Problem 30 - College Algebra with Corequisite Support

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 solutions are \(x = 0.2\) and \(y = 1.5\).

Step 1: Write the System in Matrix Form

First, represent the system of equations as a matrix equation. The system \(5x - 4y = -5\) and \(4x + y = 2.3\) can be written as: \[ A \mathbf{x} = \mathbf{b} \] where:\[ A = \begin{pmatrix} 5 & -4 \ 4 & 1 \end{pmatrix} \], \( \mathbf{x} = \begin{pmatrix} x \ y \end{pmatrix} \), and \( \mathbf{b} = \begin{pmatrix} -5 \ 2.3 \end{pmatrix} \).

Step 2: Find the Inverse of Matrix A

The inverse of a matrix \(A\) is given by \(A^{-1} = \frac{1}{ad-bc} \begin{pmatrix} d & -b \ -c & a \end{pmatrix} \). For matrix \(A = \begin{pmatrix} 5 & -4 \ 4 & 1 \end{pmatrix}\), compute the determinant: \(det(A) = (5)(1) - (-4)(4) = 5 + 16 = 21\). The inverse is then:\[ A^{-1} = \frac{1}{21} \begin{pmatrix} 1 & 4 \ -4 & 5 \end{pmatrix} \].

Step 3: Solve for x by Multiplying by the Inverse

To solve the equation \(A\mathbf{x} = \mathbf{b}\), multiply both sides by the inverse of \(A\):\[ \mathbf{x} = A^{-1} \mathbf{b} \]. Substituting the known values, we have:\[ \mathbf{x} = \frac{1}{21} \begin{pmatrix} 1 & 4 \ -4 & 5 \end{pmatrix} \begin{pmatrix} -5 \ 2.3 \end{pmatrix} \].

Step 4: Perform the Matrix Multiplication

Calculate the product \(A^{-1} \mathbf{b}\):\[ \frac{1}{21} \begin{pmatrix} 1 & 4 \ -4 & 5 \end{pmatrix} \begin{pmatrix} -5 \ 2.3 \end{pmatrix} = \frac{1}{21} \begin{pmatrix} 1(-5) + 4(2.3) \ -4(-5) + 5(2.3) \end{pmatrix} = \frac{1}{21} \begin{pmatrix} -5 + 9.2 \ 20 + 11.5 \end{pmatrix} = \frac{1}{21} \begin{pmatrix} 4.2 \ 31.5 \end{pmatrix} \].

Step 5: Simplify the Result to Find x and y

Divide each component of the resulting vector by 21 to find \(x\) and \(y\):\[ \mathbf{x} = \begin{pmatrix} \frac{4.2}{21} \ \frac{31.5}{21} \end{pmatrix} = \begin{pmatrix} 0.2 \ 1.5 \end{pmatrix} \]. Thus, \(x = 0.2\) and \(y = 1.5\).

Key Concepts

Matrix Inverse

In matrix algebra, the inverse of a matrix is like the reciprocal of a number. It is used to solve systems of equations, especially when represented as matrix equations. If we have a matrix equation of the form \( A\mathbf{x} = \mathbf{b} \), finding the inverse of matrix \( A \), denoted as \( A^{-1} \), allows us to solve for \( \mathbf{x} \) by rearranging the equation to \( \mathbf{x} = A^{-1}\mathbf{b} \).
For a \(2 \times 2\) matrix like \( A = \begin{pmatrix} a & b \ c & d \end{pmatrix} \), the inverse is calculated using the formula:

• \( A^{-1} = \frac{1}{ad-bc} \begin{pmatrix} d & -b \ -c & a \end{pmatrix} \)

This formula only works if the determinant \( ad-bc eq 0 \); otherwise, the matrix has no inverse. The inverse essentially "undoes" the multiplication by the original matrix when solving equations.

System of Equations

A system of equations is a collection of two or more equations with the same set of unknowns. In this scenario, we are dealing with two equations and two unknowns: \(5x - 4y = -5\) and \(4x + y = 2.3\). Systems of equations are commonly used in various fields to determine the values of variables that satisfy all given equations simultaneously.

The general method involves expressing these equations as a single matrix equation \( A\mathbf{x} = \mathbf{b} \), where \( A \) is a coefficient matrix, \( \mathbf{x} \) is a vector of the variables, and \( \mathbf{b} \) is the result vector. Solving this involves finding matrix \( A\)'s inverse and multiplying it by \( \mathbf{b} \) to find the values of the variables.

Determinants

The determinant is a special number that can be calculated from a square matrix. It plays a crucial role in matrix algebra, particularly in finding inverses. The determinant for a \(2 \times 2\) matrix \( A = \begin{pmatrix} a & b \ c & d \end{pmatrix} \) is calculated as \( ad-bc \).

• If the determinant is not zero, the matrix has an inverse, and thus a unique solution for the system of equations.
• If it is zero, the matrix is singular, which means it does not have an inverse, and the system may have infinitely many solutions or no solution at all.

In our problem, the determinant of matrix \( A \) is calculated as \( (5)(1) - (-4)(4) = 5 + 16 = 21 \), which is non-zero. Therefore, matrix \( A \) has an inverse, enabling us to solve the system of equations.

Matrix Multiplication

Matrix multiplication is an essential operation in linear algebra and is used extensively to solve matrix equations. When multiplying matrices, the number of columns in the first matrix must match the number of rows in the second matrix. The resulting matrix has the dimensions of the outer dimensions of the two multiplied matrices.

• For vectors, the multiplication involves taking each element of the first row of the first matrix and multiplying them by the corresponding element of the column vector.
• The results are then summed to yield the elements of the resulting vector.

In our solution, once we have the inverse matrix \( A^{-1} \) and the vector \( \mathbf{b} \), we perform the matrix multiplication \( A^{-1}\mathbf{b} \) to find the solution vector \( \mathbf{x} \). The calculation simplifies to \( \begin{pmatrix} 0.2 \ 1.5 \end{pmatrix} \), indicating \( x = 0.2 \) and \( y = 1.5 \), which are the values that satisfy the original system of equations.