<s>[INST] Finding the inverse of $A$ 
Find the inverse  of $$A =\left[\begin{matrix}0 & 1 & 0 & 0\\ 0& 0 & 1 & 0\\ 0 & 0 & 0 & 1\\ a & b & c & d\end{matrix}\right]$$

My attempts: 
$$A^{-1} = \frac {\operatorname{adj}A}{\det A}$$
As  I can  find the  $\det A$ that is $\det A =  -b$. Here,  how  can I find the  inverse? Is  there  any easy  method/ or easy procedure  to find the  inverse of $A$?
 [/INST] You're wrong: expanding by the first column, you obtain at once $\;\det A=-a$. We'll suppose $a\ne0$.
To find the inverse you can perform column reduction:
\begin{align}
\begin{bmatrix}
\begin{array}{cccc|cccc}
0&1&0&0&1&0&0&0 \\
0&0&1&0 &0&1&0&0\\
0&0&0&1&0&0&1&0\\
a&b&c&d&0&0&0&1
\end{array}
\end{bmatrix}&\xrightarrow[\scriptstyle\to\,\text{(c$_2$,c$_3$,c$_4$, c$_1$)}]{\text{(c$_1$,c$_2$,c$_3$,c$_4$)}}
\begin{bmatrix}
\begin{array}{cccc|cccc}
1&0&0&0&0&0&0&1 \\
0&1&0&0 &1&0&0&0\\
0&0&1&0&0&1&0&0\\
b&c&d&a&0&0&1&0
\end{array}
\end{bmatrix}\xrightarrow{\scriptstyle\tfrac1a\text{c}_4\,\to\,\text{c}_4}\\[1ex]
\begin{bmatrix}
\begin{array}{cccc|cccc}
1&0&0&0&0&0&0&\frac1a \\
0&1&0&0 &1&0&0&0\\
0&0&1&0&0&1&0&0\\
b&c&d&1&0&0&1&0
\end{array}
\end{bmatrix}&\xrightarrow[\scriptstyle\text{(c}_2-c\,\text{c}_4,\text{c}_3-d\,\text{c}_4)]{(\text{c}_1,\text{c}_2,\text{c}_3)\,\to\,(\text{c}_1-b\,\text{c}_4,\:}
\begin{bmatrix}
\begin{array}{cccc|cccc}
\!\!1&0 &0&0&-\frac ba&-\frac ca& -\frac da&\frac1a \\
0&1&0&0 &1&0&0&0\\
0&0&1&0&0&1&0&0\\
0&0&0&1&0&0&1&0
\end{array}
\end{bmatrix}
\end{align}
</s>