- #1
Buffu
- 849
- 146
Let ##(AB)_j## be the jth column of ##AB##, then ##\displaystyle (AB)_j = \sum^n_{r= 1} B_{rj} \alpha_r## where ##\alpha_r## is the rth column of ##A##.
Also ##(BA)_j = B \alpha_j \implies A(BA)_j = \alpha_j## susbtituting this in the sum ##\displaystyle (AB)_j = \sum^n_{r = 1} B_{rj}A(BA)_r##
Now if we multiply any matrix by some column of ##I## we get that column of the matrix as the result,
So ##\displaystyle B_j = BA \sum^n_{r= 1} B_{rj} (BA)_r##
Also ## \displaystyle [(BA)B]_j = \sum^n_{r= 1} B_{rj}(BA)_r##
therefore ##\displaystyle B_j = (BA) [(BA)B]_j = (BA)[B(AB)]_j = (BA)B_j## Therefore ##BA = I##
Is this correct ? Is there a easy way to do this ?
Also ##(BA)_j = B \alpha_j \implies A(BA)_j = \alpha_j## susbtituting this in the sum ##\displaystyle (AB)_j = \sum^n_{r = 1} B_{rj}A(BA)_r##
Now if we multiply any matrix by some column of ##I## we get that column of the matrix as the result,
So ##\displaystyle B_j = BA \sum^n_{r= 1} B_{rj} (BA)_r##
Also ## \displaystyle [(BA)B]_j = \sum^n_{r= 1} B_{rj}(BA)_r##
therefore ##\displaystyle B_j = (BA) [(BA)B]_j = (BA)[B(AB)]_j = (BA)B_j## Therefore ##BA = I##
Is this correct ? Is there a easy way to do this ?