Problem 53 - Precalculus with Limits

Question

In Exercises 51-58, use an inverse matrix to solve (if possible) the system of linear equations. \(\begin{cases} -0.4x + 0.8y = 1.6 \\ 2x - 4y = 5 \end{cases}\)

Step-by-Step Solution

Answer

This system of equations cannot be solved using an inverse matrix since the determinant of the corresponding matrix is 0, which means the matrix does not have an inverse.

Step 1: Set Up The Matrices

First, the system of equations could be converted into equation AX = B. Where: A = \(\begin{bmatrix} -0.4 & 0.8 \ 2 & -4 \end{bmatrix}\)B = \(\begin{bmatrix} 1.6 \ 5 \end{bmatrix}\)and X = \(\begin{bmatrix} x \ y \end{bmatrix}\)

Step 2: Find the Inverse of Matrix A

The inverse of a 2x2 matrix A = \(\begin{bmatrix} a & b \ c & d \end{bmatrix}\)could be found out using the formula A^-1 = \( \frac{1}{{ad - bc}}\begin{bmatrix} d & -b \ -c & a \end{bmatrix}\).For the given 2x2 matrix A = \(\begin{bmatrix} -0.4 & 0.8 \ 2 & -4 \end{bmatrix}\)We first compute the determinant, ad - bc = (-0.4*-4) - (0.8*2) = 0. When the determinant is zero, it implies that the matrix doesn't have an inverse, thus this system of equations cannot be solved using an inverse matrix.

Step 3: Conclusion

If it was possible to find an inverse for the matrix A, the next step would have been to multiply that inverse with the matrix B to find the values of x and y. However, in this case, the determinant is 0, meaning matrix A does not have an inverse and the system of equations cannot be solved by this method.

Key Concepts

Linear Equations

Linear equations form the bedrock of algebra. They express relationships between variables through constant coefficients. In the given problem, we are dealing with a system of linear equations:

• \(-0.4x + 0.8y = 1.6\)
• \(2x - 4y = 5\)

These equations represent lines on a graphical plane, where each solution \(x, y\) corresponds to a point where these lines intersect. This system can be visualized as searching for coordinates that satisfy both lines simultaneously.

Such systems can be solved using various methods like substitution, elimination, or, as intended here, through matrices. Specifically, the aim is to use inverse matrices—a method that requires converting the system into a single matrix equation.

Determinant

The determinant is a scalar value that can be computed from the elements of a square matrix. It plays a crucial role in matrix mathematics—including in calculating matrix inverses. For a 2x2 matrix \(A = \begin{bmatrix} a & b \ c & d \end{bmatrix}\), the determinant \(det(A)\) can be found as:

• \(det(A) = ad - bc\)

In our original exercise, the matrix \(A = \begin{bmatrix} -0.4 & 0.8 \ 2 & -4 \end{bmatrix}\) has a determinant calculated as:

• \(det(A) = (-0.4)(-4) - (0.8)(2) = 0\)

A zero determinant indicates a singular matrix, which means the matrix does not have an inverse. Without an inverse, it's impossible to solve the linear system using matrix inversion.

This scenario typically suggests that the system of equations may be dependent (infinitely many solutions) or inconsistent (no solutions). Understanding the determinant helps in quickly assessing such properties.

Matrix Operations

Matrix operations are essential for manipulating linear equations in a compact form. In our exercise, converting the system of linear equations into a matrix equation is the initial step. The matrix equation is expressed as \(AX = B\):

• \(A = \begin{bmatrix} -0.4 & 0.8 \ 2 & -4 \end{bmatrix}\) – coefficients matrix
• \(X = \begin{bmatrix} x \ y \end{bmatrix}\) – variables matrix
• \(B = \begin{bmatrix} 1.6 \ 5 \end{bmatrix}\) – constants matrix

An inverse matrix \(A^{-1}\) would allow us to solve for \(X\) by multiplying \(A^{-1}\) with \(B\), provided \(A^{-1}\) exists.

Since our calculation shows the matrix \(A\) has no inverse due to a zero determinant, this operation cannot proceed. This highlights the importance of matrix operations like finding inverses and determinants in solving linear systems. Mastery of these concepts enables rigorous tackling of more complex algebraic problems.