Problem 17 - Precalculus: Mathematical for Calculus

Question

The system of linear equations has a unique solution. Find the solution using Gaussian elimination or Gauss-Jordan elimination.$$\left\\{\begin{aligned} x+y+z &=2 \\ 2 x-3 y+2 z &=4 \\ 4 x+y-3 z &=1 \end{aligned}\right.$$

Step-by-Step Solution

Answer

x = 1, y = 0, z = 1.

Step 1: Write the augmented matrix

First, express the system of linear equations as an augmented matrix. The given system is: \(\begin{aligned} x+y+z &=2 \ 2x-3y+2z &=4 \ 4x+y-3z &=1 \end{aligned}\). This converts to the matrix: \(\left[ \begin{array}{ccc|c} 1 & 1 & 1 & 2 \ 2 & -3 & 2 & 4 \ 4 & 1 & -3 & 1 \end{array} \right]\).

Step 2: Perform row operations to get upper triangular form

Use row operations to simplify the matrix into upper triangular form (i.e., zeros below the leading 1 in the first column). \( R_2 = R_2 - 2R_1 \) and \( R_3 = R_3 - 4R_1 \). This gives:\[\begin{array}{ccc|c} 1 & 1 & 1 & 2 \ 0 & -5 & 0 & 0 \ 0 & -3 & -7 & -7 \end{array}\].

Step 3: Simplify the matrix further

Next, eliminate the second column below the leading 1 of the second column. Perform the operation \( R_3 = R_3 + \frac{3}{5}R_2 \) to obtain:\[\begin{array}{ccc|c} 1 & 1 & 1 & 2 \ 0 & -5 & 0 & 0 \ 0 & 0 & -7 & -7 \end{array}\].

Step 4: Back substitution to find solution

Since the lower triangular portion is now zero, use back substitution to solve for the variables: Solve for \( z \) from the third row: \(-7z = -7\) implies \(z = 1\). Substitute \( z = 1 \) into the second row: \(-5y = 0\) implies \(y = 0\). Substitute \( y = 0 \) and \( z = 1\) into the first row: \( x + 1 = 2\) implies \(x = 1\).

Step 5: State the solution

The solution to the system of equations is \( x = 1 \), \( y = 0 \), and \( z = 1 \).

Key Concepts

System of Linear Equations

A system of linear equations consists of two or more equations with common variables. The main goal is to find the specific values of these variables that satisfy all the equations simultaneously. For example, consider the system:

• \(x + y + z = 2\)
• \(2x - 3y + 2z = 4\)
• \(4x + y - 3z = 1\)

Each equation represents a plane in three-dimensional space. The solution to this system is the point where all these planes intersect. In simple terms, you are finding a single set of \(x, y, \) and \(z\) values that make all the equations true at the same time. This set is called a consistent and independent solution, indicating that there is a unique solution for the variables involved.

Gauss-Jordan Elimination

Gauss-Jordan elimination is an efficient method to solve systems of linear equations. It transforms the system’s augmented matrix into reduced row-echelon form through a series of row operations. This makes it easy to read off the solutions directly.
Start by writing the system as an augmented matrix. For example, the provided system converts to:

• \[ \left[ \begin{array}{ccc|c} 1 & 1 & 1 & 2 \ 2 & -3 & 2 & 4 \ 4 & 1 & -3 & 1 \end{array} \right] \]

Use row operations to achieve leading 1s (pivot elements) in each row and zeros below and above them. The sequence involves:

• Exchange rows if necessary.
• Scale rows to ensure the leading coefficient becomes 1.
• Subtract multiples of the pivot row from the rows beneath to eliminate lower coefficients.
• Repeat the process for each row to eventually achieve \[ \begin{array}{ccc|c} 1 & 0 & 0 & a \ 0 & 1 & 0 & b \ 0 & 0 & 1 & c \end{array} \]

This matrix form provides the values \(x = a\), \(y = b\), and \(z = c\). It’s a straightforward way to see solutions, especially when dealing with multiple variables.

Back Substitution

Back substitution is the final step in solving a system of linear equations after applying Gaussian or Gauss-Jordan elimination. Once the system is in an upper triangular form, it allows us to easily solve for each variable starting from the last equation.
In the upper triangular form, the matrix looks something like:

• \[ \begin{array}{ccc|c} 1 & 1 & 1 & 2 \ 0 & -5 & 0 & 0 \ 0 & 0 & -7 & -7 \end{array} \]

Back substitution entails:

• First, solve for \(z\) in the last row: if \(-7z = -7\), then \(z = 1\).
• Next, substitute \(z = 1\) into the second row \((0 - 5y + 0 = 0)\). This leads to \(y = 0\).
• Finally, use both \(y = 0\) and \(z = 1\) in the first row \((x + 1 = 2)\) to find \(x = 1\).

Back substitution provides the final values of each variable systematically, ensuring accuracy in the solution of the system. It is a powerful method because it helps untangle complex systems step by step, ensuring a clear path to the solution.