Problem 30 - College Algebra

Question

For the following exercises, solve the system using the inverse of a \(2 \times 2\) matrix. $$\begin{array}{l}{5 x-4 y=-5} \\ {4 x+y=2.3}\end{array}$$

Step-by-Step Solution

Answer

The solution is \(x = 0.2\) and \(y = 1.5\).

Step 1: Write the system in matrix form

First, express the given system of equations in matrix form. The system\[\begin{align*}5x - 4y &= -5 \4x + y &= 2.3\end{align*}\]can be written as\[A \begin{pmatrix} x \ y \end{pmatrix} = B\]where\[A = \begin{pmatrix} 5 & -4 \ 4 & 1 \end{pmatrix}, \quad B = \begin{pmatrix} -5 \ 2.3 \end{pmatrix}\]

Step 2: Find the inverse of matrix A

To solve the system using the matrix inverse method, we first need to find the inverse of matrix A. The inverse of a 2x2 matrix \[A = \begin{pmatrix} a & b \ c & d \end{pmatrix}\]is given by\[A^{-1} = \frac{1}{ad - bc} \begin{pmatrix} d & -b \ -c & a \end{pmatrix}\]For our matrix A,\[a = 5, \ b = -4, \ c = 4, \ d = 1\]Calculate the determinant: \[det(A) = (5)(1) - (-4)(4) = 5 + 16 = 21\]Then, the inverse is:\[A^{-1} = \frac{1}{21} \begin{pmatrix} 1 & 4 \ -4 & 5 \end{pmatrix}\]

Step 3: Multiply inverse of A by B

Now that we have the inverse of matrix A, we can find the solution vector by computing \[\begin{pmatrix} x \ y \end{pmatrix} = A^{-1}B\]Substitute in the values:\[A^{-1} = \frac{1}{21} \begin{pmatrix} 1 & 4 \ -4 & 5 \end{pmatrix}, \quad B = \begin{pmatrix} -5 \ 2.3 \end{pmatrix}\]\[\begin{pmatrix} x \ y \end{pmatrix} = \frac{1}{21} \begin{pmatrix} 1 & 4 \ -4 & 5 \end{pmatrix} \begin{pmatrix} -5 \ 2.3 \end{pmatrix}\]Perform the matrix multiplication:\[= \frac{1}{21} \begin{pmatrix} (1)(-5) + (4)(2.3) \ (-4)(-5) + (5)(2.3) \end{pmatrix}= \frac{1}{21} \begin{pmatrix} -5 + 9.2 \ 20 + 11.5 \end{pmatrix}= \frac{1}{21} \begin{pmatrix} 4.2 \ 31.5 \end{pmatrix}\]

Step 4: Simplify the solution

Simplify the resulting vector to find the values of \(x\) and \(y\):\[\begin{pmatrix} x \ y \end{pmatrix} = \begin{pmatrix} \frac{4.2}{21} \ \frac{31.5}{21} \end{pmatrix}= \begin{pmatrix} 0.2 \ 1.5 \end{pmatrix}\]Therefore, the solution is \(x = 0.2\) and \(y = 1.5\).

Key Concepts

2x2 Matrix

A 2x2 matrix is a simple, square matrix consisting of two rows and two columns, which makes it perfect for solving systems of two linear equations. In a mathematical sense, when we write a matrix, we show it as an array of numbers. If you have two equations, for instance:

• Equation 1: \(5x - 4y = -5\)
• Equation 2: \(4x + y = 2.3\)

You can represent these equations with a matrix containing coefficients of the variables \(x\) and \(y\). This is the coefficient matrix, often noted as \(A\). In this case:\[ A = \begin{pmatrix} 5 & -4 \ 4 & 1 \end{pmatrix} \]Learning how to express equations in terms of matrices standardizes the problem into a form that is more easily manipulated mathematically, particularly for finding solutions using the inverse method.

Matrix Multiplication

Matrix multiplication is a core concept when using the inverse method to solve equations. It involves the multiplication of rows by columns; however, it differs from arithmetic multiplication. When multiplying two matrices, such as a matrix\( \vec{A}\) and a vector \( \vec{B} \), you multiply the rows of the first matrix by the columns of the second matrix, then sum the results. For solving the system of equations using the inverse matrix \(A^{-1}\) and vector \(B\), represented as:\[ A^{-1}B = \begin{pmatrix} x \ y \end{pmatrix} \]You first find the product of the first row of \(A^{-1}\) with the column vector \(B\):

• Row 1: \((1)(-5) + (4)(2.3) = -5 + 9.2 = 4.2\)
• Row 2: \((-4)(-5) + (5)(2.3) = 20 + 11.5 = 31.5\)

This multiplication gives the vector \(\begin{pmatrix} 4.2 \ 31.5 \end{pmatrix}\). It's a systematic method for finding unknown solutions in linear algebra problems.

Determinant of a Matrix

The determinant of a matrix is a special scalar value that can provide useful information about the matrix properties, and is crucial in finding the inverse of a matrix. For a 2x2 matrix \(A = \begin{pmatrix} a & b \ c & d \end{pmatrix}\), the determinant is calculated as \(det(A) = ad - bc\). This value should not be zero for the matrix to have an inverse.In our example, we computed the determinant by using:\[ det(A) = (5)(1) - (-4)(4) = 5 + 16 = 21 \]Since the determinant is not zero, we can compute the inverse of the matrix.If the determinant were zero, the system would not have a unique solution and the inverse matrix method wouldn't be applicable. The determinant helps check if the matrix is invertible, allowing us to proceed with solving the linear equations.

Linear Equations

Linear equations are mathematical expressions that represent straight lines when plotted on a graph. Each equation depicts a relationship between variables without any exponents or powers, keeping things straightforward. In a system of linear equations, such as:

• \(5x - 4y = -5\)
• \(4x + y = 2.3\)

Our goal is to find the values of \(x\) and \(y\) that simultaneously satisfy both equations. This type of system can be effectively solved using the inverse matrix method.By expressing the system as a matrix equation \(AX = B\) where \(A\) is the coefficient matrix and \(B\) is the constant matrix, we transform the problem-solving approach. Using inverse matrices is a powerful technique because it provides precise solutions to the equations, allowing us to determine specific values for variables such as \(x = 0.2\) and \(y = 1.5\). This approach highlights the beauty and efficiency of linear algebra in solving real-world challenges.