
#1
Nov3012, 06:01 PM

P: 13

Hi guys
I wonder if you know any linear algebra formalism or something to solve the following question systematically? Give A,B with A=n1xn2 matrix and B=n1xm2 matrix. How do we get C=n2xm2 matrix such that A*C=B. A simple example if A=(1,2)^t, B=(2,4)^t, then C=2 The question is how to solve C given A and B. Thanks 



#2
Dec112, 09:09 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,885

That depends upon how many dimensions A and B have. In the example you give the simplest method is the write out the components and solve the resulting system of equations. Since A is "1 by 2" and B is "2 by 1", C must be a 1 by 1 matrix (a single number) and AC= B becomes
[tex]\begin{bmatrix}1 \\ 2\end{bmatrix}\begin{bmatrix}c\end{bmatrix}= \begin{bmatrix}c \\ 2c\end{bmatrix}= \begin{bmatrix}2 \\ 4\end{bmatrix}[/tex] which gives the two equations c= 2 and 2c= 4 so that c= 2. Of course, if B has NOT been a simple multiple of A, there would not have been any solution. 



#3
Dec112, 11:28 AM

Engineering
Sci Advisor
HW Helper
Thanks
P: 6,348

If you think about how matrix multiplication AC is defined, the i'th column of C only affects the i'th column of B.
So you only need to consider the simpler problem where C and B are vectors. This is just a set of linear equations, which may have zero, one, or more than one solution depending on the row and column dimensions of A and the rank of A. The details should be covered in any course on linear algebra, or numerical methods for solving linear equations. 



#4
Dec312, 10:37 AM

P: 13

Solve Linear equations
Thanks AlephZero and HallsofIvy.
I guess the example is too simple. I like to consider the general case. Here is the way I solve it. Want to solve C with AC=B. Multiply A^t on both sides A^t.A.C=A^t.B Now A^t.A^t is a square matrix so I can calculate its inverse (if it is not singular) Then I get C=(A^t.A)^1.A^t.B (1) However if I calculate C in this way, something goes wrong! For example, Say A=[1,2]^t, B=[[1,0,2],[2,1,0]] By(1) C=1/5[5,2,2] But A.C=1/5[[5,2,2],[10,4,4]]\=B. So I wonder what is wrong with (1). Why does (1) get correct C such that A.C=B? 


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