systems of equations

Homogeneous System of Equations

A system of equations is called homoegenous if each of the contstant terms is equal to 0. A homogeneous system therefore has the form:

a_{11} x_{1} + a_{12} x_{2} + \dots + a_{1 n} x_{n} & = 0 a_{12} x_{1} + a_{22} x_{2} + \dots + a_{2 n} x_{n} & = 0 ⋮ a_{m 1} x_{1} + a_{m 2} x_{2} + \dots + a_{m n} x_{n} & = 0

where

a_{i j}

are coefficients and

x_{j}

are variables

Encoding a system of Equations into Matrix Vector Multiplication

The solution set of the following system of equations is the same as the solution set for

A x = b

Solution space of a homogeneous system of equations

Given a homogenous system of equations embedded in

A x = 0

, where

A

is an

m \times n

matrix. We Define the solution space to be

S = {x \in ℝ^{n} : A x = 0}

The solution space of a homogeneous system of equations forms a subspace

ℝ^{n}

Basis for a Solution Space

Find a basis for the solution space of the following matrix encoded system of equations

[\begin{matrix} 2 & 4 & - 2 & 0 \\ 3 & 6 & - 3 & 0 \\ 5 & 8 & - 8 & 1 \end{matrix}] \vec{v} = \vec{0}

The matrix can be row reduced to the following form:

[\begin{matrix} 1 & 0 & - 4 & 1 \\ 0 & 1 & \frac{3}{2} & - \frac{1}{2} \\ 0 & 0 & 0 & 0 \end{matrix}]

Which represents the following system of equations

1 x + 0 y + - 4 z + 1 w & = 0 0 x + 1 y + \frac{3}{2} z - \frac{1}{2} w & = 0 0 x + 0 y + 0 z + 0 w & = 0

Since $z$ and $w$ are free variables, we let $s, t \in ℝ$ and set $z = s$ and $w = t$ , now using back substitution, we can deduce the following

0 x + 1 y + \frac{3}{2} z - \frac{1}{2} t = 0 & ⟹ y = \frac{1}{2} t - \frac{3}{2} s 1 x + 0 y + - 4 z + 1 w = 0 & ⟹ x = 4 s - t

Therefore the solution set is given by

{[\begin{matrix} 4 s - t \\ - \frac{3}{2} s + \frac{1}{2} t \\ s \\ t \end{matrix}] : s, t \in ℝ} = {s [\begin{matrix} 4 \\ - \frac{3}{2} \\ 1 \\ 0 \end{matrix}] + t [\begin{matrix} - 1 \\ \frac{1}{2} \\ 0 \\ 1 \end{matrix}] : s, t \in ℝ}

It can be confirmed that the two vectors $\vec{v_{1}} = [\begin{matrix} - 1 \\ \frac{1}{2} \\ 0 \\ 1 \end{matrix}]$ and $\vec{v_{2}} = [\begin{matrix} 4 \\ - \frac{3}{2} \\ 1 \\ 0 \end{matrix}]$ are linearly independent due to the non-matching zero in the third component, thus the solution space can be written as $span ({\vec{v_{1}}, \vec{v_{2}}})$ , therefore a possible basis for the solution set is $(\vec{v_{1}}, \vec{v_{2}})$ .

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