Example of use of the math::linearalgebra package to solve a question posed on [

1]

This text copied directly from the above reference:

Example:
A man buys 3 fish and 2 chips for £2.80
A woman buys 1 fish and 4 chips for £2.60
How much are the fish and how much are the chips?

Use the linear algebra package thus:

package require math::linearalgebra
set mt [math::linearalgebra::mkMatrix 2 2 0.0] ;# there are 2 transactions buying 2 items
set vc [math::linearalgebra::mkVector 2 0.] ;# 2 transactions
math::linearalgebra::setelem mt 0 0 3 ;# 3 fish in transaction 1...
math::linearalgebra::setelem mt 0 1 2 ;# 2 chips
math::linearalgebra::setelem mt 1 0 1 ;# 1 fish in transaction 2
math::linearalgebra::setelem mt 1 1 4 ;# 4 chips (this is a rather unbalanced meal)
math::linearalgebra::setelem vc 0 2.8 ;# cost of transaction 1
math::linearalgebra::setelem vc 1 2.6 ;# and transaction 2
math::linearalgebra::solveGauss $mt $vc ;# calculate the result

And the result:

0.6 0.500000000001

or the fish cost 0.60 (rather cheap) and the chips 0.50. As given in the reference above.

An even shorter version is:

package require math::linearalgebra
set mt {{3 2} {1 4}}
set vc {2.8 2.6}
math::linearalgebra::solveGauss $mt $vc

Do the simultaneous equations have a solution? Only if the

Matrix determinant is not zero. Can we have a determinant function added to the linearalgebra package please!

Here is a matrix with a zero determinant:

set mt {{3 2} {6 4}}
math::linearalgebra::solveGauss $mt $vc

the stdout output is:

divide by zero

- not very helpful but the equations do not allow a unique solution. The matrix is equivalent to:

3 x + 2 y = 2.8
6 x + 4 y = 2.6

Twice the first equation minus the second equation leaves zero = 3.