anonymous
  • anonymous
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?
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
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amistre64
  • amistre64
set up some systems of equations is my first thought
anonymous
  • anonymous
if it helps, P,Q e M2x2 (Z5)
amistre64
  • 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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anonymous
  • anonymous
rows
experimentX
  • experimentX
AQ = P inverse
anonymous
  • anonymous
yes that does seem like the way to go, I remember one of the theorems being something like that
anonymous
  • anonymous
how do we proceed though?
amistre64
  • 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
  • amistre64
its definantly not the way to type it up on internet explorer tho
anonymous
  • anonymous
right, this seems a bit long, but we can maybe find a pattern or something
experimentX
  • experimentX
8 unknown and 4 equations ..
amistre64
  • 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
  • anonymous
what can we use by the fact that P and Q are invertible, and everything is 2x2?
amistre64
  • 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
  • amistre64
since D is th eidentity matrix there it kinda just drops out doesnt it
anonymous
  • anonymous
yup
amistre64
  • 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
  • amistre64
dP and dQ
anonymous
  • anonymous
A=B^-1C^-1...where the inverses B and C are made of elementary matrices
anonymous
  • anonymous
QP=A^-1
anonymous
  • anonymous
there must be something that says that P and Q must be the inverse of A
anonymous
  • anonymous
or no, not true
amistre64
  • 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
  • amistre64
ahh, back on the chrome, might be a little easier to bear now
anonymous
  • 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
  • amistre64
that would explain it, im only versed in the first year beginning stuff :)
amistre64
  • amistre64
AQ = P-1 D AQ = P-1
anonymous
  • anonymous
ok as always I appreciate your help :), I think I have an answer now...
amistre64
  • amistre64
A = P-1 Q-1
amistre64
  • amistre64
P P-1 Q-1 Q = I
anonymous
  • anonymous
A=P^-1Q^-1, and AQ=P^-1, so Q=In
amistre64
  • amistre64
since D = I
amistre64
  • amistre64
sounds like a winner to me :)
anonymous
  • anonymous
yup haha nice, so P will be A inverse then right?
amistre64
  • amistre64
if we try to uncover P PAQ = D, but D=I PAQQ' = IQ' PA = Q' PAA' = Q'A' P = Q'A'
amistre64
  • amistre64
QP = QQ'A' QP = A' would seem more like it to me
anonymous
  • anonymous
oh right, mine was wrong, damn
amistre64
  • 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
  • anonymous
then we have to figure out Q and P still, is there just one way to construct it?
amistre64
  • amistre64
if we can get some equations set up with the aid of A and D it seems doable to me
anonymous
  • anonymous
It's in Z5, which means that if we multiply each entry in the inverse by 5, we get the inverse again?
anonymous
  • anonymous
nope, it's if you add 5
amistre64
  • 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
  • anonymous
thanks, I'm sorry, D=1 I guess, so PAQ=1
amistre64
  • 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
  • anonymous
oh wow you got an answer? well D is sppsed to be the Identity rank of A in Z
amistre64
  • amistre64
i got something, not an answer perse since 20 30 aint invertible
amistre64
  • 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
  • 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
  • anonymous
ok thanks for your help
amistre64
  • 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
  • anonymous
could you check the rank of A? I got 1, but it could be 2
amistre64
  • 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
  • anonymous
how did you get to there?
amistre64
  • 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
  • amistre64
how? by doing that mess up above
anonymous
  • anonymous
hmm, ok i follow everything up to Q=A'P', but how did you get those answers
amistre64
  • 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
  • amistre64
or something along those lines, then i subbed inwhat i could till i got a system of 4 eqs in 4 unknowns
amistre64
  • amistre64
stillhaving troubles in the end tho at any rate, when we know p we can form q
anonymous
  • anonymous
I think I'll ask the prof on this one, thanks for your insight
amistre64
  • amistre64
good luck :)

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