Question and Answers Forum

All Questions      Topic List

UNKNOWN Questions

Previous in All Question      Next in All Question      

Previous in UNKNOWN      Next in UNKNOWN      

Question Number 107330 by saorey0202 last updated on 10/Aug/20

If a+b+c=0 one root of determinant (((a−x),( c),( b)),(( c),(b−x),( a)),(( b),( a),(c−x)))=0 is

Ifa+b+c=0onerootof|a−xcbcb−xabac−x|=0is

Answered by som(math1967) last updated on 10/Aug/20

determinant (((a+b+c−x),(a+b+c−x),(a+b+c−x)),(c,(b−x),a),(b,a,(c−x)))=0 [R_1 ^′ →R_1 +R_2 +R_3 ] (a+b+c−x) determinant ((1,1,1),(c,(b−x),a),(b,(a ),(c−x)))=0 (a+b+c−x) determinant ((0,0,1),((c−b+x),(b−x−a),a),((b−a),(a−c+x),(c−x)))=0★ (a+b+c−x) determinant (((c−b+x),(b−x−a)),((b−a),(a−c+x)))=0 (a+b+c−x)(x^2 −a^2 −b^2 −c^2 +ab+bc+ca)=0 ⇒x=(a+b+c)=0 x=±(√(a^2 +b^2 +c^2 −ab−bc−ca)) ★C_1 ^′ →C_1 −C_2 C_2 ^′ →C_2 −C_3

|a+b+c−xa+b+c−xa+b+c−xcb−xabac−x|=0[R1′→R1+R2+R3](a+b+c−x)|111cb−xabac−x|=0(a+b+c−x)|001c−b+xb−x−aab−aa−c+xc−x|=0★(a+b+c−x)|c−b+xb−x−ab−aa−c+x|=0(a+b+c−x)(x2−a2−b2−c2+ab+bc+ca)=0⇒x=(a+b+c)=0x=±a2+b2+c2−ab−bc−ca★C1′→C1−C2C2′→C2−C3

Answered by 1549442205PVT last updated on 10/Aug/20

We have determinant (((a−x),( c),( b)),(( c),(b−x),( a)),(( b),( a),(c−x)))=0 ⇔(a−x)(b−x)(c−x)+2abc−b^2 (b−x)−c^2 (c−x)−a^2 (a−x)=0 ⇔−x^3 +(a+b+c)x^2 +(a^2 +b^2 +c^2 −ab−bc−ca)x −(a^3 +b^3 +c^3 −2abc)+abc=0 ⇔x^3 −(a+b+c)x^2 −(a^2 +b^2 +c^2 −ab−bc−ca)x +(a^3 +b^3 +c^3 −3abc)=0(∗) Apply the identity a^3 +b^3 +c^3 −3abc=(a+b+c)(a^2 +b^2 +c^2 −ab−bc−ca)(1)we have (∗)⇔x^2 [x−(a+b+c)]−(a^2 +b^2 +c^2 −ab−bc−ca)x +(a+b+c)(a^2 +b^2 +c^2 −ab−bc−ca)=0 ⇔(∗)⇔x^2 [x−(a+b+c)]−[x−(a+b+c)](a^2 +b^2 +c^2 −ab−bc−ca)=0 ⇔[x−(a+b+c)].[x^2 −(a^2 +b^2 +c^2 −ab−bc−ca)]=0 This show that x=a+b+c is a root of the equation (∗)(q.e.d) In adition,we see that the given has two another real roots that are x=±(√(a^2 +b^2 +c^2 −ab−bc−ca)) .The under root expression is non−negative number since we have always a^2 +b^2 +c^2 −ab−bc−ca= (1/2)[(a−b)^2 +(b−c)^2 +(c−a)^2 ]≥0∀a,b,c∈R

Wehave|a−xcbcb−xabac−x|=0⇔(a−x)(b−x)(c−x)+2abc−b2(b−x)−c2(c−x)−a2(a−x)=0⇔−x3+(a+b+c)x2+(a2+b2+c2−ab−bc−ca)x−(a3+b3+c3−2abc)+abc=0⇔x3−(a+b+c)x2−(a2+b2+c2−ab−bc−ca)x+(a3+b3+c3−3abc)=0(∗)Applytheidentitya3+b3+c3−3abc=(a+b+c)(a2+b2+c2−ab−bc−ca)(1)wehave(∗)⇔x2[x−(a+b+c)]−(a2+b2+c2−ab−bc−ca)x+(a+b+c)(a2+b2+c2−ab−bc−ca)=0⇔(∗)⇔x2[x−(a+b+c)]−[x−(a+b+c)](a2+b2+c2−ab−bc−ca)=0⇔[x−(a+b+c)].[x2−(a2+b2+c2−ab−bc−ca)]=0Thisshowthatx=a+b+cisarootoftheequation(∗)(q.e.d)Inadition,weseethatthegivenhastwoanotherrealrootsthatarex=±a2+b2+c2−ab−bc−ca.Theunderrootexpressionisnon−negativenumbersincewehavealwaysa2+b2+c2−ab−bc−ca=12[(a−b)2+(b−c)2+(c−a)2]⩾0∀a,b,c∈R

Terms of Service

Privacy Policy

Contact: info@tinkutara.com