find-determinant-a-2-b-2-c-c-c-a-b-2-c-2-a-a-b-b-a-2-c-2-b-

Question Number 118090 by bemath last updated on 15/Oct/20

find | \begin{matrix} \frac{a^{2} + b^{2}}{c} c c \\ a \frac{b^{2} + c^{2}}{a} a \\ b b \frac{a^{2} + c^{2}}{b} \end{matrix} | = ?

Commented by bemath last updated on 15/Oct/20

yes . . thank you all

Answered by MJS_new last updated on 15/Oct/20

just calculate it ?!

= 4 a b c

Answered by som(math1967) last updated on 15/Oct/20

\frac{1}{abc} | \begin{matrix} a^{2} + b^{2} & c^{2} & c^{2} \\ a^{2} & b^{2} + c^{2} & a^{2} \\ b^{2} & b^{2} & c^{2} + a^{2} \end{matrix} |

R_{1} \to R_{1} - R_{2} - R_{3}

\frac{1}{abc} | \begin{matrix} 0 & - {2 b}^{2} & - {2 a}^{2} \\ a^{2} & b^{2} + c^{2} & a^{2} \\ b^{2} & b^{2} & c^{2} + a^{2} \end{matrix} |

\frac{- 2}{abc} | \begin{matrix} 0 & b^{2} & a^{2} \\ a^{2} & b^{2} + c^{2} & a^{2} \\ b^{2} & b^{2} & c^{2} + a^{2} \end{matrix} |

R_{2} \to R_{2} - R_{1}, R_{3} \to R_{3} - R_{1}

\frac{- 2}{abc} | \begin{matrix} 0 & b^{2} & a^{2} \\ a^{2} & c^{2} & 0 \\ b^{2} & 0 & c^{2} \end{matrix} |

\frac{- 2}{abc} [0 - b^{2} (a^{2} c^{2} - 0) + a^{2} (0 - b^{2} c^{2})]

\frac{- 2 \times - {2 a}^{2} b^{2} c^{2}}{abc} = 4 abc ans

Answered by FelipeLz last updated on 15/Oct/20

\frac{a^{2} + b^{2}}{c} [\frac{a^{2} b^{2} + a^{2} c^{2} + b^{2} c^{2} + c^{4}}{a b} - a b] - c [\frac{a^{3} + {a c}^{2}}{b} - a b] + c [a b - \frac{b^{3} - {b c}^{2}}{a}]

\frac{a^{2} + b^{2}}{c} [\frac{a^{2} b^{2} + a^{2} c^{2} + b^{2} c^{2} + c^{4}}{a b} - a b] \frac{a b}{a b} + c [2 a b - \frac{b^{3} + {b c}^{2}}{a} - \frac{a^{3} + {a c}^{2}}{b}] \frac{a b}{a b}

\frac{(a^{2} + b^{2}) c}{a b} [a^{2} + b^{2} + c^{2}] + \frac{c}{a b} [2 a^{2} b^{2} - b^{4} - b^{2} c^{2} - a^{4} - a^{2} c^{2}]

\frac{c}{a b} [(a^{2} + b^{2}) (a^{2} + b^{2} + c^{2}) - {(a^{2} - b^{2})}^{2} - c^{2} (a^{2} + b^{2})]

\frac{c}{a b} [(a^{2} + b^{2}) (a^{2} + b^{2}) - {(a^{2} - b^{2})}^{2}]

\frac{c}{a b} [a^{4} + 2 a^{2} b^{2} + b^{4} - a^{4} + 2 a^{2} b^{2} - b^{4}]

\frac{4 a^{2} b^{2} c}{a b} = 4 a b c

Answered by 1549442205PVT last updated on 15/Oct/20

By rule to calculate value of a

determinant of degree 3 we have :

Δ = \frac{(a^{2} + b^{2}) (b^{2} + c^{2}) (c^{2} + a^{2})}{abc} + 2 abc

- \frac{bc (b^{2} + c^{2})}{a} - \frac{ac (a^{2} + c^{2})}{b} - \frac{ab (a^{2} + b^{2})}{c}

= \frac{1}{abc} [(a^{2} + b^{2}) (b^{2} + c^{2}) (c^{2} + a^{2})

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

- {(ab)}^{2} (a^{2} + b^{2})] + 2 abc

\frac{1}{abc} [a^{4} (b^{2} + c^{2}) + b^{4} (c^{2} + a^{2}) + c^{4} (a^{2} + b^{2})

+ {2 a}^{2} b^{2} c^{2} - b^{4} (c^{2} + a^{2}) - c^{4} (a^{2} + b^{2})

- a^{4} (b^{2} + c^{2})] + 2 abc

= \frac{1}{abc} . 2 {(abc)}^{2} + 2 abc = 4 abc