Let $A = [a_{ij}]_{3\times3}$ be a matrix such that $a_{ij} = \dfrac{i-j}{i+j}$, then which of the following statements is/are TRUE?

A. A is a symmetric matrix

B. A is a skew symmetric matrix

C. $|A| = 0$

D. $|A| \neq 0$

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If $A = [a_{ij}]_{3 \times 3}$ and $a_{ij} = i - j$, then which of the following statements are TRUE?

(A). $A$ is scalar matrix

(B). $A$ is skew symmetric matrix

(C). $|A| = 0$

(D). $a_{21} + a_{12} = 0$

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If $x \neq y \neq z$ then $\begin{vmatrix} 1 & x & x^2 \\ 1 & y & y^2 \\ 1 & z & z^2 \end{vmatrix}$ is equal to

If A and B are symmetric matrices, then AB - BA is

Let A be a square matrix of order 3 then |3A| is equal to

If $A = \begin{bmatrix}1 & 2\\3 & 4\end{bmatrix}$ such that $A^2 + \beta A + \gamma I = 0$ then $\beta + \gamma$ is equal to:

(where $I$ is an identity matrix of order 2)

If $A = [a_{ij}]_{3 \times 2}$ where $a_{ij} = i + j$, then

(A) A is a square matrix

(B) $a_{21} + a_{32} = 8$

(C) Number of elements in A is 6

(D) Transpose of $A = \begin{bmatrix}2 & 3 \\ 3 & 4 \\ 4 & 5\end{bmatrix}$

Choose the correct answer from the options given below: