Problem 43 - Advanced Engineering Mathematics

Question

Use an inverse matrix to solve the given system of equations. $$ \begin{aligned} x_{1}+x_{2} &=4 \\ 2 x_{1}-x_{2} &=14 \end{aligned} $$

Step-by-Step Solution

Answer

The solution is \(x_1 = 6\) and \(x_2 = -2\).

Step 1: Write the system of equations in matrix form

The given system of equations can be written in the matrix form as follows:\[A \mathbf{x} = \mathbf{b}\]where \(A\) is the coefficient matrix, \(\mathbf{x}\) is the column vector of variables, and \(\mathbf{b}\) is the column vector of constants. Thus, \[A = \begin{bmatrix} 1 & 1 \ 2 & -1 \end{bmatrix}, \quad \mathbf{x} = \begin{bmatrix} x_1 \ x_2 \end{bmatrix}, \quad \mathbf{b} = \begin{bmatrix} 4 \ 14 \end{bmatrix}\]

Step 2: Find the inverse of the coefficient matrix A

To solve for \(\mathbf{x}\) using the inverse, we need the inverse of matrix \(A\), denoted as \(A^{-1}\). The inverse of a 2x2 matrix \(\begin{bmatrix} a & b \ c & d \end{bmatrix}\) is given by:\[A^{-1} = \frac{1}{ad-bc} \begin{bmatrix} d & -b \ -c & a \end{bmatrix}\]For our coefficient matrix \(A\), with \(a=1\), \(b=1\), \(c=2\), \(d=-1\), the determinant is:\[ad-bc = (1)(-1) - (2)(1) = -1 - 2 = -3\]Thus,\[A^{-1} = \frac{1}{-3}\begin{bmatrix} -1 & -1 \ -2 & 1 \end{bmatrix} = \begin{bmatrix} \frac{1}{3} & \frac{1}{3} \ \frac{2}{3} & -\frac{1}{3} \end{bmatrix}\]

Step 3: Multiply the inverse matrix by the constant vector

Now, we need to multiply \(A^{-1}\) by \(\mathbf{b}\) to find \(\mathbf{x}\):\[\mathbf{x} = A^{-1} \mathbf{b}\]\[\begin{bmatrix} x_1 \ x_2 \end{bmatrix} = \begin{bmatrix} \frac{1}{3} & \frac{1}{3} \ \frac{2}{3} & -\frac{1}{3} \end{bmatrix} \begin{bmatrix} 4 \ 14 \end{bmatrix}\]Performing the matrix multiplication:\[\begin{bmatrix} x_1 \ x_2 \end{bmatrix} = \begin{bmatrix} \frac{1}{3}(4) + \frac{1}{3}(14) \ \frac{2}{3}(4) - \frac{1}{3}(14) \end{bmatrix} = \begin{bmatrix} \frac{18}{3} \ \frac{8}{3} - \frac{14}{3} \end{bmatrix} = \begin{bmatrix} 6 \ -2 \end{bmatrix}\]

Step 4: Interpret the solution

The solution \(\begin{bmatrix} 6 \ -2 \end{bmatrix}\) tells us the values of the variables. Therefore, \(x_1 = 6\) and \(x_2 = -2\).

Key Concepts

System of Equations

A system of equations involves finding values for variables that satisfy multiple equations at the same time. In our example, we worked with two linear equations:

• Equation 1: \(x_1 + x_2 = 4\)
• Equation 2: \(2x_1 - x_2 = 14\)

Systems like these can be solved using different methods, such as substitution, elimination, or matrix approaches.
Matrix methods become very useful for handling multiple equations, especially when technology is used for calculations due to its ability to simplify and streamline processes.

Matrix Multiplication

Matrix multiplication is a fundamental concept for manipulating matrices and solving systems of equations. In matrix multiplication, the elements from the rows of the first matrix are multiplied with corresponding elements from the columns of the second matrix.
This involves summing these products to produce a single number in the resulting matrix.

• For our exercise, we used the multiplication: \[\begin{bmatrix} \frac{1}{3} & \frac{1}{3} \\frac{2}{3} & -\frac{1}{3} \end{bmatrix} \begin{bmatrix} 4 \14 \end{bmatrix} \]
• By performing the calculations identified in each row and column, we arrive at: \[\begin{bmatrix} 6 \-2 \end{bmatrix} \]

Matrix multiplication must conform to size rules where the number of columns in the first matrix corresponds to the number of rows in the second.
Understanding these basics ensures smoother computations.

Determinant of a Matrix

The determinant is a special number that can be calculated from a square matrix. It provides important properties about the matrix, like whether an inverse exists.
For a 2x2 matrix, the determinant can be found as follows: \[ ad - bc \] where \(a\), \(b\), \(c\), and \(d\) are elements of the matrix.

• In our example, the matrix was: \[\begin{bmatrix} 1 & 1 \2 & -1 \end{bmatrix} \]
• The determinant was calculated as: \(-3 = (1)(-1) - (2)(1)\)

A non-zero determinant indicates that the matrix has an inverse, essential in solving systems with this method.

Solving Linear Equations

Solving linear equations with the help of inverse matrices is an efficient approach. The solution involves finding a matrix that reverses the effects of another matrix, simplifying the task of solving a system of equations.

• First, the system of equations is expressed in matrix form as \(A \mathbf{x} = \mathbf{b}\).
• The inverse of matrix \(A\), denoted as \(A^{-1}\), helps isolate the variable vector \(\mathbf{x}\).
• By computing \(A^{-1} \mathbf{b}\), the values of the variables are obtained directly.

In our problem, multiplication with the inverse matrix allowed us to derive that \(x_1 = 6\) and \(x_2 = -2\).
Understanding how each step connects in this process aids in efficient problem solving and builds a strong foundation for mastering algebraic operations.