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Question Number 15194 by arnabpapu550@gmail.com last updated on 08/Jun/17

The equation  determinant (((x−a),(x−b),(x−c)),((x−b),(x−c),(x−a)),((x−c),(x−a),(x−b)))=0,  where a, b, c are different, is satisfied by

$$\mathrm{The}\:\mathrm{equation}\:\begin{vmatrix}{{x}−{a}}&{{x}−{b}}&{{x}−{c}}\\{{x}−{b}}&{{x}−{c}}&{{x}−{a}}\\{{x}−{c}}&{{x}−{a}}&{{x}−{b}}\end{vmatrix}=\mathrm{0}, \\ $$$$\mathrm{where}\:{a},\:{b},\:{c}\:\mathrm{are}\:\mathrm{different},\:\mathrm{is}\:\mathrm{satisfied}\:\mathrm{by} \\ $$

Commented by prakash jain last updated on 08/Jun/17

  determinant (((x−a),(x−b),(x−c)),((x−b),(x−c),(x−a)),((x−c),(x−a),(x−b)))=0,  R1=R1+R2+R3    determinant (((3x−a−b−c),(3x−a−b−c),(3x−c−b−a)),((x−b),(x−c),(x−a)),((x−c),(x−a),(x−b)))=0,   (3x−a−b−c) determinant ((1,1,1),((x−b),(x−c),(x−a)),((x−c),(x−a),(x−b)))=0,  C1=C1−C2,C2=C2−C3   (3x−a−b−c) determinant ((0,0,1),((c−b),(a−c),(x−a)),((a−c),(b−a),(x−b)))=0,  3x−a−b−c=0⇒x=((a+b+c)/3)

$$\:\begin{vmatrix}{{x}−{a}}&{{x}−{b}}&{{x}−{c}}\\{{x}−{b}}&{{x}−{c}}&{{x}−{a}}\\{{x}−{c}}&{{x}−{a}}&{{x}−{b}}\end{vmatrix}=\mathrm{0}, \\ $$$$\mathrm{R1}=\mathrm{R1}+\mathrm{R2}+\mathrm{R3} \\ $$$$\:\begin{vmatrix}{\mathrm{3}{x}−{a}−{b}−{c}}&{\mathrm{3}{x}−{a}−{b}−{c}}&{\mathrm{3}{x}−{c}−{b}−{a}}\\{{x}−{b}}&{{x}−{c}}&{{x}−{a}}\\{{x}−{c}}&{{x}−{a}}&{{x}−{b}}\end{vmatrix}=\mathrm{0}, \\ $$$$\:\left(\mathrm{3}{x}−{a}−{b}−{c}\right)\begin{vmatrix}{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}\\{{x}−{b}}&{{x}−{c}}&{{x}−{a}}\\{{x}−{c}}&{{x}−{a}}&{{x}−{b}}\end{vmatrix}=\mathrm{0}, \\ $$$$\mathrm{C1}=\mathrm{C1}−\mathrm{C2},\mathrm{C2}=\mathrm{C2}−\mathrm{C3} \\ $$$$\:\left(\mathrm{3}{x}−{a}−{b}−{c}\right)\begin{vmatrix}{\mathrm{0}}&{\mathrm{0}}&{\mathrm{1}}\\{{c}−{b}}&{{a}−{c}}&{{x}−{a}}\\{{a}−{c}}&{{b}−{a}}&{{x}−{b}}\end{vmatrix}=\mathrm{0}, \\ $$$$\mathrm{3}{x}−{a}−{b}−{c}=\mathrm{0}\Rightarrow{x}=\frac{{a}+{b}+{c}}{\mathrm{3}} \\ $$

Commented by arnabpapu550@gmail.com last updated on 08/Jun/17

excellent!

$$\mathrm{excellent}! \\ $$$$ \\ $$

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