Give examples of matrices A,B and C, such that AB=AC Where A is non-zero matrix and B is not equal to C.
Answers
Step-by-step explanation:
Lots of answers here, but I think there are still some more things worth saying.
It has been noted that AB=ACAB=AC is equivalent to A(B−C)=[0]A(B−C)=[0] . Of course, BB and CC can be anything if A=[0]A=[0] , so let’s assume AA is not the zero matrix.
We do not assume that the matrices in question are square!
An easy situation in which we can conclude B=CB=C is if AA has a left inverse LL . Since AA is not assumed square, left inverse and right inverse are not necessarily the same thing and we don’t dare write L=A−1L=A−1 ! Still, applying LL to both sides of A(B−C)=[0]A(B−C)=[0] gives B−C=[0]B−C=[0] and so B=CB=C . AA having a left inverse is equivalent to the columns of AA being linearly independent, also equivalent to the null space of AA being zero, in which case one of the infinitely many left inverses is L=(ATA)−1AT.L=(ATA)−1AT. ( ATAATA is square and so capable of being invertible, but you have to check that ATAATA is non-singular if the columns of AA are linearly independent…)
So lets suppose the columns of AA are not linearly independent, or equivalently that AA does not have a left inverse, or that the null space of AA is not trivial.
Another way of saying A(B−C)=[0]A(B−C)=[0] is that the column space of B−CB−C is contained in the now non-trivial null space of AA , written Col(B−C)⊆Null(A)Col(B−C)⊆Null(A) .
In this situation, we can turn things around and describe (in principle) how to create all possible examples of AB=ACAB=AC with B≠CB≠C . Let AA be m×nm×n . The equality AB=ACAB=AC tells us that both BB and CC have to have nn rows, and that BB and CC have to have the same number of columns (how many is up to us). So lets say that we’ll concoct BB and CC with kk columns each, which means they will both be n×k.n×k.
The condition Col(B−C)⊆Null(A)Col(B−C)⊆Null(A) says that the kk columns of B−CB−C have to span a subspace of Null(A)Null(A) . So take n1,n2,…,nk∈Null(A)n1,n2,…,nk∈Null(A) and let NN be the matrix whose columns are n1,n2,…,nkn1,n2,…,nk . Let CC be an arbitrary n×kn×k matrix and then set B=C+NB=C+N . Then obviously Col(B−C)=Col(N)=Span{n1,n2,…,nk}⊆Null(A)Col(B−C)=Col(N)=Span{n1,n2,…,nk}⊆Null(A) , and so A(B−C)=[0]A(B−C)=[0] as desired.
More concisely, A(C+N)=AC+AN=AC+[0]=ACA(C+N)=AC+AN=AC+[0]=AC , but C+N≠C.C+N≠C. Every example of AB=ACAB=AC with B≠CB≠C is of this type
To give a pretty trivial example of all this, take
A=[100100].A=[100010].
Then
Null(A)=Span⎡⎣⎢001⎤⎦⎥.Null(A)=Span[001].
So lets take
n1=⎡⎣⎢001⎤⎦⎥ and n2=⎡⎣⎢002⎤⎦⎥, so that N=⎡⎣⎢001002⎤⎦⎥.n1=[001] and n2=[002], so that N=[000012].
Now take
C=⎡⎣⎢135246⎤⎦⎥, in which case B=C+N=⎡⎣⎢136248⎤⎦⎥,C=[123456], in which case B=C+N=[123468],
and it is obvious that AB=ACAB=AC but B≠CB≠C .
Reading back through this, we can say that in order for AA to be “left cancellable,” meaning that AB=AC⟹B=CAB=AC⟹B=C for all possible matrices BB and CC , it is necessary and sufficient that one of the following equivalent conditions hold:
The columns of AA are linearly independent.
The null space of AA is trivial.
AA has left inverses, (one of which is (ATA)−1AT(ATA)−1AT )