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Question Number 57186 by Tawa1 last updated on 31/Mar/19
Commented by kaivan.ahmadi last updated on 31/Mar/19
determinant (((1 1 1)),((a b c)),((a^2 b^2 c^2 )))→_(−a^2 R_1 +R_3 ) ^(−aR_1 +R_2 ) and Sarrus Rule determinant (((1 1 1)),((0 b−a c−a)),((0 b^2 −a^2 c^2 −a^2 ))) determinant (((1 1 1)),((0 b−a c−a)),((0 b^2 −a^2 c^2 −a^2 )))= [(b−a)(c^2 −a^2 )+0+0]−[0+0+(b^2 −a^2 )(c−a)]= (b−a)(c−a)(c+a)−(b−a)(b+a)(c−a)= (b−a)(c−a)[ (c+a)−(b+a)]= (b−a)(c−a)(c−b)
$$\begin{vmatrix}{\mathrm{1}\:\:\:\:\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}}\\{{a}\:\:\:\:\:\:\:\:\:\:{b}\:\:\:\:\:\:\:\:\:\:\:{c}}\\{{a}^{\mathrm{2}} \:\:\:\:\:\:\:\:{b}^{\mathrm{2}} \:\:\:\:\:\:{c}^{\mathrm{2}} }\end{vmatrix}\underset{−{a}^{\mathrm{2}} {R}_{\mathrm{1}} +{R}_{\mathrm{3}} } {\overset{−{aR}_{\mathrm{1}} +{R}_{\mathrm{2}} } {\rightarrow}}\:{and}\:{Sarrus}\:{Rule} \\ $$$$\begin{vmatrix}{\mathrm{1}\:\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}}\\{\mathrm{0}\:\:\:\:\:\:{b}−{a}\:\:\:\:\:\:{c}−{a}}\\{\mathrm{0}\:\:\:\:{b}^{\mathrm{2}} −{a}^{\mathrm{2}} \:\:\:\:\:{c}^{\mathrm{2}} −{a}^{\mathrm{2}} }\end{vmatrix}\begin{vmatrix}{\mathrm{1}\:\:\:\:\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}}\\{\mathrm{0}\:\:\:\:{b}−{a}\:\:\:\:\:\:\:\:\:{c}−{a}}\\{\mathrm{0}\:\:\:\:{b}^{\mathrm{2}} −{a}^{\mathrm{2}} \:\:\:\:\:{c}^{\mathrm{2}} −{a}^{\mathrm{2}} }\end{vmatrix}= \\ $$$$\left[\left({b}−{a}\right)\left({c}^{\mathrm{2}} −{a}^{\mathrm{2}} \right)+\mathrm{0}+\mathrm{0}\right]−\left[\mathrm{0}+\mathrm{0}+\left({b}^{\mathrm{2}} −{a}^{\mathrm{2}} \right)\left({c}−{a}\right)\right]= \\ $$$$\left({b}−{a}\right)\left({c}−{a}\right)\left({c}+{a}\right)−\left({b}−{a}\right)\left({b}+{a}\right)\left({c}−{a}\right)= \\ $$$$\left({b}−{a}\right)\left({c}−{a}\right)\left[\:\left({c}+{a}\right)−\left({b}+{a}\right)\right]= \\ $$$$\left({b}−{a}\right)\left({c}−{a}\right)\left({c}−{b}\right) \\ $$$$ \\ $$
Commented by kaivan.ahmadi last updated on 31/Mar/19
this determinant is vandermlnd. easly i change a=a_1 ,b=a_2 and c=a_3
$${this}\:{determinant}\:{is}\:{vandermlnd}. \\ $$$${easly}\:{i}\:{change}\:{a}={a}_{\mathrm{1}} ,{b}={a}_{\mathrm{2}} \:{and}\:{c}={a}_{\mathrm{3}} \\ $$
Commented by Tawa1 last updated on 31/Mar/19
God bless you sir
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 31/Mar/19
∣0 0 1∣ ∣a_1 −a_3 a_2 −a_3 a_3 ∣ ∣(a_1 +a_3 )(a_1 −a_3 ) (a_2 +a_3 )(a_2 −a_3 ) a_3 ^2 ∣ (a_1 −a_3 )(a_2 −a_3 )∣0 0 1∣ ∣1 1 a_3 ∣ ∣a_1 +a_3 a_2 +a_3 a_3 ^2 ∣ (a_1 −a_3 )(a_2 −a_3 )(a_2 +a_3 −a_1 −a_3 ) =(a_2 −a_1 )(a_3 −a_1 )(a_3 −a_2 ) printing error in problem element printed (a_1 ^2 )_(3×3) but should be(a_3 ^2 )_(3×3)
$$\mid\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}\mid \\ $$$$\mid{a}_{\mathrm{1}} −{a}_{\mathrm{3}} \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{a}_{\mathrm{2}} −{a}_{\mathrm{3}} \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{a}_{\mathrm{3}} \mid \\ $$$$\mid\left({a}_{\mathrm{1}} +{a}_{\mathrm{3}} \right)\left({a}_{\mathrm{1}} −{a}_{\mathrm{3}} \right)\:\:\left({a}_{\mathrm{2}} +{a}_{\mathrm{3}} \right)\left({a}_{\mathrm{2}} −{a}_{\mathrm{3}} \right)\:\:\:\:\:\:\:{a}_{\mathrm{3}} ^{\mathrm{2}} \:\mid \\ $$$$ \\ $$$$\left({a}_{\mathrm{1}} −{a}_{\mathrm{3}} \right)\left({a}_{\mathrm{2}} −{a}_{\mathrm{3}} \right)\mid\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}\mid \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mid\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{a}_{\mathrm{3}} \mid \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mid{a}_{\mathrm{1}} +{a}_{\mathrm{3}} \:\:\:\:\:{a}_{\mathrm{2}} +{a}_{\mathrm{3}} \:\:\:\:\:\:\:{a}_{\mathrm{3}} ^{\mathrm{2}} \:\:\mid \\ $$$$\left({a}_{\mathrm{1}} −{a}_{\mathrm{3}} \right)\left({a}_{\mathrm{2}} −{a}_{\mathrm{3}} \right)\left({a}_{\mathrm{2}} +{a}_{\mathrm{3}} −{a}_{\mathrm{1}} −{a}_{\mathrm{3}} \right) \\ $$$$=\left({a}_{\mathrm{2}} −{a}_{\mathrm{1}} \right)\left({a}_{\mathrm{3}} −{a}_{\mathrm{1}} \right)\left({a}_{\mathrm{3}} −{a}_{\mathrm{2}} \right) \\ $$$${printing}\:{error}\:{in}\:{problem}\:{element}\: \\ $$$${printed}\:\left({a}_{\mathrm{1}} ^{\mathrm{2}} \right)_{\mathrm{3}×\mathrm{3}} {but}\:{should}\:{be}\left({a}_{\mathrm{3}} ^{\mathrm{2}} \right)_{\mathrm{3}×\mathrm{3}} \\ $$
Commented by Tawa1 last updated on 31/Mar/19
God bless you sir
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

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