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 = (\begin{matrix} a b c \\ b c a \\ c a b \end{matrix}) and a, b, c > 0

such that abc = 1 and A^{T} . A = I

find a^{3} + b^{3} + c^{3} - 3 abc .

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

A . A^{T} = I

(\begin{matrix} a & b & c \\ b & c & a \\ c & a & b \end{matrix}) (\begin{matrix} a & b & c \\ b & c & a \\ c & a & b \end{matrix}) = (\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix})

(\begin{matrix} a^{2} + b^{2} + c^{2} & a b + b c + c a & a b + b c + c a \\ a b + b c + c a & a^{2} + b^{2} + c^{2} & a b + b c + c a \\ a b + b c + c a & a b + b c + c a & a^{2} + b^{2} + c^{2} \end{matrix})

= (\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix})

∴ a^{2} + b^{2} + c^{2} = 1

a b + b c + c a = 0

{(a + b + c)}^{2} = a^{2} + b^{2} + c^{2} + 2 (a b + b c + c a)

(a + b + c) = \sqrt{1 + 0} = 1 [a, b, c > 0]

a^{3} + b^{3} + c^{3} - 3 a b c

(a + b + c) (a^{2} + b^{2} + c^{2} - a b - b c - c a)

= 1 \times 1 = 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}) = \pm 1

3 abc - (a^{3} + b^{3} + c^{3}) = \pm 1

\begin{matrix} a^{3} + b^{3} + c^{3} - 3 abc = \mp 1 \end{matrix}

Commented by manxsol last updated on 14/Jul/23

m u y b i e n

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