If-a-b-c-0-one-root-of-determinant-a-x-c-b-c-b-x-a-b-a-c-x-0-is- Tinku Tara June 14, 2023 None 0 Comments FacebookTweetPin Question Number 107330 by saorey0202 last updated on 10/Aug/20 Ifa+b+c=0onerootof|a−xcbcb−xabac−x|=0is Answered by som(math1967) last updated on 10/Aug/20 |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 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 FacebookTweetPin Post navigation Previous Previous post: If-A-is-an-invertible-matrix-then-det-A-1-is-equal-to-Next Next post: If-are-the-roots-of-the-equation-ax-2-bx-c-0-then-the-value-of-the-determinant-determinant-1-cos-cos-cos-1-cos-cos-cos-1-is- Leave a Reply Cancel replyYour email address will not be published. Required fields are marked *Comment * Name * Save my name, email, and website in this browser for the next time I comment.
![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](https://www.tinkutara.com/question/Q107361.png)
![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](https://www.tinkutara.com/question/Q107391.png)