If `P = [[cos frac(pi)(6), sin frac(pi)(6) ],[-sinfrac(pi)(6),cosfrac(pi)(6)]], A = [[1,1],[0,1]] then `P+ A`

Explanation

To solve the question, we need to find the sum of the matrices \( P \) and \( A \). Let's break it down step by step.

### Step 1: Define the matrices \( P \) and \( A \)

The matrix \( P \) is defined as:
\[ P = \begin{bmatrix} \cos\left(\frac{\pi}{6}\right) & \sin\left(\frac{\pi}{6}\right) \\ -\sin\left(\frac{\pi}{6}\right) & \cos\left(\frac{\pi}{6}\right) \end{bmatrix} \]

The matrix \( A \) is defined as:
\[ A = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} \]

### Step 2: Calculate the trigonometric values

We need to calculate \( \cos\left(\frac{\pi}{6}\right) \) and \( \sin\left(\frac{\pi}{6}\right) \):
- \( \cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2} \)
- \( \sin\left(\frac{\pi}{6}\right) = \frac{1}{2} \)

### Step 3: Substitute the trigonometric values into matrix \( P \)

Now we can substitute these values into the matrix \( P \):
\[ P = \begin{bmatrix} \frac{\sqrt{3}}{2} & \frac{1}{2} \\ -\frac{1}{2} & \frac{\sqrt{3}}{2} \end{bmatrix} \]

### Step 4: Add the matrices \( P \) and \( A \)

Now we will add the matrices \( P \) and \( A \):
\[ P + A = \begin{bmatrix} \frac{\sqrt{3}}{2} & \frac{1}{2} \\ -\frac{1}{2} & \frac{\sqrt{3}}{2} \end{bmatrix} + \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} \]

To add the matrices, we add the corresponding elements:
\[ P + A = \begin{bmatrix} \frac{\sqrt{3}}{2} + 1 & \frac{1}{2} + 1 \\ -\frac{1}{2} + 0 & \frac{\sqrt{3}}{2} + 1 \end{bmatrix} \]

### Step 5: Simplify the resulting matrix

Now we simplify each element:
1. \( \frac{\sqrt{3}}{2} + 1 = \frac{\sqrt{3} + 2}{2} \)
2. \( \frac{1}{2} + 1 = \frac{1 + 2}{2} = \frac{3}{2} \)
3. \( -\frac{1}{2} + 0 = -\frac{1}{2} \)
4. \( \frac{\sqrt{3}}{2} + 1 = \frac{\sqrt{3} + 2}{2} \)

Thus, the final result is:
\[ P + A = \begin{bmatrix} \frac{\sqrt{3} + 2}{2} & \frac{3}{2} \\ -\frac{1}{2} & \frac{\sqrt{3} + 2}{2} \end{bmatrix} \]

### Final Answer
\[ P + A = \begin{bmatrix} \frac{\sqrt{3} + 2}{2} & \frac{3}{2} \\ -\frac{1}{2} & \frac{\sqrt{3} + 2}{2} \end{bmatrix} \]