Problem 9 - Calculus for Biology and Medicine

Question

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

Step-by-Step Solution

Answer

The solution is \( x = \frac{2}{5} \) and \( y = -\frac{11}{5} \).

Step 1: Write the System in Matrix Form

The system of linear equations can be represented using an augmented matrix. For the given system, the corresponding augmented matrix is: \[\begin{bmatrix} 2 & -1 & | & 3 \ 1 & -3 & | & 7 \end{bmatrix}\]

Step 2: Eliminate the First Column of the Second Row

To create a zero below the leading 2 in the first column, multiply the first row by 1 and subtract it from the second row. Perform the operation: Row 2 = Row 2 - (1/2) * Row 1 The new augmented matrix becomes: \[\begin{bmatrix} 2 & -1 & | & 3 \ 0 & -5/2 & | & 11/2 \end{bmatrix}\]

Step 3: Convert to Upper Triangular Form

The matrix is now in upper triangular form, as there are zeros below the leading entry in the first column. The system can be reverted back to equations for easier solving:\[\begin{align*}2x - y &= 3 \-\frac{5}{2}y &= \frac{11}{2}\end{align*}\]

Step 4: Solve for the Second Variable, y

Use the second equation to solve for \( y \): \[-\frac{5}{2}y = \frac{11}{2}\]Multiply both sides by \(-\frac{2}{5}\) to solve for \( y \):\[y = -\frac{11}{5}\]

Step 5: Substitute y Back into the First Equation

Substitute \( y = -\frac{11}{5} \) back into the first equation to solve for \( x \):\[2x - \left( -\frac{11}{5} \right) = 3\]Simplify and solve for \( x \):\[2x + \frac{11}{5} = 3\]Multiply the entire equation by 5 to clear the fraction, and simplify:\[10x + 11 = 15\]\[10x = 4\]Divide both sides by 10:\[x = \frac{2}{5}\]

Step 6: Solution of the System

The solutions for the system of equations are \( x = \frac{2}{5} \) and \( y = -\frac{11}{5} \). The system is consistent and has a unique solution.

Key Concepts

System of Linear Equations

A system of linear equations is a collection of two or more linear equations involving the same set of variables. Linear equations are expressions where each term is either a constant or the product of a constant and a single variable. For example, in the given problem, we have:

• \(2x - y = 3\)
• \(x - 3y = 7\)

The aim of solving a system of linear equations is to find values for the variables that satisfy all equations simultaneously. There are three possible outcomes when solving these systems:

• No solution, when the lines are parallel.
• One unique solution, when the lines intersect at exactly one point.
• Infinitely many solutions, when the lines are identical.

In our exercise, the system has a unique solution, where the values of \(x\) and \(y\) satisfy both equations simultaneously.

Augmented Matrix

An augmented matrix is an essential tool when solving systems of linear equations, especially for larger systems. It is a compact representation of the system that combines the coefficients and the constants of the equations into a matrix form. For the problem at hand, the system of equations:\[\begin{array}{c}2x - y = 3 \x - 3y = 7\end{array}\]is represented as the augmented matrix:\[\begin{bmatrix}2 & -1 & | & 3 \1 & -3 & | & 7\end{bmatrix}\]The vertical line in the matrix separates the coefficients of the variables from the constants of the equations. By operating on this matrix, we can manipulate the equations simultaneously to simplify the process of finding the solution.

Upper Triangular Form

Converting a matrix to upper triangular form is a technique used to simplify the process of solving systems of linear equations. An upper triangular matrix is one where all entries below the main diagonal are zero. This format makes it easier to apply back-substitution to find solutions.In the exercise, we perform row operations to produce zeros below the pivot (the leading coefficient) in the first column. Here is how the transformation in the exercise happens:- Start with the original augmented matrix: \[\begin{bmatrix}2 & -1 & | & 3 \ 1 & -3 & | & 7\end{bmatrix}\]- Subtract \((\frac{1}{2})\) the first row from the second row to get: \[\begin{bmatrix}2 & -1 & | & 3 \0 & -\frac{5}{2} & | & \frac{11}{2}\end{bmatrix}\]The resulting matrix is in upper triangular form, allowing us to systematically solve for the variables starting from the bottom. With \(-\frac{5}{2}y = \frac{11}{2}\) from the second row, we solve for \(y\) first, then substitute back into the first equation to find \(x\). This stepwise approach is crucial to solving systems efficiently and correctly.