- #1
franky2727
- 132
- 0
given A={(1,2,1),(2,4,2),(3,6,3)} findr and invertable matrices Q and P such that Q-1AP={(Ir,0),(0,0)} where each zero denotes a matrix of zeros not necessarily the same size
paying special attension to the order of the vectors write down the bases of R3 with respect to which Q-1AP represents the mapping x->Ax
i think i can do the first part getting row opps of r3-3r1 and r2-2r1 and then column opps of c2-2c1 and c3-c1 giving me {(1,0,0),(0,0,0),(0,0,0) and therefore r=1
then i do I3 with the same row opps giving Q-1={(1,0,0),(-2,1,0),(-3,0,1)) giving Q=(Q-1)-1 = {(1,0,0),(1/2,1,0),(1/3,0,1)
same with the column opps on P gives me {1.-2.-1),(0,1,0),(0,0,1)}=P
i believe this is right but have not done it in a while and may be messing up the method so a check wouldn't go a miss, also i don't know how to do the second part of the question, it looks slightly familia with the getting vectors in the right order but i can't remember where to start so help here would be aprichiated thanks. on a side note this is revision not homework so feal free to splurt it all out :P
paying special attension to the order of the vectors write down the bases of R3 with respect to which Q-1AP represents the mapping x->Ax
i think i can do the first part getting row opps of r3-3r1 and r2-2r1 and then column opps of c2-2c1 and c3-c1 giving me {(1,0,0),(0,0,0),(0,0,0) and therefore r=1
then i do I3 with the same row opps giving Q-1={(1,0,0),(-2,1,0),(-3,0,1)) giving Q=(Q-1)-1 = {(1,0,0),(1/2,1,0),(1/3,0,1)
same with the column opps on P gives me {1.-2.-1),(0,1,0),(0,0,1)}=P
i believe this is right but have not done it in a while and may be messing up the method so a check wouldn't go a miss, also i don't know how to do the second part of the question, it looks slightly familia with the getting vectors in the right order but i can't remember where to start so help here would be aprichiated thanks. on a side note this is revision not homework so feal free to splurt it all out :P
Last edited: