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 $A$?
 A: I think that the formula that you mentioned is the best way for this matrix with lots of zeros. Note that you made a mistake: $\det A=-a$, not $-b$. And$$\operatorname{Adj}A=\begin{bmatrix}b & c & d & -1 \\ -a & 0 & 0 & 0 \\ 0 & -a & 0 & 0 \\ 0 & 0 & -a & 0\end{bmatrix}.$$Therefore$$A^{-1}=\begin{bmatrix}-\frac{b}{a} & -\frac{c}{a} & -\frac{d}{a} &   \frac{1}{a} \\ 1 & 0 & 0 & 0 \\
 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0\end{bmatrix}.$$
A: Let
$$\left[\begin{array}{c|ccc} 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1\\ \hline a & b & c & d\end{array}\right]^{-1} := \left[\begin{array}{c|c} \mathrm p_{11}^\top & p_{12}\\ \hline \mathrm P_{21} & \mathrm p_{22}\end{array}\right]$$
Let $\mathrm r^\top := \begin{bmatrix} b & c & d\end{bmatrix}$. From
$$\left[\begin{array}{c|c} 0_3 & \mathrm I_3\\ \hline a & \mathrm r^\top\end{array}\right] \left[\begin{array}{c|c} \mathrm p_{11}^\top & p_{12}\\ \hline \mathrm P_{21} & \mathrm p_{22}\end{array}\right] = \left[\begin{array}{c|c} \mathrm I_3 & 0_3\\ \hline 0_3^\top & 1\end{array}\right]$$
we obtain $\mathrm P_{21} = \mathrm I_3$, then $\mathrm p_{22} = 0_3$, then $\mathrm p_{11}^\top = -\frac{1}{a} \mathrm r^\top$ and, lastly, $p_{12} = \frac 1a$. Thus,
$$\left[\begin{array}{c|ccc} 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1\\ \hline a & b & c & d\end{array}\right]^{-1} = \color{blue}{\left[\begin{array}{ccc|c} - \frac{b}{a} & - \frac{c}{a} & - \frac{d}{a} & \frac{1}{a}\\ \hline1 & 0 & 0 & 0\\0 & 1 & 0 & 0\\0 & 0 & 1 & 0\end{array}\right]}$$
A: I suggest Gauss-Jordan method
$$[A|I]\to [I|A^{-1}]$$
Refer also to Method for Finding Matrix-Inverse Through Gauss-Jordan?
A: 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}
A: Let
$$\mathrm M := \begin{bmatrix} 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1\\ a & b & c & d\end{bmatrix}$$
which is the companion matrix to the following (characteristic) polynomial
$$p (s) := s^4 - d s^3 - c s^2 - b s - a$$
Using the Cayley-Hamilton theorem,
$$\mathrm M^4 - d \,\mathrm M^3 - c \,\mathrm M^2 - b\,\mathrm M - a \,\mathrm I_4 = \mathrm O_4$$
Multiplying both sides by $\mathrm M^{-1}$, we obtain
$$\mathrm M^3 - d \,\mathrm M^2 - c \,\mathrm M - b\,\mathrm I_4 - a \mathrm M^{-1} = \mathrm O_4$$
Assuming that $a \neq 0$,
$$\mathrm M^{-1} = \frac{1}{a} \, \mathrm M^3 - \frac{d}{a} \,\mathrm M^2 - \frac{c}{a} \,\mathrm M - \frac{b}{a}\,\mathrm I_4$$
Using SymPy:
>>> a, b, c, d = symbols('a b c d')
>>> M = Matrix([[0,1,0,0],
                [0,0,1,0],
                [0,0,0,1],
                [a,b,c,d]])
>>> simplify((1/a)*M**3 - (d/a)*M**2 - (c/a)*M - (b/a)*eye(4))
Matrix([[-b/a, -c/a, -d/a, 1/a],
        [   1,    0,    0,   0],
        [   0,    1,    0,   0],
        [   0,    0,    1,   0]])

