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Question Number 2564 by Rasheed Soomro last updated on 22/Nov/15

A= [(1,2),(2,5) ],B= [(3,(−1)),(4,(+2)) ]  determine A^B

$${A}=\begin{bmatrix}{\mathrm{1}}&{\mathrm{2}}\\{\mathrm{2}}&{\mathrm{5}}\end{bmatrix},{B}=\begin{bmatrix}{\mathrm{3}}&{−\mathrm{1}}\\{\mathrm{4}}&{+\mathrm{2}}\end{bmatrix} \\ $$$${determine}\:{A}^{{B}} \\ $$

Answered by Yozzi last updated on 22/Nov/15

∣A∣=5−4=1≠0 ⇒A is non−singular.  A^B =e^((lnA)B) =Σ_(k=0) ^∞ (1/(k!)){(lnA)B}^k   For eigenvalues k of A  ∣A−kI∣=0  ∴ determinant (((1−k   2)),((2    5−k)))=0  (1−k)(5−k)−4=0  5−6k+k^2 −4=0  k^2 −6k+1=0  (k−3)^2 −9+1=0  (k−3)^2 =8⇒k=3±2(√2)  For k=3+2(√2),    (∵ Ae=ke⇒(A−kI)e=0)   [((1−3−2(√2)          2)),((2                   5−3−2(√2))) ] [(x),(y) ]= [(0),(0) ]  (−2−2(√2))x+2y=0⇒y=(1+(√2))x.....(i)    2x+(2−2(√2))y=0  x+(1−(√2))y=0......(ii)  y from (i) leads to  x+(1−(√2))(1+(√2))x=0  x+(1−2)x=0  x−x=0 (true)  ∴ We can choose eigenvector e_1 = [(1),((1+(√2))) ]   for k=3+2(√2)    For k=3−2(√2)   [((−2+2(√2)      2)),((2                 2+2(√2))) ] [(x),(y) ]= [(0),(0) ]  ⇒(−1+(√2))x+y=0⇒y=(1−(√2))x ..............(iii)  and 2x+(2+2(√2))y=0⇒x+(1+(√2))y=0........(iv)  y from (iii) into (iv) gives  x+(1−2)x=0⇒x−x=0 (true)  So let e_2 = [(1),((1−(√2))) ] for k=3−2(√2).  ∴ Let V=[e_1    e_2 ]= [((1              1)),((1+(√2)    1−(√2))) ]  and A^′ = [((3+2(√2)     0)),((0     3−2(√2))) ]⇒lnA^′ = [((ln(3+2(√2))    0)),((0                       ln(3−2(√2)))) ]  V^(−1) =(1/(∣V∣))adj(V)  adj(V)= [((1−(√2)     −1)),((−1−(√2)     1)) ]  ∣V∣=1−(√2)−1−(√2)=−2(√2)  V^(−1) =((−1)/(2(√2))) [((1−(√2)   −1)),((−1−(√2)    1)) ]  ∴  lnA=V(lnA^′ )V^(−1)   lnA=((−1)/(2(√2))) [((1              1)),((1+(√2)     1−(√2))) ] [((ln(3+2(√2))   0)),((0    ln(3−2(√2)))) ] [((1−(√2)    −1)),((−1−(√2)     1)) ]  lnA=((−1)/(2(√2))) [((ln(3+2(√2))                   ln(3−2(√2)))),(((1+(√2))ln(3+2(√2))   (1−(√2))ln(3−2(√2)))) ] [((1−(√2)         −1)),((−1−(√2)        1)) ]  lnA=((−1)/(2(√2))) [(((1−(√2))ln(3+2(√2))−(1+(√2))ln(3−2(√2))                      ln(((3−2(√2))/(3+2(√2))))           )),((                         ln(((3−2(√2))/(3+2(√2))))                                 (1−(√2))ln(3−2(√2))−(1+(√2))ln(3+2(√2)))) ]  ∴(lnA)B=((−1)/(2(√2))) [(((1−(√2))ln(3+2(√2))−(1+(√2))ln(3−2(√2))                           ln(((3−2(√2))/(3+2(√2)))))),((                          ln(((3−2(√2))/(3+2(√2))))                              (1−(√2))ln(3−2(√2))−(1+(√2))ln(3+2(√2)))) ]×B

