Problem 22 - Algebra 2

Question

Use a matrix equation to solve each system of equations. \(x+2 y=8\) \(3 x+2 y=6\)

Step-by-Step Solution

Answer

\(x = -1\), \(y = 4.5\).

Step 1: Write the system of equations in matrix form

We are given the system of equations:\(x + 2y = 8\)\(3x + 2y = 6\).We can express this system in the form of a matrix equation \(A\mathbf{x} = \mathbf{b}\), where \(A\) is the coefficient matrix, \(\mathbf{x}\) is the column vector of variables, and \(\mathbf{b}\) is the column vector of constants.Thus, \(A = \begin{bmatrix} 1 & 2 \ 3 & 2 \end{bmatrix}\), \(\mathbf{x} = \begin{bmatrix} x \ y \end{bmatrix}\), and \(\mathbf{b} = \begin{bmatrix} 8 \ 6 \end{bmatrix}\). So the matrix equation is:\[\begin{bmatrix} 1 & 2 \ 3 & 2 \end{bmatrix} \begin{bmatrix} x \ y \end{bmatrix} = \begin{bmatrix} 8 \ 6 \end{bmatrix}\]

Step 2: Find the inverse of the coefficient matrix

To solve the matrix equation \(A\mathbf{x} = \mathbf{b}\), we need to find the inverse of matrix \(A\), denoted as \(A^{-1}\). The formula for finding the inverse of a 2x2 matrix \(\begin{bmatrix} a & b \ c & d \end{bmatrix}\) is: \[A^{-1} = \frac{1}{ad - bc}\begin{bmatrix} d & -b \ -c & a \end{bmatrix}\]For \(A = \begin{bmatrix} 1 & 2 \ 3 & 2 \end{bmatrix}\), compute the determinant: \(ad - bc = (1)(2) - (2)(3) = 2 - 6 = -4\).Thus, the inverse is:\[A^{-1} = \frac{1}{-4} \begin{bmatrix} 2 & -2 \ -3 & 1 \end{bmatrix} = \begin{bmatrix} -0.5 & 0.5 \ 0.75 & -0.25 \end{bmatrix}\]

Step 3: Solve for the variable vector

Using the formula \(\mathbf{x} = A^{-1}\mathbf{b}\), we substitute the matrices:\[\mathbf{x} = \begin{bmatrix} -0.5 & 0.5 \ 0.75 & -0.25 \end{bmatrix} \begin{bmatrix} 8 \ 6 \end{bmatrix}\] Now perform the matrix multiplication:\[\begin{bmatrix} -0.5 \times 8 + 0.5 \times 6 \ 0.75 \times 8 - 0.25 \times 6 \end{bmatrix} = \begin{bmatrix} -4 + 3 \ 6 - 1.5 \end{bmatrix} = \begin{bmatrix} -1 \ 4.5 \end{bmatrix}\].

Step 4: Interpret the solution

The solution \(\begin{bmatrix} x \ y \end{bmatrix} = \begin{bmatrix} -1 \ 4.5 \end{bmatrix}\) means \(x = -1\) and \(y = 4.5\).Substitute \(x = -1\) and \(y = 4.5\) back into the original equations to verify that both equations are satisfied, confirming that the solution is correct.

Key Concepts

Systems of Equations

A system of equations is a set of two or more equations with the same variables. Here, we have:

• Equation 1: \( x + 2y = 8 \)
• Equation 2: \( 3x + 2y = 6 \)

The goal in solving these equations is to find values for \( x \) and \( y \) that satisfy both equations simultaneously. Representing a system in a matrix form, as \( A\mathbf{x} = \mathbf{b} \), introduces a neat and structured way to handle such problems. We express the coefficients in a matrix, the variables in a vector, and the constants in another vector. This translation from equations to matrices sets the stage for using matrix operations to solve the system efficiently.

Inverse of a Matrix

Finding the inverse of a matrix is akin to finding a reciprocal of a number in basic arithmetic. Only square matrices (same number of rows and columns) can have inverses. The inverse, when multiplied by the original matrix, gives the identity matrix, much like how multiplying a number by its reciprocal gives 1. For a 2x2 matrix \( \begin{bmatrix} a & b \ c & d \end{bmatrix} \), the inverse is given by \[ A^{-1} = \frac{1}{ad-bc} \begin{bmatrix} d & -b \ -c & a \end{bmatrix} \].In our exercise, the matrix \( A = \begin{bmatrix} 1 & 2 \ 3 & 2 \end{bmatrix} \) has a determinant of \( -4 \). Computing its inverse gives \( A^{-1} = \begin{bmatrix} -0.5 & 0.5 \ 0.75 & -0.25 \end{bmatrix} \). This step is crucial as applying the inverse matrix to the constant vector \( \mathbf{b} \) allows us to find the variables \( \mathbf{x} \).

Matrix Multiplication

Matrix multiplication is a fundamental operation that combines matrices in a specific way. When multiplying a matrix by a vector, each entry in the resulting vector is a dot product of the corresponding row of the matrix and the column vector.To solve \( A\mathbf{x} = \mathbf{b} \), we use the inverse matrix we calculated and multiply it by \( \mathbf{b} \). This results in:\[\begin{bmatrix} -0.5 & 0.5 \ 0.75 & -0.25 \end{bmatrix} \begin{bmatrix} 8 \ 6 \end{bmatrix} = \begin{bmatrix} -1 \ 4.5 \end{bmatrix}\]The first row calculation: \(-0.5 \times 8 + 0.5 \times 6 = -1\), and the second row: \(0.75 \times 8 - 0.25 \times 6 = 4.5\). These calculations give us the values for \( x \) and \( y \), thus solving the system.

Solution Verification

Solution verification ensures that the calculated solution satisfies the original system of equations. It is like checking your work for accuracy.Using the values \( x = -1 \) and \( y = 4.5 \) from our solution:

• Substitute into Equation 1: \( -1 + 2(4.5) = 8 \), and indeed it equals 8.
• Substitute into Equation 2: \( 3(-1) + 2(4.5) = 6 \), which simplifies correctly to 6.

Both equations are satisfied, thus verifying the solution is correct. This step is important as it confirms the reliability of our algebraic work using matrices.