Simple Matrix Question 1

[haynewp]

Why do you have to put in a matrix to the power of (-1) and multiply with another matrix in order to divide?

For instance, I am trying to divide a vector with 1 column and 3 rows by a matrix with 3 columns and 3 rows and I get the error that these array dimensions don't match. But I am able to do this on my calculator and get the right answer. In Mathcad I have to multiply the matrices and put one to the power of (-1) in order to divide.

 

[davidbeach]

Basic rules of matrix arithmetic.

 

[pstuckey]

It has to with the fact that matrix multiplication is not commutative.

for real numbers x,y:
x/y = x*(y^-1) = (y^-1)*x

this is not true for matrixes
x*(y^-1) does NOT equal (y^-1)*x

so there is a potential ambiguity when talking about division with matrices as to whether x/y is x*(y^-1) or (y^-1)*x.

textbooks generally avoid this ambiguity by always writing matrix division as multiplication (using the inverse). some math software define two divison operators for matrices '/' and '\' to handle the two cases above seperately.

Mcad apparently uses the definition x/y = x*(y^-1), and in your case i believe the quantity you are solving for is (y^-1)*x, which in Mcad requires multiplication as you described.

Peter

 

[haynewp]

Thanks, I haven't worked with matrices in 10 years and have forgotten the basics. I found in my HP49 manual describing the way it support "division" of matrices.