Problem 3 - Algebra A Combined Function

Question

Use matrices to solve each system of linear equations. See Example 1. $$ \left\\{\begin{array}{l} x+3 y=2 \\ x+2 y=0 \end{array}\right. $$

Step-by-Step Solution

Answer

x = -4, y = 2.

Step 1: Write the System in Matrix Form

The given system of equations is: \( \begin{align*} x + 3y &= 2, \ x + 2y &= 0 \end{align*} \). To express this system in matrix form, we set up a matrix equation \( A\mathbf{x} = \mathbf{b} \), where \( A \) is the coefficient matrix, \( \mathbf{x} \) is the variable matrix, and \( \mathbf{b} \) is the constant matrix. Thus, we have: \[ A = \begin{pmatrix} 1 & 3 \ 1 & 2 \end{pmatrix}, \quad \mathbf{x} = \begin{pmatrix} x \ y \end{pmatrix}, \quad \mathbf{b} = \begin{pmatrix} 2 \ 0 \end{pmatrix}\]

Step 2: Set Up the Equation to Solve for \( \mathbf{x} \)

We need to solve the matrix equation \( A\mathbf{x} = \mathbf{b} \). This requires finding the inverse of the matrix \( A \), if it exists, to obtain \( \mathbf{x} = A^{-1}\mathbf{b} \).

Step 3: Find the Inverse of Matrix \( A \)

The inverse of a 2x2 matrix \( \begin{pmatrix} a & b \ c & d \end{pmatrix} \) is given by:\[A^{-1} = \frac{1}{ad - bc} \begin{pmatrix} d & -b \ -c & a \end{pmatrix}\]For our matrix \( A \), \( a = 1, b = 3, c = 1, d = 2 \), so we calculate:\[det(A) = ad - bc = 1 \times 2 - 3 \times 1 = 2 - 3 = -1\]The inverse is:\[A^{-1} = \frac{1}{-1} \begin{pmatrix} 2 & -3 \ -1 & 1 \end{pmatrix} = \begin{pmatrix} -2 & 3 \ 1 & -1 \end{pmatrix}\]

Step 4: Multiply the Inverse by Matrix \( \mathbf{b} \)

Using \( A^{-1} \), we multiply it by \( \mathbf{b} \) to find \( \mathbf{x} \).\[\mathbf{x} = A^{-1}\mathbf{b} = \begin{pmatrix} -2 & 3 \ 1 & -1 \end{pmatrix} \begin{pmatrix} 2 \ 0 \end{pmatrix} \]Carrying out the multiplication:\[\begin{pmatrix} -2 \times 2 + 3 \times 0 \ 1 \times 2 + (-1) \times 0 \end{pmatrix} = \begin{pmatrix} -4 \ 2 \end{pmatrix}\]Thus, \( x = -4 \) and \( y = 2 \).

Step 5: Verify the Solution

Substitute \( x = -4 \) and \( y = 2 \) back into the original equations to ensure they satisfy both.- For \( x + 3y = 2 \): \[ -4 + 3(2) = -4 + 6 = 2 \] which is correct.- For \( x + 2y = 0 \): \[ -4 + 2(2) = -4 + 4 = 0 \] which is also correct.Both equations are satisfied by \( x = -4 \) and \( y = 2 \).

Key Concepts

System of Linear Equations

A system of linear equations is a set of two or more equations with the same variables. These equations can be worked on together to find common solutions. For example, if we have equations involving variables \( x \) and \( y \), our goal is to find such values for these variables that satisfy each equation in the system simultaneously.

In the given exercise, the system consists of two linear equations:

• \( x + 3y = 2 \)
• \( x + 2y = 0 \)

The solution to a system like this involves finding the values of \( x \) and \( y \) that make both equations true when substituted back into them. One common method to solve these is by using matrices, which can simplify the process, especially for larger systems involving more variables and equations.

Matrix Inverse

To solve a system of linear equations using matrices, finding the inverse of a matrix is a critical step. In our exercise, the goal was to find the inverse of the coefficient matrix \( A \). The coefficient matrix \( A \) for this system is:

• \( A = \begin{pmatrix} 1 & 3 \ 1 & 2 \end{pmatrix} \)

For a 2x2 matrix \( \begin{pmatrix} a & b \ c & d \end{pmatrix} \), the inverse is calculated as:

• \( A^{-1} = \frac{1}{ad - bc} \begin{pmatrix} d & -b \ -c & a \end{pmatrix} \)

In this exercise, the determinant \( ad - bc \) is \(-1\), making the inverse:

• \( A^{-1} = \begin{pmatrix} -2 & 3 \ 1 & -1 \end{pmatrix} \)

The inverse matrix, when it exists, helps transform the matrix equation into a solvable form. By multiplying the inverse of the coefficient matrix by the constants matrix, we can find the solution for the variables in the system.

Matrix Multiplication

Matrix multiplication is the process used to derive solutions from matrix equations. After finding the inverse of the matrix \( A \), it's crucial to multiply it by the constants vector \( \mathbf{b} \) to find the solution \( \mathbf{x} \). Given:

• \( A^{-1} = \begin{pmatrix} -2 & 3 \ 1 & -1 \end{pmatrix} \)
• \( \mathbf{b} = \begin{pmatrix} 2 \ 0 \end{pmatrix} \)

The multiplication is carried out as follows:

• \( \mathbf{x} = A^{-1}\mathbf{b} = \begin{pmatrix} -2 & 3 \ 1 & -1 \end{pmatrix} \begin{pmatrix} 2 \ 0 \end{pmatrix} \)
• Perform the calculation: \( \begin{pmatrix} -2 \times 2 + 3 \times 0 \ 1 \times 2 + (-1) \times 0 \end{pmatrix} = \begin{pmatrix} -4 \ 2 \end{pmatrix} \)

This means \( x = -4 \) and \( y = 2 \). This step is crucial as it directly leads to the solution of the system of equations. Each product in the multiplication contributes to specific solutions for the variables \( x \) and \( y \).