Problem 48 - College Algebra with Modeling and Visualization

Question

Complete the following. (A) Write the system in the form \(A X=B\). (B) Solve the system by finding \(A^{-1}\) and then using the equation \(\boldsymbol{X}=\boldsymbol{A}^{-1} \boldsymbol{B}\). (Hint: Some of your answers from Exercises \(15-28\) may be helpful.) $$ \begin{array}{l} 2 x+y=4 \\ -x+2 y=-1 \end{array} $$

Step-by-Step Solution

Answer

The solution is \(x = \frac{9}{5}\) and \(y = \frac{2}{5}\).

Step 1: Write the system in matrix form

The given system of equations is: \[\begin{align*} 2x + y &= 4 \ -x + 2y &= -1 \end{align*}\] Convert this into the matrix form \(AX = B\). Identify the coefficient matrix \(A\), the variable matrix \(X\), and the constant matrix \(B\).\[A = \begin{bmatrix} 2 & 1 \ -1 & 2 \end{bmatrix}, \quad X = \begin{bmatrix} x \ y \end{bmatrix}, \quad B = \begin{bmatrix} 4 \ -1 \end{bmatrix}\]

Step 2: Find the inverse of matrix A

To solve for \(X\) using \(A^{-1}\), we first need to find the inverse of matrix \(A\). The inverse of a 2x2 matrix \(A = \begin{bmatrix} a & b \ c & d \end{bmatrix}\) is given by \[A^{-1} = \frac{1}{ad-bc}\begin{bmatrix} d & -b \ -c & a \end{bmatrix}\]. For \(A = \begin{bmatrix} 2 & 1 \ -1 & 2 \end{bmatrix}\), calculate the determinant: \[ad-bc = (2)(2) - (-1)(1) = 4 + 1 = 5\] Thus, \[A^{-1} = \frac{1}{5}\begin{bmatrix} 2 & -1 \ 1 & 2 \end{bmatrix}\].

Step 3: Use the inverse to solve for X

With \(A^{-1}\) computed, solve \(X = A^{-1}B\):\[\begin{aligned} X = \frac{1}{5}\begin{bmatrix} 2 & -1 \ 1 & 2 \end{bmatrix} \begin{bmatrix} 4 \ -1 \end{bmatrix} &= \frac{1}{5} \begin{bmatrix} (2)(4) + (-1)(-1) \ (1)(4) + (2)(-1) \end{bmatrix} \ &= \frac{1}{5} \begin{bmatrix} 8 + 1 \ 4 - 2 \end{bmatrix} = \frac{1}{5} \begin{bmatrix} 9 \ 2 \end{bmatrix} = \begin{bmatrix} \frac{9}{5} \ \frac{2}{5} \end{bmatrix}. \end{aligned}\]

Step 4: Interpret the solution

The solution to the system of equations provides the values of \(x\) and \(y\). Thus, the values are \(x = \frac{9}{5}\) and \(y = \frac{2}{5}\).

Key Concepts

System of Equations

A system of equations is a set of two or more equations with the same variables. The goal is to find values for the variables that satisfy all the equations simultaneously. In this exercise, we're dealing with two equations: \(2x + y = 4\) and \(-x + 2y = -1\). Here, the variables are \(x\) and \(y\).

This particular system can be solved using different methods, such as substitution, elimination, or matrix algebra. Matrix algebra is convenient when dealing with multiple equations and unknowns. It allows us to systematically manipulate the equations using matrices. The matrix form of a system is expressed as \(AX = B\), where \(A\) is the coefficient matrix, \(X\) is the variable matrix, and \(B\) is the constant matrix. By organizing the system in this way, it becomes easier to apply matrix operations to find the solution.

Matrix Inverse

The inverse of a matrix is analogous to the reciprocal of a number. It is used to solve matrix equations, much like how we use division to solve arithmetic equations. Not every matrix has an inverse, but those that do are called invertible or non-singular.

To find the inverse of a 2x2 matrix \(A\), you can use the formula: \[A^{-1} = \frac{1}{ad-bc}\begin{bmatrix} d & -b \ -c & a \end{bmatrix}\] Here, \(ad-bc\) is the determinant. If this value is zero, the matrix does not have an inverse. In the exercise, the matrix \(A\) is \(\begin{bmatrix} 2 & 1 \ -1 & 2 \end{bmatrix}\), and the determinant is 5, which means \(A\) is invertible. After calculating, we found the inverse to be \(\frac{1}{5}\begin{bmatrix} 2 & -1 \ 1 & 2 \end{bmatrix}\). Applying the inverse can then help us solve the system of equations by isolating the variable matrix \(X\).

Determinants

Determinants play a crucial role in matrix algebra, especially for finding inverse matrices. A determinant is a special number calculated from a square matrix. For a 2x2 matrix, the determinant is calculated as \(ad-bc\).

In this case, the matrix \(A = \begin{bmatrix} 2 & 1 \ -1 & 2 \end{bmatrix}\) has a determinant of 5. The value of the determinant tells us if a matrix has an inverse or not. If the determinant is zero, the matrix is singular and does not have an inverse. Otherwise, the matrix is invertible, and we can proceed with the inverse calculation.

The determinant also provides a sense of the linear independence of rows or columns in the matrix. If the determinant is not zero, the matrix's rows or columns are linearly independent, meaning no row or column can be written as a combination of others.

Matrix Multiplication

Matrix multiplication is a fundamental operation in matrix algebra. It involves multiplying corresponding elements and summing the products. For two matrices to be multiplied, the number of columns in the first matrix must equal the number of rows in the second matrix.

In solving systems of equations, matrix multiplication is used to combine the inverse matrix \(A^{-1}\) with the constant matrix \(B\) to find the variable matrix \(X\). This is expressed as \(X = A^{-1}B\).

In our exercise, we found that\[\frac{1}{5}\begin{bmatrix} 2 & -1 \ 1 & 2 \end{bmatrix} \begin{bmatrix} 4 \ -1 \end{bmatrix} = \begin{bmatrix} \frac{9}{5} \ \frac{2}{5} \end{bmatrix}\]This operation reveals the values of \(x\) and \(y\) that solve the original system of equations. Understanding matrix multiplication is essential for using matrices to solve complex algebraic problems efficiently.