Problem 24 - Intermediate Algebra

Question

Describe how to use matrices to solve the system $$ \left(\begin{array}{r} x-2 y=5 \\ 2 x+7 y=9 \end{array}\right) $$

Step-by-Step Solution

Answer

\(x = \frac{53}{11}\), \(y = -\frac{1}{11}\)

Step 1: Write the system of equations as a matrix equation

Convert the given system of linear equations into a matrix equation of the form \[ A \vec{x} = \vec{b} \]where \[ A = \begin{pmatrix} 1 & -2 \ 2 & 7 \end{pmatrix}, \quad \vec{x} = \begin{pmatrix} x \ y \end{pmatrix}, \quad \text{and} \quad \vec{b} = \begin{pmatrix} 5 \ 9 \end{pmatrix}. \] The matrix equation becomes:\[\begin{pmatrix} 1 & -2 \ 2 & 7 \end{pmatrix} \begin{pmatrix} x \ y \end{pmatrix} = \begin{pmatrix} 5 \ 9 \end{pmatrix}\]

Step 2: Determine the inverse of matrix A

To solve for \(\vec{x}\), we need the inverse of matrix \(A\). The inverse \(A^{-1}\) 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 \(A = \begin{pmatrix} 1 & -2 \ 2 & 7 \end{pmatrix}\), calculate the determinant:\[ \text{det}(A) = 1 \cdot 7 - (-2) \cdot 2 = 11.\]Then, the inverse is:\[A^{-1} = \frac{1}{11}\begin{pmatrix} 7 & 2 \ -2 & 1 \end{pmatrix}.\]

Step 3: Multiply the inverse of A with vector b

To find \(\vec{x}\), perform the matrix multiplication \(A^{-1} \vec{b}\):\[ \vec{x} = A^{-1} \vec{b} = \frac{1}{11} \begin{pmatrix} 7 & 2 \ -2 & 1 \end{pmatrix} \begin{pmatrix} 5 \ 9 \end{pmatrix}.\]Compute the multiplication:\[\vec{x} = \frac{1}{11} \begin{pmatrix} 7 \times 5 + 2 \times 9 \ -2 \times 5 + 1 \times 9 \end{pmatrix} = \frac{1}{11} \begin{pmatrix} 35 + 18 \ -10 + 9 \end{pmatrix} = \frac{1}{11} \begin{pmatrix} 53 \ -1 \end{pmatrix}.\]Thus, \(\vec{x} = \begin{pmatrix} \frac{53}{11} \ -\frac{1}{11} \end{pmatrix}.\)

Step 4: Interpret the result

From the solution vector \(\vec{x} = \begin{pmatrix} \frac{53}{11} \ -\frac{1}{11} \end{pmatrix}\), interpret the individual values for \(x\) and \(y\). Therefore, we have:\[ x = \frac{53}{11} \approx 4.818 \quad \text{and} \quad y = -\frac{1}{11} \approx -0.091. \]

Key Concepts

Matrix Equation

To solve a system of linear equations using matrices, the first step is to convert the equations into a matrix form known as a matrix equation. This matrix equation is expressed as \( A \vec{x} = \vec{b} \). Here:

• \( A \) is the coefficient matrix. It includes the coefficients of the variables from the system of equations.
• \( \vec{x} \) is the variable vector. It consists of the variables you're solving for, such as \( x \) and \( y \) in our example.
• \( \vec{b} \) is the constant vector, which holds the constants on the right side of the equations.

For instance, given the equations \( x - 2y = 5 \) and \( 2x + 7y = 9 \), we create matrix \( A = \begin{pmatrix} 1 & -2 \ 2 & 7 \end{pmatrix} \), vector \( \vec{x} = \begin{pmatrix} x \ y \end{pmatrix} \), and vector \( \vec{b} = \begin{pmatrix} 5 \ 9 \end{pmatrix} \). The matrix equation is then \( \begin{pmatrix} 1 & -2 \ 2 & 7 \end{pmatrix} \begin{pmatrix} x \ y \end{pmatrix} = \begin{pmatrix} 5 \ 9 \end{pmatrix} \). This form simplifies the process of solving the system by using matrix operations.

Inverse Matrix

Once the matrix equation is formulated, the next task is determining the inverse of the coefficient matrix \( A \). The inverse of a matrix \( A \) is denoted as \( A^{-1} \) and is used to find the solution vector \( \vec{x} \) by multiplying it with \( \vec{b} \). For a 2x2 matrix \( \begin{pmatrix} a & b \ c & d \end{pmatrix} \), the formula for the inverse is:\[A^{-1} = \frac{1}{ad - bc}\begin{pmatrix} d & -b \ -c & a \end{pmatrix}\]However, the inverse only exists if the determinant \( ad - bc eq 0 \). In our example with \( A = \begin{pmatrix} 1 & -2 \ 2 & 7 \end{pmatrix} \), we compute the determinant as \( 11 \). Thus:\[A^{-1} = \frac{1}{11}\begin{pmatrix} 7 & 2 \ -2 & 1 \end{pmatrix}\]The next step involves utilizing this inverse matrix to solve the system of equations.

Determinant Calculation

The calculation of the determinant is crucial when working with matrix inverses. It helps determine whether a matrix is invertible. For a 2x2 matrix \( \begin{pmatrix} a & b \ c & d \end{pmatrix} \), the determinant is calculated as \( ad - bc \). If the determinant equals zero, the matrix lacks an inverse, and alternative methods would be necessary to solve the system.For the exercise example, the matrix is \( \begin{pmatrix} 1 & -2 \ 2 & 7 \end{pmatrix} \). Calculating the determinant involves:

• Multiplying the diagonal elements: \( 1 \times 7 = 7 \)
• Subtracting the product of the off-diagonal elements: \( -2 \times 2 = -4 \)
• Adding them together yields a determinant of \( 11 \).

Since 11 is not zero, \( A \) is invertible. With this determinant, the inverse can be effectively computed, leading to the matrix solution \( \vec{x} = A^{-1}\vec{b} \).