Question

Linear algebra,

Can you please help me solve this ?
https://www.dropbox.com/s/am92go4cs84z3rd/Screenshot%202013-11-15%2022.53.41.jpg

Answer 1

plug in a dummy matrix for B and work the 2 operations AB and BA; then equate the elements to solve the dummy variables

Answer 2

"dummy" as in ?can you please show me what you mean ? :(

Answer 3

a c
b d


a+2b c+2d
0+ b 0+d



a 2a+c
b 2b+d


a = a+2b
b = b

c+2d = 2a+c
d = 2b+d

Answer 4

since yo have no idea what B needs to be, fill it with variables ... dummy variables is just another way of saying it

Answer 5

so thats AB a+2b c+2d
0+ b 0+d

Answer 6

and thats BA a 2a+c
b 2b+d

Answer 7

yes, then equate them part for part

Answer 8

whats this a = a+2b
b = b

Answer 9

if AB = BA .. then the elements have to be the same

Answer 10

ab11 = ba11

ab12 = ba12

ab21 = ba 21

ab22 = ba22

Answer 11

so a = d and c = 0

Answer 12

and then what ?

Answer 13

a = a+2b; b=0, a=a
b = b

c+2d = 2a+c
d = 2b+d d=d

c+2d = 2a+c; c any value

d=a

B = a c
0 a

for any real value a and c

Answer 14

can you please tell me how did you came up with this B = a c
0 a

Answer 15

whats the first step ?

Answer 16

i worked it all out already ....

Answer 17

a = a+2b; only if b=0
b = 0

c+2d = 2a+c ; subtract c to see that d = a
d = a

Answer 18

if a=1, c not equal 0 to conform to the rules

and if a=c=0 is not allowed either, but other than that ...

Answer 19

ok please tell mw why this equation a = a+2b

Answer 20

from where is this equation ?

Answer 21

form equating AB with BA of course

Answer 22

*from

Answer 23

ok

Answer 24

so which elements are we equating here ?

Answer 25

i already went over this above ....

\[AB=\begin{pmatrix}ab_{11}&ab_{12}\\ab_{21}&ab_{22}\end{pmatrix}\]
\[BA=\begin{pmatrix}ba_{11}&ba_{12}\\ba_{21}&ba_{22}\end{pmatrix}\]
if AB = BA, then they have to have the same parts .....\[ab_{11}=ba_{11}\\ab_{12}=ba_{12}\\ab_{21}=ba_{21}\\ab_{22}=ba_{22}\]

Answer 26

ok but the thing that confuses me is that no A nor B has this as a one of it's parts: "a+2b"

Answer 27

of course it does ...

let B = a c and work the operations AB and BA
b d

Answer 28

https://www.dropbox.com/s/26cirzqtzych6y5/Screenshot%202013-11-15%2023.55.41.jpg

Answer 29

https://www.dropbox.com/s/dz6duusm3int5e2/Screenshot%202013-11-15%2023.55.49.jpg

Answer 30

then we are not using the same B setup ... you placed the dummy variables in a different way so naturally your setup will "look" different :)

Answer 31

oh wait wait -_-

Answer 32

super confused now :D

Answer 33

sec

Answer 34

from yours, I can say AB = BA --> \(B^{-1}AB = B^{-1}BA\) or \(B^{-1}AB = A . that means you have to find out diagonalized matrix of A.

Answer 35

but eigenvalues is 0 double roots and it gives out a Jordan one, which is \(B= \left[\begin{matrix}0&1\\0&0\end{matrix}\right]\), let check, you have AB =BA

Answer 36

neeeverrmind I got it all sorted out luckily, I understand whats going on now , thanks a lot ! = )

Answer 37

for me its c = 0 and a = d

Answer 38

but same thing

Answer 39

yeah; we should both end up with:\[B=\begin{pmatrix}R_a&R_b\\0&R_a\end{pmatrix}\]

Answer 40

R is a real number in this case ...

Answer 41

yep I got 0 in the exact same spot

Answer 42

since there are countless real numbers; the solution set is infinite

Answer 43

so I can put ANY combination of numbers in a b d

Answer 44

no, the instructions say that we avoid all zeros, and identitiy

Answer 45

if Ra = Rb = 0 its against the rules

if Ra = 1 and Rb = 0 its against the rules

otherwise its fair game

Answer 46

in other words, they wanted the nontrivial solutions

Answer 47

I see! ty !

Answer 48

youre welcome