Problem 28 - Intermediate Algebra

Question

Use matrices to solve each system of equations. $$ \left\\{\begin{array}{l} 5 x-4 y=10 \\ x-7 y=2 \end{array}\right. $$

Step-by-Step Solution

Answer

The solution is \(x = 2\) and \(y = 0\).

Step 1: Write the System in Matrix Form

First, express the given system of equations as a matrix equation of the form \( A\mathbf{x} = \mathbf{b} \). This involves separating the coefficients, variables, and constants.The system given:\[\begin{align*}&5x - 4y = 10 \&x - 7y = 2\end{align*}\]In matrix form, it can be written as:\[\begin{bmatrix} 5 & -4 \ 1 & -7 \end{bmatrix} \begin{bmatrix} x \ y \end{bmatrix} = \begin{bmatrix} 10 \ 2 \end{bmatrix}\]Here, \( A = \begin{bmatrix} 5 & -4 \ 1 & -7 \end{bmatrix} \), \( \mathbf{x} = \begin{bmatrix} x \ y \end{bmatrix} \), and \( \mathbf{b} = \begin{bmatrix} 10 \ 2 \end{bmatrix} \).

Step 2: Calculate the Inverse of Matrix A

To solve for \( \mathbf{x} \), you need the inverse of matrix \( A \). The inverse of a 2x2 matrix \( \begin{bmatrix} a & b \ c & d \end{bmatrix} \) is given by:\[\frac{1}{ad-bc} \begin{bmatrix} d & -b \ -c & a \end{bmatrix}\]For matrix \( A = \begin{bmatrix} 5 & -4 \ 1 & -7 \end{bmatrix} \), first calculate the determinant:\[ad - bc = (5)(-7) - (-4)(1) = -35 + 4 = -31\]The inverse is calculated as:\[A^{-1} = \begin{bmatrix} -7 & 4 \ -1 & 5 \end{bmatrix} \cdot \frac{1}{-31} = \begin{bmatrix} \frac{7}{31} & -\frac{4}{31} \ \frac{1}{31} & -\frac{5}{31} \end{bmatrix}\]

Step 3: Multiply the Inverse by Vector b

Now, solve for the vector \( \mathbf{x} \) by multiplying \( A^{-1} \) with \( \mathbf{b} \).\[\begin{bmatrix} x \ y \end{bmatrix} = A^{-1} \begin{bmatrix} 10 \ 2 \end{bmatrix} = \begin{bmatrix} \frac{7}{31} & -\frac{4}{31} \ \frac{1}{31} & -\frac{5}{31} \end{bmatrix} \begin{bmatrix} 10 \ 2 \end{bmatrix}\]Compute the product:\[\begin{align*}x &= \frac{7 \times 10}{31} + \frac{-4 \times 2}{31} = \frac{70 - 8}{31} = \frac{62}{31} = 2 \y &= \frac{1 \times 10}{31} + \frac{-5 \times 2}{31} = \frac{10 - 10}{31} = 0\end{align*}\]Thus, the solution to the system is \( x = 2 \), \( y = 0 \).

Step 4: Verify the Solution

Lastly, substitute \( x = 2 \) and \( y = 0 \) back into the original equations to verify the solution.1. Substitute into the first equation:\[5(2) - 4(0) = 10 \, \text{True}\]2. Substitute into the second equation:\[2 - 7(0) = 2 \, \text{True}\]Since both equations hold true, \( x = 2 \) and \( y = 0 \) is the correct solution.

Key Concepts

Systems of Equations

To start with, it's important to know what a system of equations is. It's simply a set of two or more equations that share the same variables. The challenge is to find the values for these variables that satisfy all the equations simultaneously.

For example, consider the system: \[ \begin{align*} 5x - 4y & = 10 \ x - 7y & = 2 \end{align*} \] Here, we need to find values for \(x\) and \(y\) that make both equations true at the same time. To solve such systems, there are several methods like substitution, elimination, and matrix algebra. Using matrices is a powerful approach, especially for larger systems. This method involves expressing the system in a matrix form \( A\mathbf{x} = \mathbf{b} \), where \(A\) is the coefficient matrix, \(\mathbf{x}\) is the variable matrix, and \(\mathbf{b}\) is the constants matrix.

Matrix Inversion

Matrix inversion is a key operation in matrix algebra that allows us to solve systems of equations efficiently. When the matrix \(A\) is invertible, we can find its inverse and use it to solve the equation \( A\mathbf{x} = \mathbf{b} \) by calculating \( \mathbf{x} = A^{-1}\mathbf{b} \).

For a 2x2 matrix \( \begin{bmatrix} a & b \ c & d \end{bmatrix} \), the inverse is given by:

• First, compute the determinant: \( ad - bc \).
• If the determinant is non-zero, the inverse exists and is: \( \frac{1}{ad-bc} \begin{bmatrix} d & -b \ -c & a \end{bmatrix} \).

In our example, the determinant of \( A = \begin{bmatrix} 5 & -4 \ 1 & -7 \end{bmatrix} \) is \(-31\), so the inverse is\( \begin{bmatrix} \frac{7}{31} & -\frac{4}{31} \ \frac{1}{31} & -\frac{5}{31} \end{bmatrix} \). This inverse allows us to solve for \( \mathbf{x} \).

Determinants

Determinants have a crucial role in matrix algebra. Specifically, they determine if a matrix is invertible. For a 2x2 matrix \( \begin{bmatrix} a & b \ c & d \end{bmatrix} \), the determinant is calculated as \( ad - bc \).

If the determinant is not zero, the matrix is invertible, meaning we can find its inverse. This is important because using the inverse matrix, we can solve the equation \( A\mathbf{x} = \mathbf{b} \). If the determinant is zero, the matrix is singular, and you cannot find its inverse. In our example problem, the determinant \(-31\) is not zero, indicating that the inverse of matrix \(A\) can be found and used to solve the system of equations.

Verification of Solutions

Once we have a solution, it's important to verify it to ensure accuracy. Verification means plugging the obtained values of the variables back into the original equations to check if they hold true.

For the system \[ \begin{align*} 5x - 4y & = 10 \ x - 7y & = 2 \end{align*} \], we found \( x = 2 \) and \( y = 0 \). Let's verify:

• Substitute \( x = 2 \) and \( y = 0 \) into the first equation: \( 5(2) - 4(0) = 10 \), which is true.
• Substitute into the second equation: \( 2 - 7(0) = 2 \), which is also true.

Both equations check out, confirming that our solution is indeed correct. This step is essential as it verifies the solution's validity and ensures we didn't make any mistakes during calculations.