Question

If the maximum and minimum values of the determinant

$\begin{vmatrix} 1+\text{sin}^{2}x & \text{cos}^{2}x & \text{sin 2x} \\ \text{sin}^{2}x & 1+\text{cos}^{2}\text{x} & \text{sin 2x} \\ \text{sin}^{2}\text{x} & \text{cos}^{2}\text{x} & 1+\text{sin 2x} \end{vmatrix}$ are $\alpha $ and $\beta $ respectively, then $\alpha + 2 \beta $ is equal to

Answer 1

Given determinant is
$\Delta = \begin{vmatrix} 1+\text{sin}^{2}x & \text{cos}^{2}x & \text{sin 2x} \\ \text{sin}^{2}x & 1+\text{cos}^{2}\text{x} & \text{sin 2x} \\ \text{sin}^{2}\text{x} & \text{cos}^{2}\text{x} & 1+\text{sin 2x} \end{vmatrix}$
Applying the operation $\text{C}_{1} \rightarrow \text{C}_{1} + \text{C}_{2}$ , we get
$\Delta = \begin{vmatrix} 2 & \text{cos}^{2}x & \text{sin 2x} \\ 2 & 1+\text{cos}^{2}x & \text{sin 2x} \\ 1 & \text{cos}^{2}x & 1+\text{sin 2x} \end{vmatrix}$
Applying the operations $\text{R}_{2} \rightarrow \text{R}_{2} - \text{R}_{1}$ and then $\text{R}_{3} \rightarrow \text{R}_{3} - \text{R}_{1}$ , we get
$\Delta = \begin{vmatrix} 2 & \text{cos}^{2}x & \text{sin 2x} \\ 0 & 1 & 0 \\ -1 & 0 & 1 \end{vmatrix}$
Now, expanding the above determinant along $R_{2}$ , we get
$\Delta = - 0 + \left(\right. 2 + \text{sin 2x} \left.\right) - 0 = 2 + \text{sin 2x}$
Since, the maximum value of $\text{sin} \text{2x}$ is $1$ and minimum value of $\text{sin} \text{2x}$ is $-1$ .
Therefore, $\alpha =$ maximum value of $\Delta = 2 + 1 = 3$ and $\beta =$ minimum value of $\Delta = 2 - 1 = 1$ .
$\Rightarrow \alpha = 3$ and $\beta = 1$
$\therefore \alpha + 2 \beta = 3 + 2 = 5$