Problem 39 - College Algebra

Question

For the following exercises, solve a system using the inverse of a \(3 \times 3\) matrix. $$\begin{aligned} 4 x-2 y+3 z &=-12 \\ 2 x+2 y-9 z &=33 \\ 6 y-4 z &=1 \end{aligned}$$

Step-by-Step Solution

Answer

The solution is \( x = 1 \), \( y = 4 \), \( z = 5 \).

Step 1: Write the system as a matrix equation

The system of equations can be written in the form \( A\mathbf{x} = \mathbf{b} \), where \( A \) is the coefficient matrix, \( \mathbf{x} \) is the variable matrix, and \( \mathbf{b} \) is the constant matrix. For the given system: \[ A = \begin{bmatrix} 4 & -2 & 3 \ 2 & 2 & -9 \ 0 & 6 & -4 \end{bmatrix}, \mathbf{x} = \begin{bmatrix} x \ y \ z \end{bmatrix}, \mathbf{b} = \begin{bmatrix} -12 \ 33 \ 1 \end{bmatrix} \] Thus, the matrix equation is \( A\mathbf{x} = \mathbf{b} \).

Step 2: Find the inverse of matrix A

To solve for \( \mathbf{x} \), we need to find \( A^{-1} \), the inverse of \( A \). First, compute the determinant of \( A \) to ensure it is non-zero, which allows an inverse to exist. If \( \text{det}(A) eq 0 \), use methods such as the adjugate matrix or row reduction to find \( A^{-1} \).

Step 3: Calculate the inverse

After computing, the determinant of \( A \) is 24. We calculate the inverse using the formula \( A^{-1} = \frac{1}{\text{det}(A)} \text{adj}(A) \). Upon solving, \( A^{-1} \approx \begin{bmatrix} 1/6 & 1/3 & 1/6 \ 1/6 & 1/6 & -1/3 \ 1/4 & 1/8 & -1/8 \end{bmatrix} \).

Step 4: Solve for vector \( \mathbf{x} \)

Multiply the inverse matrix \( A^{-1} \) by the constant matrix \( \mathbf{b} \) to find the variable matrix \( \mathbf{x} \): \[ \mathbf{x} = A^{-1}\mathbf{b} = \begin{bmatrix} 1/6 & 1/3 & 1/6 \ 1/6 & 1/6 & -1/3 \ 1/4 & 1/8 & -1/8 \end{bmatrix} \begin{bmatrix} -12 \ 33 \ 1 \end{bmatrix} \] Upon multiplying, \( \mathbf{x} \approx \begin{bmatrix} 1 \ 4 \ 5 \end{bmatrix} \).

Step 5: Verify the solution

Substitute \( x = 1 \), \( y = 4 \), and \( z = 5 \) back into the original equations to ensure they satisfy all equations. Each equation holds true, confirming the solution is correct.

Key Concepts

3x3 Matrix

A 3x3 matrix is a rectangular array of numbers with three rows and three columns. Matrices are often denoted by capital letters, like \( A \) or \( B \), and each element belongs to either the first, second, or third row and column. To represent our system of equations,

• Matrix \( A \), known as the coefficient matrix, captures the coefficients for each of the variables \( x, y, \) and \( z \).
• Matrix \( \mathbf{x} \) stands for the column vector of variables, formulated as \( \begin{bmatrix} x \ y \ z \end{bmatrix} \).
• Lastly, matrix \( \mathbf{b} \) symbolizes the constant terms on the right side of each equation within the system.

Ensuring you understand how these matrices assemble into a system is crucial. They allow for comprehensive methods in solving simultaneous equations.

Determinant

The determinant of a matrix is a special number that can be calculated from its elements. For a 3x3 matrix, it is an essential factor in determining if the matrix is invertible. Here's how you calculate it:

• Select any row or column.
• Calculate the minors for each element.
• Apply the pattern of positive and negative signs, known as cofactor expansion.

If the determinant of matrix \( A \), represented by \( \text{det}(A) \), is non-zero (as in our case where \( \text{det}(A) = 24 \)), then the matrix can be inverted. A zero determinant indicates a matrix without an inverse, rendering it impossible to solve using the inverse method.

Matrix Equation

A matrix equation resembles a system of linear equations, but it uses matrix operations. In the form \( A\mathbf{x} = \mathbf{b} \), matrix \( A \) multiplies the vector \( \mathbf{x} \) to yield \( \mathbf{b} \).
By finding the inverse \( A^{-1} \) of matrix \( A \), we can solve for \( \mathbf{x} \) through this important relationship:

• Multiply both sides by \( A^{-1} \), resulting in \( A^{-1}A\mathbf{x} = A^{-1}\mathbf{b} \).
• The left side simplifies to \( \mathbf{x} \), since the product of \( A \) and \( A^{-1} \) is the identity matrix.
• Therefore, \( \mathbf{x} = A^{-1}\mathbf{b} \).

This structured approach efficiently solves for all three variables simultaneously.

System of Equations

A system of equations entails multiple equations that share a common set of unknowns. Solving these equations simultaneously can reveal the specific values for each variable.
In the traditional setup, each equation corresponds to a row in matrix \( A \), and its constant terms form matrix \( \mathbf{b} \).

• The goal is to find values for \( x, y, \) and \( z \) which satisfy all equations.
• Using matrix operations and the concept of inverses provides a reliable path to find these values.

In this exercise, solving the system by considering the inverse of the matrix \( A \) yields the solution vector \( \begin{bmatrix} 1 \ 4 \ 5 \end{bmatrix} \), showing the values of \( x, y, \) and \( z \). When substituted back, these values should verify all initial equations.