Problem 9 - Calculus for Biology and Medicine

Question

In Problems 9-16, reduce the system of equations to upper triangular form and find all the solutions. $$ \begin{array}{r} 2 x-y=3 \\ x-3 y=7 \end{array} $$

Step-by-Step Solution

Answer

The solution is \( x = 0.4 \) and \( y = -2.2 \).

Step 1: Write the System of Equations in Matrix Form

First, express the system of linear equations as an augmented matrix. The given system is \( \begin{array}{r} 2x - y = 3 \ x - 3y = 7 \end{array} \). This translates to the augmented matrix \( \begin{bmatrix} 2 & -1 & | & 3 \ 1 & -3 & | & 7 \end{bmatrix} \).

Step 2: Eliminate the x-coefficient in the Second Row

We aim to make the first column of the second row zero to achieve an upper triangular form. Subtract the first row, multiplied by 0.5 (or \( \frac{1}{2} \)), from the second row. The operation is \( R_2 = R_2 - 0.5 R_1 \). Calculate: \( 1 - 0.5 \cdot 2 = 0 \), \( -3 - 0.5 \cdot (-1) = -2.5 \), and \( 7 - 0.5 \cdot 3 = 5.5 \). This gives the new matrix: \( \begin{bmatrix} 2 & -1 & | & 3 \ 0 & -2.5 & | & 5.5 \end{bmatrix} \).

Step 3: Solve the System by Back Substitution

Now that we have an upper triangular matrix, use back substitution to solve for \( y \) in the second equation \( -2.5y = 5.5 \), giving \( y = -\frac{5.5}{2.5} = -2.2 \). Then substitute \( y \) back into the first equation: \( 2x - (-2.2) = 3 \). Solving gives \( 2x + 2.2 = 3 \), or \( 2x = 0.8 \). Thus, \( x = 0.4 \).

Step 4: Verify the Solution

Substitute \( x = 0.4 \) and \( y = -2.2 \) back into the original equations to verify correctness. For \( 2x - y = 3 \): \( 2(0.4) - (-2.2) = 0.8 + 2.2 = 3 \). For \( x - 3y = 7 \): \( 0.4 - 3(-2.2) = 0.4 + 6.6 = 7 \). Both check out, confirming the solution is correct.

Key Concepts

Systems of Equations

A system of equations consists of two or more equations with multiple variables. In linear algebra, solving these systems helps to find values for the variables that satisfy all equations simultaneously.
In the given exercise, we have two equations: one that looks like a line plotted on a graph, and another. Our goal is to find the point where both lines intersect. This is the solution to the system.
For this problem, our equations are:

• \(2x - y = 3\)
• \(x - 3y = 7\)

This is a simple system with two equations and two variables. The goal is to find the values of \(x\) and \(y\) that satisfy both equations at the same time. This foundational concept is crucial in many areas of mathematics, physics, and engineering.

Matrix Form

To efficiently solve systems of equations, we often convert them into matrix form. A matrix is a rectangular array of numbers arranged in rows and columns. This makes it easier to handle plans like row operations, which simplify the system.
In our exercise, the system \(2x - y = 3\) and \(x - 3y = 7\) becomes:

• Matrix: \(\begin{bmatrix} 2 & -1 & | & 3 \ 1 & -3 & | & 7 \end{bmatrix}\)

Here, each row in the matrix represents an equation, and the vertical bar separates the coefficients of the variables from the constants. This approach helps change the equations systematically and lead us on to the next phase of solving—which includes operations like row elimination. The matrix form thus becomes a vital step for transforming the problem into something more accessible and calculable.

Back Substitution

Once a system has been converted into an upper triangular matrix form, the final step is to use back substitution to find the solution of the system. In upper triangular form, each row represents an equation, with each row below it having zero coefficients for the first variables.
From the exercise, we reach:

• \(\begin{bmatrix} 2 & -1 & | & 3 \ 0 & -2.5 & | & 5.5 \end{bmatrix}\)

Back substitution involves starting from the bottom row and solving for one variable at a time. Here, we first solve for \(y\) using the second row equation: \(-2.5y = 5.5\), which gives us \(y = -2.2\). Next, we substitute this value into the first row's equation to find \(x\).
This method is crucial in linear algebra as it allows for solving systems efficiently and is especially useful when dealing with larger systems of equations.