Problem 44 - Advanced Engineering Mathematics

Question

Use an inverse matrix to solve the given system of equations. $$ \begin{gathered} x_{1}-x_{2}=2 \\ 2 x_{1}+4 x_{2}=-5 \end{gathered} $$

Step-by-Step Solution

Answer

The solution is \( x_1 = \frac{1}{2} \) and \( x_2 = -\frac{4}{3} \).

Step 1: Write the system in matrix form

First, express the system of equations as a matrix equation in the form \( AX = B \). For the given system: \[A = \begin{bmatrix} 1 & -1 \ 2 & 4 \end{bmatrix}, \quad X = \begin{bmatrix} x_1 \ x_2 \end{bmatrix}, \quad B = \begin{bmatrix} 2 \ -5 \end{bmatrix}\] So, the matrix equation can be written as: \[ AX = B \]

Step 2: Find the inverse of matrix A

To solve for \( X \), we need the inverse of \( A \), denoted \( A^{-1} \). For the matrix \( A \), the inverse is found using the formula:\[ A^{-1} = \frac{1}{\text{det}(A)} \text{adj}(A) \]First, calculate the determinant of \( A \):\[ \text{det}(A) = (1)(4) - (-1)(2) = 4 + 2 = 6 \]Next, find the adjugate of \( A \):\[ \text{adj}(A) = \begin{bmatrix} 4 & 1 \ -2 & 1 \end{bmatrix} \]Thus, the inverse of \( A \) is:\[ A^{-1} = \frac{1}{6} \begin{bmatrix} 4 & 1 \ -2 & 1 \end{bmatrix} = \begin{bmatrix} \frac{2}{3} & \frac{1}{6} \ -\frac{1}{3} & \frac{1}{6} \end{bmatrix} \]

Step 3: Compute X by multiplying A^{-1} and B

Multiply the inverse matrix \( A^{-1} \) by matrix \( B \) to find matrix \( X \):\[X = A^{-1}B = \begin{bmatrix} \frac{2}{3} & \frac{1}{6} \ -\frac{1}{3} & \frac{1}{6} \end{bmatrix} \begin{bmatrix} 2 \ -5 \end{bmatrix}\]Calculate the multiplication:\[\begin{array}{ll}x_1 = \frac{2}{3} \times 2 + \frac{1}{6} \times (-5) = \frac{4}{3} - \frac{5}{6} = \frac{3}{6} = \frac{1}{2} \x_2 = -\frac{1}{3} \times 2 + \frac{1}{6} \times (-5) = -\frac{2}{3} - \frac{5}{6} = -\frac{4}{3}\end{array}\]

Step 4: Interpret the solution

The resulting matrix \( X \) provides the solution to the system of equations, which is:\[x_1 = \frac{1}{2}, \quad x_2 = -\frac{4}{3}\]This means that the values \( x_1 = \frac{1}{2} \) and \( x_2 = -\frac{4}{3} \) satisfy both equations given in the problem.

Key Concepts

System of Equations

A system of equations is a collection of two or more equations that share the same set of unknowns. These equations are solved simultaneously to find the values of the unknowns that satisfy all the equations in the system. In the given exercise, we have a simple system of two linear equations involving variables \(x_1\) and \(x_2\):\[ \begin{aligned} x_1 - x_2 &= 2, \ 2x_1 + 4x_2 &= -5. \end{aligned} \]
To solve this system, we look for values of \(x_1\) and \(x_2\) that make both equations true at the same time. There are several methods to solve such systems, including substitution, elimination, and using matrices. Here, we are interested in using matrices, specifically an inverse matrix, to find the solution efficiently.

Matrix Equation

When dealing with systems of equations, matrix representation provides a clean and organized way to handle the equations using linear algebra. A matrix equation organizes these equations into a concise form \(AX = B\), where \(A\) is the coefficient matrix containing the constants multiplying the variables, \(X\) is the column matrix of the variables, and \(B\) is the constant matrix.
In our example, we have:

• \(A = \begin{bmatrix} 1 & -1 \ 2 & 4 \end{bmatrix}\)
• \(X = \begin{bmatrix} x_1 \ x_2 \end{bmatrix}\)
• \(B = \begin{bmatrix} 2 \ -5 \end{bmatrix}\)

So the matrix equation \(AX = B\) succinctly represents the system. Solving \(AX = B\) for \(X\) involves using matrix techniques such as finding the inverse of \(A\), provided \(A\) is invertible.

Adjugate Matrix

The adjugate matrix, sometimes called the adjoint, is a crucial component in finding the inverse of a matrix. For a 2x2 matrix \(A\):\[ A = \begin{bmatrix} a & b \ c & d \end{bmatrix} \]
The adjugate, often denoted as \(\text{adj}(A)\), is constructed by swapping the elements on the main diagonal and changing the signs of the off-diagonal elements. Hence, the adjugate is given by:

• \(\text{adj}(A) = \begin{bmatrix} d & -b \ -c & a \end{bmatrix}\)

In our example, matrix \(A = \begin{bmatrix} 1 & -1 \ 2 & 4 \end{bmatrix}\), so the adjugate matrix is:\[ \text{adj}(A) = \begin{bmatrix} 4 & 1 \ -2 & 1 \end{bmatrix} \]This adjugate matrix is essential for calculating the inverse of \(A\), which in turn, is critical to solving the matrix equation \(AX = B\).

Determinant of a Matrix

The determinant of a matrix is a special number that can be calculated from its elements and is useful in a variety of applications in linear algebra. For a 2x2 matrix \(A = \begin{bmatrix} a & b \ c & d \end{bmatrix}\), the determinant, denoted as \(\text{det}(A)\), is calculated as:
\[ \text{det}(A) = ad - bc \]
In the given exercise, the matrix \(A\) is \(\begin{bmatrix} 1 & -1 \ 2 & 4 \end{bmatrix}\), and its determinant is:

• \(\text{det}(A) = (1 \times 4) - (-1 \times 2) = 4 + 2 = 6\)

The determinant helps determine if a matrix is invertible. If the determinant is zero, the matrix does not have an inverse. Since \(\text{det}(A)\) is 6 in this case, matrix \(A\) is invertible, allowing us to proceed with calculating its inverse and solving the system of equations.