Linear Equations in Linear Algebra

1. System of Linear Equations

A system of linear equations consists of two or more linear equations involving the same set of variables. A solution to the system is a set of values for the variables that satisfy all equations simultaneously.

General Form

A system of \(m\) linear equations in \(n\) variables can be written as:

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

where:

• \(a_{ij}\) are the coefficients of the variables \(x_1, x_2, \ldots, x_n\).
• \(b_i\) are the constants.

Example

Solve the system of equations:

\[ \begin{aligned} 2x + 3y &= 8 \\ 4x - y &= 2 \end{aligned} \]

Solution

1. Multiply the second equation by 3 to eliminate \(y\):

\[ 12x - 3y = 6 \]

1. Add to the first equation:

\[ (2x + 3y) + (12x - 3y) = 8 + 6 \\ 14x = 14 \\ x = 1 \]

1. Substitute \(x = 1\) into the first equation:

\[ 2(1) + 3y = 8 \\ 3y = 6 \\ y = 2 \]

Thus, the solution is \(x = 1, y = 2\).

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2. Row Reduction and Echelon Forms

Row reduction transforms a matrix into a simpler form to solve linear systems. The two key forms are:

Row Echelon Form (REF)

1. Each leading entry in a row is to the right of the leading entry in the row above.
2. All entries below a leading entry are zero.

Reduced Row Echelon Form (RREF)

1. The matrix is in REF.
2. Each leading entry is 1.
3. Each leading 1 is the only nonzero entry in its column.

Steps for Gaussian Elimination

1. Write the augmented matrix of the system.
2. Use row operations to achieve REF.
3. (Optional) Continue to RREF for simplicity.

Example

Solve the system:

\[ \begin{aligned} x + y + z &= 6 \\ 2y + 5z &= -4 \\ 2x + 5y - z &= 27 \end{aligned} \]

Augmented Matrix:

\[ \begin{bmatrix} 1 & 1 & 1 & | & 6 \\ 0 & 2 & 5 & | & -4 \\ 2 & 5 & -1 & | & 27 \end{bmatrix} \]

Perform row reduction to solve:

\[ \begin{bmatrix} 1 & 1 & 1 & | & 6 \\ 0 & 2 & 5 & | & -4 \\ 0 & 3 & -3 & | & 15 \end{bmatrix} \rightarrow \begin{bmatrix} 1 & 1 & 1 & | & 6 \\ 0 & 1 & 5/2 & | & -2 \\ 0 & 0 & -15/2 & | & 21 \end{bmatrix} \rightarrow \begin{bmatrix} 1 & 0 & 0 & | & 1 \\ 0 & 1 & 0 & | & 2 \\ 0 & 0 & 1 & | & -1 \end{bmatrix} \]

Solution: \(x = 1, y = 2, z = -1\).

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3. Vector Equations

A vector equation is an equation involving vectors and their linear combinations.

General Form

\[ c_1\mathbf{v}_1 + c_2\mathbf{v}_2 + \cdots + c_n\mathbf{v}_n = \mathbf{b} \]

Example

Given:

\[ \mathbf{v}_1 = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}, \mathbf{v}_2 = \begin{bmatrix} 4 \\ 5 \\ 6 \end{bmatrix}, \mathbf{b} = \begin{bmatrix} 7 \\ 8 \\ 9 \end{bmatrix} \]

Find scalars \(c_1\) and \(c_2\) such that:

\[ c_1\mathbf{v}_1 + c_2\mathbf{v}_2 = \mathbf{b} \]

Write as a system:

\[ \begin{aligned} 1c_1 + 4c_2 &= 7 \\ 2c_1 + 5c_2 &= 8 \\ 3c_1 + 6c_2 &= 9 \end{aligned} \]

Solve using row reduction or substitution to find \(c_1\) and \(c_2\).

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4. The Matrix Equation \(A\mathbf{x} = \mathbf{b}\)

The matrix equation represents a system of linear equations compactly:

\[ A\mathbf{x} = \mathbf{b} \]

where:

• \(A\) is the coefficient matrix.
• \(\mathbf{x}\) is the vector of variables.
• \(\mathbf{b}\) is the vector of constants.

Example

\[ A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}, \mathbf{x} = \begin{bmatrix} x_1 \\ x_2 \end{bmatrix}, \mathbf{b} = \begin{bmatrix} 5 \\ 6 \end{bmatrix} \]

\[ \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}\begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} 5 \\ 6 \end{bmatrix} \]

Solve using matrix operations:

\[ \mathbf{x} = A^{-1}\mathbf{b} \]

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5. Applications of Linear Systems

1. Physics: Modeling forces and motion.
2. Economics: Solving input-output models.
3. Computer Graphics: Transformations and projections.
4. Engineering: Circuit analysis.

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6. Linear Independence

Vectors \(\mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_n\) are linearly independent if the only solution to:

\[ c_1\mathbf{v}_1 + c_2\mathbf{v}_2 + \cdots + c_n\mathbf{v}_n = \mathbf{0} \]

is \(c_1 = c_2 = \cdots = c_n = 0\).

Example

Are the vectors:

\[ \mathbf{v}_1 = \begin{bmatrix} 1 \\ 0 \end{bmatrix}, \mathbf{v}_2 = \begin{bmatrix} 0 \\ 1 \end{bmatrix} \]

linearly independent?

Solution:

\[ c_1\begin{bmatrix} 1 \\ 0 \end{bmatrix} + c_2\begin{bmatrix} 0 \\ 1 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \]

Results in \(c_1 = 0, c_2 = 0\). Hence, the vectors are independent.

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Code Example

import numpy as np

# Solve Ax = b
A = np.array([[1, 2], [3, 4]])
b = np.array([5, 6])

x = np.linalg.solve(A, b)
print("Solution:", x)

Solution: [-4. 4.5]