SOLUTION: show that the jth column of the matrix product AB is equal to the matrix product ABj, where Bj is the jth column of B

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WITH USUAL NOTATION.CAPITAL LETTERS A,B,C,D FOR MATRICES AND SMALL LETTERS a,b,c,d FOR THEIR ELEMENTS.
LET A BE (m,n)MATRIX,B BE (n,p)MATRIX AND LET AB = C (m,p) MATRIX.HENCE WE HAVE
c(m,p)=SIGMA a(m,r)*b(r,p)WITH r VARYING FROM 1 TO n.
HENCE JTH COLUMN OF C IS GIVEN BY ..
c(m,j)= SIGMA a(m,r)*b(r,j)WITH r VARYING FROM 1 TO n.THAT IS
c(1,j)= SIGMA a(1,r)*b(r,j)WITH r VARYING FROM 1 TO n.
c(2,j)= SIGMA a(2,r)*b(r,j)WITH r VARYING FROM 1 TO n.
ETC...
NOW WE WERE GIVEN THAT MATRIX BJ IS (n,1)MATRIX WITH
b(n,j)= bj(n,1) AND
LET A*BJ = D (m,1) MATRIX. HENCE
d(m,1)=SIGMA a(m,r)*bj(r,1)WITH r VARYING FROM 1 TO n.THAT IS
d(1,1)=SIGMA a(1,r)*bj(r,1)WITH r VARYING FROM 1 TO n
d(1,1)=SIGMA a(2,r)*bj(r,1)WITH r VARYING FROM 1 TO n
ETC...
BUT SINCE bj(n,1)=b(n,j) WE GET
d(m,1)=SIGMA a(m,r)*bj(r,1)=SIGMA a(m,r)*b(r,j) WITH r VARYING FROM 1 TO n
= c(m,j)...HENCE WE HAVE D=A*BJ = CJ WHERE CJ IS THE J TH. COLUMN MATRIX OF AB.