07. Solution Sets of Linear Systems

Dated: 12-03-2025

Solution Set

A solution of a linear system¹ is an assignment of values to the variables \(x_1, x_2, \ldots, x_n\) such that each of the equations in the linear system is satisfied.
The set of all possible solutions is called the solution set.

Homogeneous Linear System

A linear system¹ is homogeneous if it can be written in the form \(A \vec x = \vec 0\) where \(A\) is a \(m \times n\) matrix² and \(\vec 0\) is a null vector³ in \(\mathbb R^m\).

Trivial Solution

A homogeneous system \(A\vec x = \vec 0\) always has at least one solution, namely, \(\vec x = \vec 0\).
This zero solution is usually called the trivial solution of the homogeneous system.

Non Trivial Solution

A solution other than trivial solution is called non trivial solution.
Meaning the solution to \(A \vec x = \vec 0\) such that \(\vec x \ne \vec 0\) is called non trivial solution.

Existence and Uniqueness Theorem

The homogeneous equation \(A \vec x = \vec 0\) has a nontrivial solution if and only if the equation has at least one free variable.

Geometric Interpretation

Geometrically, the solution set is a line⁴ through \(\vec 0\) in \(\mathbb R^3\).

Parametric Vector Form of the Solution

The equation

\[\vec x = s \vec u + t \vec v \quad (s, t \in \mathbb R)\]

is called a parametric vector equation of the plane.⁵

Solution of Non Homogeneous Systems

If a non homogeneous system has many solutions, the general solution can be written in the parametric vector form as the sum of

One particular solution vector³
Any arbitrary linear combination⁶ of vectors³ which satisfy the corresponding homogeneous system.

\[\vec x = \vec p + t \vec v\]

Example

Describe all solutions of \(A \vec x = \vec b\), where

\[ A = \begin{bmatrix} 3 & 5 & -4 \\ -3 & -2 & 4 \\ 6 & 1 & -8 \end{bmatrix} \]

\[ \vec b = \begin{bmatrix} 7 \\ -1 \\ -4 \end{bmatrix} \]

Solution

Take the augmented matrix² \(\begin{bmatrix}A & \vec b\end{bmatrix}\).

\[ \begin{bmatrix} 3 & 5 & -4 & 7 \\ -3 & -2 & 4 & -1 \\ 6 & 1 & -8 & -4 \end{bmatrix} \]

\[ \Big \downarrow R_2 = R_1 + R_2 \]

\[ \Big \downarrow R_3 = -2R_1 + R_3 \]

\[ \sim \begin{bmatrix} 3 & 5 & -4 & 7 \\ 0 & 3 & 0 & 6 \\ 0 & -9 & 0 & -18 \end{bmatrix} \]

\[ \Big \downarrow R_3 = 3R_2 + R_3 \]

\[ \Big \downarrow R_2 = \frac 1 3 R_2 \]

\[ \sim \begin{bmatrix} 3 & 5 & -4 & 7 \\ 0 & 1 & 0 & 2 \\ 0 & 0 & 0 & 0 \end{bmatrix} \]

\[ \Big \downarrow R_1 = -5R_2 + R_1 \]

\[ \Big \downarrow R_1 = \frac 1 3 R_1 \]

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

\[ \begin{array}{cccccc} &x_1 & &- \frac 4 3 x_3 &= &-1 \\ & &x_2 & &= &2 \\ & & & 0 &= &0 \end{array} \]

Thus \(x_1 = -1 + \frac 4 3 x_3, x_2 = 2\) and \(x_3\) is free.

\[ x = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} -1 + \frac{4}{3}x_3 \\ 2 \\ x_3 \end{bmatrix} = \begin{bmatrix} -1 \\ 2 \\ 0 \end{bmatrix} + \begin{bmatrix} \frac{4}{3}x_3 \\ 0 \\ x_3 \end{bmatrix} = 1 \begin{bmatrix} -1 \\ 2 \\ 0 \end{bmatrix} + x_3 \begin{bmatrix} \frac 4 3 \\ 0 \\ 1 \end{bmatrix} \]

Therefore,

\[\vec x = \vec p + x_3 \vec v\]

Theorem

Suppose \(A \vec x = \vec b\) is consistent¹ for some given \(\vec b\) and let \(\vec p\) be a solution.
Then the solution set for \(A \vec x = \vec b\) is set⁶ of all vectors³ of the form

\[\vec w = \vec p + \vec {v_h}\]

where \(\vec{v_h}\) is the solution of the associated homogeneous equation \(A \vec x = \vec 0\).

Steps of Writing a Solution Set (of a Consistent System) in a Parametric Vector Form

Row reduces the augmented matrix² to reduced echelon form.²
Express each basic variable in terms of any free variables appearing in an equation.
Write a typical solution \(x\) as a vector³ whose entries depend on the free variables if any.
Decompose \(x\) into a linear combination¹ of vectors³ (with numeric entries) using the free variables as parameters.