Problem 29 - College Algebra

Question

For the following exercises, solve the system using the inverse of a \(2 \times 2\) matrix. $$\begin{array}{l}{3 x-2 y=6} \\ {-x+5 y=-2}\end{array}$$

Step-by-Step Solution

Answer

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

Step 1: Write the System as a Matrix Equation

The given system can be written as a matrix equation of the form \( AX = B \), where \( A \) is the matrix of coefficients, \( X \) is the column matrix of variables, and \( B \) is the column matrix of constants. Here:\[ A = \begin{bmatrix} 3 & -2 \ -1 & 5 \end{bmatrix}, \quad X = \begin{bmatrix} x \ y \end{bmatrix}, \quad B = \begin{bmatrix} 6 \ -2 \end{bmatrix} \]

Step 2: Find the Inverse of Matrix A

To solve the system using the inverse, first find the inverse of matrix \( A \). The inverse of a 2x2 matrix \( \begin{bmatrix} a & b \ c & d \end{bmatrix} \) is given by:\[A^{-1} = \frac{1}{ad-bc} \begin{bmatrix} d & -b \ -c & a \end{bmatrix}\]For matrix \( A \), we have:\( a = 3 \), \( b = -2 \), \( c = -1 \), \( d = 5 \). Thus, the determinant is:\( ad - bc = (3)(5) - (-1)(-2) = 15 - 2 = 13 \).So, the inverse is:\[A^{-1} = \frac{1}{13} \begin{bmatrix} 5 & 2 \ 1 & 3 \end{bmatrix}\]

Step 3: Multiply the Inverse of A by B

Now, solve for \( X \) using the inverse of \( A \) by multiplying it with \( B \):\[X = A^{-1}B\]Substituting the values:\[X = \frac{1}{13} \begin{bmatrix} 5 & 2 \ 1 & 3 \end{bmatrix} \begin{bmatrix} 6 \ -2 \end{bmatrix}\]Perform the matrix multiplication:\[= \frac{1}{13} \begin{bmatrix} (5)(6) + (2)(-2) \ (1)(6) + (3)(-2) \end{bmatrix}\]\[= \frac{1}{13} \begin{bmatrix} 30 - 4 \ 6 - 6 \end{bmatrix}\]\[= \frac{1}{13} \begin{bmatrix} 26 \ 0 \end{bmatrix}\]\[= \begin{bmatrix} 2 \ 0 \end{bmatrix}\]

Step 4: Interpret the Result

From the result of the matrix multiplication, we have:\( X = \begin{bmatrix} 2 \ 0 \end{bmatrix} \).This means \( x = 2 \) and \( y = 0 \) is the solution to the system of equations.

Key Concepts

Inverse Matrix

The concept of an inverse matrix is crucial in matrix algebra, especially when solving systems of equations. An inverse of a matrix, denoted as \( A^{-1} \), is essentially the matrix that 'undoes' what the original matrix \( A \) does. Anything you multiply by its inverse returns the original identity matrix, matrix-wise.For a 2x2 matrix \( A = \begin{bmatrix} a & b \ c & d \end{bmatrix} \), the inverse is given by:\[A^{-1} = \frac{1}{ad-bc} \begin{bmatrix} d & -b \ -c & a \end{bmatrix}\] Here are some key points to remember about inverse matrices:

• The determinant \( ad-bc \) should not be zero. If it is zero, the matrix does not have an inverse.
• Finding the inverse is crucial for certain algebraic manipulations, such as solving matrix equations.

In the context of solving a system of equations, finding the inverse allows us to effectively "divide" by a matrix—a concept that doesn't exist straightforwardly in matrix terms. By multiplying both sides of a matrix equation by the inverse, we can isolate the variable matrix.

2x2 Matrix

A 2x2 matrix is a fundamental building block in matrix algebra, consisting of two rows and two columns. Understanding how to manipulate and interpret these matrices is essential for solving simpler systems of linear equations. A general 2x2 matrix looks like this:\[ A = \begin{bmatrix} a & b \ c & d \end{bmatrix} \]Here are some important features and operations you can perform with 2x2 matrices:

• Determinant: This is a scalar value calculated as \( ad - bc \). It dictates many properties of the matrix, such as whether an inverse exists.
• Matrix Multiplication: Multiplying matrices involves combining rows of the first matrix with columns of the second, resulting in another matrix.
• Transpose: Swapping rows and columns provides the transpose, which can be useful in certain algebraic operations.

The 2x2 format is widespread due to its simplicity, yet it provides a solid foundation for moving onto more complex matrices and learning how matrices behave.

System of Equations

A system of equations involves multiple equations that share variables. Solving these systems is about finding the values of these shared variables that satisfy all the equations at once. Matrix algebra provides a powerful way to address systems of equations efficiently.There are several ways to solve systems of equations:

• Substitution and Elimination: Traditional methods that focus on algebraically solving one equation and substituting it into another.
• Matrix Methods: Using matrix algebra, you can express a system of equations in the form \( AX = B \), where \( A \) is a matrix of coefficients, \( X \) is a column of the variables, and \( B \) is a column of constants.

The solution is obtained by manipulating these matrices, often through finding the inverse of \( A \) and solving for \( X \) as \( X = A^{-1}B \). This method is particularly powerful for larger systems, although we've used it here for a simple 2x2 example.Understanding how systems of equations work with matrices not only deepens our understanding of algebra but also equips us with methods applicable in higher dimensions and more complex scenarios.