Dependent and Independent Linear Systems • Oscar's Blog

Dependent and Independent Systems

\begin{array}{cccc} % cccc specifies that there are four centered columns % First Matrix \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 3 & 2 & 3 \end{bmatrix} & % Second Matrix \begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & 2 \\ 0 & 0 & -1 \end{bmatrix} & % Third Matrix \begin{bmatrix} 1 & 1 & 1 \\ 0 & 2 & 2 \\ 0 & 0 & 3 \end{bmatrix} & % Fourth Matrix \begin{bmatrix} 1 & 2 & 5 \\ 0 & 3 & -2 \\ 2 & 4 & 10 \end{bmatrix} \\ \\ % Titles 3r{1} + 2r_{2} = r_{3} & r{1} - r_{2} = r_{3} & \textbf{No relation} & 2r_{1} = r_{3} \\ \text{Dependent} & \text{Dependent} & \text{Independent} & \text{Dependent} \\ \text{Singular} & \text{Singular} & \text{Non Singular} & \text{Singular} \\ \end{array}

Singular: Infinite or No solutions.

Singular: Redundant or Contradictory.

Non Singular: Unique solution / Complete system.

U = \begin{pmatrix} \textcolor{magenta}{a_{11}} & a_{12} & a_{13} & \cdots & a_{1n} \\ 0 & \textcolor{magenta}{a_{22}} & a_{23} & \cdots & a_{2n} \\ 0 & 0 & \textcolor{magenta}{a_{33}} & \cdots & a_{3n} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & \textcolor{magenta}{a_{nn}} \end{pmatrix}

Its determinant is defined as the product of its diagonal elements.

\det(U) = \begin{vmatrix} \textcolor{magenta}{a_{11}} & a_{12} & a_{13} & \cdots & a_{1n} \\ 0 & \textcolor{magenta}{a_{22}} & a_{23} & \cdots & a_{2n} \\ 0 & 0 & \textcolor{magenta}{a_{33}} & \cdots & a_{3n} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & \textcolor{magenta}{a_{nn}} \end{vmatrix} = a_{11} \cdot a_{22} \cdot a_{33} \cdot \ldots \cdot a_{nn} = \prod_{i=1}^{n} a_{ii}

For the lower triangular matrix, the determinant is the same as the upper triangular matrix.

\begin{equation} \left\{ \begin{array}{rcl} -x_{1} + 3x_{2} & = & 7 \\ 3x_{1} + 2x_{2} & = & 1 \end{array} \right. \end{equation}

A = \begin{bmatrix} -1 & 3 \\ 3 & 2 \end{bmatrix} , b = \begin{bmatrix} 7 \\ 1 \end{bmatrix}

A = np.array([
    [-1, 3],
    [3, 2]
], dtype=np.dtype(float))
b = np.array([7, 1], dtype=np.dtype(float))
x = np.linalg.solve(A, b)

Output (Only has one solution):

[-1.  2.]

np.linalg is a numpy module related to linear algebra and matrix operations.

The matrix $A$ is non-singular because its determinant is not zero.

\det(A) = -1 \cdot 2 - 3 \cdot 3 = -2 - 9 = -11 \ne 0

det_A = np.linalg.det(A)
print(f"Determinant of matrix A: {d:.2f}")

Output:

Determinant of matrix A: -11.00