Choose the correct answer. If x, y, z are nonzero real numbers,

Answer: A

$A=\left|\begin{array}{lll}x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z\end{array}\right|$

$\therefore|A|=x(y z-0)=x y z \neq 0$

Now, $A_{11}=y z, A_{12}=0, A_{13}=0$

$A_{21}=0, A_{22}=x z, A_{23}=0$

$A_{31}=0, A_{32}=0, A_{33}=x y$

$\therefore \operatorname{adj} A=\left[\begin{array}{lll}y z & 0 & 0 \\ 0 & x z & 0 \\ 0 & 0 & x y\end{array}\right]$

$\therefore A^{-1}=\frac{1}{|A|}$ adjA

$=\frac{1}{x y z}\left[\begin{array}{lll}y z & 0 & 0 \\ 0 & x z & 0 \\ 0 & 0 & x y\end{array}\right]$

$=\left[\begin{array}{ccc}\frac{y z}{x y z} & 0 & 0 \\ 0 & \frac{x z}{x y z} & 0 \\ 0 & 0 & \frac{x y}{x y z}\end{array}\right]$

$=\left[\begin{array}{ccc}\frac{1}{x} & 0 & 0 \\ 0 & \frac{1}{y} & 0 \\ 0 & 0 & \frac{1}{z}\end{array}\right]=\left[\begin{array}{lll}x^{-1} & 0 & 0 \\ 0 & y^{-1} & 0 \\ 0 & 0 & z^{-1}\end{array}\right]$

The correct answer is A.