$$\mid{A}\mid=\mathrm{5}−\mathrm{4}=\mathrm{1}\neq\mathrm{0}\:\Rightarrow{A}\:{is}\:{non}−{singular}. \\ $$$${A}^{{B}} ={e}^{\left({lnA}\right){B}} =\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{k}!}\left\{\left({lnA}\right){B}\right\}^{{k}} \\ $$$${For}\:{eigenvalues}\:{k}\:{of}\:{A} \\ $$$$\mid{A}−{kI}\mid=\mathrm{0} \\ $$$$\therefore\begin{vmatrix}{\mathrm{1}−{k}\:\:\:\mathrm{2}}\\{\mathrm{2}\:\:\:\:\mathrm{5}−{k}}\end{vmatrix}=\mathrm{0} \\ $$$$\left(\mathrm{1}−{k}\right)\left(\mathrm{5}−{k}\right)−\mathrm{4}=\mathrm{0} \\ $$$$\mathrm{5}−\mathrm{6}{k}+{k}^{\mathrm{2}} −\mathrm{4}=\mathrm{0} \\ $$$${k}^{\mathrm{2}} −\mathrm{6}{k}+\mathrm{1}=\mathrm{0} \\ $$$$\left({k}−\mathrm{3}\right)^{\mathrm{2}} −\mathrm{9}+\mathrm{1}=\mathrm{0} \\ $$$$\left({k}−\mathrm{3}\right)^{\mathrm{2}} =\mathrm{8}\Rightarrow{k}=\mathrm{3}\pm\mathrm{2}\sqrt{\mathrm{2}} \\ $$$${For}\:{k}=\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}},\:\:\:\:\left(\because\:{A}\boldsymbol{{e}}={k}\boldsymbol{{e}}\Rightarrow\left({A}−{kI}\right)\boldsymbol{{e}}=\mathrm{0}\right) \\ $$$$\begin{bmatrix}{\mathrm{1}−\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}\:\:\:\:\:\:\:\:\:\:\mathrm{2}}\\{\mathrm{2}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{5}−\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}}\end{bmatrix}\begin{bmatrix}{{x}}\\{{y}}\end{bmatrix}=\begin{bmatrix}{\mathrm{0}}\\{\mathrm{0}}\end{bmatrix} \\ $$$$\left(−\mathrm{2}−\mathrm{2}\sqrt{\mathrm{2}}\right){x}+\mathrm{2}{y}=\mathrm{0}\Rightarrow{y}=\left(\mathrm{1}+\sqrt{\mathrm{2}}\right){x}.....\left({i}\right) \\ $$$$\:\:\mathrm{2}{x}+\left(\mathrm{2}−\mathrm{2}\sqrt{\mathrm{2}}\right){y}=\mathrm{0} \\ $$$${x}+\left(\mathrm{1}−\sqrt{\mathrm{2}}\right){y}=\mathrm{0}......\left({ii}\right) \\ $$$${y}\:{from}\:\left({i}\right)\:{leads}\:{to} \\ $$$${x}+\left(\mathrm{1}−\sqrt{\mathrm{2}}\right)\left(\mathrm{1}+\sqrt{\mathrm{2}}\right){x}=\mathrm{0} \\ $$$${x}+\left(\mathrm{1}−\mathrm{2}\right){x}=\mathrm{0} \\ $$$${x}−{x}=\mathrm{0}\:\left({true}\right) \\ $$$$\therefore\:{We}\:{can}\:{choose}\:{eigenvector}\:\boldsymbol{{e}}_{\mathrm{1}} =\begin{bmatrix}{\mathrm{1}}\\{\mathrm{1}+\sqrt{\mathrm{2}}}\end{bmatrix}\: \\ $$$${for}\:{k}=\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}} \\ $$$$ \\ $$$${For}\:{k}=\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}} \\ $$$$\begin{bmatrix}{−\mathrm{2}+\mathrm{2}\sqrt{\mathrm{2}}\:\:\:\:\:\:\mathrm{2}}\\{\mathrm{2}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}+\mathrm{2}\sqrt{\mathrm{2}}}\end{bmatrix}\begin{bmatrix}{{x}}\\{{y}}\end{bmatrix}=\begin{bmatrix}{\mathrm{0}}\\{\mathrm{0}}\end{bmatrix} \\ $$$$\Rightarrow\left(−\mathrm{1}+\sqrt{\mathrm{2}}\right){x}+{y}=\mathrm{0}\Rightarrow{y}=\left(\mathrm{1}−\sqrt{\mathrm{2}}\right){x}\:..............\left({iii}\right) \\ $$$${and}\:\mathrm{2}{x}+\left(\mathrm{2}+\mathrm{2}\sqrt{\mathrm{2}}\right){y}=\mathrm{0}\Rightarrow{x}+\left(\mathrm{1}+\sqrt{\mathrm{2}}\right){y}=\mathrm{0}........\left({iv}\right) \\ $$$${y}\:{from}\:\left({iii}\right)\:{into}\:\left({iv}\right)\:{gives} \\ $$$${x}+\left(\mathrm{1}−\mathrm{2}\right){x}=\mathrm{0}\Rightarrow{x}−{x}=\mathrm{0}\:\left({true}\right) \\ $$$${So}\:{let}\:\boldsymbol{{e}}_{\mathrm{2}} =\begin{bmatrix}{\mathrm{1}}\\{\mathrm{1}−\sqrt{\mathrm{2}}}\end{bmatrix}\:{for}\:{k}=\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}. \\ $$$$\therefore\:{Let}\:{V}=\left[\boldsymbol{{e}}_{\mathrm{1}} \:\:\:\boldsymbol{{e}}_{\mathrm{2}} \right]=\begin{bmatrix}{\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}}\\{\mathrm{1}+\sqrt{\mathrm{2}}\:\:\:\:\mathrm{1}−\sqrt{\mathrm{2}}}\end{bmatrix} \\ $$$${and}\:{A}^{'} =\begin{bmatrix}{\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}\:\:\:\:\:\mathrm{0}}\\{\mathrm{0}\:\:\:\:\:\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}}\end{bmatrix}\Rightarrow{lnA}^{'} =\begin{bmatrix}{{ln}\left(\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}\right)\:\:\:\:\mathrm{0}}\\{\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{ln}\left(\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}\right)}\end{bmatrix} \\ $$$${V}^{−\mathrm{1}} =\frac{\mathrm{1}}{\mid{V}\mid}{adj}\left({V}\right) \\ $$$${adj}\left({V}\right)=\begin{bmatrix}{\mathrm{1}−\sqrt{\mathrm{2}}\:\:\:\:\:−\mathrm{1}}\\{−\mathrm{1}−\sqrt{\mathrm{2}}\:\:\:\:\:\mathrm{1}}\end{bmatrix}\:\:\mid{V}\mid=\mathrm{1}−\sqrt{\mathrm{2}}−\mathrm{1}−\sqrt{\mathrm{2}}=−\mathrm{2}\sqrt{\mathrm{2}} \\ $$$${V}^{−\mathrm{1}} =\frac{−\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\begin{bmatrix}{\mathrm{1}−\sqrt{\mathrm{2}}\:\:\:−\mathrm{1}}\\{−\mathrm{1}−\sqrt{\mathrm{2}}\:\:\:\:\mathrm{1}}\end{bmatrix} \\ $$$$\therefore\:\:{lnA}={V}\left({lnA}^{'} \right){V}^{−\mathrm{1}} \\ $$$${lnA}=\frac{−\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\begin{bmatrix}{\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}}\\{\mathrm{1}+\sqrt{\mathrm{2}}\:\:\:\:\:\mathrm{1}−\sqrt{\mathrm{2}}}\end{bmatrix}\begin{bmatrix}{{ln}\left(\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}\right)\:\:\:\mathrm{0}}\\{\mathrm{0}\:\:\:\:{ln}\left(\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}\right)}\end{bmatrix}\begin{bmatrix}{\mathrm{1}−\sqrt{\mathrm{2}}\:\:\:\:−\mathrm{1}}\\{−\mathrm{1}−\sqrt{\mathrm{2}}\:\:\:\:\:\mathrm{1}}\end{bmatrix} \\ $$$${lnA}=\frac{−\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\begin{bmatrix}{{ln}\left(\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{ln}\left(\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}\right)}\\{\left(\mathrm{1}+\sqrt{\mathrm{2}}\right){ln}\left(\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}\right)\:\:\:\left(\mathrm{1}−\sqrt{\mathrm{2}}\right){ln}\left(\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}\right)}\end{bmatrix}\begin{bmatrix}{\mathrm{1}−\sqrt{\mathrm{2}}\:\:\:\:\:\:\:\:\:−\mathrm{1}}\\{−\mathrm{1}−\sqrt{\mathrm{2}}\:\:\:\:\:\:\:\:\mathrm{1}}\end{bmatrix} \\ $$$${lnA}=\frac{−\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\begin{bmatrix}{\left(\mathrm{1}−\sqrt{\mathrm{2}}\right){ln}\left(\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}\right)−\left(\mathrm{1}+\sqrt{\mathrm{2}}\right){ln}\left(\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{ln}\left(\frac{\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}}\right)\:\:\:\:\:\:\:\:\:\:\:}\\{\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{ln}\left(\frac{\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{1}−\sqrt{\mathrm{2}}\right){ln}\left(\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}\right)−\left(\mathrm{1}+\sqrt{\mathrm{2}}\right){ln}\left(\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}\right)}\end{bmatrix} \\ $$$$\therefore\left({lnA}\right){B}=\frac{−\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\begin{bmatrix}{\left(\mathrm{1}−\sqrt{\mathrm{2}}\right){ln}\left(\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}\right)−\left(\mathrm{1}+\sqrt{\mathrm{2}}\right){ln}\left(\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{ln}\left(\frac{\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}}\right)}\\{\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{ln}\left(\frac{\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{1}−\sqrt{\mathrm{2}}\right){ln}\left(\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}\right)−\left(\mathrm{1}+\sqrt{\mathrm{2}}\right){ln}\left(\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}\right)}\end{bmatrix}×{B} \\ $$$$ \\ $$$$ \\ $$

