Problem 46 - Advanced Engineering Mathematics

Question

Use an inverse matrix to solve the given system of equations. $$ \begin{aligned} x_{1}+2 x_{2} &=4 \\ 3 x_{1}+4 x_{2} &=-3 \end{aligned} $$

Step-by-Step Solution

Answer

\(x_1 = -11\) and \(x_2 = 7.5\).

Step 1: Write the System in Matrix Form

First, express the system of equations in the form \(A\mathbf{x} = \mathbf{b}\). The given equations can be written as:\[\begin{bmatrix} 1 & 2 \ 3 & 4 \end{bmatrix} \begin{bmatrix} x_1 \ x_2 \end{bmatrix} = \begin{bmatrix} 4 \ -3 \end{bmatrix}\]

Step 2: Determine the Inverse of Matrix A

Calculate the inverse of matrix \(A\), which is \(\begin{bmatrix} 1 & 2 \ 3 & 4 \end{bmatrix}\). The formula for the inverse of a 2x2 matrix \(\begin{bmatrix} a & b \ c & d \end{bmatrix}\) is:\[A^{-1} = \frac{1}{ad-bc} \begin{bmatrix} d & -b \ -c & a \end{bmatrix}\]Substitute \(a=1\), \(b=2\), \(c=3\), \(d=4\) into the formula:\[A^{-1} = \frac{1}{1 \times 4 - 2 \times 3} \begin{bmatrix} 4 & -2 \ -3 & 1 \end{bmatrix} = \frac{1}{-2} \begin{bmatrix} 4 & -2 \ -3 & 1 \end{bmatrix}\]Which simplifies to:\[A^{-1} = \begin{bmatrix} -2 & 1 \ 1.5 & -0.5 \end{bmatrix}\]

Step 3: Multiply the Inverse Matrix by Vector b

Now, find \(\mathbf{x}\) by calculating \(\mathbf{x} = A^{-1}\mathbf{b}\), where \(\mathbf{b} = \begin{bmatrix} 4 \ -3 \end{bmatrix}\).Multiply:\[\begin{bmatrix} -2 & 1 \ 1.5 & -0.5 \end{bmatrix} \begin{bmatrix} 4 \ -3 \end{bmatrix} = \begin{bmatrix} (-2)(4) + (1)(-3) \ (1.5)(4) + (-0.5)(-3) \end{bmatrix} = \begin{bmatrix} -8 - 3 \ 6 + 1.5 \end{bmatrix} = \begin{bmatrix} -11 \ 7.5 \end{bmatrix}\]

Step 4: Solution Interpretation

The solution vector \(\mathbf{x}\) gives the values for \(x_1\) and \(x_2\):\(x_1 = -11\) and \(x_2 = 7.5\). Therefore, the values for \(x_1\) and \(x_2\) that satisfy the system of equations are \(-11\) and \(7.5\), respectively.

Key Concepts

System of Linear Equations

A system of linear equations consists of two or more equations that together impose restrictions on a set of variables. In this case, we have two equations with two unknowns, which we can write as follows:
\[\begin{aligned}x_{1} + 2x_{2} &= 4 \3x_{1} + 4x_{2} &= -3\end{aligned}\]
This means we are seeking values for \(x_1\) and \(x_2\) that render both equations true simultaneously. Systems like this are common in algebra and can often be solved using different methods. Here, we'll explore the inverse matrix method.

This method is particularly useful for solving systems where there is the same number of equations as unknowns. It allows us to transform the system into a matrix equation of the form \(A\mathbf{x} = \mathbf{b}\), where \(A\) is the coefficient matrix, \(\mathbf{x}\) is the column matrix containing variables, and \(\mathbf{b}\) is the constant matrix. This elegant structure helps in methodically finding solutions by leveraging matrix operations.

Matrix Inverse

The matrix inverse plays a pivotal role in solving equations like \(A\mathbf{x} = \mathbf{b}\), by converting it into a form where the solution \(\mathbf{x}\) can be obtained directly. Just like the inverse of a number in arithmetic (where the inverse of 5 is 1/5) allows us to solve equations by multiplying both sides, the inverse of our matrix \(A\) can help solve matrix equations.

To find the inverse of a 2x2 matrix \(\begin{bmatrix} a & b \ c & d \end{bmatrix}\), we use:
\[A^{-1} = \frac{1}{ad-bc} \begin{bmatrix} d & -b \ -c & a \end{bmatrix}\]
First, calculate the determinant \(ad - bc\). If the determinant is not zero, the inverse exists. In our example, the determinant is \((-2)\). Since it's not zero, the inverse of \(\begin{bmatrix} 1 & 2 \ 3 & 4 \end{bmatrix}\) can be computed:
\[A^{-1} = \begin{bmatrix} -2 & 1 \ 1.5 & -0.5 \end{bmatrix}\]
This inverse is the critical factor in unlocking the system's solution.

2x2 Matrix Solution

The solution process beyond finding the inverse matrix involves utilizing it to determine the values of \(x_1\) and \(x_2\) from our original system of equations. Once we have \(A^{-1}\), we can solve \(A\mathbf{x} = \mathbf{b}\) by multiplying both sides by \(A^{-1}\) to get \(\mathbf{x} = A^{-1}\mathbf{b}\).

So, given \(A^{-1} = \begin{bmatrix} -2 & 1 \ 1.5 & -0.5 \end{bmatrix}\) and \(\mathbf{b} = \begin{bmatrix} 4 \ -3 \end{bmatrix}\), the calculation becomes:
\[\mathbf{x} = \begin{bmatrix} -2 & 1 \ 1.5 & -0.5 \end{bmatrix} \begin{bmatrix} 4 \ -3 \end{bmatrix} = \begin{bmatrix} -11 \ 7.5 \end{bmatrix}\]
The result \(\mathbf{x} = \begin{bmatrix} -11 \ 7.5 \end{bmatrix}\) provides the solution \(x_1 = -11\) and \(x_2 = 7.5\).

This process demonstrates how matrices offer a streamlined way of handling multiple linear equations, providing precise solutions and showcasing the power of linear algebra.