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Question Number 56090 by Tawa1 last updated on 10/Mar/19

Commented by math1967 last updated on 10/Mar/19

pls check I think it should be   determinant (((bc),(ca),(ab)),(a,b,c),(a^2 ,b^2 ,c^2 ))

$${pls}\:{check}\:{I}\:{think}\:{it}\:{should}\:{be} \\ $$$$\begin{vmatrix}{{bc}}&{{ca}}&{{ab}}\\{{a}}&{{b}}&{{c}}\\{{a}^{\mathrm{2}} }&{{b}^{\mathrm{2}} }&{{c}^{\mathrm{2}} }\end{vmatrix} \\ $$

Commented by Tawa1 last updated on 10/Mar/19

Please help me do it like this

$$\mathrm{Please}\:\mathrm{help}\:\mathrm{me}\:\mathrm{do}\:\mathrm{it}\:\mathrm{like}\:\mathrm{this}\: \\ $$

Commented by Tawa1 last updated on 10/Mar/19

But that is the question.

$$\mathrm{But}\:\mathrm{that}\:\mathrm{is}\:\mathrm{the}\:\mathrm{question}. \\ $$

Commented by math1967 last updated on 10/Mar/19

 determinant (((bc−ca),(ca−ab),(ab)),((a−b),(b−c),c),((a^2 −b^2 ),(b^2 −c^(2 ) ),(c^(2 )   )))c_(1  ) −c_2 ,c_(2 ) −c_(3 )   (a−b)(b−c) determinant (((−c),(−a),(ab)),(1,1,c),((a+b),(b+c),c^2 ))  =(a−b)(b−c) determinant (((−c),(c−a),(ab)),(1,0,c),((a+b ),(c−a),c^2 ))c_2 −c_1   =(a−b)(b−c)(c−a) determinant (((−c),1,(ab)),(1,0,c),((a+b),1,c^2 ))  =(a−b)(b−c)(c−a){c^2 −c^2 +ca+bc+ab}  =(b−a)(c−b)(c−a)(ab+bc+ca)

$$\begin{vmatrix}{{bc}−{ca}}&{{ca}−{ab}}&{{ab}}\\{{a}−{b}}&{{b}−{c}}&{{c}}\\{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }&{{b}^{\mathrm{2}} −{c}^{\mathrm{2}\:} }&{{c}^{\mathrm{2}\:} \:\:}\end{vmatrix}{c}_{\mathrm{1}\:\:} −{c}_{\mathrm{2}} ,{c}_{\mathrm{2}\:} −{c}_{\mathrm{3}\:} \\ $$$$\left({a}−{b}\right)\left({b}−{c}\right)\begin{vmatrix}{−{c}}&{−{a}}&{{ab}}\\{\mathrm{1}}&{\mathrm{1}}&{{c}}\\{{a}+{b}}&{{b}+{c}}&{{c}^{\mathrm{2}} }\end{vmatrix} \\ $$$$=\left({a}−{b}\right)\left({b}−{c}\right)\begin{vmatrix}{−{c}}&{{c}−{a}}&{{ab}}\\{\mathrm{1}}&{\mathrm{0}}&{{c}}\\{{a}+{b}\:}&{{c}−{a}}&{{c}^{\mathrm{2}} }\end{vmatrix}{c}_{\mathrm{2}} −{c}_{\mathrm{1}} \\ $$$$=\left({a}−{b}\right)\left({b}−{c}\right)\left({c}−{a}\right)\begin{vmatrix}{−{c}}&{\mathrm{1}}&{{ab}}\\{\mathrm{1}}&{\mathrm{0}}&{{c}}\\{{a}+{b}}&{\mathrm{1}}&{{c}^{\mathrm{2}} }\end{vmatrix} \\ $$$$=\left({a}−{b}\right)\left({b}−{c}\right)\left({c}−{a}\right)\left\{{c}^{\mathrm{2}} −{c}^{\mathrm{2}} +{ca}+{bc}+{ab}\right\} \\ $$$$=\left({b}−{a}\right)\left({c}−{b}\right)\left({c}−{a}\right)\left({ab}+{bc}+{ca}\right) \\ $$

Commented by Tawa1 last updated on 11/Mar/19

God bless you sir

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

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