let `P_(1)=[(1,0,0),(0,1,0),(0,0,1)],P_(2)=[(1,0,0),(0,0,1),(0,1,0)],P_(3)=[(0,1,0),(1,0,0),(0,0,1)],P_(4)=[(0,1,0),(0,0,1),(1,0,0)],P_(5)=[(0,0,1),(1,0,0),(0,1,0)],P_(6)=[(0,0,1),(0,1,0),(1,0,0)]` and `X=sum_(k=1)^(6) P_(k)[(2,1,3),(1,0,2),(3,2,1)]P_(k)^(T)` Where `P_(k)^(T)` is transpose of matrix `P_(k)`. Then which of the following options is/are correct?

Options

A. X is a symmetric matrix
B. if `X=[(1),(1),(1)]=alpha[(1),(1),(1)]`, then `alpha=30`
C. X-30I is an invertible matrix
D. The sum of diagonal entries of X is 18.

Explanation

To solve the problem step by step, we will analyze the given matrices \( P_k \) and the expression for \( X \).

### Step 1: Understand the given matrices

The matrices \( P_1, P_2, P_3, P_4, P_5, P_6 \) are defined as follows:

\[ P_1 = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}, \quad P_2 = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix}, \quad P_3 = \begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix} \]
\[ P_4 = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{pmatrix}, \quad P_5 = \begin{pmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix}, \quad P_6 = \begin{pmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{pmatrix} \]

### Step 2: Define the matrix \( A \)

The matrix \( A \) is given by:

\[ A = \begin{pmatrix} 2 & 1 & 3 \\ 1 & 0 & 2 \\ 3 & 2 & 1 \end{pmatrix} \]

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

The expression for \( X \) is:

\[ X = \sum_{k=1}^{6} P_k A P_k^T \]

Here, \( P_k^T \) is the transpose of the matrix \( P_k \).

### Step 4: Verify the symmetry of \( X \)

To show that \( X \) is symmetric, we need to demonstrate that \( X^T = X \).

Using the property of transpose:

\[ X^T = \left( \sum_{k=1}^{6} P_k A P_k^T \right)^T = \sum_{k=1}^{6} (P_k A P_k^T)^T = \sum_{k=1}^{6} P_k^T A^T P_k \]

Since \( P_k^T = P_k \) (as shown in the video), we have:

\[ X^T = \sum_{k=1}^{6} P_k A P_k = X \]

Thus, \( X \) is symmetric.

### Step 5: Calculate \( X \cdot \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} \)

Let \( B = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} \).

We need to compute \( X B \):

\[ X B = \sum_{k=1}^{6} P_k A P_k^T B \]

Since \( P_k^T B = P_k B \) and \( P_k B \) results in a vector that sums the entries of \( B \):

\[ P_k B = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} \text{ (for all k)} \]

Thus,

\[ X B = \sum_{k=1}^{6} P_k A \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} \]

Calculating \( A \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} \):

\[ A \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 2 + 1 + 3 \\ 1 + 0 + 2 \\ 3 + 2 + 1 \end{pmatrix} = \begin{pmatrix} 6 \\ 3 \\ 6 \end{pmatrix} \]

Thus,

\[ X B = \sum_{k=1}^{6} P_k \begin{pmatrix} 6 \\ 3 \\ 6 \end{pmatrix} \]

This results in:

\[ X B = 6 \cdot \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} + 3 \cdot \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} + 6 \cdot \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} = 30 \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} \]

Thus, \( \alpha = 30 \).

### Step 6: Check if \( X - 30I \) is invertible

Since \( X B = 30 B \), we have:

\[ (X - 30I) B = 0 \]

This implies that \( X - 30I \) is not invertible.

### Step 7: Calculate the trace of \( X \)

The trace of \( X \) is given by:

\[ \text{trace}(X) = \text{trace}(P_1 A P_1^T) + \text{trace}(P_2 A P_2^T) + \ldots + \text{trace}(P_6 A P_6^T) \]

Each \( P_k A P_k^T \) contributes to the trace, and since the trace of a product is invariant under cyclic permutations:

\[ \text{trace}(P_k A P_k^T) = \text{trace}(A) \]

Calculating the trace of \( A \):

\[ \text{trace}(A) = 2 + 0 + 1 = 3 \]

Thus,

\[ \text{trace}(X) = 6 \cdot 3 = 18 \]

### Conclusion

The correct options are:
1. \( X \) is a symmetric matrix.
2. \( \alpha = 30 \).
3. The sum of diagonal entries of \( X \) is 18.