Once a matrix has been decomposed into lower and upper triangular parts it is possible to obtain the solution to in a direct way. The procedure can be summarised as follows
- Given , find and so that . Hence .
- Let so that . Solve this triangular system for .
- Finally solve the triangular system for .
The benefit of this approach is that we only ever need to solve triangular systems. The cost is that we have to solve two of them.
[Here we solve only small systems; a large system is presented in Engineering Example 1 on page 62.]
Find the solution of of the system
The first step is to calculate the
decomposition of the coefficient matrix on the left-hand side. In this case that job has already been done since this is the matrix we considered earlier. We found that
The next step is to solve
for the vector
. That is we consider
which can be solved by forward substitution . From the top equation we see that . The middle equation states that and hence . Finally the bottom line says that from which we see that .
Now that we have found
we finish the procedure by solving
. That is we solve
by using back substitution . Starting with the bottom equation we see that so clearly . The middle equation implies that and it follows that . The top equation states that and consequently .
Therefore we have found that the solution to the system of simultaneous equations
Use the decomposition you found earlier in the last Task (page 24) to solve
We found earlier that the coefficient matrix is equal to .
First we solve for , we have
The top line implies that . The middle line states that and therefore . The last line tells us that and therefore .
Finally we solve for , we have
The bottom line shows that . The middle line then shows that , and then the top line gives us that . The required solution is .