The inverse of the matrix $A = \begin{bmatrix}\frac{1}{2}&0&0\\0&\frac{1}{4}&0\\0&0&\frac{1}{8}\end{bmatrix}$ is

$\begin{bmatrix}2&0&0\\0&4&0\\0&0&8\end{bmatrix}$

The correct answer is Option (3) → $\begin{bmatrix}2&0&0\\0&4&0\\0&0&8\end{bmatrix}$

$A=\begin{bmatrix}\frac{1}{2}&0&0\\[4pt]0&\frac{1}{4}&0\\[4pt]0&0&\frac{1}{8}\end{bmatrix}$

Since $A$ is diagonal, its inverse is obtained by taking the reciprocal of each diagonal entry.

$\frac{1}{\frac{1}{2}}=2,\quad \frac{1}{\frac{1}{4}}=4,\quad \frac{1}{\frac{1}{8}}=8$

Therefore,

$A^{-1}=\begin{bmatrix}2&0&0\\[4pt]0&4&0\\[4pt]0&0&8\end{bmatrix}$

The inverse matrix is $\begin{bmatrix}2&0&0\\0&4&0\\0&0&8\end{bmatrix}$.