If the matrix `({:(cos theta, sin theta, 0),(sin theta, cos theta, 0),(0,0,1):})` is singular, then what is one of the value of `theta` ?

Options

A. `(pi)/(2)`
B. `(pi)/(4)`
C. `pi`
D. 0

Answer

To determine the value of \( \theta \) for which the matrix

\[ \begin{pmatrix} \cos \theta & \sin \theta & 0 \\ \sin \theta & \cos \theta & 0 \\ 0 & 0 & 1 \end{pmatrix} \]

is singular, we need to find the determinant of the matrix and set it equal to zero.

### Step 1: Write down the determinant of the matrix.

The determinant of a 3x3 matrix

\[ \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix} \]

is given by the formula:

\[ \text{det} = a(ei - fh) - b(di - fg) + c(dh - eg) \]

For our matrix, we have:

- \( a = \cos \theta \)
- \( b = \sin \theta \)
- \( c = 0 \)
- \( d = \sin \theta \)
- \( e = \cos \theta \)
- \( f = 0 \)
- \( g = 0 \)
- \( h = 0 \)
- \( i = 1 \)

### Step 2: Calculate the determinant.

Using the determinant formula:

\[ \text{det} = \cos \theta \cdot (\cos \theta \cdot 1 - 0 \cdot 0) - \sin \theta \cdot (\sin \theta \cdot 1 - 0 \cdot 0) + 0 \]

This simplifies to:

\[ \text{det} = \cos \theta \cdot \cos \theta - \sin \theta \cdot \sin \theta \]

\[ \text{det} = \cos^2 \theta - \sin^2 \theta \]

### Step 3: Set the determinant equal to zero.

For the matrix to be singular, we set the determinant equal to zero:

\[ \cos^2 \theta - \sin^2 \theta = 0 \]

### Step 4: Use the identity for cosine.

We can rewrite the equation using the identity \( \cos^2 \theta - \sin^2 \theta = \cos(2\theta) \):

\[ \cos(2\theta) = 0 \]

### Step 5: Solve for \( \theta \).

The cosine function equals zero at:

\[ 2\theta = \frac{\pi}{2} + n\pi \quad \text{for } n \in \mathbb{Z} \]

Dividing by 2 gives:

\[ \theta = \frac{\pi}{4} + \frac{n\pi}{2} \]

### Step 6: Find one specific value of \( \theta \).

One specific solution is when \( n = 0 \):

\[ \theta = \frac{\pi}{4} \]

### Conclusion

Thus, one of the values of \( \theta \) for which the matrix is singular is:

\[ \theta = \frac{\pi}{4} \]