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Solving Systems of Equations Using the Row Reduction Method

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Understanding how to solve systems of equations is a foundational skill in algebra and linear algebra. One of the most powerful methods for solving these systems is row reduction, which leverages matrices to streamline the solution process. In this post, we will cover the step-by-step process of solving systems of linear equations using the row reduction method and provide detailed examples to ensure clarity.

What is Row Reduction?

Row reduction is a technique that transforms a system of linear equations into a simpler form, making it easier to solve. This method involves writing the system as an augmented matrix and performing row operations to achieve a simplified form, such as row echelon form or reduced row echelon form.

Steps to Solve a System Using Row Reduction

  1. Write the Augmented Matrix: Convert the system of equations into an augmented matrix, where the coefficients and constants are written in a compact form.
  2. Apply Row Operations: Use these three types of row operations to simplify the matrix:
    • Swap two rows.
    • Multiply a row by a nonzero scalar.
    • Add or subtract a multiple of one row from another row.
  3. Reach Row Echelon Form: Simplify the matrix so that all entries below the main diagonal are zeros.
  4. (Optional) Reach Reduced Row Echelon Form: Further simplify the matrix so that all entries above and below the main diagonal are zeros, leaving only 1s on the diagonal.
  5. Solve for Variables: Interpret the simplified matrix to find the solution to the system of equations.

Example 1: Solving a 2×2 System

Consider the system of equations:

    \[\begin{cases} x + y = 5 \\2x - y = 1\end{cases}\]

Step 1: Write the Augmented Matrix

    \[\begin{bmatrix}1 & 1 & 5 \\2 & -1 & 1\end{bmatrix}\]

Step 2: Apply Row Operations

  • Subtract 2 \times \text{Row 1} from Row 2:

    \[\begin{bmatrix}1 & 1 & 5 \\0 & -3 & -9\end{bmatrix}\]

  • Divide Row 2 by -3:

    \[\begin{bmatrix}1 & 1 & 5 \\0 & 1 & 3\end{bmatrix}\]

  • Subtract Row 2 from Row 1:

    \[\begin{bmatrix}1 & 0 & 2 \\0 & 1 & 3\end{bmatrix}\]

Step 3: Solve for Variables

The matrix corresponds to:

    \[x = 2, \quad y = 3\]

Example 2: Solving a 3×3 System

Solve the system:

    \[\begin{cases}x + y + z = 6 \\2x - y + z = 3 \\x +4y + 3z = 14\end{cases}\]

Step 1: Write the Augmented Matrix

    \[\begin{bmatrix}1 & 1 & 1 & 6 \\2 & -1 & 1 & 3 \\1 & 4 & 3 & 14\end{bmatrix}\]

Step 2: Apply Row Operations

  1. Subtract 2 \times \text{Row 1} from Row 2 and subtract Row 1 from Row 3:

    \[\begin{bmatrix}1 & 1 & 1 & 6 \\0 & -3 & -1 & -9 \\0 & 3 & 2 & 8\end{bmatrix}\]

  1. Add Row 2 to Row 3:

    \[\begin{bmatrix}1 & 1 & 1 & 6 \\0 & -3 & -1 & -9 \\0 & 0 & 1 & -1\end{bmatrix}\]

  1. Divide Row 2 by -3:

    \[\begin{bmatrix}1 & 1 & 1 & 6 \\0 & 1 & \frac{1}{3} & 3 \\0 & 0 & 1 & -1\end{bmatrix}\]

  1. Eliminate the z-term from Row 2 by dividing Row 3 by 3 and subtracting from row 2:

    \[\begin{bmatrix}1 & 1 & 0 & 7 \\0 & 1 & 0 & \frac{10}{3} \\0 & 0 & 1 & -1\end{bmatrix}\]

  1. Eliminate the y-term from Row 1 by subtracting Row 2 from Row 1:

    \[\begin{bmatrix}1 & 0 & 0 & \frac{11}{3} \\0 & 1 & 0 & \frac{10}{3} \\0 & 0 & 1 & -1\end{bmatrix}\]

Step 3: Solve for Variables

The matrix corresponds to:

    \[x = \frac{11}{3}, \quad y = \frac{10}{3} , \quad z = -1\]

Example 3: Special Case – No Solution

Consider the system:

    \[\begin{cases}x + y = 4 \\2x + 2y = 10\end{cases}\]

Step 1: Write the Augmented Matrix

    \[\begin{bmatrix}1 & 1 & 4 \\2 & 2 & 10\end{bmatrix}\]

Step 2: Apply Row Operations

  • Subtract 2 \times \text{Row 1} from Row 2:

    \[\begin{bmatrix}1 & 1 & 4 \\0 & 0 & 2\end{bmatrix}\]

Analysis: The second row translates to 0 = 2, which is a contradiction. Therefore, the system has no solution.

Example 4: Special Case – Infinitely Many Solutions

Consider the system:

    \[\begin{cases}x + y + z = 5 \\2x + 2y + 2z = 10\end{cases}\]

Step 1: Write the Augmented Matrix

    \[\begin{bmatrix}1 & 1 & 1 & 5 \\2 & 2 & 2 & 10\end{bmatrix}\]

Step 2: Apply Row Operations

  • Subtract 2 \times \text{Row 1} from Row 2:

    \[\begin{bmatrix}1 & 1 & 1 & 5 \\0 & 0 & 0 & 0\end{bmatrix}\]

Analysis: The second row translates to 0 = 0, which means there are infinitely many solutions. Let z = t, where t is a parameter. Then:

    \[x + y + t = 5 \implies y = 5 - x - t\]

The solution is:

    \[x = x, \quad y = 5 - x - t, \quad z = t\]

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