Question

Find invertible matrics P,Q such that PAQ=D where A=[1 2; 3 1]? D=[1 0; 0 1]...How do I go about solving this?

Answer 1

set up some systems of equations is my first thought

Answer 2

if it helps, P,Q e M2x2 (Z5)

Answer 3

\[\begin{vmatrix}a&b\\c&d\end{vmatrix} A\begin{vmatrix}p&q\\ s&t\end{vmatrix}=D\]
are those columns or rows in your post for A and D ?

Answer 4

rows

Answer 5

AQ = P inverse

Answer 6

yes that does seem like the way to go, I remember one of the theorems being something like that

Answer 7

how do we proceed though?

Answer 8

\[\begin{vmatrix}a&b\\c&d\end{vmatrix} \begin{vmatrix}1&2\\3&1\end{vmatrix}\begin{vmatrix}p&q\\ s&t\end{vmatrix}=\begin{vmatrix}1&0\\0&1\end{vmatrix}\]
\[\begin{vmatrix}a+3b&2a+b\\c+3d&2c+d\end{vmatrix}\begin{vmatrix}p&q\\ s&t\end{vmatrix}=\begin{vmatrix}1&0\\0&1\end{vmatrix}\]
\[\begin{vmatrix}p(a+3b)+s(2a+b)&q(a+3b)+t(2a+b)\\p(c+3d)+s(2c+d)&q(c+3d)+t(2a+b)\end{vmatrix}=\begin{vmatrix}1&0\\0&1\end{vmatrix}\]

\[\begin{array}\ p(a+3b)+s(2a+b)=1\\q(a+3b)+t(2a+b)=0\\p(c+3d)+s(2c+d)=0\\q(c+3d)+t(2a+b)=1\end{array}\]
this might be the long way, but it may be fruitful

Answer 9

its definantly not the way to type it up on internet explorer tho

Answer 10

right, this seems a bit long, but we can maybe find a pattern or something

Answer 11

8 unknown and 4 equations ..

Answer 12

the next thing I would try would be to assume an inverse can be found and leave the determinant open for interpretation; P-1 D Q-1

Answer 13

what can we use by the fact that P and Q are invertible, and everything is 2x2?

Answer 14

\[ \begin{vmatrix}1&2\\3&1\end{vmatrix}=\begin{vmatrix}d&-b\\-c&a\end{vmatrix}\begin{vmatrix}1&0\\0&1\end{vmatrix}\begin{vmatrix}t&-q\\ -s&p\end{vmatrix}\]
\[ \begin{vmatrix}1&2\\3&1\end{vmatrix}=\begin{vmatrix}d&-b\\-c&a\end{vmatrix}\begin{vmatrix}t&-q\\ -s&p\end{vmatrix}\]

Answer 15

since D is th eidentity matrix there it kinda just drops out doesnt it

Answer 16

yup

Answer 17

\[\begin{vmatrix}1&2\\3&1\end{vmatrix}=\begin{vmatrix}td+sb&-qd-pb\\-tc-sa&qc+pa\end{vmatrix}\]
\[\begin{array}\ td+sb=1\\-qd-pb=2\\-tc-sa=3\\qc+pa=1\end{array}\]

and there is a matter of keeping the determinants in mind

Answer 18

dP and dQ

Answer 19

A=B^-1C^-1...where the inverses B and C are made of elementary matrices

Answer 20

QP=A^-1

Answer 21

there must be something that says that P and Q must be the inverse of A

Answer 22

or no, not true

Answer 23

im sure theres "something" to say one thing or another, but without knowing what your material has covered, this is just a shot in the dark :)

Answer 24

ahh, back on the chrome, might be a little easier to bear now

Answer 25

true...its 2nd year advanced lin algebra, we should be looking for consequences of things like invertibility and matrix multiplication to answer it. something elegant and short

Answer 26

that would explain it, im only versed in the first year beginning stuff :)

Answer 27

AQ = P-1 D

AQ = P-1

Answer 28

ok as always I appreciate your help :), I think I have an answer now...

Answer 29

A = P-1 Q-1

Answer 30

P P-1 Q-1 Q = I

Answer 31

A=P^-1Q^-1, and AQ=P^-1, so Q=In

Answer 32

since D = I

Answer 33

sounds like a winner to me :)

Answer 34

yup haha nice, so P will be A inverse then right?

Answer 35

if we try to uncover P

PAQ = D, but D=I

PAQQ' = IQ'

PA = Q'

PAA' = Q'A'

P = Q'A'

Answer 36

QP = QQ'A'

QP = A' would seem more like it to me

Answer 37

oh right, mine was wrong, damn

Answer 38

pa+sb = -1/5
qa+tb = 3/5
pc+sd = 2/5
qc+td = -1/5

is what i get for that set

Answer 39

then we have to figure out Q and P still, is there just one way to construct it?

