Answer
$\begin{bmatrix}1/a &0 &0 &0 \\0 &1/b &0 &0 \\0 &0 &1/c &0 \\0 &0 &0 &1/d \end{bmatrix} $
Work Step by Step
Step 1. Write the matrix together with an identify as:
$\begin{array}( \\A|I= \\ \\ \end{array}
\begin{bmatrix} a & 0 &0 &0 | &1 &0 &0 &0 \\0&b & 0 &0| &0 &1 &0 &0 \\0 & 0 &c &0 | &0 &0 &1 &0 \\0&0 & 0 &d| &0 &0 &0 &1 \end{bmatrix} \begin{array}( R_1/a\to R_1\\R_2/b\to R_2 \\ R_3/c\to R_3\\R_4/d\to R_4 \end{array}$
Step 2. Use row operations to transform the left half into a reduced row-echelon:
$\begin{array}( \\A|I= \\ \\ \end{array}
\begin{bmatrix} 1 & 0 &0 &0 | &1/a &0 &0 &0 \\0&1 & 0 &0| &0 &1/b &0 &0 \\0 & 0 &1 &0 | &0 &0 &1/c &0 \\0&0 & 0 &1| &0 &0 &0 &1/d \end{bmatrix} $
Step 3. The inverse of the original matrix is:
$\begin{array}( \\A^{-1}= \\ \\ \end{array}
\begin{bmatrix}1/a &0 &0 &0 \\0 &1/b &0 &0 \\0 &0 &1/c &0 \\0 &0 &0 &1/d \end{bmatrix} $
