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) $x ≠ y ≠ z$

Choose the correct answer from the options given below:

(C) and (D) only

The correct answer is Option (3) → (C) and (D) only

$\det A= \begin{vmatrix} x&3\\ 2&4\end{vmatrix}=4x-6=0 \Rightarrow x=\frac{3}{2}$

$\det B= \begin{vmatrix} 2&3\\ y&3\end{vmatrix}=6-3y=0 \Rightarrow y=2$

$\det C= \begin{vmatrix} z&1\\ 8&2\end{vmatrix}=2z-8=0 \Rightarrow z=4$

$\Rightarrow z>y>x\;(4>2>\frac{3}{2})$

Correct: (C) $z>x$ and (D) $x\ne y\ne z$