Question Number 118090 by bemath last updated on 15/Oct/20
$$\mathrm{find}\:\begin{vmatrix}{\frac{\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} }{\mathrm{c}}\:\:\:\:\:\:\:\:\mathrm{c}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{c}}\\{\:\:\:\:\:\mathrm{a}\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{b}^{\mathrm{2}} +\mathrm{c}^{\mathrm{2}} }{\mathrm{a}}\:\:\:\:\:\:\:\mathrm{a}}\\{\:\:\:\:\:\mathrm{b}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{b}\:\:\:\:\:\:\:\:\frac{\mathrm{a}^{\mathrm{2}} +\mathrm{c}^{\mathrm{2}} }{\mathrm{b}}}\end{vmatrix}=?\: \\ $$
Commented by bemath last updated on 15/Oct/20
$$\mathrm{yes}..\mathrm{thank}\:\mathrm{you}\:\mathrm{all} \\ $$
Answered by MJS_new last updated on 15/Oct/20
$$\mathrm{just}\:\mathrm{calculate}\:\mathrm{it}?! \\ $$$$=\mathrm{4}{abc} \\ $$
Answered by som(math1967) last updated on 15/Oct/20
![(1/(abc)) determinant (((a^2 +b^2 ),c^2 ,c^2 ),(a^2 ,(b^2 +c^2 ),a^2 ),(b^2 ,b^2 ,(c^2 +a^2 ))) R_1 →R_1 −R_2 −R_3 (1/(abc)) determinant ((0,(−2b^2 ),(−2a^2 )),(a^2 ,(b^2 +c^2 ),a^2 ),(b^(2 ) ,b^2 ,(c^2 +a^2 ))) ((−2)/(abc)) determinant ((0,b^2 ,a^2 ),(a^2 ,(b^2 +c^2 ),a^2 ),(b^(2 ) ,b^2 ,(c^2 +a^2 ))) R_2 →R_2 −R_1 ,R_3 →R_3 −R_1 ((−2)/(abc)) determinant ((0,b^2 ,a^2 ),(a^2 ,c^2 ,0),(b^2 ,0,c^2 )) ((−2)/(abc))[0−b^2 (a^2 c^2 −0)+a^2 (0−b^2 c^2 )] ((−2×−2a^2 b^2 c^2 )/(abc))=4abc ans](https://www.tinkutara.com/question/Q118103.png)
$$\frac{\mathrm{1}}{\mathrm{abc}}\begin{vmatrix}{\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} }&{\mathrm{c}^{\mathrm{2}} }&{\mathrm{c}^{\mathrm{2}} }\\{\mathrm{a}^{\mathrm{2}} }&{\mathrm{b}^{\mathrm{2}} +\mathrm{c}^{\mathrm{2}} }&{\mathrm{a}^{\mathrm{2}} }\\{\mathrm{b}^{\mathrm{2}} }&{\mathrm{b}^{\mathrm{2}} }&{\mathrm{c}^{\mathrm{2}} +\mathrm{a}^{\mathrm{2}} }\end{vmatrix} \\ $$$$\mathrm{R}_{\mathrm{1}} \rightarrow\mathrm{R}_{\mathrm{1}} −\mathrm{R}_{\mathrm{2}} −\mathrm{R}_{\mathrm{3}} \\ $$$$\frac{\mathrm{1}}{\mathrm{abc}}\begin{vmatrix}{\mathrm{0}}&{−\mathrm{2b}^{\mathrm{2}} }&{−\mathrm{2a}^{\mathrm{2}} }\\{\mathrm{a}^{\mathrm{2}} }&{\mathrm{b}^{\mathrm{2}} +\mathrm{c}^{\mathrm{2}} }&{\mathrm{a}^{\mathrm{2}} }\\{\mathrm{b}^{\mathrm{2}\:} }&{\mathrm{b}^{\mathrm{2}} }&{\mathrm{c}^{\mathrm{2}} +\mathrm{a}^{\mathrm{2}} }\end{vmatrix} \\ $$$$\frac{−\mathrm{2}}{\mathrm{abc}}\begin{vmatrix}{\mathrm{0}}&{\mathrm{b}^{\mathrm{2}} }&{\mathrm{a}^{\mathrm{2}} }\\{\mathrm{a}^{\mathrm{2}} }&{\mathrm{b}^{\mathrm{2}} +\mathrm{c}^{\mathrm{2}} }&{\mathrm{a}^{\mathrm{2}} }\\{\mathrm{b}^{\mathrm{2}\:} }&{\mathrm{b}^{\mathrm{2}} }&{\mathrm{c}^{\mathrm{2}} +\mathrm{a}^{\mathrm{2}} }\end{vmatrix} \\ $$$$\mathrm{R}_{\mathrm{2}} \rightarrow\mathrm{R}_{\mathrm{2}} −\mathrm{R}_{\mathrm{1}} ,\mathrm{R}_{\mathrm{3}} \rightarrow\mathrm{R}_{\mathrm{3}} −\mathrm{R}_{\mathrm{1}} \\ $$$$\frac{−\mathrm{2}}{\mathrm{abc}}\begin{vmatrix}{\mathrm{0}}&{\mathrm{b}^{\mathrm{2}} }&{\mathrm{a}^{\mathrm{2}} }\\{\mathrm{a}^{\mathrm{2}} }&{\mathrm{c}^{\mathrm{2}} }&{\mathrm{0}}\\{\mathrm{b}^{\mathrm{2}} }&{\mathrm{0}}&{\mathrm{c}^{\mathrm{2}} }\end{vmatrix} \\ $$$$\frac{−\mathrm{2}}{\mathrm{abc}}\left[\mathrm{0}−\mathrm{b}^{\mathrm{2}} \left(\mathrm{a}^{\mathrm{2}} \mathrm{c}^{\mathrm{2}} −\mathrm{0}\right)+\mathrm{a}^{\mathrm{2}} \left(\mathrm{0}−\mathrm{b}^{\mathrm{2}} \mathrm{c}^{\mathrm{2}} \right)\right] \\ $$$$\frac{−\mathrm{2}×−\mathrm{2a}^{\mathrm{2}} \mathrm{b}^{\mathrm{2}} \mathrm{c}^{\mathrm{2}} }{\mathrm{abc}}=\mathrm{4abc}\:\mathrm{ans} \\ $$$$ \\ $$
Answered by FelipeLz last updated on 15/Oct/20
![((a^2 +b^2 )/c)[((a^2 b^2 +a^2 c^2 +b^2 c^2 +c^4 )/(ab))−ab]−c[((a^3 +ac^2 )/b)−ab]+c[ab−((b^3 −bc^2 )/a)] ((a^2 +b^2 )/c)[((a^2 b^2 +a^2 c^2 +b^2 c^2 +c^4 )/(ab))−ab]((ab)/(ab))+c[2ab−((b^3 +bc^2 )/a)−((a^3 +ac^2 )/b)]((ab)/(ab)) (((a^2 +b^2 )c)/(ab))[a^2 +b^2 +c^2 ]+(c/(ab))[2a^2 b^2 −b^4 −b^2 c^2 −a^4 −a^2 c^2 ] (c/(ab))[(a^2 +b^2 )(a^2 +b^2 +c^2 )−(a^2 −b^2 )^2 −c^2 (a^2 +b^2 )] (c/(ab))[(a^2 +b^2 )(a^2 +b^2 )−(a^2 −b^2 )^2 ] (c/(ab))[a^4 +2a^2 b^2 +b^4 −a^4 +2a^2 b^2 −b^4 ] ((4a^2 b^2 c)/(ab)) = 4abc](https://www.tinkutara.com/question/Q118107.png)
$$\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }{{c}}\left[\frac{{a}^{\mathrm{2}} {b}^{\mathrm{2}} +{a}^{\mathrm{2}} {c}^{\mathrm{2}} +{b}^{\mathrm{2}} {c}^{\mathrm{2}} +{c}^{\mathrm{4}} }{{ab}}−{ab}\right]−{c}\left[\frac{{a}^{\mathrm{3}} +{ac}^{\mathrm{2}} }{{b}}−{ab}\right]+{c}\left[{ab}−\frac{{b}^{\mathrm{3}} −{bc}^{\mathrm{2}} }{{a}}\right] \\ $$$$\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }{{c}}\left[\frac{{a}^{\mathrm{2}} {b}^{\mathrm{2}} +{a}^{\mathrm{2}} {c}^{\mathrm{2}} +{b}^{\mathrm{2}} {c}^{\mathrm{2}} +{c}^{\mathrm{4}} }{{ab}}−{ab}\right]\frac{{ab}}{{ab}}+{c}\left[\mathrm{2}{ab}−\frac{{b}^{\mathrm{3}} +{bc}^{\mathrm{2}} }{{a}}−\frac{{a}^{\mathrm{3}} +{ac}^{\mathrm{2}} }{{b}}\right]\frac{{ab}}{{ab}} \\ $$$$\frac{\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right){c}}{{ab}}\left[{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right]+\frac{{c}}{{ab}}\left[\mathrm{2}{a}^{\mathrm{2}} {b}^{\mathrm{2}} −{b}^{\mathrm{4}} −{b}^{\mathrm{2}} {c}^{\mathrm{2}} −{a}^{\mathrm{4}} −{a}^{\mathrm{2}} {c}^{\mathrm{2}} \right] \\ $$$$\frac{{c}}{{ab}}\left[\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)−\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)^{\mathrm{2}} −{c}^{\mathrm{2}} \left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)\right] \\ $$$$\frac{{c}}{{ab}}\left[\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)−\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)^{\mathrm{2}} \right] \\ $$$$\frac{{c}}{{ab}}\left[{a}^{\mathrm{4}} +\mathrm{2}{a}^{\mathrm{2}} {b}^{\mathrm{2}} +{b}^{\mathrm{4}} −{a}^{\mathrm{4}} +\mathrm{2}{a}^{\mathrm{2}} {b}^{\mathrm{2}} −{b}^{\mathrm{4}} \right] \\ $$$$\frac{\mathrm{4}{a}^{\mathrm{2}} {b}^{\mathrm{2}} {c}}{{ab}}\:=\:\mathrm{4}{abc} \\ $$
