Question

If $\left|\begin{array}{c}x+\alpha \\ \alpha \\ \alpha \end{array}\begin{array}{c}\beta \\ x+\beta \\ \beta \end{array}\begin{array}{c}\gamma \\ \gamma \\ x+\gamma \end{array}\right|=0$, then $x$ is equal to

• A. $0,-(\alpha +\beta +\gamma )$
• B. $0,\alpha +\beta +\gamma$
• C. $1,\alpha +\beta +\gamma$
• D. $0,{\alpha }^2+{\beta }^2+{\gamma }^2$

Answer 1

The correct option is A $0,-(\alpha +\beta +\gamma )$

Expand the above determinant along $R_1$ as shown below:

$\Delta =(x+\alpha )[(x+\beta )(x+\gamma )-\left(\beta \right)\left(\gamma \right)]-\beta [\left(\alpha \right)(x+\gamma )-\left(\alpha \right)\left(\gamma \right)]+\gamma [\left(\alpha \right)\left(\beta \right)-\left(\alpha \right)(x+\beta )]=0$

$\Delta =(x+\alpha )[x^2+x\gamma +x\beta +\beta \gamma -\beta \gamma ]-\beta [\alpha x+\alpha y-\alpha y]+\gamma [\alpha \beta -\alpha x-\alpha \beta ]=0$

$\Delta =(x+\alpha )[x^2+x\gamma +x\beta ]-\beta \left[\alpha x\right]+\gamma [-\alpha x]=0$

$\Delta =(x+\alpha )[x^2+x\gamma +x\beta ]-\alpha x\beta -\gamma \alpha x=0$

$\Delta =x^3+x^2\gamma +x^2\beta +x^2\alpha +\alpha x\beta +\alpha x\gamma +x\beta -\alpha x\beta -\gamma \alpha x=0$

$\Delta =x^3+x^2(\alpha +\beta +\gamma )=0$

$\Delta =x^2(x+\alpha +\beta +\gamma )=0$

either $x^2=0$ or $(x+\alpha +\beta +\gamma )=0$

$\Rightarrow$$x=0$ or $x=-(\alpha +\beta +\gamma )$

Hence option $A$ is correct.