Question

I have a matrix A(Mxn) and I need to prove that for all j 1<=j<=n there is a solution for Ax=A(column j)
I don't even know where to start. anyone can explain plz?

Answer 1

Assume the equation is AX=B where X and B are column vectors of length n.
If you expand the matrix multiplication, you'll find that
bi=sum(aij.xj) j=1 to n
If X is such that xj=1 when j=k, and xj=0 otherwise, then bi=aik, which means that the RHS is simply column k.

Answer 2

\[b _{i} = \sum_{j=1}^{n} a _{ij}x _{j}\]
\[x _{j} = 1\] when j = k and
\[x _{j} = 0\] otherwise, then
\[b _{i} = a _{ik}\]
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ok, so it took me a while staring at the answer to get it. But I think I got it! You just added me 20 points to my assignment. Thanks!

Answer 3

Sorry if it was not very clear. But I am glad that you got it in the end.