Register to reply

Solve Linear equations

by phynewb
Tags: equations, linear, solve
Share this thread:
Nov30-12, 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.
Phys.Org News Partner Science news on
What lit up the universe?
Sheepdogs use just two simple rules to round up large herds of sheep
Animals first flex their muscles
Dec1-12, 09:09 AM
Sci Advisor
PF Gold
P: 39,533
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.
Dec1-12, 11:28 AM
Sci Advisor
HW Helper
P: 7,150
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.

Dec3-12, 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?

Register to reply

Related Discussions
Solve linear equations using simplex method Calculus 2
Using Characteristic equations to solve 2nd order linear DEQ's Calculus & Beyond Homework 1
Non linear system of 4 equations, how to solve it? Calculus & Beyond Homework 11
Using Linear Algebra to solve systems of non-linear equations Calculus 9