# Staircase Linear Equation Systems

A linear equation system is called a **staircase system** (echelon form) if it can be written in this format.

The number of unknowns in each equation decreases progressively from the first to the last equation of the system.

The **number of steps** in the linear system is equal to the matrix rank.

This approach simplifies the calculation of the rank, even when the system consists of many equations and variables.

**Why are staircase systems useful?**

Staircase linear systems are particularly easy to solve because the values of the unknowns can be determined starting from the last equation.

If a linear system is not in staircase form, it can be transformed into one using the Gauss-Jordan elimination method.

**Note**. For a system to be a staircase, each step must descend by one row. If the matrix descends by two or more rows, it is not a staircase matrix.

If a step spans two variables (columns), one of the variables should be converted into a parameter of the linear system.

## A Practical Example

**Example 1**

This linear equation system is a staircase

The system has 4 steps and n=4 unknown variables.

The complete matrix of the system has a rank equal to r=4 because the system has 4 steps.

The coefficient matrix also has a rank of 4 for the same reason.

Therefore, according to the **Rouché-Capelli Theorem**, the system admits one or more solutions.

In this case, it admits one solution because the rank (r=4) is equal to the number of variables (n=4).

$$ \infty^{n-r} = \infty^{4-4} = \infty^0 = 1 $$

To solve it, start with the last equation 2x_{4}=0, which is immediately resolvable. It's clear that x_{4}=0.

Substitute x_{4}=0 into the third equation to get 4x_{3}=4, hence x_{3}=1.

Substitute x_{3}=1 and x_{4}=0 into the second equation to get 2x_{2}-6·1-5·0=-2, hence x_{2}=2.

Finally, substitute x_{2}=2, x_{3}=1, and x_{4}=0 into the first equation to get x_{1}+2·2+3·1+4·0=1, hence x_{1}=-6.

The system is solved in a few steps.

The solutions of the system are x_{1}=-6, x_{2}=2, x_{3}=1, and x_{4}=0.

**Note**. This is why it's very useful to transform a linear equation system into a staircase system. Solving the system becomes much simpler, even if it consists of many equations and unknown variables.

**Example 2**

This linear equation system is also in a stepwise form.

The full matrix of the system has a rank equal to r=3 because the system has 3 steps.

The coefficient matrix also has a rank equal to r=3 for the same reason.

According to the **Rouché–Capelli Theorem**, the system has one or infinitely many solutions.

In this case, the system has infinitely many solutions because the rank (r=3) is less than the number of variables (n=4).

$$ \infty^{n-r} = \infty^{4-3} = \infty^1 = \infty $$

To find the solutions, I must **parameterize the system**.

For example, in the last equation, I express variable x_{3} using variable x_{4} as the system's parameter.

Then, I substitute x_{3}=2-x_{4} in the second equation and find x_{2}=-1.

Finally, I substitute x_{2}=-1 and x_{3}=2-x_{4} in the first equation and find x_{1}=1.

The solutions of the system of equations are x_{1}=1, x_{2}=-1, and x_{3}=2-x_{4}, considering x_{4} as a parameter k=x_{4} with k∈R.

$$ \begin{cases} x_1 = 1 \\ x_2 = -1 \\ x_3 = 2 - k \end{cases} $$

Therefore, the system has infinitely many solutions.

## Observations

Some useful observations about stepwise linear systems:

**All stepwise matrices are upper triangular matrices**. However, the converse is not true. Not all upper triangular matrices are also stepwise matrices.**The determinant of a square stepwise matrix is equal to the product of the elements on the main diagonal**. Being a triangular matrix, the determinant is simply calculated by multiplying the elements on the main diagonal because the others are zero. For this reason, calculating the determinant in square stepwise matrices is very straightforward.

And so on.