If `[(cos alpha,-sinalpha),(sinalpha,cos alpha)]` is identity matrix then what is the value of `alpha`?

Options

A. 0
B. `pi/2`
C. `pi/4`
D. `(3pi)/2`

Answer

To solve the problem, we need to determine the value of \( \alpha \) such that the matrix

\[ \begin{pmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{pmatrix} \]

is equal to the identity matrix

\[ \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}. \]

### Step 1: Set up the equations

We can equate the elements of the two matrices:

1. \( \cos \alpha = 1 \)
2. \( -\sin \alpha = 0 \)
3. \( \sin \alpha = 0 \)
4. \( \cos \alpha = 1 \)

From these equations, we can simplify our problem.

### Step 2: Solve the equations

From the first equation \( \cos \alpha = 1 \):

- The cosine function equals 1 at \( \alpha = 0 + 2k\pi \) where \( k \) is any integer.

From the second equation \( -\sin \alpha = 0 \):

- This implies \( \sin \alpha = 0 \), which is true for \( \alpha = n\pi \) where \( n \) is any integer.

### Step 3: Find common solutions

The values of \( \alpha \) that satisfy both conditions are:

- \( \alpha = 0 \) (which is \( 0 + 2k\pi \))
- \( \alpha = \pi \) (which is \( n\pi \) for \( n=1 \))

However, since \( \cos \alpha = 1 \) only holds true at \( \alpha = 0 \) in the range of \( [0, 2\pi) \), we can conclude that:

\[ \alpha = 0 \]

### Final Answer

Thus, the value of \( \alpha \) is

\[ \boxed{0}. \]

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