Lecture 05: Solving triangular systems

Lecture 05: Solving triangular systems#

A general lower triangular system of equations has \(a_{ij} = 0\) for \(j > i\) and takes the form:

\[\begin{split} \begin{pmatrix} a_{11} & 0 & 0 & \cdots & 0 \\ a_{21} & a_{22} & 0 & \cdots & 0 \\ a_{31} & a_{32} & a_{33} & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ a_{n1} & a_{n2} & a_{n3} & \cdots & a_{nn} \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3 \\ \vdots \\ x_n \end{pmatrix} = \begin{pmatrix} b_1 \\ b_2 \\ b_3 \\ \vdots \\ b_n \end{pmatrix}. \end{split}\]

Note the first equation is

\[ a_{11} x_1 = b_1. \]

The \(x_i\) can be found by calculating

\[ x_i = \frac{1}{a_{ii}} \left(b_i - \sum_{j=1}^{i-1} a_{ij} x_j \right) \]

for each row \(i = 1, 2, \ldots, n\) in turn.

  • Each calculation requires only previously computed values \(x_j\) (and the sum gives a loop for \(j < i\).

  • The matrix \(A\) must have nonzero diagonal entries
    i.e. \(a_{ii} \neq 0\) for \(i = 1, 2, \ldots, n\).

  • Upper triangular systems of equations can be solved in a similar manner.

Example 1#

Solve the lower triangular system of equations given by

\[\begin{split} \begin{aligned} 2 x_1 && && &= 2 \\ x_1 &+& 2 x_2 && &= 7 \\ 2 x_1 &+& 4 x_2 &+& 6 x_3 &= 26 \end{aligned} \end{split}\]

or, equivalently,

\[\begin{split} \begin{pmatrix} 2 & 0 & 0 \\ 1 & 2 & 0 \\ 2 & 4 & 6 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = \begin{pmatrix} 2 \\ 7 \\ 26 \end{pmatrix}. \end{split}\]

Example 1: solution#

The solution can be calculated systematically from

\[\begin{split} \begin{aligned} x_1 &= \frac{b_1}{a_{11}} = \frac{2}{2} = 1 \\ x_2 &= \frac{b_2 - a_{21} x_1}{a_{22}} = \frac{7 - 1 \times 1}{2} = \frac{6}{2} = 3 \\ x_3 &= \frac{b_3 - a_{31} x_1 - a_{32} x_2}{a_33} = \frac{26 - 2 \times 1 - 4 \times 3}{6} = \frac{12}{6} = 2 \end{aligned} \end{split}\]

which gives the solution \(\vec{x} = (1, 3, 2)^T\).

Example 2 (homework)#

Solve the upper triangular linear system given by

\[\begin{split} \begin{aligned} 2 x_1 &+& x_2 &+& 4 x_3 &=& 12 \\ && 1.5 x_2 && &=& 3 \\ && && 2 x_3 &=& 4 \end{aligned}. \end{split}\]

Notes#

  • It is simple to solve a lower (upper) triangular system of equations (provided the diagonal is nonzero).

  • This process is often referred to as forward (backward) substitution.

  • A general system of equations (i.e. a full matrix \(A\)) can be solved rapidly once it has been reduced to upper triangular form.

  • This will be the topic of the next section…

Further reading#

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