Answer 40

if we can get some equations set up with the aid of A and D it seems doable to me

Answer 41

It's in Z5, which means that if we multiply each entry in the inverse by 5, we get the inverse again?

Answer 42

nope, it's if you add 5

Answer 43

PAQ=D

P'PAQ = P'D

AQ = P' ; |p|' = 1/detP for shorthand

p+2s = d|p|'
3p+ s = -c|p|'
q+2t = -b|p|'
3q+ t = a|p|'

pa+sb = -1/5
qa+tb = 3/5
pc+sd = 2/5
qc+td = -1/5


pa+sb = -1/5
pc+sd = 2/5

p ' s
a -1/5 b
c 2/5 d

p = -d/5 -2b/5
----------
ad-bc

s = 2a/5-c/5
--------
ad-bc

qa+tb = 3/5
qc+td = -1/5

q ' t
a 3/5 b
c -1/5 d

q = 3d/5 + b/5
----------
ad-bc

t = -a/5 -3c/5
----------
ad-bc


\[p=\frac{-d-2b}{5(ad-bc)}\]
\[s=\frac{2a-c}{5(ad-bc)}\]
\[q=\frac{3d+b}{5(ad-bc)}\]
\[t=\frac{-a-3c}{5(ad-bc)}\]

\[ \frac{-d-2b+4a-2c}{5(ad-bc)} = \frac{d}{detP}\]\[\frac{-3d-6b+2a-c}{5(ad-bc)} =\frac{-c}{detP}\]\[\frac{3d+b-2a-6c}{5(ad-bc)}= \frac{-b}{detP}\]\[\frac{9d+3b-a-3c}{5(ad-bc)} = \frac{a}{detP} \]

but ad-bc = detP, if i see that right; so if we multiply it all by detP those vanish


\[ \frac{-d-2b+4a-2c}{5} = \frac{d}{}\]\[\frac{-3d-6b+2a-c}{5} =\frac{-c}{}\]\[\frac{3d+b-2a-6c}{5}= \frac{-b}{}\]\[\frac{9d+3b-a-3c}{5} = \frac{a}{} \]


\[-d-2b+4a-2c =5d\]\[-3d-6b+2a-c =-5c\]\[3d+b-2a-6c=-5b\]\[9d+3b-a-3c = 5a \]


\[-6d-2b+4a-2c =0\]\[-3d-6b+2a-4c =0\]\[3d+6b-2a-6c=0\]\[9d+3b-6a-3c =0 \]

with any luck; this is a rref-able matrix to find the entries of P = ab,cd

Answer 44

thanks, I'm sorry, D=1 I guess, so PAQ=1

Answer 45

rref{{-6,-2,4,-2},{-3,-6,2,-4},{3,6,-2,-6},{9,3,-6,-3}}

a 2/3
b = c. 0
c 1
d 0

http://www.wolframalpha.com/input/?i=rref%7B%7B-6%2C-2%2C4%2C-2%7D%2C%7B-3%2C-6%2C2%2C-4%7D%2C%7B3%2C6%2C-2%2C-6%7D%2C%7B9%2C3%2C-6%2C-3%7D%7D

D = "one" or indentity?

Answer 46

oh wow you got an answer? well D is sppsed to be the Identity rank of A in Z

Answer 47

i got something, not an answer perse since

20
30 aint invertible

Answer 48

id have to sit down and try it on paper; this typing to code it up and math at the same time is brain wracking

Answer 49

ill see what i can do later and try to post something later on since i apparently need alot of practice at this :)

Answer 50

ok thanks for your help

Answer 51

i think i see what i did, towards the end i dint bother to line up the abcds in columns to pull off the matrix ....

Answer 52

could you check the rank of A? I got 1, but it could be 2

Answer 53

well, on second look they kinda just lined them selves up, but i dint apply them correctly afterwards

d 2
b = c. 0
a 3
c 0

30
02 = P if im lucky

Answer 54

how did you get to there?

Answer 55

i believe :
PAQ=D

P'PAQ = P'D

AQ = P'

A'AQ=A'P'

Q= A'P'

A' = 1 -2 P' = 2 0 Q = 2 -6
-3 1 0 3 -6 3

to test out

Answer 56

how? by doing that mess up above

Answer 57

hmm, ok i follow everything up to Q=A'P', but how did you get those answers

Answer 58

i created 8 equations based on 2 scenarios: AQ=P' and i think thats was QP=A'

paq=d

p'paq=p'd

aq=p'

a'aq=a'p'

q=a'p'

qp=a'p'p

qp=a'

Answer 59

or something along those lines, then i subbed inwhat i could till i got a system of 4 eqs in 4 unknowns

Answer 60

stillhaving troubles in the end tho

at any rate, when we know p we can form q

Answer 61

I think I'll ask the prof on this one, thanks for your insight

Answer 62

good luck :)