If-A-a-b-c-b-c-a-c-a-b-and-a-b-c-gt-0-such-that-abc-1-and-A-T-A-I-find-a-3-b-3-c-3-3abc-

Question Number 194642 by horsebrand11 last updated on 12/Jul/23

If A= (((a b c)),((b c a)),((c a b)) ) and a,b,c >0 such that abc=1 and A^T .A=I find a^3 +b^3 +c^3 −3abc .

$$\:\mathrm{If}\:\mathrm{A}=\begin{pmatrix}{\mathrm{a}\:\:\:\:\mathrm{b}\:\:\:\:\:\:\mathrm{c}}\\{\mathrm{b}\:\:\:\:\mathrm{c}\:\:\:\:\:\:\mathrm{a}}\\{\mathrm{c}\:\:\:\:\:\mathrm{a}\:\:\:\:\:\:\mathrm{b}}\end{pmatrix}\:\mathrm{and}\:\mathrm{a},\mathrm{b},\mathrm{c}\:>\mathrm{0} \\ $$$$\:\:\mathrm{such}\:\mathrm{that}\:\mathrm{abc}=\mathrm{1}\:\mathrm{and}\:\mathrm{A}^{\mathrm{T}} .\mathrm{A}=\mathrm{I} \\ $$$$\:\mathrm{find}\:\mathrm{a}^{\mathrm{3}} +\mathrm{b}^{\mathrm{3}} +\mathrm{c}^{\mathrm{3}} −\mathrm{3abc}\:. \\ $$

Answered by som(math1967) last updated on 12/Jul/23

A.A^T =I ((a,b,c),(b,c,a),(c,a,b) ) ((a,b,c),(b,c,a),(c,a,b) )= ((1,0,0),(0,1,0),(0,0,1) ) (((a^2 +b^2 +c^2 ),(ab+bc+ca),(ab+bc+ca)),((ab+bc+ca),(a^2 +b^2 +c^2 ),(ab+bc+ca)),((ab+bc+ca),(ab+bc+ca),(a^2 +b^2 +c^2 )) ) = ((1,0,0),(0,1,0),(0,0,1) ) ∴a^2 +b^2 +c^2 =1 ab+bc+ca=0 (a+b+c)^2 =a^2 +b^2 +c^2 +2(ab+bc+ca) (a+b+c)=(√(1+0))=1 [a,b,c>0] a^3 +b^3 +c^3 −3abc (a+b+c)(a^2 +b^2 +c^2 −ab−bc−ca) =1×1=1

$$\:{A}.{A}^{{T}} ={I} \\ $$$$\begin{pmatrix}{{a}}&{{b}}&{{c}}\\{{b}}&{{c}}&{{a}}\\{{c}}&{{a}}&{{b}}\end{pmatrix}\begin{pmatrix}{{a}}&{{b}}&{{c}}\\{{b}}&{{c}}&{{a}}\\{{c}}&{{a}}&{{b}}\end{pmatrix}=\begin{pmatrix}{\mathrm{1}}&{\mathrm{0}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{1}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{0}}&{\mathrm{1}}\end{pmatrix} \\ $$$$\begin{pmatrix}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }&{{ab}+{bc}+{ca}}&{{ab}+{bc}+{ca}}\\{{ab}+{bc}+{ca}}&{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }&{{ab}+{bc}+{ca}}\\{{ab}+{bc}+{ca}}&{{ab}+{bc}+{ca}}&{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }\end{pmatrix} \\ $$$$=\begin{pmatrix}{\mathrm{1}}&{\mathrm{0}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{1}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{0}}&{\mathrm{1}}\end{pmatrix} \\ $$$$\therefore{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} =\mathrm{1} \\ $$$${ab}+{bc}+{ca}=\mathrm{0} \\ $$$$\left({a}+{b}+{c}\right)^{\mathrm{2}} ={a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +\mathrm{2}\left({ab}+{bc}+{ca}\right) \\ $$$$\left({a}+{b}+{c}\right)=\sqrt{\mathrm{1}+\mathrm{0}}=\mathrm{1}\:\:\:\:\:\left[{a},{b},{c}>\mathrm{0}\right] \\ $$$${a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} −\mathrm{3}{abc} \\ $$$$\left({a}+{b}+{c}\right)\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{ab}−{bc}−{ca}\right) \\ $$$$=\mathrm{1}×\mathrm{1}=\mathrm{1} \\ $$

Answered by cortano12 last updated on 12/Jul/23

A^T A=I ∣A∣^2 = 1 ∣A∣=a(bc−a^2 )−b(b^2 −ac)+c(ab−c^2 )=±1 3abc−(a^3 +b^3 +c^3 )= ± 1 determinant (((a^3 +b^3 +c^3 −3abc=∓ 1)))

$$\:\:\mathrm{A}^{\mathrm{T}} \mathrm{A}=\mathrm{I} \\ $$$$\:\mid\mathrm{A}\mid^{\mathrm{2}} =\:\mathrm{1} \\ $$$$\:\mid\mathrm{A}\mid=\mathrm{a}\left(\mathrm{bc}−\mathrm{a}^{\mathrm{2}} \right)−\mathrm{b}\left(\mathrm{b}^{\mathrm{2}} −\mathrm{ac}\right)+\mathrm{c}\left(\mathrm{ab}−\mathrm{c}^{\mathrm{2}} \right)=\pm\mathrm{1} \\ $$$$\:\mathrm{3abc}−\left(\mathrm{a}^{\mathrm{3}} +\mathrm{b}^{\mathrm{3}} +\mathrm{c}^{\mathrm{3}} \right)=\:\pm\:\mathrm{1} \\ $$$$\:\begin{array}{|c|}{\mathrm{a}^{\mathrm{3}} +\mathrm{b}^{\mathrm{3}} +\mathrm{c}^{\mathrm{3}} −\mathrm{3abc}=\mp\:\mathrm{1}}\\\hline\end{array} \\ $$

Commented by manxsol last updated on 14/Jul/23

$${muy}\:{bien} \\ $$

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