Commented by RasheedAhmad last updated on 22/Nov/15

I′ve learnt something from you  that I didn′t know   before!

$$\mathcal{I}'{ve}\:{learnt}\:{something}\:{from}\:{you} \\ $$$${that}\:{I}\:{didn}'{t}\:{know}\:\:\:{before}! \\ $$

Commented by Yozzi last updated on 22/Nov/15

Honestly, I hadn′t known how to find  A^(B )  until today. Sometime ago  I looked at Wikipedia to obtain information  on solving a matrix differential  equation of the form (dr/dt)+Ar=o   where it was suggested that a   suitable matrix integrating factor  be employed. In (incorrectly) guessing that the  required factor is of the form e^A   I was met with a cool idea concerning  the expression e^(X )  for X being a matrix  and e being Euler′s constant. So  when I saw your question on A^B  I thought  about the equivalent expression,  which wiki corrected me on, e^((lnA)B) .  We have B_A ≡e^(B(lnA))  due to the general  non−commutative nature of matrices.  Luckily I studied some linear algebra  before so applying the method I  found on Wikipedia wasn′t too hard.

$${Honestly},\:{I}\:{hadn}'{t}\:{known}\:{how}\:{to}\:{find} \\ $$$${A}^{{B}\:} \:{until}\:{today}.\:{Sometime}\:{ago} \\ $$$${I}\:{looked}\:{at}\:{Wikipedia}\:{to}\:{obtain}\:{information} \\ $$$${on}\:{solving}\:{a}\:{matrix}\:{differential} \\ $$$${equation}\:{of}\:{the}\:{form}\:\frac{{d}\boldsymbol{{r}}}{{dt}}+{A}\boldsymbol{{r}}=\boldsymbol{{o}}\: \\ $$$${where}\:{it}\:{was}\:{suggested}\:{that}\:{a}\: \\ $$$${suitable}\:{matrix}\:{integrating}\:{factor} \\ $$$${be}\:{employed}.\:{In}\:\left({incorrectly}\right)\:{guessing}\:{that}\:{the} \\ $$$${required}\:{factor}\:{is}\:{of}\:{the}\:{form}\:{e}^{{A}} \\ $$$${I}\:{was}\:{met}\:{with}\:{a}\:{cool}\:{idea}\:{concerning} \\ $$$${the}\:{expression}\:{e}^{{X}\:} \:{for}\:{X}\:{being}\:{a}\:{matrix} \\ $$$${and}\:{e}\:{being}\:{Euler}'{s}\:{constant}.\:{So} \\ $$$${when}\:{I}\:{saw}\:{your}\:{question}\:{on}\:{A}^{{B}} \:{I}\:{thought} \\ $$$${about}\:{the}\:{equivalent}\:{expression}, \\ $$$${which}\:{wiki}\:{corrected}\:{me}\:{on},\:{e}^{\left({lnA}\right){B}} . \\ $$$${We}\:{have}\:{B}_{{A}} \equiv{e}^{{B}\left({lnA}\right)} \:{due}\:{to}\:{the}\:{general} \\ $$$${non}−{commutative}\:{nature}\:{of}\:{matrices}. \\ $$$${Luckily}\:{I}\:{studied}\:{some}\:{linear}\:{algebra} \\ $$$${before}\:{so}\:{applying}\:{the}\:{method}\:{I} \\ $$$${found}\:{on}\:{Wikipedia}\:{wasn}'{t}\:{too}\:{hard}. \\ $$

