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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-
$0$
$-1$
$1$
$2$
Solution
Condition for non-trivial solution is
$\left|\begin{array}{ccc}{a-1} & {a} & {a} \\ {b} & {b-1} & {b} \\ {c} & {c} & {c-1}\end{array}\right|=0$
$\Rightarrow(a+b+c-1)\left|\begin{array}{ccc}{1} & {1} & {1} \\ {b} & {b-1} & {b} \\ {c} & {c} & {c-1}\end{array}\right|=0$
$\mathrm{C}_{2} \rightarrow \mathrm{C}_{2}-\mathrm{C}_{1}$
$\mathrm{C}_{3} \rightarrow \mathrm{C}_{3}-\mathrm{C}_{1}$
$\Rightarrow a+b+c=1$
Similar Questions
Let $\alpha, \beta$ and $\gamma$ be real numbers. consider the following system of linear equations
$x+2 y+z=7$
$x+\alpha z=11$
$2 x-3 y+\beta z=\gamma$
Match each entry in List – $I$ to the correct entries in List-$II$
| List – $I$ | List – $II$ |
| ($P$) If $\beta=\frac{1}{2}(7 \alpha-3)$ and $\gamma=28$, then the system has | ($1$) a unique solution |
| ($Q$) If $\beta=\frac{1}{2}(7 \alpha-3)$ and $\gamma \neq 28$, then the system has | ($2$) no solution |
|
($R$) If $\beta \neq \frac{1}{2}(7 \alpha-3)$ where $\alpha=1$ and $\gamma \neq 28$, then the system has |
($3$) infinitely many solutions |
| ($S$) If $\beta \neq \frac{1}{2}(7 \alpha-3)$ where $\alpha=1$ and $\gamma=28$, then the system has | ($4$) $x=11, y=-2$ and $z=0$ as a solution |
| ($5$) $x=-15, y=4$ and $z=0$ as a solution |
Then the system has
