Given system of equation a(x + y + z)=x,b(x + y + z) = y, c(x + y + z) = z where a,b,c a

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3 and 4 .Determinants and Matrices

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Given the system of equation $a(x + y + z)=x,b(x + y + z) = y, c(x + y + z) = z$ where $a,b,c$ are non-zero real numbers. If the real numbers $x,y,z$ are such that $xyz \neq 0,$ then $(a + b + c)$ is equal to-

A

$0$

B

$-1$

C

$1$

D

$2$

Solution

Condition for non-trivial solution is

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

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

$\mathrm{C}_{2} \rightarrow \mathrm{C}_{2}-\mathrm{C}_{1}$

$\mathrm{C}_{3} \rightarrow \mathrm{C}_{3}-\mathrm{C}_{1}$

$\Rightarrow a+b+c=1$

Standard 12

Mathematics

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