Commented by Rasheed Soomro last updated on 23/Nov/15

THaNkSsss for Sharing thoughts with open mind!  It can be said that my question has become cause to  increase your knowledge and interest AND your   Sharing−Knowledge has become cause to increase  my knowledge!  Actually the question is result of my imagination,  otherwise my knowledge regarding Matrices is of  elementary level.  Your comment contains an expression ′ B_( A) ≡e^(B(ln A)) ...due to  non−commutative nature...′.  I would like to know  the meaning of B_( A) . Considering non−commutative  nature ...shouldn′t we write   B^A  as non−commutative  part of A^B  ?

$$\mathcal{TH}{a}\mathcal{N}{k}\mathcal{S}{sss}\:{for}\:\mathcal{S}{haring}\:{thoughts}\:{with}\:{open}\:{mind}! \\ $$$$\mathcal{I}{t}\:{can}\:{be}\:{said}\:{that}\:{my}\:{question}\:{has}\:{become}\:{cause}\:{to} \\ $$$${increase}\:{your}\:{knowledge}\:{and}\:{interest}\:\mathcal{AND}\:{your}\: \\ $$$$\mathcal{S}{haring}−\mathcal{K}{nowledge}\:{has}\:{become}\:{cause}\:{to}\:{increase} \\ $$$${my}\:{knowledge}! \\ $$$$\mathcal{A}{ctually}\:{the}\:{question}\:{is}\:{result}\:{of}\:{my}\:{imagination}, \\ $$$${otherwise}\:{my}\:{knowledge}\:{regarding}\:\mathcal{M}{atrices}\:{is}\:{of} \\ $$$${elementary}\:{level}. \\ $$$$\mathcal{Y}{our}\:{comment}\:{contains}\:{an}\:{expression}\:'\:{B}_{\:{A}} \equiv{e}^{{B}\left({ln}\:{A}\right)} ...{due}\:{to} \\ $$$${non}−{commutative}\:{nature}...'.\:\:\mathcal{I}\:{would}\:{like}\:{to}\:{know} \\ $$$${the}\:{meaning}\:{of}\:{B}_{\:{A}} .\:{Considering}\:{non}−{commutative} \\ $$$${nature}\:...{shouldn}'{t}\:{we}\:{write}\:\:\:{B}^{{A}} \:{as}\:{non}−{commutative} \\ $$$${part}\:{of}\:{A}^{{B}} \:? \\ $$

Commented by Yozzi last updated on 23/Nov/15

The wiki indicated that while  A^B =e^((lnA)B) , ^A B=e^(B(lnA)) .  I just figured out how to obtain   left−hand exponents. I was trying  to do so by writting B_A ...

$${The}\:{wiki}\:{indicated}\:{that}\:{while} \\ $$$${A}^{{B}} ={e}^{\left({lnA}\right){B}} ,\:\:^{{A}} {B}={e}^{{B}\left({lnA}\right)} . \\ $$$${I}\:{just}\:{figured}\:{out}\:{how}\:{to}\:{obtain}\: \\ $$$${left}−{hand}\:{exponents}.\:{I}\:{was}\:{trying} \\ $$$${to}\:{do}\:{so}\:{by}\:{writting}\:{B}_{{A}} ... \\ $$

Commented by Rasheed Soomro last updated on 23/Nov/15

THANKS !

$$\mathbb{THANKS}\:! \\ $$

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