If $A = \begin{bmatrix} x & 3 \\ 2 & 4 \end{bmatrix}$, $B = \begin{bmatrix} 2 & 3 \\ y & 3 \end{bmatrix}$ and $C = \begin{bmatrix} z & 1 \\ 8 & 2 \end{bmatrix}$ are singular matrices then:

(A) $x > y$

(B) $y > z$

(C) $z > x$

(D) $x \neq y \neq z$

Choose the correct answer from the options given below:

A singular matrix has determinant equal to 0.

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For matrix $A = \begin{bmatrix} x & 3 \\ 2 & 4 \end{bmatrix}$:

$\det(A) = (x)(4) - (3)(2) = 0$

$4x - 6 = 0$

$4x = 6$

$x = \frac{6}{4} = 1.5$

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For matrix $B = \begin{bmatrix} 2 & 3 \\ y & 3 \end{bmatrix}$:

$\det(B) = (2)(3) - (3)(y) = 0$

$6 - 3y = 0$

$3y = 6$

$y = 2$

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For matrix $C = \begin{bmatrix} z & 1 \\ 8 & 2 \end{bmatrix}$:

$\det(C) = (z)(2) - (1)(8) = 0$

$2z - 8 = 0$

$2z = 8$

$z = 4$

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With $x = 1.5$, $y = 2$, and $z = 4$:

(A) $x > y$: Is $1.5 > 2$? False

(B) $y > z$: Is $2 > 4$? False

(C) $z > x$: Is $4 > 1.5$? True

(D) $x \neq y \neq z$: Are $1.5$, $2$, and $4$ all different? True

Both (C) and (D) are correct.