Matrix Algebra

Date: 06/01/98 at 06:32:02
From: Andrea Quintiliani
Subject: a problem in matrix algebra

Hello, I've been trying to help my son with his school work, and have 
come upon a problem that I've not been able to solve. I'm a graduate 
in physics and have asked other colleagues, but nobody seems to come 
up with a correct solution (everybody got rusty since University!). I 
also searched your database with no success.

The problem is this:

Let A, B, C, X be matrices, for instance 2x2 matrices. Calculate X in 
the expression:

   A*X - X*B = C

For example:

   [1  2] * X - X * [2  2] = [2  4]
   [2  1]           [-1 0]   [0 -1]

If it were:

   A*X + B*X = C

then (A + B) * X = C, and setting A + B = D, D!= inverse matrix of D,
we have, D!*D*X = D!*C. Thus:

   I*X = X = D!*C     where I = identity matrix

But, given also the non-commutability of matrix products, I guess I 
cannot apply this grouping technique to the problem at hand. I am 
really curious about finding a solution to this problem and hope that 
somebody will find the time to help me.

Thank you very much,
Andrea Quintiliani

Date: 06/01/98 at 11:19:47
From: Doctor Anthony
Subject: Re: a problem in matrix algebra

Using your values in the equation A*X - X*B = C  we get:

   [1   2]*[a   b]  -  [a   b]*[ 2   2] =  [2    4]
   [2   1] [c   d]     [c   d] [-1   0]    [0   -1]

   [a+2c    b+2d] - [2a-b   2a]  =  [2   4]
   [2a+c    2b+d]   [2c-d   2c]     [0   1]

   [-a+b+2c   -2a+b+2d]  =  [2   4]
   [ 2a-c+d    2b-2c+d]     [0   1]

So we get four equations:

    -a +   b +  2c + 0*d = 2
   -2a +   b + 0*c + 2*d = 4
    2a + 0*b -   c +   d = 0
   0*a +  2b -  2c +   d = 1

And we obtain  a = -2/5, b = 2/5, c = 3/5, d = 7/5; so the matrix X 
is:

   X =  (1/5)[-2    2]
             [ 3    7]   

-Doctor Anthony, The Math Forum
Check out our web site! http://mathforum.org/dr.math/