Answered by 1549442205PVT last updated on 15/Oct/20
![By rule to calculate value of a determinant of degree 3 we have: Δ=(((a^2 +b^2 )(b^2 +c^2 )(c^2 +a^2 ))/(abc))+2abc −((bc(b^2 +c^2 ))/a)−((ac(a^2 +c^2 ))/b)−((ab(a^2 +b^2 ))/c) =(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 )]+2abc (1/(abc))[a^4 (b^2 +c^2 )+b^4 (c^2 +a^2 )+c^4 (a^2 +b^2 ) +2a^2 b^2 c^2 −b^4 (c^2 +a^2 )−c^4 (a^2 +b^2 ) −a^4 (b^2 +c^2 )]+2abc =(1/(abc)).2(abc)^2 +2abc=4abc](https://www.tinkutara.com/question/Q118112.png)
$$\mathrm{By}\:\mathrm{rule}\:\mathrm{to}\:\mathrm{calculate}\:\mathrm{value}\:\mathrm{of}\:\mathrm{a}\: \\ $$$$\:\mathrm{determinant}\:\mathrm{of}\:\mathrm{degree}\:\mathrm{3}\:\mathrm{we}\:\mathrm{have}: \\ $$$$\Delta=\frac{\left(\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} \right)\left(\mathrm{b}^{\mathrm{2}} +\mathrm{c}^{\mathrm{2}} \right)\left(\mathrm{c}^{\mathrm{2}} +\mathrm{a}^{\mathrm{2}} \right)}{\mathrm{abc}}+\mathrm{2abc} \\ $$$$−\frac{\mathrm{bc}\left(\mathrm{b}^{\mathrm{2}} +\mathrm{c}^{\mathrm{2}} \right)}{\mathrm{a}}−\frac{\mathrm{ac}\left(\mathrm{a}^{\mathrm{2}} +\mathrm{c}^{\mathrm{2}} \right)}{\mathrm{b}}−\frac{\mathrm{ab}\left(\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} \right)}{\mathrm{c}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{abc}}\left[\left(\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} \right)\left(\mathrm{b}^{\mathrm{2}} +\mathrm{c}^{\mathrm{2}} \right)\left(\mathrm{c}^{\mathrm{2}} +\mathrm{a}^{\mathrm{2}} \right)\right. \\ $$$$−\mathrm{b}^{\mathrm{2}} \mathrm{c}^{\mathrm{2}} \left(\mathrm{b}^{\mathrm{2}} +\mathrm{c}^{\mathrm{2}} \right)−\left(\mathrm{ca}\right)^{\mathrm{2}} \left(\mathrm{c}^{\mathrm{2}} +\mathrm{a}^{\mathrm{2}} \right) \\ $$$$\left.−\left(\mathrm{ab}\right)^{\mathrm{2}} \left(\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} \right)\right]+\mathrm{2abc} \\ $$$$\frac{\mathrm{1}}{\mathrm{abc}}\left[\mathrm{a}^{\mathrm{4}} \left(\mathrm{b}^{\mathrm{2}} +\mathrm{c}^{\mathrm{2}} \right)+\mathrm{b}^{\mathrm{4}} \left(\mathrm{c}^{\mathrm{2}} +\mathrm{a}^{\mathrm{2}} \right)+\mathrm{c}^{\mathrm{4}} \left(\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} \right)\right. \\ $$$$+\mathrm{2a}^{\mathrm{2}} \mathrm{b}^{\mathrm{2}} \mathrm{c}^{\mathrm{2}} −\mathrm{b}^{\mathrm{4}} \left(\mathrm{c}^{\mathrm{2}} +\mathrm{a}^{\mathrm{2}} \right)−\mathrm{c}^{\mathrm{4}} \left(\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} \right) \\ $$$$\left.−\mathrm{a}^{\mathrm{4}} \left(\mathrm{b}^{\mathrm{2}} +\mathrm{c}^{\mathrm{2}} \right)\right]+\mathrm{2abc} \\ $$$$=\frac{\mathrm{1}}{\mathrm{abc}}.\mathrm{2}\left(\mathrm{abc}\right)^{\mathrm{2}} +\mathrm{2abc}=\mathrm{4abc} \\ $$