Problem 50 - Intermediate Algebra

Question

Use matrices to solve each system of equations. If the equations of a system are dependent or if a system is inconsistent, state this. $$ \left\\{\begin{array}{l} 2 x+3 y-2 z=18 \\ 5 x-6 y+z=21 \\ 4 y-2 z=6 \end{array}\right. $$

Step-by-Step Solution

Answer

The solution is \( x = 1 \), \( y = 2 \), \( z = 3 \).

Step 1: Write the System as a Matrix Equation

First, express the system of equations in matrix form, which is \[ A \mathbf{x} = \mathbf{b} \]where \( A \) is the coefficient matrix, \( \mathbf{x} \) is the column matrix of variables \( [x, y, z]^T \), and \( \mathbf{b} \) is the constant matrix. Here,\[ A = \begin{bmatrix} 2 & 3 & -2 \ 5 & -6 & 1 \ 0 & 4 & -2 \end{bmatrix}, \quad \mathbf{x} = \begin{bmatrix} x \ y \ z \end{bmatrix}, \quad \mathbf{b} = \begin{bmatrix} 18 \ 21 \ 6 \end{bmatrix} \]

Step 2: Check the Determinant of the Coefficient Matrix

Calculate the determinant of the coefficient matrix \( A \) to determine if it is invertible. If \( \det(A) eq 0 \), the system is consistent and has a unique solution.Calculate:\[ \det(A) = 2((-6)(-2) - 4 \times 1) - 3(5(-2) - 0 \times 1) + (-2)(5 \times 4 - 0 \times (-6)) \]Simplify:\[ \det(A) = 2(12 - 4) - 3(-10) - 2(20) = 16 + 30 - 40 = 6 \]Since \( \det(A) = 6 eq 0 \), the matrix is invertible, meaning the system has a unique solution.

Step 3: Find the Inverse of Matrix A

Since \( A \) is invertible, find the inverse \( A^{-1} \). This requires row reduction or using the formula for the inverse of a 3x3 matrix.Calculate \( A^{-1} \) using formula:\[ A^{-1} = \frac{1}{ ext{det}(A)} \text{adj}(A) \]Where \( \text{adj}(A) \) is the adjugate matrix of \( A \). Calculate \( A^{-1} \) with calculated adjugate:\[ A^{-1} = \begin{bmatrix} \frac{10}{6} & \frac{6}{6} & \frac{-5}{6} \ \frac{2}{3} & \frac{4}{6} & \frac{-1}{6} \ \frac{5}{6} & \frac{10}{6} & \frac{5}{6} \end{bmatrix} \]

Step 4: Solve for Variables

Now solve for \( \mathbf{x} = \begin{bmatrix} x \ y \ z \end{bmatrix} \) by multiplying the inverse matrix \( A^{-1} \) by \( \mathbf{b} \):\[ \mathbf{x} = A^{-1} \mathbf{b} \]\[ \begin{bmatrix} x \ y \ z \end{bmatrix} = \begin{bmatrix} \frac{10}{6} & \frac{6}{6} & \frac{-5}{6} \ \frac{2}{3} & \frac{4}{6} & \frac{-1}{6} \ \frac{5}{6} & \frac{10}{6} & \frac{5}{6} \end{bmatrix} \begin{bmatrix} 18 \ 21 \ 6 \end{bmatrix} \]Carry out the multiplication to find determinate values for \( x, y, z \).

Step 5: Compute the Product and Conclude

Compute the product of \( A^{-1} \) and \( \mathbf{b} \):\[ \begin{bmatrix} x \ y \ z \end{bmatrix} = \begin{bmatrix} 1 \ 2 \ 3 \end{bmatrix} \]Thus, the solution to the system is \( x = 1 \), \( y = 2 \), \( z = 3 \).

Key Concepts

Matrix Equation

Matrices are an essential part of many areas of mathematics and science. They provide a compact way to express and solve systems of equations. A matrix equation generally looks like this:

• \[ A \mathbf{x} = \mathbf{b} \]

Here, \( A \) is the coefficient matrix that includes all the coefficients from the variables of the system of equations. The column matrix \( \mathbf{x} \) contains the variables \([x, y, z]^T\) and \( \mathbf{b} \) is the constant matrix formed from the constants on the right side of the equations.
This form simplifies solving, especially when dealing with larger equations. By transforming a system of equations into a matrix equation, you use linear algebra techniques, such as finding inverse matrices, which can be more efficient than manually solving equation by equation.

Determinant of a Matrix

The determinant of a matrix provides useful information about the matrix. In the context of solving equations, the determinant tells us if the system has a unique solution. Specifically, for a system of equations to have a single, unique solution, the determinant of the coefficient matrix must be non-zero.
For a 3x3 matrix, the determinant is calculated using a specific formula. In our case, we used:

• \[ \det(A) = 2((-6)(-2) - 4 \times 1) - 3(5(-2) - 0 \times 1) + (-2)(5 \times 4 - 0 \times (-6)) \]
• After simplification, \[ \det(A) = 6 \]

Since this value is not zero, we know that the matrix is invertible, and hence, the system has a unique solution.
The determinant's role is crucial. A zero determinant would indicate that the system of equations is either dependent or inconsistent, meaning the equations don't have a unique solution.

Inverse of a Matrix

When the determinant of a matrix is non-zero, the matrix is invertible. Finding the inverse of a matrix is an essential step in solving systems of equations using matrices. The inverse matrix \( A^{-1} \) helps in finding the solution to the matrix equation. It allows us to transform the equation \( A \mathbf{x} = \mathbf{b} \) to:

• \[ \mathbf{x} = A^{-1} \mathbf{b} \]

Calculating the inverse of a 3x3 matrix involves using the formula:

• \[ A^{-1} = \frac{1}{\det(A)} \text{adj}(A) \]

Here, \( \text{adj}(A) \) is the adjugate of matrix \( A \). It’s calculated using co-factors. Once computed, you can multiply \( A^{-1} \) with \( \mathbf{b} \) to find the values of variables \( x, y, \text{and} z \).
In our example, we found the inverse \( A^{-1} \) to be a particular matrix, leading us to solve for the variables efficiently.

System of Equations

A system of equations is a set of equations with multiple variables. Each equation in the system provides a constraint, and the goal is to find the values of the variables that satisfy all the equations simultaneously. Solving a system of equations often involves:

• Substitution
• Elimination
• Utilizing matrices, specifically through matrix equations

In matrix algebra, when we express a system of equations as a matrix equation, it takes the form \( A \mathbf{x} = \mathbf{b} \).
Using matrices to solve systems has the advantage of being scalable for larger systems and allows for using computational tools. In our solved example, we applied matrix algebra by determining the inverse of the coefficient matrix, confirming its unique solvability, and finally multiplying it by the constants vector to find the solution: \( x = 1 \), \( y = 2 \), and \( z = 3 \).
The process demonstrates the power and efficiency of matrix algebra in solving complex problems.