If `A=[[costheta,sintheta],[-sintheta,costheta]]` and `beta=A^4+A` then determinant of `beta=`

Options

A. `2-2cos3theta`
B. `2+3cos2theta`
C. `3+cos2theta`
D. `2+2cos3theta`

Answer

To solve the problem, we need to find the determinant of the matrix \(\beta = A^4 + A\), where \(A = \begin{bmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix}\).

### Step 1: Calculate \(A^2\)

First, we calculate \(A^2\):

\[ A^2 = A \cdot A = \begin{bmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix} \cdot \begin{bmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix} \]

Calculating the elements:

- First row, first column: \(\cos^2 \theta + \sin^2 \theta = 1\)
- First row, second column: \(\cos \theta \sin \theta + \sin \theta \cos \theta = 2 \cos \theta \sin \theta\)
- Second row, first column: \(-\sin \theta \cos \theta + \cos \theta \sin \theta = 0\)
- Second row, second column: \(-\sin^2 \theta + \cos^2 \theta = \cos 2\theta\)

So, we have:

\[ A^2 = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I \]

### Step 2: Calculate \(A^4\)

Next, we calculate \(A^4\):

\[ A^4 = (A^2)^2 = I^2 = I \]

### Step 3: Calculate \(\beta\)

Now we can find \(\beta\):

\[ \beta = A^4 + A = I + A = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} + \begin{bmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix} \]

Calculating the elements:

\[ \beta = \begin{bmatrix} 1 + \cos \theta & \sin \theta \\ -\sin \theta & 1 + \cos \theta \end{bmatrix} \]

### Step 4: Calculate the determinant of \(\beta\)

Now we calculate the determinant of \(\beta\):

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

Calculating this gives:

\[ \text{det}(\beta) = (1 + \cos \theta)^2 + \sin^2 \theta \]

Using the identity \(\sin^2 \theta + \cos^2 \theta = 1\):

\[ \text{det}(\beta) = (1 + \cos \theta)^2 + (1 - \cos^2 \theta) = (1 + \cos \theta)^2 + 1 - \cos^2 \theta \]

Expanding \((1 + \cos \theta)^2\):

\[ = 1 + 2\cos \theta + \cos^2 \theta + 1 - \cos^2 \theta = 2 + 2\cos \theta \]

Thus, the determinant of \(\beta\) is:

\[ \text{det}(\beta) = 2(1 + \cos \theta) \]

### Final Answer

The determinant of \(\beta\) is \(2(1 + \cos \theta)\).