For $\alpha, \beta \in R$ and a natural number $n$,let $A_r = \begin{vmatrix} r & 1 & \frac{n^2}{2} + \alpha \\ 2r & 2 & n^2 - \beta \\ 3r - 2 & 3 & \frac{n(3n - 1)}{2} \end{vmatrix}$. Then $2A_{10} - A_8$ is equal to:

If $A$ and $B$ are two matrices such that $AB = B$ and $BA = A$, then $A^{2} + B^{2}$ equals

If $\Delta=\left|\begin{array}{lll}1 & 5 & 6 \\ 0 & 1 & 7 \\ 0 & 0 & 1\end{array}\right|$ and $\Delta^{\prime}=\left|\begin{array}{ccc}1 & 0 & 1 \\ 3 & 0 & 3 \\ 4 & 6 & 100\end{array}\right|$, then

$A = [a_{ij}]$ is a $3 \times 3$ matrix with positive integers as its elements. Elements of $A$ are such that the sum of all the elements of each row is equal to $6$ and $a_{22} = 2$. If $a_{ii} = \begin{cases} a_{ij} + a_{ji}, & j = i + 1 \text{ when } i < 3 \\ a_{ij} + a_{ji}, & j = 4 - i \text{ when } i = 3 \end{cases}$ for $i = 1, 2, 3$, then $|A| = $

Let $ABC = I$. Then $tr(ABC + BCA + CAB)$ is (where the order of matrices $A, B, C$ is $3 \times 3$ and $tr(A)$ is the sum of the diagonal elements in $A$).

Let $I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$ and $P = \begin{bmatrix} 2 & 0 \\ 0 & 3 \end{bmatrix}$. Let $Q = \begin{bmatrix} x & y \\ z & 4 \end{bmatrix}$ for some non-zero real numbers $x, y$,and $z$,for which there exists a $2 \times 2$ matrix $R$ with all entries being non-zero real numbers,such that $QR = RP$. Then which of the following statements is (are) true?
$(A)$ $|Q-2I| = 0$
$(B)$ $|Q-6I| = 12$
$(C)$ $|Q-3I| = 15$
$(D)$ $yz = 2$