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Question Number 175110 by nadovic last updated on 19/Aug/22
Prove that determinant ((1,1,1),(a,b,c),(a^2 ,b^2 ,c^2 ))= (a−b)(b−c)(c−a)
$$\mathrm{Prove}\:\mathrm{that}\begin{vmatrix}{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}\\{{a}}&{{b}}&{{c}}\\{{a}^{\mathrm{2}} }&{{b}^{\mathrm{2}} }&{{c}^{\mathrm{2}} }\end{vmatrix}=\:\left({a}−{b}\right)\left({b}−{c}\right)\left({c}−{a}\right) \\ $$
Commented by kaivan.ahmadi last updated on 19/Aug/22
determinant (((1 1 1)),((a b c)),((a^2 b^2 c^2 ))) determinant (((1 1 1)),((a b c)),((a^2 b^2 c^2 )))= (bc^2 +ca^2 +ab^2 )−(ba^2 +ac^2 +cb^2 ) =c(bc+a^2 −ac−b^2 )−ab(a−b) =c(c(b−a)+(a−b)(a+b))−ab(a−b) =(a−b)(−c^2 +ac+bc−ab) =(a−b)(c(a−c)+b(c−a)) =(a−b)((c−a)(b−c))= (a−b)(b−c)(c−a) note that this is a Vandermonde determinant. In general case determinant (((1 1 ... 1)),((a_(1 ) a_(2 ) ... a_n )),((⋮ ⋮ ⋮)),((a_1 ^(n−1) ... a_n ^(n−1) )))=Π_(1≤i<j≤n) (a_j −a_i )
$$\begin{vmatrix}{\mathrm{1}\:\:\:\:\:\mathrm{1}\:\:\:\:\:\mathrm{1}}\\{{a}\:\:\:\:\:{b}\:\:\:\:\:{c}}\\{{a}^{\mathrm{2}} \:\:\:{b}^{\mathrm{2}} \:\:\:{c}^{\mathrm{2}} }\end{vmatrix}\begin{vmatrix}{\mathrm{1}\:\:\:\:\:\mathrm{1}\:\:\:\:\:\mathrm{1}}\\{{a}\:\:\:\:\:{b}\:\:\:\:\:{c}}\\{{a}^{\mathrm{2}} \:\:{b}^{\mathrm{2}} \:\:\:{c}^{\mathrm{2}} }\end{vmatrix}= \\ $$$$\left({bc}^{\mathrm{2}} +{ca}^{\mathrm{2}} +{ab}^{\mathrm{2}} \right)−\left({ba}^{\mathrm{2}} +{ac}^{\mathrm{2}} +{cb}^{\mathrm{2}} \right) \\ $$$$={c}\left({bc}+{a}^{\mathrm{2}} −{ac}−{b}^{\mathrm{2}} \right)−{ab}\left({a}−{b}\right) \\ $$$$={c}\left({c}\left({b}−{a}\right)+\left({a}−{b}\right)\left({a}+{b}\right)\right)−{ab}\left({a}−{b}\right) \\ $$$$=\left({a}−{b}\right)\left(−{c}^{\mathrm{2}} +{ac}+{bc}−{ab}\right) \\ $$$$=\left({a}−{b}\right)\left({c}\left({a}−{c}\right)+{b}\left({c}−{a}\right)\right) \\ $$$$=\left({a}−{b}\right)\left(\left({c}−{a}\right)\left({b}−{c}\right)\right)= \\ $$$$\left({a}−{b}\right)\left({b}−{c}\right)\left({c}−{a}\right) \\ $$$${note}\:{that}\:{this}\:{is}\:{a}\:{Vandermonde} \\ $$$${determinant}. \\ $$$${In}\:{general}\:{case} \\ $$$$\begin{vmatrix}{\mathrm{1}\:\:\:\:\:\:\mathrm{1}\:\:\:\:…\:\:\:\:\mathrm{1}}\\{{a}_{\mathrm{1}\:\:\:} \:{a}_{\mathrm{2}\:\:} …\:\:\:\:\:{a}_{{n}} }\\{\vdots\:\:\:\vdots\:\:\:\:\vdots}\\{{a}_{\mathrm{1}} ^{{n}−\mathrm{1}} \:\:…\:\:\:\:\:{a}_{{n}} ^{{n}−\mathrm{1}} \:}\end{vmatrix}=\underset{\mathrm{1}\leqslant{i}<{j}\leqslant{n}} {\prod}\left({a}_{{j}} −{a}_{{i}} \right) \\ $$
Commented by Tawa11 last updated on 19/Aug/22
Great sir.
