3x3 matrix is simply a matrix with three rows and three columns. These matrices are common when dealing with systems of three equations, like the one in this exercise. The matrix \( A \) in our problem is a 3x3 matrix composed of the coefficients of the variables in the equations:\[ A = \begin{pmatrix} \frac{1}{10} & -\frac{1}{5} & 4 \ \frac{1}{5} & -20 & \frac{2}{5} \ \frac{3}{10} & 4 & -\frac{3}{10} \end{pmatrix} \]Each row of the matrix \( A \) corresponds to one equation in the system. Similarly, each column corresponds to the coefficients of one of the variables \( x \), \( y \), or \( z \). When the inverse of this 3x3 matrix is correctly calculated, it becomes a powerful tool to solve the system of equations.

Matrix Multiplication

Matrix multiplication is an essential operation in matrix algebra. When multiplying matrices, the rows from the first matrix are combined with the columns of the second matrix. The element in the resulting matrix is obtained by calculating the sum of products of corresponding elements.For instance, to solve the system of equations using the inverse matrix, we perform the multiplication \( A^{-1}\mathbf{b} \). Here's a simple breakdown:

• Each element of the resulting vector \( \mathbf{x} \) is calculated by multiplying the rows of \( A^{-1} \) with the column vector \( \mathbf{b} \).
• This multiplication gives a new vector containing the values of \( x \), \( y \), and \( z \) that satisfy the original system.

Keep in mind that matrix multiplication is not commutative, meaning \( A \times B \) is not necessarily equal to \( B \times A \). This property is important to consider while solving problems involving matrices.