Systems of Equations and Gaussian Elimination

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This tutorial provides a detailed guide on systems of equations, their representation, solutions, and the Gaussian elimination method used to solve them.

1. Systems of Equations: Definition and Representation

A system of linear equations consists of multiple linear equations that share variables. A general system with $n$ variables and $m$ equations can be written as:

$\begin{aligned} a_{11}x_1 + a_{12}x_2 + \dots + a_{1n}x_n &= b_1 a_{21}x_1 + a_{22}x_2 + \dots + a_{2n}x_n &= b_2 \vdots & a_{m1}x_1 + a_{m2}x_2 + \dots + a_{mn}x_n &= b_m \end{aligned}$

where:

$x_1, x_2, \dots, x_n$ are the unknown variables.
$a_{ij}$ are the coefficients.
$b_1, b_2, \dots, b_m$ are the constants on the right-hand side.

Matrix Representation

A system of equations can be represented using matrices:

$A \mathbf{x} = \mathbf{b}$

where:

$A$ is the coefficient matrix:

$A = \begin{bmatrix} a_{11} & a_{12} & \dots & a_{1n} a_{21} & a_{22} & \dots & a_{2n} \vdots & \vdots & \ddots & \vdots a_{m1} & a_{m2} & \dots & a_{mn} \end{bmatrix}$
$\mathbf{x}$ is the variable column vector:

$\mathbf{x} = \begin{bmatrix} x_1 x_2 \vdots x_n \end{bmatrix}$
$\mathbf{b}$ is the constant column vector:

$\mathbf{b} = \begin{bmatrix} b_1 b_2 \vdots b_m \end{bmatrix}$

The augmented matrix is written as:

$[A | \mathbf{b}$ ]

Example:
$\begin{aligned} 2x + 3y &= 8 5x - y &= 3 \end{aligned}$

Matrix representation:
$\begin{bmatrix} 2 & 3 5 & -1 \end{bmatrix} \begin{bmatrix} x y \end{bmatrix} = \begin{bmatrix} 8 3 \end{bmatrix}$

Augmented matrix:
$\left[ \begin{array}{cc|c} 2 & 3 & 8 5 & -1 & 3 \end{array} \right$ ]

2. Gaussian Elimination Method

Gaussian elimination is a systematic method for solving systems of equations by transforming the augmented matrix into row echelon form (REF) and then solving for the variables using back-substitution.

Steps of Gaussian Elimination

Convert the augmented matrix into an upper triangular (row echelon) form by using row operations:
- Swap rows if needed.
- Multiply a row by a nonzero constant.
- Add or subtract a multiple of one row from another.
Back-substitution to find the solution.

Example 1: Solving a System using Gaussian Elimination

Solve the system:
$\begin{aligned} 2x + y - z &= 3 4x - 6y &= 2 -2x + 7y + 2z &= 5 \end{aligned}$

Step 1: Convert to Augmented Matrix

$\left[ \begin{array}{ccc|c} 2 & 1 & -1 & 3 4 & -6 & 0 & 2 -2 & 7 & 2 & 5 \end{array} \right$ ]

Step 2: Make the First Pivot 1

Divide row 1 by 2: $\left[ \begin{array}{ccc|c} 1 & 0.5 & -0.5 & 1.5 4 & -6 & 0 & 2 -2 & 7 & 2 & 5 \end{array} \right$ ]

Step 3: Eliminate First Column Below Pivot

Replace row 2 by subtracting 4 times row 1:
Replace row 3 by adding 2 times row 1:

$\left[ \begin{array}{ccc|c} 1 & 0.5 & -0.5 & 1.5 0 & -8 & 2 & -4 0 & 8 & 1 & 8 \end{array} \right$ ]

Step 4: Make Second Pivot 1

Divide row 2 by -8:

$\left[ \begin{array}{ccc|c} 1 & 0.5 & -0.5 & 1.5 0 & 1 & -0.25 & 0.5 0 & 8 & 1 & 8 \end{array} \right$ ]

Step 5: Eliminate Second Column Below Pivot

Replace row 3 by subtracting 8 times row 2:

$\left[ \begin{array}{ccc|c} 1 & 0.5 & -0.5 & 1.5 0 & 1 & -0.25 & 0.5 0 & 0 & 3 & 4 \end{array} \right$ ]

Step 6: Back Substitution

Solve from bottom to top:

$3z = 4 \Rightarrow z = \frac{4}{3}$
$y - 0.25z = 0.5 \Rightarrow y = 0.5 + 0.25(4/3) = \frac{7}{6}$
$x + 0.5y - 0.5z = 1.5 \Rightarrow x = 1.5 - 0.5(7/6) + 0.5(4/3) = \frac{11}{6}$

Solution: $x = \frac{11}{6}, \quad y = \frac{7}{6}, \quad z = \frac{4}{3}$

3. Solution Types

Unique Solution: A single solution for $x_1, x_2, \dots$ .
Infinite Solutions: Many solutions (system is dependent).
No Solution: Inconsistent system.

4. Homogeneous vs. Non-Homogeneous Systems

Homogeneous system: Ax=0
- Always has at least the trivial solution $\mathbf{x} = 0$ .
- May have infinitely many solutions if the determinant of $A$ is zero.
Non-homogeneous system: Ax=b
- May have unique, infinite, or no solutions.

Conclusion

Gaussian elimination is a powerful method to solve systems.
The number of solutions depends on the rank of the matrix.
Homogeneous systems always have a trivial solution.

Let me know if you need more examples or explanations! 🚀

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