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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 Question 25010: 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 Answer by venugopalramana(3286)   (Show Source): You can put this solution on YOUR website!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.