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](https://www.tinkutara.com/question/Q2564.png)
$${A}=\begin{bmatrix}{\mathrm{1}}&{\mathrm{2}}\\{\mathrm{2}}&{\mathrm{5}}\end{bmatrix},{B}=\begin{bmatrix}{\mathrm{3}}&{−\mathrm{1}}\\{\mathrm{4}}&{+\mathrm{2}}\end{bmatrix} \\ $$$${determine}\:{A}^{{B}} \\ $$
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](https://www.tinkutara.com/question/Q2572.png)
$$\mid{A}\mid=\mathrm{5}−\mathrm{4}=\mathrm{1}\neq\mathrm{0}\:\Rightarrow{A}\:{is}\:{non}−{singular}. \\ $$$${A}^{{B}} ={e}^{\left({lnA}\right){B}} =\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{k}!}\left\{\left({lnA}\right){B}\right\}^{{k}} \\ $$$${For}\:{eigenvalues}\:{k}\:{of}\:{A} \\ $$$$\mid{A}−{kI}\mid=\mathrm{0} \\ $$$$\therefore\begin{vmatrix}{\mathrm{1}−{k}\:\:\:\mathrm{2}}\\{\mathrm{2}\:\:\:\:\mathrm{5}−{k}}\end{vmatrix}=\mathrm{0} \\ $$$$\left(\mathrm{1}−{k}\right)\left(\mathrm{5}−{k}\right)−\mathrm{4}=\mathrm{0} \\ $$$$\mathrm{5}−\mathrm{6}{k}+{k}^{\mathrm{2}} −\mathrm{4}=\mathrm{0} \\ $$$${k}^{\mathrm{2}} −\mathrm{6}{k}+\mathrm{1}=\mathrm{0} \\ $$$$\left({k}−\mathrm{3}\right)^{\mathrm{2}} −\mathrm{9}+\mathrm{1}=\mathrm{0} \\ $$$$\left({k}−\mathrm{3}\right)^{\mathrm{2}} =\mathrm{8}\Rightarrow{k}=\mathrm{3}\pm\mathrm{2}\sqrt{\mathrm{2}} \\ $$$${For}\:{k}=\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}},\:\:\:\:\left(\because\:{A}\boldsymbol{{e}}={k}\boldsymbol{{e}}\Rightarrow\left({A}−{kI}\right)\boldsymbol{{e}}=\mathrm{0}\right) \\ $$$$\begin{bmatrix}{\mathrm{1}−\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}\:\:\:\:\:\:\:\:\:\:\mathrm{2}}\\{\mathrm{2}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{5}−\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}}\end{bmatrix}\begin{bmatrix}{{x}}\\{{y}}\end{bmatrix}=\begin{bmatrix}{\mathrm{0}}\\{\mathrm{0}}\end{bmatrix} \\ $$$$\left(−\mathrm{2}−\mathrm{2}\sqrt{\mathrm{2}}\right){x}+\mathrm{2}{y}=\mathrm{0}\Rightarrow{y}=\left(\mathrm{1}+\sqrt{\mathrm{2}}\right){x}…..\left({i}\right) \\ $$$$\:\:\mathrm{2}{x}+\left(\mathrm{2}−\mathrm{2}\sqrt{\mathrm{2}}\right){y}=\mathrm{0} \\ $$$${x}+\left(\mathrm{1}−\sqrt{\mathrm{2}}\right){y}=\mathrm{0}……\left({ii}\right) \\ $$$${y}\:{from}\:\left({i}\right)\:{leads}\:{to} \\ $$$${x}+\left(\mathrm{1}−\sqrt{\mathrm{2}}\right)\left(\mathrm{1}+\sqrt{\mathrm{2}}\right){x}=\mathrm{0} \\ $$$${x}+\left(\mathrm{1}−\mathrm{2}\right){x}=\mathrm{0} \\ $$$${x}−{x}=\mathrm{0}\:\left({true}\right) \\ $$$$\therefore\:{We}\:{can}\:{choose}\:{eigenvector}\:\boldsymbol{{e}}_{\mathrm{1}} =\begin{bmatrix}{\mathrm{1}}\\{\mathrm{1}+\sqrt{\mathrm{2}}}\end{bmatrix}\: \\ $$$${for}\:{k}=\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}} \\ $$$$ \\ $$$${For}\:{k}=\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}} \\ $$$$\begin{bmatrix}{−\mathrm{2}+\mathrm{2}\sqrt{\mathrm{2}}\:\:\:\:\:\:\mathrm{2}}\\{\mathrm{2}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}+\mathrm{2}\sqrt{\mathrm{2}}}\end{bmatrix}\begin{bmatrix}{{x}}\\{{y}}\end{bmatrix}=\begin{bmatrix}{\mathrm{0}}\\{\mathrm{0}}\end{bmatrix} \\ $$$$\Rightarrow\left(−\mathrm{1}+\sqrt{\mathrm{2}}\right){x}+{y}=\mathrm{0}\Rightarrow{y}=\left(\mathrm{1}−\sqrt{\mathrm{2}}\right){x}\:…………..\left({iii}\right) \\ $$$${and}\:\mathrm{2}{x}+\left(\mathrm{2}+\mathrm{2}\sqrt{\mathrm{2}}\right){y}=\mathrm{0}\Rightarrow{x}+\left(\mathrm{1}+\sqrt{\mathrm{2}}\right){y}=\mathrm{0}……..\left({iv}\right) \\ $$$${y}\:{from}\:\left({iii}\right)\:{into}\:\left({iv}\right)\:{gives} \\ $$$${x}+\left(\mathrm{1}−\mathrm{2}\right){x}=\mathrm{0}\Rightarrow{x}−{x}=\mathrm{0}\:\left({true}\right) \\ $$$${So}\:{let}\:\boldsymbol{{e}}_{\mathrm{2}} =\begin{bmatrix}{\mathrm{1}}\\{\mathrm{1}−\sqrt{\mathrm{2}}}\end{bmatrix}\:{for}\:{k}=\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}. \\ $$$$\therefore\:{Let}\:{V}=\left[\boldsymbol{{e}}_{\mathrm{1}} \:\:\:\boldsymbol{{e}}_{\mathrm{2}} \right]=\begin{bmatrix}{\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}}\\{\mathrm{1}+\sqrt{\mathrm{2}}\:\:\:\:\mathrm{1}−\sqrt{\mathrm{2}}}\end{bmatrix} \\ $$$${and}\:{A}^{'} =\begin{bmatrix}{\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}\:\:\:\:\:\mathrm{0}}\\{\mathrm{0}\:\:\:\:\:\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}}\end{bmatrix}\Rightarrow{lnA}^{'} =\begin{bmatrix}{{ln}\left(\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}\right)\:\:\:\:\mathrm{0}}\\{\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{ln}\left(\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}\right)}\end{bmatrix} \\ $$$${V}^{−\mathrm{1}} =\frac{\mathrm{1}}{\mid{V}\mid}{adj}\left({V}\right) \\ $$$${adj}\left({V}\right)=\begin{bmatrix}{\mathrm{1}−\sqrt{\mathrm{2}}\:\:\:\:\:−\mathrm{1}}\\{−\mathrm{1}−\sqrt{\mathrm{2}}\:\:\:\:\:\mathrm{1}}\end{bmatrix}\:\:\mid{V}\mid=\mathrm{1}−\sqrt{\mathrm{2}}−\mathrm{1}−\sqrt{\mathrm{2}}=−\mathrm{2}\sqrt{\mathrm{2}} \\ $$$${V}^{−\mathrm{1}} =\frac{−\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\begin{bmatrix}{\mathrm{1}−\sqrt{\mathrm{2}}\:\:\:−\mathrm{1}}\\{−\mathrm{1}−\sqrt{\mathrm{2}}\:\:\:\:\mathrm{1}}\end{bmatrix} \\ $$$$\therefore\:\:{lnA}={V}\left({lnA}^{'} \right){V}^{−\mathrm{1}} \\ $$$${lnA}=\frac{−\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\begin{bmatrix}{\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}}\\{\mathrm{1}+\sqrt{\mathrm{2}}\:\:\:\:\:\mathrm{1}−\sqrt{\mathrm{2}}}\end{bmatrix}\begin{bmatrix}{{ln}\left(\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}\right)\:\:\:\mathrm{0}}\\{\mathrm{0}\:\:\:\:{ln}\left(\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}\right)}\end{bmatrix}\begin{bmatrix}{\mathrm{1}−\sqrt{\mathrm{2}}\:\:\:\:−\mathrm{1}}\\{−\mathrm{1}−\sqrt{\mathrm{2}}\:\:\:\:\:\mathrm{1}}\end{bmatrix} \\ $$$${lnA}=\frac{−\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\begin{bmatrix}{{ln}\left(\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{ln}\left(\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}\right)}\\{\left(\mathrm{1}+\sqrt{\mathrm{2}}\right){ln}\left(\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}\right)\:\:\:\left(\mathrm{1}−\sqrt{\mathrm{2}}\right){ln}\left(\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}\right)}\end{bmatrix}\begin{bmatrix}{\mathrm{1}−\sqrt{\mathrm{2}}\:\:\:\:\:\:\:\:\:−\mathrm{1}}\\{−\mathrm{1}−\sqrt{\mathrm{2}}\:\:\:\:\:\:\:\:\mathrm{1}}\end{bmatrix} \\ $$$${lnA}=\frac{−\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\begin{bmatrix}{\left(\mathrm{1}−\sqrt{\mathrm{2}}\right){ln}\left(\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}\right)−\left(\mathrm{1}+\sqrt{\mathrm{2}}\right){ln}\left(\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{ln}\left(\frac{\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}}\right)\:\:\:\:\:\:\:\:\:\:\:}\\{\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{ln}\left(\frac{\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{1}−\sqrt{\mathrm{2}}\right){ln}\left(\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}\right)−\left(\mathrm{1}+\sqrt{\mathrm{2}}\right){ln}\left(\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}\right)}\end{bmatrix} \\ $$$$\therefore\left({lnA}\right){B}=\frac{−\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\begin{bmatrix}{\left(\mathrm{1}−\sqrt{\mathrm{2}}\right){ln}\left(\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}\right)−\left(\mathrm{1}+\sqrt{\mathrm{2}}\right){ln}\left(\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{ln}\left(\frac{\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}}\right)}\\{\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{ln}\left(\frac{\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{1}−\sqrt{\mathrm{2}}\right){ln}\left(\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}\right)−\left(\mathrm{1}+\sqrt{\mathrm{2}}\right){ln}\left(\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}\right)}\end{bmatrix}×{B} \\ $$$$ \\ $$$$ \\ $$
Commented by RasheedAhmad last updated on 22/Nov/15
$$\mathcal{I}'{ve}\:{learnt}\:{something}\:{from}\:{you} \\ $$$${that}\:{I}\:{didn}'{t}\:{know}\:\:\:{before}! \\ $$
Commented by Yozzi last updated on 22/Nov/15
