Question and Answers Forum

All Questions      Topic List

Geometry Questions

Previous in All Question      Next in All Question      

Previous in Geometry      Next in Geometry      

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

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

Terms of Service

Privacy Policy

Contact: info@tinkutara.com