Problem 76 - Precalculus

Question

Solve the system of linear equations using Gauss-Jordan elimination. $$\begin{array}{rr} x+2 y-z= & 6 \\ 2 x-y+3 z= & -13 \\ 3 x-2 y+3 z= & -16 \end{array}$$

Step-by-Step Solution

Answer

The solution is \(x = -1\), \(y = 2\), \(z = -3\).

Step 1: Write the Augmented Matrix

First, convert the given system of equations into an augmented matrix:\[\begin{bmatrix}1 & 2 & -1 & | & 6 \2 & -1 & 3 & | & -13 \3 & -2 & 3 & | & -16 \end{bmatrix}\]

Step 2: Make Leading Coefficient of Row 1 Equal to 1

The leading coefficient of the first row is already 1, so we can proceed to the next step without changes.

Step 3: Eliminate First Column Below the First Row

We need to eliminate the first element in the second and third rows below the leading 1. Use row operations: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 matrix becomes:\[\begin{bmatrix}1 & 2 & -1 & | & 6 \0 & -5 & 5 & | & -25 \0 & -8 & 6 & | & -34\end{bmatrix}\]

Step 4: Make Leading Coefficient of Row 2 Equal to 1

Divide the entire second row by -5 to make the leading coefficient 1:\[R_2 = \frac{1}{-5} R_2\]The matrix becomes:\[\begin{bmatrix}1 & 2 & -1 & | & 6 \0 & 1 & -1 & | & 5 \0 & -8 & 6 & | & -34\end{bmatrix}\]

Step 5: Eliminate Second Column Elements in Other Rows

To eliminate the second column elements in the first and third rows, perform the following operations:First Row:Subtract 2 times the second row from the first row:\[R_1 = R_1 - 2R_2\]Third Row:Add 8 times the second row to the third row:\[R_3 = R_3 + 8R_2\]The matrix becomes:\[\begin{bmatrix}1 & 0 & 1 & | & -4 \0 & 1 & -1 & | & 5 \0 & 0 & -2 & | & 6\end{bmatrix}\]

Step 6: Make Leading Coefficient of Row 3 Equal to 1

Divide the third row by -2 to make the leading coefficient 1:\[R_3 = \frac{1}{-2} R_3\]The matrix becomes:\[\begin{bmatrix}1 & 0 & 1 & | & -4 \0 & 1 & -1 & | & 5 \0 & 0 & 1 & | & -3\end{bmatrix}\]

Step 7: Back Substitution to Find Other Variables

Now that the matrix is in reduced row-echelon form, use back substitution.For the first row, subtract the third column element from the constant:\[x + 1(-3) = -4 \rightarrow x = -1\]For the second row, substitute 1 for the third column element:\[y - 1(-3) = 5 \rightarrow y = 2\]The solution is:\[x = -1, \, y = 2, \, z = -3\]

Step 8: Verify the Solution

Substitute the values \(x = -1\), \(y = 2\), and \(z = -3\) back into the original equations to verify:1. \( x + 2y - z = 6 \rightarrow -1 + 4 + 3 = 6 \)2. \( 2x - y + 3z = -13 \rightarrow -2 - 2 - 9 = -13 \)3. \( 3x - 2y + 3z = -16 \rightarrow -3 - 4 - 9 = -16 \)The solution satisfies all equations.

Key Concepts

Systems of Linear Equations

A system of linear equations consists of multiple equations that need to be solved together, as they share common variables. In our case, the system has three equations with three variables: \(x\), \(y\), and \(z\). These types of systems can be classified based on their solutions into:

• Consistent: Having at least one set of solutions.
• Inconsistent: Having no possible solutions.
• Dependent: Where all equations represent the same line, resulting in infinitely many solutions.
Each equation in the system describes a plane in a three-dimensional space, and the solution to the system—the values of \(x\), \(y\), and \(z\) that satisfy all equations—can be seen as the point at which these planes intersect.
Many methods exist for solving these systems, including substitution and elimination. However, Gauss-Jordan elimination is a systematic algorithm that simplifies the process and can be applied to any system of linear equations.
It involves transforming the system into a simpler form where solutions can easily be identified.

Augmented Matrix

An augmented matrix is a powerful tool used to represent a system of linear equations in a concise form. This matrix combines the coefficient matrix with a column for the constants from each equation. For example, for our system:
The equations \[\begin{array}{rrr|r}1 & 2 & -1 & 6 \2 & -1 & 3 & -13 \3 & -2 & 3 & -16\end{array}\]reflect the coefficients of \(x\), \(y\), \(z\), respectively, with a separator for the constants (right-hand side of the equations).
The augmented matrix allows us to employ row reduction operations effectively, helping to transform the matrix into a form from which we can easily read off the solutions.
Converting a system of equations into an augmented matrix is a crucial first step in applying techniques like Gauss-Jordan elimination or Gaussian elimination.
This approach provides a more structured method to approach complex systems, compared to direct algebraic manipulation of equations.

Row Operations

Row operations are essential elementary operations that we perform on matrices to simplify them directly and systematically. The three permissible row operations are:

• Row Replacement: Adding or subtracting the multiple of one row to another row.
• Row Swap: Exchanging two rows.
• Scalar Multiplication: Multiplying a row by a non-zero constant.

In Gauss-Jordan elimination, these operations are used strategically to convert a matrix to its reduced row-echelon form (RREF), where each leading entry (or pivot) in a row is 1, and it is the only non-zero number in its column.
The process is systematic: eliminate variables column by column and achieve zeros below and above pivots. Once in RREF, the matrix directly reveals the solutions to the system of equations.
Performing these operations requires careful tracking to ensure that no errors creep into the process, fundamentally altering the solution the operations aim to achieve.
Practicing these operations on smaller matrices helps grasp the underlying principles, making it easier to handle more complex systems.