If $A =\begin{bmatrix}x+z&2&-3\\x&0&4\\3&x-y&0\end{bmatrix}$ is a skew-symmetric matrix, then which of the following are true?

(A) $y >z>x$
(B) $x > y$
(C) $x + y + z>0$
(D) $z>x$

Choose the correct answer from the options given below:

(C) and (D) only

The correct answer is Option (4) → (C) and (D) only

Given matrix: $A = \begin{bmatrix} x+z & 2 & -3 \\ x & 0 & 4 \\ 3 & x-y & 0 \end{bmatrix}$

For a skew-symmetric matrix: $A^T = -A \Rightarrow a_{ii} = 0$ and $a_{ij} = -a_{ji}$

Check diagonal elements:

$a_{11} = x+z = 0 \Rightarrow x + z = 0 \Rightarrow z = -x$

$a_{22} = 0$

$a_{33} = 0$

Check off-diagonal elements:

$a_{12} = 2$, $a_{21} = x \Rightarrow x = -2$

Then $z = -x = 2$

$a_{23} = 4$, $a_{32} = x - y = -2 - y \Rightarrow 4 = -(-2 - y) = 2 + y \Rightarrow y = 2$

Now values: $x = -2, y = 2, z = 2$

Check options:

(A) y > z > x → 2 > 2 > -2 → false (y ≯ z)

(B) x > y → -2 > 2 → false

(C) x + y + z = -2 + 2 + 2 = 2 > 0 → true

(D) z > x → 2 > -2 → true

Correct statements: (C), (D)