# Free and Basic Variables of a Matrix - Examples with Solutions

   

## Free and Basic Variables of a Matrix Representing a System of Equations

Definition For a system of equations in row echelon form , which of course may be represented by the augmented matrix, a variable whose coefficient is a leading 1 ( pivot ) is called a basic variable and a varibale without pivot is called a free variable. Let us consider the following system of equations in row echelon form
\quad \quad \begin{align*} x_1 + 2x_2 - x_3 + x_4 - 3 x_5 = 0 \\ x_3 - 2 x_5 = 0 \\ x_4 - x_5 = 0 \end{align*}
The augmented matrix in row echelon of the above system is as follows
$$\quad \quad \begin{bmatrix} \color{red}{1} & 2 & - 1 & 1 & -3 & 0\\ 0 & 0 & \color{red}{1} & 0 & -2 & 0\\ 0 & 0 & 0 & \color{red}{1} & -1 & 0 \end{bmatrix}$$

According fo the above definition
$$\quad \quad x_1, x_3$$ and $$x_4$$ are the basic variables and $$x_2$$ and $$x_5$$ are the free variables.
When we solve the above system, we express the basic variables in terms of the free variables
The third equation in the system gives
$$\quad \quad x_4 = x_5$$
The second equation gives
$$\quad \quad x_3 = 2 x_5$$
The first equation gives
$$\quad \quad x_1 = - 2 x_2 + x_3 - x_4 + 3 x_5$$
Substitute the basic variables on the right
$$\quad \quad x_1 = - 2 x_2 + 2 x_5 - x_5 + 3 x_5$$
Simplify
$$\quad \quad x_1 = - 2 x_2 + 4 x_5$$
The solution is written as
$$\quad \quad x_1 = - 2 x_2 + 4 x_5$$
$$\quad \quad x_3 = 2 x_5$$
$$\quad \quad x_4 = x_5$$
where $$x_2$$ and $$x_5$$ can be any real numbers hence their names as "free variables".
Definition The use of free variables helps us to write an explicit formula for the solutions of our system.

## Questions with Solution

For each of the following augmented matrices in row echelon form, which are basic variables and which are free variables?
1. $$\begin{bmatrix} 1 & 4 & 3 & 0 & 0\\ 0 & 0 & 1 & -3 & 1\\ 0 & 0 & 0 & 0 & 0 \\ \end{bmatrix}$$

2. $$\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ \end{bmatrix}$$

3. $$\begin{bmatrix} 1 & 2 & 0 & -1 & -2 & 0\\ 0 & 1 & 1 & 0 & 3 & 0\\ 0 & 0 & 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{bmatrix}$$

4. $$\begin{bmatrix} 1 & 2 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 0\\ \end{bmatrix}$$

Solutions to the Above Questions
Being augmented matrices, the number of variables is equal to the number of columns of the given matrix -1.
For examples, for a matrix of 5 columns, the number of variables is 5 - 1 = 4, named as $$x_1$$, $$x_2$$, $$x_3$$ and $$x_4$$.

1. Matrix 1 is has two pivots and 4 variables.
The first pivot at row 1 column 1; hence $$x_1$$ is a basic variable.
The second pivot is at row 2 column 3; hence $$x_3$$ is also a basic variable.
The remaining variables: $$x_2$$ and $$x_4$$ are free variables.

2. Matrix 2 is has two Pivots and 2 variables.
The first pivot is at row 1 column 1; hence hence $$x_1$$ is a basic variable.
The second pivot is at row 2 column 2; hence $$x_2$$ is a basic variable.
There are no free variables.

3. Matrix 3 has 3 pivots and 5 variables.
The first pivot at row 1 column 1; hence $$x_1$$ is a basic variable.
The second pivot is at row 2 column 2; hence $$x_2$$ is a basic variable.
The third pivot is at row 3 column 4; hence $$x_4$$ is a basic variable.
The remaining variables: $$x_3$$ and $$x_5$$ are free variables.

4. Matrix 4 has two Pivots and 3 variables.
The first pivot is at row 1 column 1; hence hence $$x_1$$ is a basic variable.
The second pivot is at row 2 column 3; hence $$x_3$$ is a basic variable.
The remaining variables: $$x_2$$ is a free variables.