Question

If the system of equations $x-\alpha y-\alpha z=0,\beta x-y+\beta z=0,\gamma x+\gamma y-z=0$, where $\alpha ,\beta ,\gamma \ne -1$ have only non-trivial solutions, $\frac{1}{1+\alpha }+\frac{1}{1+\beta }+\frac{1}{1+\gamma }=$

• A. $2$
• B. $1$
• C. $-1$
• D. $-2$

Answer 1

Answer: (A)

$2$

We have,

$x-\alpha (y+z)=0\Rightarrow (1+\alpha )x-\alpha (x+y+z)=0$ ...(1)

$y-\beta (x+z)=0(1+\beta )y-\beta (x+y+z)=0$ ...(2)

$z-\gamma (x+y)=0(1+\gamma )z-\gamma (x+y+z)=0$ ...(3)

Since the given system of equation have only non-trivial solutions, $\therefore (x+y+z)\ne 0$ $[\because$ if $x+y+z=0$, then in view of (1),(2) and (3), we have $x=0,y=0,z=0,$ a trivial solution$]$

$\therefore \frac{\alpha }{1+\alpha }=\frac{x}{x+y+z},\frac{\beta }{1+\beta }=\frac{y}{x+y+z}$

and, $\frac{\gamma }{1+\gamma }=\frac{z}{x+y+z}$

On adding, we get $\frac{\alpha }{1+\alpha }+\frac{\beta }{1+\beta }+\frac{\gamma }{1+\gamma }=1$

$\Rightarrow (1-\frac{\alpha }{1+\alpha })+(1-\frac{\beta }{1+\beta })+(1-\frac{\gamma }{1+\gamma })=3-1$

$\Rightarrow \frac{1}{1+\alpha }+\frac{1}{1+\beta }+\frac{1}{1+\gamma }=2.$