Problem 44 - Advanced Engineering Mathematics

Question

In Problems \(43-50\), use an inverse matrix to solve the given system of equations. $$ \begin{aligned} x_{1}-x_{2} &=2 \\ 2 x_{1}+4 x_{2} &=-5 \end{aligned} $$

Step-by-Step Solution

Answer

\(x_1 = \frac{1}{2}\), \(x_2 = -\frac{3}{2}\).

Step 1: Write the system of equations in matrix form

First, we rewrite the given system of equations in the form \[ A \mathbf{x} = \mathbf{b} \]where\[ A = \begin{bmatrix} 1 & -1 \ 2 & 4 \end{bmatrix}, \quad \mathbf{x} = \begin{bmatrix} x_1 \ x_2 \end{bmatrix}, \quad \mathbf{b} = \begin{bmatrix} 2 \ -5 \end{bmatrix} \]

Step 2: Find the inverse of matrix A

To find the inverse of matrix \(A\), use the formula for the inverse of a 2x2 matrix:\[ A^{-1} = \frac{1}{ad-bc} \begin{bmatrix} d & -b \ -c & a \end{bmatrix} \]where in matrix \(A = \begin{bmatrix} 1 & -1 \ 2 & 4 \end{bmatrix}\), \(a = 1\), \(b = -1\), \(c = 2\), \(d = 4\).Calculate the determinant:\[ ad - bc = 1(4) - (-1)(2) = 4 + 2 = 6 \]The inverse is:\[ A^{-1} = \frac{1}{6} \begin{bmatrix} 4 & 1 \ -2 & 1 \end{bmatrix} \]

Step 3: Multiply both sides by the inverse matrix

Now, multiply both sides of the matrix equation \(A \mathbf{x} = \mathbf{b}\) by \(A^{-1}\):\[ A^{-1}A \mathbf{x} = A^{-1} \mathbf{b} \]Since \(A^{-1}A = I\) (the identity matrix), this simplifies to:\[ \mathbf{x} = A^{-1} \mathbf{b} \]

Step 4: Calculate the solution vector x

Substitute \(A^{-1}\) and \(\mathbf{b}\) into the equation to find \(\mathbf{x}\):\[ \mathbf{x} = \frac{1}{6} \begin{bmatrix} 4 & 1 \ -2 & 1 \end{bmatrix} \begin{bmatrix} 2 \ -5 \end{bmatrix} \]Perform the matrix multiplication:First row:\[ 4(2) + 1(-5) = 8 - 5 = 3 \]Second row:\[ -2(2) + 1(-5) = -4 - 5 = -9 \]Now apply the scalar multiplication:\[ \mathbf{x} = \begin{bmatrix} \frac{3}{6} \ \frac{-9}{6} \end{bmatrix} = \begin{bmatrix} \frac{1}{2} \ -\frac{3}{2} \end{bmatrix} \]Thus, \(x_1 = \frac{1}{2}\) and \(x_2 = -\frac{3}{2}\).

Key Concepts

System of Equations

A system of equations is a collection of two or more equations with the same set of variables. In the context of linear algebra, a system of linear equations involves linear expressions. Each equation represents a line, plane, or higher-dimensional plane depending on the number of variables. Solving these systems means finding the values for the variables that satisfy all equations simultaneously.
In our exercise, we have a system with two equations and two unknowns \(x_1\) and \(x_2\):

• \(x_1 - x_2 = 2\)
• \(2x_1 + 4x_2 = -5\)

By expressing this system in matrix form, we pave the way for solving using a powerful tool in linear algebra: the inverse of a matrix.

Matrix Algebra

Matrix algebra is a branch of mathematics that deals with the manipulation of matrices, which are rectangular arrays of numbers. Matrices can represent and solve systems of equations, making them a cornerstone of linear algebra.
In the problem provided, the original system of equations is written in matrix form as \(A \mathbf{x} = \mathbf{b}\), where:

• Matrix \(A\) is the coefficients matrix: \(\begin{bmatrix} 1 & -1 \ 2 & 4 \end{bmatrix}\)
• \(\mathbf{x}\) is the column matrix of the variables: \(\begin{bmatrix} x_1 \ x_2 \end{bmatrix}\)
• \(\mathbf{b}\) is the constant column matrix: \(\begin{bmatrix} 2 \ -5 \end{bmatrix}\)

By setting up the equations this way, matrix algebra allows us to use the inverse of the matrix to solve for \(\mathbf{x}\), making it a powerful solving method.

Determinant Calculation

The determinant is a special number calculated from a square matrix. It plays an essential role in matrix operations, particularly in finding the inverse matrix. For a 2x2 matrix, the determinant \(det(A)\) can be defined as \(ad - bc\) when the matrix is \(\begin{bmatrix} a & b \ c & d \end{bmatrix}\).
In this exercise:

• Matrix \(A = \begin{bmatrix} 1 & -1 \ 2 & 4 \end{bmatrix}\)
• Determinant: \(ad - bc = 1(4) - (-1)(2) = 6\)

The determinant helps us determine if the matrix is invertible. Since the determinant is non-zero \((6)\), matrix \(A\) has an inverse, enabling us to solve the system.

Matrix Multiplication

Matrix multiplication is the operation that combines two matrices to produce another matrix. It involves multiplying rows by columns. This operation is vital in solving systems of equations using inverses.
To solve for \(\mathbf{x}\) in \(\mathbf{x} = A^{-1} \mathbf{b}\), we perform matrix multiplication following these steps:

• Insert \(A^{-1} = \frac{1}{6} \begin{bmatrix} 4 & 1 \ -2 & 1 \end{bmatrix}\) and \(\mathbf{b} = \begin{bmatrix} 2 \ -5 \end{bmatrix}\).
• Multiply: \(\begin{bmatrix} 4 & 1 \ -2 & 1 \end{bmatrix} \begin{bmatrix} 2 \ -5 \end{bmatrix} = \begin{bmatrix} 3 \ -9 \end{bmatrix}\).
• Finally, apply the scalar: \(\frac{1}{6} \begin{bmatrix} 3 \ -9 \end{bmatrix} = \begin{bmatrix} \frac{1}{2} \ -\frac{3}{2} \end{bmatrix}\).

Through matrix multiplication, we find that \(x_1 = \frac{1}{2}\) and \(x_2 = -\frac{3}{2}\), solving the system effortlessly.