Let Q = `[{: (cos ""(pi)/(4) , - sin""(pi)/(4)),(sin""(pi)/(4), cos""(pi)/(4)):}] and x = [{:((2)/(sqrt(2))),((1)/(sqrt(2))):}]` then `Q^(3)` x is equal to

Options

A. `[{:(0),(1):}]`
B. `[{:(-(1)/(sqrt(2))),((1)/(sqrt(2))):}]`
C. `[{:(-1),(0):}]`
D. `[{:(-(1)/(sqrt(2))),(- (1)/(sqrt(2))):}]`

Answer

To solve the problem, we need to calculate \( Q^3 \cdot x \), where \( Q \) and \( x \) are given matrices.

### Step 1: Define the matrices
The matrix \( Q \) is defined as:
\[ Q = \begin{pmatrix} \cos\left(\frac{\pi}{4}\right) & -\sin\left(\frac{\pi}{4}\right) \\ \sin\left(\frac{\pi}{4}\right) & \cos\left(\frac{\pi}{4}\right) \end{pmatrix} \]
Using the values of sine and cosine at \( \frac{\pi}{4} \):
\[ \cos\left(\frac{\pi}{4}\right) = \sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \]
Thus, we can rewrite \( Q \) as:
\[ Q = \begin{pmatrix} \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{pmatrix} \]

The matrix \( x \) is given as:
\[ x = \begin{pmatrix} \frac{2}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \end{pmatrix} \]

### Step 2: Calculate \( Q^2 \)
To find \( Q^3 \), we first need to calculate \( Q^2 \):
\[ Q^2 = Q \cdot Q = \begin{pmatrix} \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{pmatrix} \cdot \begin{pmatrix} \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{pmatrix} \]

Calculating the elements of \( Q^2 \):
- First row, first column:
\[ \frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}} + \left(-\frac{1}{\sqrt{2}}\right) \cdot \frac{1}{\sqrt{2}} = \frac{1}{2} - \frac{1}{2} = 0 \]
- First row, second column:
\[ \frac{1}{\sqrt{2}} \cdot \left(-\frac{1}{\sqrt{2}}\right) + \left(-\frac{1}{\sqrt{2}}\right) \cdot \frac{1}{\sqrt{2}} = -\frac{1}{2} - \frac{1}{2} = -1 \]
- Second row, first column:
\[ \frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}} = \frac{1}{2} + \frac{1}{2} = 1 \]
- Second row, second column:
\[ \frac{1}{\sqrt{2}} \cdot \left(-\frac{1}{\sqrt{2}}\right) + \frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}} = -\frac{1}{2} + \frac{1}{2} = 0 \]

Thus, we have:
\[ Q^2 = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} \]

### Step 3: Calculate \( Q^3 \)
Now we calculate \( Q^3 = Q^2 \cdot Q \):
\[ Q^3 = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} \cdot \begin{pmatrix} \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{pmatrix} \]

Calculating the elements of \( Q^3 \):
- First row, first column:
\[ 0 \cdot \frac{1}{\sqrt{2}} + (-1) \cdot \frac{1}{\sqrt{2}} = -\frac{1}{\sqrt{2}} \]
- First row, second column:
\[ 0 \cdot \left(-\frac{1}{\sqrt{2}}\right) + (-1) \cdot \frac{1}{\sqrt{2}} = -\frac{1}{\sqrt{2}} \]
- Second row, first column:
\[ 1 \cdot \frac{1}{\sqrt{2}} + 0 \cdot \frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2}} \]
- Second row, second column:
\[ 1 \cdot \left(-\frac{1}{\sqrt{2}}\right) + 0 \cdot \frac{1}{\sqrt{2}} = -\frac{1}{\sqrt{2}} \]

Thus, we have:
\[ Q^3 = \begin{pmatrix} -\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \end{pmatrix} \]

### Step 4: Multiply \( Q^3 \) by \( x \)
Now we multiply \( Q^3 \) by \( x \):
\[ Q^3 \cdot x = \begin{pmatrix} -\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \end{pmatrix} \cdot \begin{pmatrix} \frac{2}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \end{pmatrix} \]

Calculating the elements:
- First row:
\[ -\frac{1}{\sqrt{2}} \cdot \frac{2}{\sqrt{2}} + (-\frac{1}{\sqrt{2}}) \cdot \frac{1}{\sqrt{2}} = -1 - \frac{1}{2} = -\frac{3}{2} \]
- Second row:
\[ \frac{1}{\sqrt{2}} \cdot \frac{2}{\sqrt{2}} + (-\frac{1}{\sqrt{2}}) \cdot \frac{1}{\sqrt{2}} = 1 - \frac{1}{2} = \frac{1}{2} \]

Thus, we have:
\[ Q^3 \cdot x = \begin{pmatrix} -\frac{3}{2} \\ \frac{1}{2} \end{pmatrix} \]

### Final Answer
The final result is:
\[ Q^3 \cdot x = \begin{pmatrix} -\frac{3}{2} \\ \frac{1}{2} \end{pmatrix} \]