Question

If the minor of three-one element (i.e. $M_{31}$) in the determinant $\begin{vmatrix} 0 & 1 & \sec \alpha \\ \tan \alpha & -\sec \alpha & \tan \alpha \\ 1 & 0 & 1 \end{vmatrix}$ is 1 then find the value of $\alpha$. $(0 \le \alpha \le \pi)$.

Answer 1

0, $\frac{3\pi}{4}$, $\pi$

Answer 2

To find the value of $\alpha$, we first need to calculate the minor of the three-one element ($M_{31}$) of the given determinant.

The given determinant is:

$$\begin{vmatrix} 0 & 1 & \sec \alpha \\ \tan \alpha & -\sec \alpha & \tan \alpha \\ 1 & 0 & 1 \end{vmatrix}$$

The element $a_{31}$ is 1. The minor $M_{31}$ is the determinant obtained by deleting the 3rd row and the 1st column:

$$M_{31} = \begin{vmatrix} 1 & \sec \alpha \\ -\sec \alpha & \tan \alpha \end{vmatrix}$$

Now, we calculate the value of this $2 \times 2$ determinant:

$$M_{31} = (1)(\tan \alpha) - (\sec \alpha)(-\sec \alpha)$$ $$M_{31} = \tan \alpha + \sec^2 \alpha$$

We are given that the minor $M_{31}$ is equal to 1. So, we set up the equation:

$$\tan \alpha + \sec^2 \alpha = 1$$

We know the trigonometric identity $\sec^2 \alpha = 1 + \tan^2 \alpha$. Substitute this into the equation:

$$\tan \alpha + (1 + \tan^2 \alpha) = 1$$ $$\tan^2 \alpha + \tan \alpha + 1 = 1$$

Subtract 1 from both sides of the equation:

$$\tan^2 \alpha + \tan \alpha = 0$$

Factor out $\tan \alpha$:

$$\tan \alpha (\tan \alpha + 1) = 0$$

This equation implies two possible cases:

1. $\tan \alpha = 0$
2. $\tan \alpha + 1 = 0 \implies \tan \alpha = -1$

We need to find the values of $\alpha$ in the given range $0 \le \alpha \le \pi$.

Case 1: $\tan \alpha = 0$

For $0 \le \alpha \le \pi$, the values of $\alpha$ for which $\tan \alpha = 0$ are $\alpha = 0$ and $\alpha = \pi$.

Case 2: $\tan \alpha = -1$

For $0 \le \alpha \le \pi$, $\tan \alpha$ is negative only in the second quadrant. The reference angle for $\tan \alpha = 1$ is $\frac{\pi}{4}$. Therefore, the angle in the second quadrant is $\alpha = \pi - \frac{\pi}{4} = \frac{3\pi}{4}$.

Thus, the possible values of $\alpha$ in the given range are $0, \frac{3\pi}{4}, \pi$.