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This example uses various linear algebra techniques to demonstrate the following Derive routines:

identity matrix; append; append columns; row reduce; dimension; column extraction; determinant; transpose; as well as more than one argument (including text) in the return statement. As well as producing a solution, the following method also produces the inverse of the co-efficient matrix.

Given a set of simultaneous linear equations, how do we find the solutions, assuming they exist? There are many techniques, some more efficient than others, and some more complex than others. Derive already has numerous in-built functions to solve such problems. The following routine primarily demonstrates how to append two matrices/vectors (i.e. join them together), as well as how to extract one or more columns from a matrix (i.e. producing another vector/matrix).

Without entering into the detailed mathematics, there is a straightforward method of calculating
a) the inverse of the matrix of co-efficients,
b) the solution vector, and
c) both! 

The method involves augmentation, which means 'adding on' or 'appending' (as in Derive-speak) two vectors or matrices. Suppose we have the following set of linear simultaneous equations:

In matrix notation, the co-efficients, variables, and solution vectors are given by:

and identity matrix (dimension 3) is given by

i(n):=identity_matrix(n)

So

The general form of the solution is given by 

    a x = c,

and to use 'augmented' methods to calculate the solution, we append the rhs onto the matrix a (written as a|c):    

We now row reduce this 3x4 matrix, to produce an identity matrix in the first 3 columns, and the solution in the last column. In Derive, to create this new 3x4 matrix, the append routine is called. However, the append function normally only applies to vectors (one-dimensional matrices), so we need to trick the function into thinking we have two vectors. We transpose the original matrix and rhs vector, then append them, then transpose the result back again:

a:= append(a`,c`)`

where the 'prime' mark "`" represents transpose.

If we now row_reduce this 3x4 matrix:

a:= row_reduce(a) and simplify: 

The first 3 columns represent a 3x3 identity matrix, and the last column is the solution vector. To extract this solution vector, we can use the col command:

solution:=a col [4], which simplifies to:

Using a similar procedure to calculate the inverse of the co-efficients, we append the identity matrix onto this matrix. The new (3x6) matrix is similarly row_reduced, and the right-hand 'half' of this new matrix is the inverse. So:

a:= append(a`,identity_matrix(3)`)`

and a:= row_reduce(a) gives

(As a check, by entering a^(-1) on the entry line, where a is the original matrix, gives the same result).  To extract the inverse, we can use 

inverse:= a col [4,...,6]

where [4,...,6] represents an inclusive range of columns, remembering to use just three ellipses:

So, maintaining the original 

    and   

 and calling the function gauss2(a,c) gives