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?

- anonymous

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- schrodinger

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- amistre64

set up some systems of equations is my first thought

- anonymous

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

- amistre64

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

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## More answers

- anonymous

rows

- experimentX

AQ = P inverse

- anonymous

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

- anonymous

how do we proceed though?

- amistre64

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

- amistre64

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

- anonymous

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

- experimentX

8 unknown and 4 equations ..

- amistre64

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

- anonymous

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

- amistre64

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

- amistre64

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

- anonymous

yup

- amistre64

\[\begin{vmatrix}1&2\\3&1\end{vmatrix}=\begin{vmatrix}td+sb&-qd-pb\\-tc-sa&qc+pa\end{vmatrix}\]
\[\begin{array}\ td+sb=1\\-qd-pb=2\\-tc-sa=3\\qc+pa=1\end{array}\]
and there is a matter of keeping the determinants in mind

- amistre64

dP and dQ

- anonymous

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

- anonymous

QP=A^-1

- anonymous

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

- anonymous

or no, not true

- amistre64

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 :)

- amistre64

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

- anonymous

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

- amistre64

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

- amistre64

AQ = P-1 D
AQ = P-1

- anonymous

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

- amistre64

A = P-1 Q-1

- amistre64

P P-1 Q-1 Q = I

- anonymous

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

- amistre64

since D = I

- amistre64

sounds like a winner to me :)

- anonymous

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

- amistre64

if we try to uncover P
PAQ = D, but D=I
PAQQ' = IQ'
PA = Q'
PAA' = Q'A'
P = Q'A'

- amistre64

QP = QQ'A'
QP = A' would seem more like it to me

- anonymous

oh right, mine was wrong, damn

- amistre64

pa+sb = -1/5
qa+tb = 3/5
pc+sd = 2/5
qc+td = -1/5
is what i get for that set

- anonymous

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

- amistre64

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

- anonymous

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

- anonymous

nope, it's if you add 5

- amistre64

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

- anonymous

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

- amistre64

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?

- anonymous

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

- amistre64

i got something, not an answer perse since
20
30 aint invertible

- amistre64

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

- amistre64

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

- anonymous

ok thanks for your help

- amistre64

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 ....

- anonymous

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

- amistre64

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

- anonymous

how did you get to there?

- amistre64

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

- amistre64

how? by doing that mess up above

- anonymous

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

- amistre64

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'

- amistre64

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

- amistre64

stillhaving troubles in the end tho
at any rate, when we know p we can form q

- anonymous

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

- amistre64

good luck :)

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