Problem 60 - Precalculus

Question

Solve the system of linear equations using Gaussian elimination with back- substitution. $$\begin{array}{rr} 2 x_{1}+x_{2}+x_{3}= & -1 \\ x_{1}+x_{2}-x_{3}= & 5 \\ 3 x_{1}-x_{2}-x_{3}= & 1 \end{array}$$

Step-by-Step Solution

Answer

The solution to the system is \(x_1 = 0\), \(x_2 = 2\), \(x_3 = -3\).

Step 1: Write the Augmented Matrix

Begin by converting the system of equations into an augmented matrix. The augmented matrix for the given system is: \[\begin{bmatrix}2 & 1 & 1 & | & -1 \1 & 1 & -1 & | & 5 \3 & -1 & -1 & | & 1\end{bmatrix}\]

Step 2: Perform Row Operations to Get Upper Triangular Form

Apply row operations to convert the matrix into an upper triangular form. First, eliminate the first entry in the second row using the first row:Replace Row 2 with Row 2 - 0.5*Row 1:\[R_2 = R_2 - 0.5 \cdot R_1 \implies \begin{bmatrix}2 & 1 & 1 & | & -1 \0 & 0.5 & -1.5 & | & 5.5 \3 & -1 & -1 & | & 1\end{bmatrix}\]Next, eliminate the first entry in the third row:Replace Row 3 with Row 3 - 1.5*Row 1:\[R_3 = R_3 - 1.5 \cdot R_1 \implies \begin{bmatrix}2 & 1 & 1 & | & -1 \0 & 0.5 & -1.5 & | & 5.5 \0 & -2.5 & -2.5 & | & 2.5\end{bmatrix}\]

Step 3: Continue Eliminating Lower Entries

Continue with the elimination process to ensure the matrix is in row-echelon form. Eliminate the entry in the third row, second column using the second row:Replace Row 3 with Row 3 + 5*Row 2:\[R_3 = R_3 + 5 \cdot R_2 \implies \begin{bmatrix}2 & 1 & 1 & | & -1 \0 & 0.5 & -1.5 & | & 5.5 \0 & 0 & -10 & | & 30\end{bmatrix}\]

Step 4: Back-Substitution

Solve for the variables starting from the last row upwards. Begin with the last row:From Row 3: \[-10x_3 = 30 \ x_3 = -3 \]Substitute \(x_3 = -3\) into Row 2:\[0.5x_2 - 1.5(-3) = 5.5 \ 0.5x_2 + 4.5 = 5.5 \ 0.5x_2 = 1 \ x_2 = 2 \]Substitute \(x_2 = 2\) and \(x_3 = -3\) into Row 1:\[2x_1 + 1\times2 + 1\times(-3) = -1 \ 2x_1 + 2 - 3 = -1 \ 2x_1 = 0 \ x_1 = 0 \]

Step 5: Verify the Solution

To ensure the solution is correct, plug \(x_1 = 0\), \(x_2 = 2\), and \(x_3 = -3\) back into the original equations:Equation 1: \[2(0) + 2 + (-3) = -1\] Equation 2: \[0 + 2 + 3 = 5\]Equation 3: \[3(0) - 2 - (-3) = 1\]All equations are satisfied.

Key Concepts

Linear Equations

Linear equations are mathematical expressions that represent straight lines. They typically consist of variables, constants, and operations of addition or subtraction:

• Variables: Symbols (like \(x_1, x_2, x_3\)) that represent unknown values.
• Constants: Fixed values, represented by numbers.
• Operations: In linear equations, only addition and subtraction are used for variables.

In the context of solving systems of linear equations like the one presented, the objective is to find the values of the variables that satisfy all equations in the system simultaneously. The solution to a system of linear equations is a set of values for the variables that make all equations true. These systems can be represented in various forms, one of the most useful being the augmented matrix. Understanding linear equations is crucial before delving deeper into methods like Gaussian elimination.

Augmented Matrix

To solve systems of linear equations using linear algebra techniques like Gaussian elimination, we first convert them into an augmented matrix form. An augmented matrix is essentially a compact way to represent a system of linear equations:

• Each row corresponds to an equation.
• Columns, except for the last, correspond to the coefficients of each variable in the equations.
• The last column represents the constants from the right side of the equations.

For instance, the system given is transformed into the augmented matrix:\[\begin{bmatrix} 2 & 1 & 1 & | & -1 \1 & 1 & -1 & | & 5 \3 & -1 & -1 & | & 1\end{bmatrix}\]This representation allows for systematic application of row operations, making it easier to manipulate the equations simultaneously to reach a solution.

Row Operations

Row operations are the tools we use to manipulate the augmented matrix in order to solve the system of linear equations. There are a few key types of row operations that you can perform:

• Swap two rows.
• Multiply a row by a non-zero scalar.
• Add or subtract a multiple of one row to/from another row.

The goal is to use these operations to transform the matrix into an upper triangular form (row-echelon form), where all the entries below the leading diagonal (from top left to bottom right) are zeros. By following a systematic sequence of row operations, as demonstrated in the given solution, the system becomes easier to solve. This transformation paves the way for the next step: back-substitution. It's crucial to carry out these operations carefully to maintain the integrity of the system and progress towards a simplified solution.

Back-Substitution

After performing row operations to achieve an upper triangular form of the augmented matrix, we proceed with back-substitution to find the values of the variables. This process involves solving for the variables starting from the bottom of the matrix and working upwards:

• The last row gives a direct equation for the last variable, which can be solved easily.
• Having obtained the last variable, substitute it into the equation above to solve for the second to last variable.
• Continue substituting the known variables into the upper equations to solve for the remaining unknowns.

In our solution:

• Start with the third equation: \(-10x_3 = 30\) yielding \(x_3 = -3\).
• Substitute \(x_3 = -3\) into the second equation: solve for \(x_2\); this gives \(x_2 = 2\).
• Finally, insert \(x_2 = 2\) and \(x_3 = -3\) into the first equation, resulting in \(x_1 = 0\).

Back-substitution completes the solution with values for all variables, ensuring you have the set \((x_1, x_2, x_3)\) that satisfies the original set of equations. This method efficiently handles even the most complex systems once they are in row-echelon form.