Problem 59 - College Algebra and Calculus: An Applied Approach

Question

Use an inverse matrix to solve (if possible) the system of linear equations. $$ \left\\{\begin{aligned} -0.4 x+0.8 y &=1.6 \\ 2 x-4 y &=5 \end{aligned}\right. $$

Step-by-Step Solution

Answer

Since the determinant of the matrix is zero, an inverse matrix does not exist and it is not possible to solve this system of equations using an inverse matrix.

Step 1 - Write the system of equations as a matrix

We can express this system of equations as a matrix multiplication. The system of equations \(-0.4x+0.8y=1.6\) and \(2x-4y=5\) can be written as \(\begin{bmatrix}-0.4 & 0.8\2 & -4\end{bmatrix} \begin{bmatrix}x\y\end{bmatrix} = \begin{bmatrix}1.6\5\end{bmatrix}\)

Step 2 - Calculate the determinant of the matrix

We calculate the determinant (denoted as |A|) using the formula |A| = ad - bc for a matrix A = \(\begin{bmatrix}a & b\c & d\end{bmatrix}\). This gives us |A| = (-0.4)*(-4) - (0.8)*2 = 1.6 - 1.6 = 0.

Step 3 - Check if an inverse matrix exists

An inverse matrix only exists if the determinant is not equal to zero. In this case, the determinant is zero, so there is no inverse. Therefore, it is not possible to solve this system of equations using an inverse matrix.

Key Concepts

Inverse Matrix

An inverse matrix is comparable to the reciprocal value of a number. It's a matrix that, when multiplied by the original matrix, results in the identity matrix, much like multiplying a number by its reciprocal yields 1. In the context of solving systems of linear equations, the inverse matrix is an invaluable tool. It essentially allows us to 'divide' by a matrix to isolate our variable of interest. However, not all matrices have an inverse.

For a matrix to have an inverse, it must be square (same number of rows and columns) and its determinant must be non-zero. Recall that the determinant gives us information about the properties of a matrix, including whether it can be inverted. When the determinant is zero, the matrix is referred to as 'singular', meaning it has no inverse. Thus, your textbook exercise demonstrated that the system of equations provided does not have a unique solution because the corresponding matrix does not have an inverse, based on its zero determinant.

Determinant Calculation

The calculation of a determinant is a crucial step in analyzing a matrix's characteristics. Generally, the determinant can tell us if a square matrix is invertible, which is key for solving systems of equations by matrix methods. For a 2x2 matrix, which is the size we are concerned with in your exercise, the determinant is found using the simple formula \( |A| = ad - bc \) where the matrix \( A \) is represented as \( \begin{bmatrix}a & b\c & d\end{bmatrix} \).

This action of finding the determinant essentially gives us the 'scaling factor' of the linear transformation that the matrix represents. When the determinant is zero, it indicates that the transformation squashes the two-dimensional space into a lower dimension, which explains why the inverse doesn't exist - there is no way back to the original two-dimensional space from the lower dimension.

Matrix Multiplication

Matrix multiplication is not as straightforward as multiplying individual elements of the matrices. Instead, it involves a process where rows of one matrix are multiplied with columns of another and sums of these products constitute the elements of the product matrix.

When we speak of solving a system of linear equations with matrices, we often refer to this multiplication process. For example, in our system represented by \( Ax = b \), the matrix \( A \) contains the coefficients of the variables, and \( x \) is a column matrix of the variables. The product of \( A \) and \( x \) should result in \( b \) - the constants from the right side of the equations. However, as seen in the exercise, this method requires an invertible matrix \( A \) which is not the case when the determinant of \( A \) is zero. As such, matrix multiplication offers a powerful way to simplify and solve linear systems provided the matrices involved meet the necessary criteria.