3 and 4 .Determinants and Matrices
normal

If $\left| {\begin{array}{*{20}{c}}   {a - b}&{b - c}&{c - a} \\    {b - c}&{c - a}&{a - b} \\    {c - a + 1}&{a - b}&{b - c}  \end{array}} \right| = 0$ ,$\left( {a,b,c \in R - \left\{ 0 \right\}} \right),$ then

A

$a$,$b$ and $c$ must be equal

B

$a$,$b$ and $c$ may not be all equal

C

$a$,$b$ and $c$ must be in $G.P$.

D

$a$,$b$ and $c$ must be distinct

Solution

$\left|\begin{array}{lll}{a-b} & {b-c} & {c-a} \\ {b-c} & {c-a} & {a-b} \\ {c-a} & {a-b} & {b-c}\end{array}\right|+\left|\begin{array}{ccc}{0} & {b-c} & {c-a} \\ {0} & {c-a} & {a-b} \\ {1} & {a-b} & {b-c}\end{array}\right|=0$

$\Rightarrow 0+(a-b)(b-c)=(c-a)^{2}$

$\Rightarrow a b-a c-b^{2}+b c=c^{2}+a^{2}-2 a c$

$\Rightarrow a^{2}+b^{2}+c^{2}=a b+b c+c a$

$\Rightarrow a=b=c$

Standard 12
Mathematics

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