Question

If $\alpha ,\beta $ and $\gamma $ are the roots of the equation $x^{3}-3x^{2}+4x+4=0$ , then the value of $\begin{vmatrix} \alpha ^{2}+1 & 1 & 1 \\ 1 & \beta ^{2}+1 & 1 \\ 1 & 1 & \gamma ^{2}+1 \end{vmatrix}$ is equal to

• A. $32$
• B. $16$
• C. $56$
• D. $64$

Answer 1

Answer: (C)

$\alpha +\beta +\gamma =3$
$\alpha \beta +\beta \gamma +\gamma \alpha =4$
$\alpha \beta \gamma =-4$
$\begin{vmatrix} \left(\alpha \right)^{2}+1 & 1 & 1 \\ 1 & \left(\beta \right)^{2}+1 & 1 \\ 1 & 1 & \left(\gamma \right)^{2}+1 \end{vmatrix}=\left(\left(\alpha \right)^{2} + 1\right)\left[\left(\left(\beta \right)^{2} + 1\right) \left(\left(\gamma \right)^{2} + 1\right) - 1\right]-1\left[\left(\gamma \right)^{2}\right]+1\left(- \left(\beta \right)^{2}\right)$
$=\left(\left(\alpha \right)^{2} + 1\right)\left[\left(\beta \right)^{2} + \left(\gamma \right)^{2} + \left(\beta \right)^{2} \left(\gamma \right)^{2}\right]-\left(\gamma \right)^{2}-\left(\beta \right)^{2}$
$=\alpha ^{2}\beta ^{2}+\alpha ^{2}\gamma ^{2}+\alpha ^{2}\beta ^{2}\gamma ^{2}+\beta ^{2}+\gamma ^{2}+\beta ^{2}\gamma ^{2}-\gamma ^{2}-\beta ^{2}$
$=\left(\alpha \right)^{2}\left(\beta \right)^{2}+\left(\beta \right)^{2}\left(\gamma \right)^{2}+\left(\gamma \right)^{2}\left(\alpha \right)^{2}+\left(\alpha \beta \gamma \right)^{2}$
$=\left(\alpha \beta + \beta \gamma + \gamma \alpha \right)^{2}-2\alpha \beta \gamma \left(\alpha + \beta + \gamma \right)+\left(\alpha \beta \gamma \right)^{2}$
$=16+8\left(3\right)+16$
$=56$