$${Honestly},\:{I}\:{hadn}'{t}\:{known}\:{how}\:{to}\:{find} \\ $$$${A}^{{B}\:} \:{until}\:{today}.\:{Sometime}\:{ago} \\ $$$${I}\:{looked}\:{at}\:{Wikipedia}\:{to}\:{obtain}\:{information} \\ $$$${on}\:{solving}\:{a}\:{matrix}\:{differential} \\ $$$${equation}\:{of}\:{the}\:{form}\:\frac{{d}\boldsymbol{{r}}}{{dt}}+{A}\boldsymbol{{r}}=\boldsymbol{{o}}\: \\ $$$${where}\:{it}\:{was}\:{suggested}\:{that}\:{a}\: \\ $$$${suitable}\:{matrix}\:{integrating}\:{factor} \\ $$$${be}\:{employed}.\:{In}\:\left({incorrectly}\right)\:{guessing}\:{that}\:{the} \\ $$$${required}\:{factor}\:{is}\:{of}\:{the}\:{form}\:{e}^{{A}} \\ $$$${I}\:{was}\:{met}\:{with}\:{a}\:{cool}\:{idea}\:{concerning} \\ $$$${the}\:{expression}\:{e}^{{X}\:} \:{for}\:{X}\:{being}\:{a}\:{matrix} \\ $$$${and}\:{e}\:{being}\:{Euler}'{s}\:{constant}.\:{So} \\ $$$${when}\:{I}\:{saw}\:{your}\:{question}\:{on}\:{A}^{{B}} \:{I}\:{thought} \\ $$$${about}\:{the}\:{equivalent}\:{expression}, \\ $$$${which}\:{wiki}\:{corrected}\:{me}\:{on},\:{e}^{\left({lnA}\right){B}} . \\ $$$${We}\:{have}\:{B}_{{A}} \equiv{e}^{{B}\left({lnA}\right)} \:{due}\:{to}\:{the}\:{general} \\ $$$${non}−{commutative}\:{nature}\:{of}\:{matrices}. \\ $$$${Luckily}\:{I}\:{studied}\:{some}\:{linear}\:{algebra} \\ $$$${before}\:{so}\:{applying}\:{the}\:{method}\:{I} \\ $$$${found}\:{on}\:{Wikipedia}\:{wasn}'{t}\:{too}\:{hard}. \\ $$
Commented by Rasheed Soomro last updated on 23/Nov/15
$$\mathcal{TH}{a}\mathcal{N}{k}\mathcal{S}{sss}\:{for}\:\mathcal{S}{haring}\:{thoughts}\:{with}\:{open}\:{mind}! \\ $$$$\mathcal{I}{t}\:{can}\:{be}\:{said}\:{that}\:{my}\:{question}\:{has}\:{become}\:{cause}\:{to} \\ $$$${increase}\:{your}\:{knowledge}\:{and}\:{interest}\:\mathcal{AND}\:{your}\: \\ $$$$\mathcal{S}{haring}−\mathcal{K}{nowledge}\:{has}\:{become}\:{cause}\:{to}\:{increase} \\ $$$${my}\:{knowledge}! \\ $$$$\mathcal{A}{ctually}\:{the}\:{question}\:{is}\:{result}\:{of}\:{my}\:{imagination}, \\ $$$${otherwise}\:{my}\:{knowledge}\:{regarding}\:\mathcal{M}{atrices}\:{is}\:{of} \\ $$$${elementary}\:{level}. \\ $$$$\mathcal{Y}{our}\:{comment}\:{contains}\:{an}\:{expression}\:'\:{B}_{\:{A}} \equiv{e}^{{B}\left({ln}\:{A}\right)} …{due}\:{to} \\ $$$${non}−{commutative}\:{nature}…'.\:\:\mathcal{I}\:{would}\:{like}\:{to}\:{know} \\ $$$${the}\:{meaning}\:{of}\:{B}_{\:{A}} .\:{Considering}\:{non}−{commutative} \\ $$$${nature}\:…{shouldn}'{t}\:{we}\:{write}\:\:\:{B}^{{A}} \:{as}\:{non}−{commutative} \\ $$$${part}\:{of}\:{A}^{{B}} \:? \\ $$
Commented by Yozzi last updated on 23/Nov/15
$${The}\:{wiki}\:{indicated}\:{that}\:{while} \\ $$$${A}^{{B}} ={e}^{\left({lnA}\right){B}} ,\:\:^{{A}} {B}={e}^{{B}\left({lnA}\right)} . \\ $$$${I}\:{just}\:{figured}\:{out}\:{how}\:{to}\:{obtain}\: \\ $$$${left}−{hand}\:{exponents}.\:{I}\:{was}\:{trying} \\ $$$${to}\:{do}\:{so}\:{by}\:{writting}\:{B}_{{A}} … \\ $$
Commented by Rasheed Soomro last updated on 23/Nov/15
$$\mathbb{THANKS}\:! \\ $$