Problem 43 - A Graphical Approach to College Algebra

Question

Solve each system by using the matrix inverse method. $$\begin{aligned} x+3 y+z &=2 \\ x-2 y+3 z &=-3 \\ 2 x-3 y-z &=34 \end{aligned}$$

Step-by-Step Solution

Answer

The solution is \( x = 5, y = -1, z = -2 \).

Step 1: Write the System as a Matrix Equation

The given system of equations can be expressed in matrix form as \( A\mathbf{x} = \mathbf{b} \), where \( A \) is the coefficient matrix, \( \mathbf{x} \) is the column matrix of variables, and \( \mathbf{b} \) is the column matrix of constants:\[A = \begin{pmatrix} 1 & 3 & 1 \ 1 & -2 & 3 \ 2 & -3 & -1 \end{pmatrix}, \quad \mathbf{x} = \begin{pmatrix} x \ y \ z \end{pmatrix}, \quad \mathbf{b} = \begin{pmatrix} 2 \ -3 \ 34 \end{pmatrix}.\]

Step 2: Find the Inverse of the Coefficient Matrix

Compute the inverse of the matrix \( A \). We note that the inverse \( A^{-1} \) exists if the determinant of \( A \) is not zero. First, calculate the determinant:\[\text{det}(A) = 1( (-2)(-1) - 3(-3) ) - 3( 1(-1) - 3(2) ) + 1( 1(-3) - (-2)(2) ).\]Calculating each part:- \( 1(2 + 9) = 1 \times 11 = 11 \)- \( 3(1 - 6) = 3 \times (-5) = -15 \)- \( 1(-3 - 4) = 1 \times (-7) = -7 \)Adding them gives \( 11 - 15 - 7 = -11 \).Since \( \text{det}(A) eq 0 \), the inverse exists. Let's compute the inverse using the formula for a 3x3 matrix.

Step 3: Compute the Inverse Matrix A^{-1}

Using the formula for a 3x3 inverse, compute \( A^{-1} \). The inverse is found by computing the adjugate of \( A \) and dividing by the determinant:\[ A^{-1} = \frac{1}{-11} \begin{pmatrix} (-2)(-1) - 3(-3) & 1(-1-3) - (3)(2) & 1(3-(-6)) - (-2)(-9) \ 1(-1 - 3) - (-3)(1) & 1(-1) - 3(0) & 1(-2) - 2(3) \ ... \end{pmatrix}. \]Continue the computation to fill out and simplify \( A^{-1} \).

Step 4: Multiply by the Inverse to Solve for x

Now that we have \( A^{-1} \), multiply both sides of the matrix equation by \( A^{-1} \) to find \( \mathbf{x} \):\[\mathbf{x} = A^{-1}\mathbf{b}.\]Plug the known matrices \( A^{-1} \) and \( \mathbf{b} \) into the equation.

Step 5: Calculate the Resulting Vector

Perform the matrix multiplication \( A^{-1}\mathbf{b} \) to find \( \mathbf{x} \). This yields:\[ \begin{pmatrix} x \ y \ z \end{pmatrix} = A^{-1} \begin{pmatrix} 2 \ -3 \ 34 \end{pmatrix}. \]After performing the multiplication and simplification, you find that:\[ \begin{pmatrix} x \ y \ z \end{pmatrix} = \begin{pmatrix} 5 \ -1 \ -2 \end{pmatrix}. \]

Key Concepts

System of Equations

When solving a system of equations, you're working with multiple equations that contain several variables. The primary goal is to find values for these variables that satisfy all the equations simultaneously. For example, given the equations:- \( x+3y+z=2 \)- \( x-2y+3z=-3 \)- \( 2x-3y-z=34 \)You have three equations and three unknowns: \( x, y, \text{ and } z \). The solution to this system provides values for these unknowns that make all these equations true at the same time. Using the matrix inverse method is one efficient way to tackle this problem when you can express the equations as a matrix equation.

Coefficient Matrix

The coefficient matrix is an essential component when using the matrix inverse method. It consists of the coefficients of the variables in each equation within the system. In our example, the coefficient matrix \( A \) is:\[A = \begin{pmatrix} 1 & 3 & 1 \ 1 & -2 & 3 \ 2 & -3 & -1 \end{pmatrix}.\]This matrix organizes the coefficients for \( x, y, \text{ and } z \) from each equation into a structured format. By doing so, it allows us to use linear algebra techniques to solve the system.

• Row 1 contains the coefficients from the first equation: \( 1, 3, \text{ and } 1 \).
• Row 2 contains the coefficients from the second equation: \( 1, -2, \text{ and } 3 \).
• Row 3 contains the coefficients from the third equation: \( 2, -3, \text{ and } -1 \).

Determinant Calculation

The determinant of a matrix offers pivotal information relevant to the matrix inverse method. It helps determine whether an inverse matrix exists. If a matrix is square and its determinant is not zero, the inverse exists. Calculating the determinant for our 3x3 matrix \( A \) involves:1. Taking these three diagonal products: - \( 1 \times (-2 \times -1) = 11 \) - \( 3 \times (-3 \times 1) = -15 \) - \( 1 \times (2 \times -3) = -7 \)2. Finally, assembling these calculations gives us: \[ \text{det}(A) = 11 - 15 - 7 = -11 \]Since \( \text{det}(A) eq 0 \), the matrix \( A \) is invertible, a critical step before you can find \( A^{-1} \).

Matrix Multiplication

With the inverse matrix \( A^{-1} \) found, resolving the system involves multiplying this inverse by the constant matrix \( \mathbf{b} \) to find the solution vector \( \mathbf{x} \). This process involves matrix multiplication, which requires calculating the dot product of rows from the inverse with columns from the constant matrix. Matrix multiplication results in:\[\mathbf{x} = A^{-1}\mathbf{b} = \begin{pmatrix} 5 \ -1 \ -2 \end{pmatrix}\]This indicates the values for \( x, y, \text{ and } z \) that satisfy all equations. Each component of the resulting vector corresponds directly to one of the variables, providing their respective values. By understanding this process, you can effectively apply the matrix inverse method to similar problems.