Strictly triangular systems of linear equations
Prerequisites
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Coefficient and augmented matrices Show
Strictly triangular systems of linear equations
Introduction
- A system of linear equations is strictly triangular if:
- It is n x n, i.e. the number of equations is equal to the number of variables.
- For equation k, the coefficients for x1 to xk - 1 are 0.
- For equation k, the coefficient for xk is not 0.
- For k in 1, ..., n
- A 3x3 strictly triangular system has the form:

- Or as an augmented matrix:

Solving strictly triangular systems
- Strictly triangular systems are very easy to solve.
- For example:

- From the 3rd equation, we know that:

- So we can substitute this into the second equation:

- And finally substitute x2 and x3 into our first equation:

- This method of solving strictly triangular systems is called back-substitution.
Converting systems to be strictly triangular
- If a system of equations has exactly one solution then it can be converted to a strictly triangular system.
- For example, given the system:

- First, we convert it into an augmented martrix:

- We start by eliminating the coefficients of x for all but the first row:
- To do this, we subtract multiples of the 1st row from the other rows.
- The row that is used for elimination is called the pivotal row.
- We subtract the 1st row from the 2nd:

- And then we subtract 3 times the 1st row from the 3rd row:

- Finally we eliminate the coefficients of y from the 3rd row:
- To do this, we use the 2nd row as our pivotal row.
- We subtract 3 times the 2nd row from the 3rd row:

Choosing a pivotal row
- Sometimes a system will be given like:

- In this case the first row can't be used as a pivotal row to eliminate the coefficients in the first column.
- Instead, it is necessary to switch round the first two rows.