let-a-gt-b-gt-c-gt-0-solve-in-R-ax-by-cz-a-bx-cy-az-b-cx-ay-bz-c-

Question Number 158724 by HongKing last updated on 08/Nov/21

let a > b > c > 0 solve in R

{\begin{cases} ax + by + cz = a \\ bx + cy + az = b \\ cx + ay + bz = c \end{cases}

Answered by ajfour last updated on 08/Nov/21

x + y + z = 1

a x + b y + c (1 - x - y) = a

\Rightarrow (a - c) x + (b - c) y = a - c

x + (\frac{b - c}{a - c}) y = 1

(a - c) x + (b - a) y

+ (c - b) (1 - x - y) = a - c

\Rightarrow (a + b - 2 c) x + (2 b - a - c) y

= a + b - 2 c

\Rightarrow x - y = 1

\Rightarrow y + 1 + (\frac{b - c}{a - c}) y = 1

\Rightarrow y = 0, x = 1, z = 0 \cdot

Commented by HongKing last updated on 08/Nov/21

thank you dear Ser cool

Answered by som(math1967) last updated on 08/Nov/21

D = | \begin{matrix} a & b & c \\ b & c & a \\ c & a & b \end{matrix} | = a (b c - a^{2}) - b (b^{2} - c a) + c (a b - c^{2})

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

D_{1} = | \begin{matrix} a & b & c \\ b & c & a \\ c & a & b \end{matrix} | = a b c - a^{3} - b^{3} + a b c + a b c - c^{3}

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

D_{2} = | \begin{matrix} a & a & c \\ b & b & a \\ c & c & b \end{matrix} | = 0 [C_{1,} C_{2} i d e n t i c a l]

D_{3} = | \begin{matrix} a & b & a \\ b & c & b \\ c & a & c \end{matrix} | = 0 [C_{1}, C_{3} i d e n t i c a l]

x = \frac{D_{1}}{D} = 1, y = \frac{D_{2}}{D} = 0, z = \frac{D_{3}}{D} = 0

a > b > c > 0 ∴ (3 a b c - a^{3} - b^{3} - c^{3}) \neq 0

Commented by HongKing last updated on 08/Nov/21

thank you dear Ser cool