To find the sum of all elements of the matrix [tex]\(X\)[/tex], let's break down the problem into detailed steps. We are given:
[tex]\[ P = \begin{pmatrix}
\frac{\sqrt{3}}{2} & \frac{1}{2} \\
\frac{-1}{2} & \frac{\sqrt{3}}{2}
\end{pmatrix} \][/tex]
[tex]\[ A = \begin{pmatrix}
1 & 1 \\
0 & 1
\end{pmatrix} \][/tex]
We need to calculate matrices step-by-step and finally find the sum of all elements in matrix [tex]\(X\)[/tex].
### Step 1: Compute [tex]\( P^\tau \)[/tex]
Since the transpose ([tex]\(\tau\)[/tex]) of matrix [tex]\(P\)[/tex] is simply swapping rows with columns,
[tex]\[ P^\tau = \begin{pmatrix}
\frac{\sqrt{3}}{2} & \frac{-1}{2} \\
\frac{1}{2} & \frac{\sqrt{3}}{2}
\end{pmatrix} \][/tex]
### Step 2: Compute [tex]\( Q = P A P^\tau \)[/tex]
We perform the matrix multiplication for [tex]\( Q \)[/tex]:
[tex]\[ Q = P A P^\tau = \begin{pmatrix}
\frac{\sqrt{3}}{2} & \frac{1}{2} \\
\frac{-1}{2} & \frac{\sqrt{3}}{2}
\end{pmatrix} \begin{pmatrix}
1 & 1 \\
0 & 1
\end{pmatrix} \begin{pmatrix}
\frac{\sqrt{3}}{2} & \frac{-1}{2} \\
\frac{1}{2} & \frac{\sqrt{3}}{2}
\end{pmatrix} \][/tex]
Upon calculation (details omitted for brevity), this results in:
[tex]\[ Q = \begin{pmatrix}
1.4330127 & 0.75 \\
-0.25 & 0.5669873
\end{pmatrix} \][/tex]
### Step 3: Compute [tex]\( Q^{2015} \)[/tex]
We now find the 2015th power of matrix [tex]\(Q\)[/tex]:
[tex]\[ Q^{2015} = \begin{pmatrix}
873.52059474 & 1511.25000074 \\
-503.75000024 & -871.52059473
\end{pmatrix} \][/tex]
### Step 4: Compute [tex]\( X = P^\tau Q^{2015} P \)[/tex]
Next, we compute [tex]\(X\)[/tex] by multiplying [tex]\(P^\tau\)[/tex], [tex]\(Q^{2015}\)[/tex], and [tex]\(P\)[/tex]:
[tex]\[ X = P^\tau Q^{2015} P \][/tex]
Upon computation (details omitted), it results in:
[tex]\[ X = \begin{pmatrix}
1.0 & 2015.0 \\
-5.57803803 \times 10^{-12} & 1.0
\end{pmatrix} \][/tex]
### Step 5: Sum of all elements in [tex]\(X\)[/tex]
Finally, we sum all the elements of [tex]\(X\)[/tex]:
[tex]\[
\text{Sum} = 1.0 + 2015.0 - 5.57803803 \times 10^{-12} + 1.0 \approx 2017.0
\][/tex]
Thus, the sum of all elements of [tex]\(X\)[/tex] is:
[tex]\[
\boxed{2017}
\][/tex]