$$\mathrm{Great}\:\mathrm{sir}. \\ $$
Commented by puissant last updated on 21/Aug/22
Vandermonde method
$${Vandermonde}\:{method} \\ $$
Answered by Rasheed.Sindhi last updated on 19/Aug/22
Prove that determinant ((1,1,1),(a,b,c),(a^2 ,b^2 ,c^2 ))= (a−b)(b−c)(c−a) ∣ (/)Through properties(/) ∣ ^(−) = determinant ((1,( 0),( 0)),(a,(b−a),(c−a)),(a^2 ,(b^2 −a^2 ),(c^2 −a^2 )))C2−C1_(C3−C1) =(b−a)(c−a) determinant ((1,( 0),( 0)),(a,( 1),( 1)),(a^2 ,(b+a),(c+a))) =(b−a)(c−a)∙1{1(c+a)−1(b+a)} =(b−a)(c−a)(c−b) =(a−b)(b−c)(c−a)
$$\mathrm{Prove}\:\mathrm{that}\begin{vmatrix}{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}\\{{a}}&{{b}}&{{c}}\\{{a}^{\mathrm{2}} }&{{b}^{\mathrm{2}} }&{{c}^{\mathrm{2}} }\end{vmatrix}=\:\left({a}−{b}\right)\left({b}−{c}\right)\left({c}−{a}\right)\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\overline {\:\:\:\:\:\:\:\:\underline{\mid}\:\underline{\frac{}{}\boldsymbol{\mathrm{Through}}\:\boldsymbol{\mathrm{properties}}\frac{}{}\:\mid}\:\:\:\:\:\:\:} \\ $$$$=\begin{vmatrix}{\mathrm{1}}&{\:\:\:\:\mathrm{0}}&{\:\:\:\mathrm{0}}\\{{a}}&{{b}−{a}}&{{c}−{a}}\\{{a}^{\mathrm{2}} }&{{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }&{{c}^{\mathrm{2}} −{a}^{\mathrm{2}} }\end{vmatrix}\underset{\mathrm{C3}−\mathrm{C1}} {\mathrm{C2}−\mathrm{C1}} \\ $$$$ \\ $$$$=\left({b}−{a}\right)\left({c}−{a}\right)\begin{vmatrix}{\mathrm{1}}&{\:\:\:\:\mathrm{0}}&{\:\:\:\mathrm{0}}\\{{a}}&{\:\:\:\:\mathrm{1}}&{\:\:\:\mathrm{1}}\\{{a}^{\mathrm{2}} }&{{b}+{a}}&{{c}+{a}}\end{vmatrix} \\ $$$$=\left({b}−{a}\right)\left({c}−{a}\right)\centerdot\mathrm{1}\left\{\mathrm{1}\left({c}+{a}\right)−\mathrm{1}\left({b}+{a}\right)\right\} \\ $$$$=\left({b}−{a}\right)\left({c}−{a}\right)\left({c}−{b}\right) \\ $$$$=\left({a}−{b}\right)\left({b}−{c}\right)\left({c}−{a}\right) \\ $$
Commented by puissant last updated on 21/Aug/22
Great
$${Great} \\ $$
Commented by Rasheed.Sindhi last updated on 21/Aug/22
Thanks sir!
$$\mathcal{T}{hanks}\:{sir}! \\ $$
Answered by Rajpurohith last updated on 10/Jun/23
Apply the operation C_1 →C_1 −C_2 and C_2 →C_2 −C_3 The determinant reduces to, determinant ((( 0 0 1)),((a−b b−c c)),((a^2 −b^2 b^2 −c^2 c^2 )))=(a−b)(b−c) determinant (((0 0 1)),((1 1 c)),((a+b b+c c^2 ))) This is by taking the common factors from C_1 & C_2 and operate C_1 →C_1 −C_2 =(a−b)(b−c) determinant (((0 0 1)),((0 1 c)),((a−c b+c c^2 ))) expanding along the first row, =(a−b)(b−c)(c−a)
$${Apply}\:{the}\:{operation}\:{C}_{\mathrm{1}} \rightarrow{C}_{\mathrm{1}} −{C}_{\mathrm{2}} \:{and}\:{C}_{\mathrm{2}} \rightarrow{C}_{\mathrm{2}} −{C}_{\mathrm{3}} \\ $$$${The}\:{determinant}\:{reduces}\:{to}, \\ $$$$\begin{vmatrix}{\:\:\:\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}}\\{{a}−{b}\:\:\:\:\:\:\:\:\:{b}−{c}\:\:\:\:\:\:\:\:{c}}\\{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} \:\:\:\:\:{b}^{\mathrm{2}} −{c}^{\mathrm{2}} \:\:\:\:\:{c}^{\mathrm{2}} }\end{vmatrix}=\left({a}−{b}\right)\left({b}−{c}\right)\begin{vmatrix}{\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{0}\:\:\:\:\:\:\mathrm{1}}\\{\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:{c}}\\{{a}+{b}\:\:\:\:{b}+{c}\:\:\:{c}^{\mathrm{2}} }\end{vmatrix} \\ $$$${This}\:{is}\:{by}\:{taking}\:{the}\:{common}\:{factors}\:{from}\:{C}_{\mathrm{1}} \:\&\:{C}_{\mathrm{2}} \\ $$$${and}\:{operate}\:{C}_{\mathrm{1}} \rightarrow{C}_{\mathrm{1}} −{C}_{\mathrm{2}} \\ $$$$=\left({a}−{b}\right)\left({b}−{c}\right)\begin{vmatrix}{\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{0}\:\:\:\:\:\:\:\:\mathrm{1}}\\{\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\:\:\:\:{c}}\\{{a}−{c}\:\:\:\:{b}+{c}\:\:\:\:{c}^{\mathrm{2}} }\end{vmatrix} \\ $$$${expanding}\:{along}\:{the}\:{first}\:{row}, \\ $$$$=\left({a}−{b}\right)\left({b}−{c}\right)\left({c}−{a}\right) \\ $$$$ \\ $$$$ \\ $$

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