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 $A$?

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,
$$\left[\begin{array}{c|ccc} 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1\\ \hline a & b & c & d\end{array}\right]^{-1} = \color{blue}{\left[\begin{array}{ccc|c} - \frac{b}{a} & - \frac{c}{a} & - \frac{d}{a} & \frac{1}{a}\\ \hline1 & 0 & 0 & 0\\0 & 1 & 0 & 0\\0 & 0 & 1 & 0\end{array}\right]}$$
