Problem 35 - Precalculus: Mathematics for Calculus

Question

The system of linear equations has a unique solution. Find the solution using Gaussian elimination or Gauss-Jordan elimination. $$\left\\{\begin{aligned} x_{1}+2 x_{2}-x_{3} &=9 \\ 2 x_{1} &-x_{3}=-2 \\ 3 x_{1}+5 x_{2}+2 x_{3} &=22 \end{aligned}\right.$$

Step-by-Step Solution

Answer

The unique solution is \(x_1 = \frac{61}{9}\), \(x_2 = \frac{130}{36}\), \(x_3 = -\frac{40}{9}\).

Step 1: Write the Augmented Matrix

The first step is to write the augmented matrix for the given system of equations. The system is: 1. \( x_{1} + 2x_{2} - x_{3} = 9 \) 2. \( 2x_{1} + 0x_{2} - x_{3} = -2 \) 3. \( 3x_{1} + 5x_{2} + 2x_{3} = 22 \) This results in the augmented matrix: \[\begin{bmatrix} 1 & 2 & -1 & | & 9 \ 2 & 0 & -1 & | & -2 \ 3 & 5 & 2 & | & 22 \end{bmatrix}\]

Step 2: Perform Row Operations to Obtain Upper Triangular Form

We need zeros below the first element of the first column. Subtract 2 times the first row from the second row: \[ R_2 = R_2 - 2R_1 \] Subtract 3 times the first row from the third row: \[ R_3 = R_3 - 3R_1 \] The new augmented matrix is: \[\begin{bmatrix} 1 & 2 & -1 & | & 9 \ 0 & -4 & 1 & | & -20 \ 0 & -1 & 5 & | & -5 \end{bmatrix}\]

Step 3: Eliminate Below the Second Pivot

To create a zero below the \(-4\) in the second column, add \(\frac{1}{4}\) of the second row to the third row: \[ R_3 = R_3 + \frac{1}{4}R_2 \] The augmented matrix becomes: \[\begin{bmatrix} 1 & 2 & -1 & | & 9 \ 0 & -4 & 1 & | & -20 \ 0 & 0 & \frac{9}{4} & | & -10 \end{bmatrix}\]

Step 4: Back Substitution

Now solve the equations starting from the bottom row. The third equation simplifies to \( \frac{9}{4}x_3 = -10 \), thus \( x_3 = -\frac{40}{9} \). Substitute \( x_3 \) into the second equation: \[ -4x_2 + 1 \left(-\frac{40}{9}\right) = -20 \] This simplifies to \[ x_2 = \frac{130}{36} \]. Substitute \( x_2 \) and \( x_3 \) into the first equation: \[ x_1 + 2 \left(\frac{130}{36}\right) - \left(-\frac{40}{9}\right) = 9 \] This simplifies to \( x_1 = 9 - \left(\frac{65}{9}\right) + \left(\frac{40}{9}\right) = \frac{61}{9} \).

Step 5: Verify the Solution

Substitute the values of \( x_1, x_2, x_3 \) back into the original equations to verify: 1. \( x_1 + 2x_2 - x_3 = 9 \) is satisfied. 2. \( 2x_1 - x_3 = -2 \) is satisfied. 3. \( 3x_1 + 5x_2 + 2x_3 = 22 \) is satisfied.This confirms the solution is correct.

Key Concepts

Augmented Matrix

When you're working with a system of linear equations like the one presented, one efficient way to organize the data is using an augmented matrix. An augmented matrix neatly arranges both the coefficients of the variables and the constants from the equations into a convenient matrix form. This structure makes it easier to apply methods like Gaussian elimination for solving.
An augmented matrix typically takes the form:

• The left side of the matrix contains the coefficients. Each row corresponds to a linear equation.
• The right part after the "|" includes the constants from each equation.

For example, the given system of equations:

• \( x_1 + 2x_2 - x_3 = 9 \)
• \( 2x_1 - x_3 = -2 \)
• \( 3x_1 + 5x_2 + 2x_3 = 22 \)

converts to the augmented matrix:

• \[\begin{bmatrix} 1 & 2 & -1 & | & 9 \ 2 & 0 & -1 & | & -2 \ 3 & 5 & 2 & | & 22 \end{bmatrix}\]

This setup allows for focused operations on transforming the system to a simpler form.

Row Operations

In Gaussian elimination, row operations are used to manipulate the augmented matrix into simpler forms, ultimately to solve the system of equations. The goal is to create zeros below the pivot positions, which are the leading coefficients in each row.
There are three types of row operations you may perform:

• Swapping two rows
• Multiplying a row by a non-zero constant
• Adding or subtracting a multiple of one row from another

For instance, the provided problem requires:

• Subtracting 2 times the first row from the second row
• Subtracting 3 times the first row from the third row

These operations yield:
\[\begin{bmatrix} 1 & 2 & -1 & | & 9 \ 0 & -4 & 1 & | & -20 \ 0 & -1 & 5 & | & -5 \end{bmatrix}\]
Such strategic row operations are key to transforming matrices towards triangular forms, making back substitution feasible.

Unique Solution

A system of linear equations may have a unique solution, infinitely many solutions, or no solution at all. A unique solution, as is the case in the given exercise, indicates a single set of values for the variables that satisfies all the equations simultaneously.
In the context of Gaussian elimination:

• The system is reduced to an upper triangular matrix where the variables can be isolated through back substitution.
• If the matrix can be shaped into a form that clearly dictates only one possible value for each variable, this confirms the uniqueness of the solution.

For example, our system reduces to the matrix:

• \[\begin{bmatrix} 1 & 2 & -1 & | & 9 \ 0 & -4 & 1 & | & -20 \ 0 & 0 & \frac{9}{4} & | & -10 \end{bmatrix}\]

Here, each equation unambiguously determines the value of one variable once back substitution is performed:

• From \( \frac{9}{4}x_3 = -10 \), we find \( x_3 \).
• Next, using \( x_3 \), we solve \(-4x_2 + x_3 = -20 \) to get \( x_2 \).
• Finally, substituting \( x_2 \) and \( x_3 \) into \( x_1 + 2x_2 - x_3 = 9 \) gives \( x_1 \).

This stepwise approach, where each operation brings us one step closer to solving for a specific variable, epitomizes finding a unique solution in a linear system.