Let

P_{1} = ∣ ∣ ∣ ∣ \begin{matrix} 1 & 0 & 0 0 & 1 & 0 0 & 0 & 1 \end{matrix} ∣ ∣ ∣ ∣,

P_{2} = ∣ ∣ ∣ ∣ \begin{matrix} 1 & 0 & 0 0 & 0 & 1 0 & 1 & 0 \end{matrix} ∣ ∣ ∣ ∣,

P_{3} = ∣ ∣ ∣ ∣ \begin{matrix} 0 & 1 & 0 1 & 0 & 0 0 & 0 & 1 \end{matrix} ∣ ∣ ∣ ∣,

P_{4} = ∣ ∣ ∣ ∣ \begin{matrix} 0 & 1 & 0 0 & 0 & 1 1 & 0 & 0 \end{matrix} ∣ ∣ ∣ ∣,

P_{5} = ∣ ∣ ∣ ∣ \begin{matrix} 0 & 0 & 1 1 & 0 & 0 0 & 1 & 0 \end{matrix} ∣ ∣ ∣ ∣,

P_{6} = ∣ ∣ ∣ ∣ \begin{matrix} 0 & 0 & 1 0 & 1 & 0 1 & 0 & 0 \end{matrix} ∣ ∣ ∣ ∣,

and

X = 6 \sum k = 1 P_{k} ⎡ ⎢ ⎣ \begin{matrix} 2 & 1 & 3 1 & 0 & 2 3 & 2 & 1 \end{matrix} ⎤ ⎥ ⎦ P_{k}^{T},

where

P_{k}^{T}

denotes the transpose of the matrix

P_{k} .

Then which of the following options is/are correct?

$X - 30 I$ is an invertible matrix
The sum of diagonal entries of $X$ is $18$
$X$ is a symmetric matrix
$If X ⎡ ⎢ ⎣ \begin{matrix} 111 \end{matrix} ⎤ ⎥ ⎦ = α ⎡ ⎢ ⎣ \begin{matrix} 111 \end{matrix} ⎤ ⎥ ⎦, then α = 30$

In the Boolean algebra, the following one is wrong1+0=11+1=10+1=10+0=1

In the Boolean algebra, the following one is wrong
$1 + 0 = 1$
$1 + 1 = 1$
$0 + 1 = 1$
$0 + 0 = 1$