$4 \times 4$ matrix and its inverse. Is my method ok? I have been struggling to derive inverse matrix of a $4 \times 4$  Lorenz matrix $\Lambda$. 
$$
\Lambda =
\begin{bmatrix}
\gamma&0&0&-\beta \gamma \\
0 & 1 & 0 & 0\\
0 & 0 & 1 & 0\\
-\beta \gamma & 0 & 0 & \gamma 
\end{bmatrix}
$$
My professor says that inverse is:
$$
\Lambda^{-1} =
\begin{bmatrix}
\gamma&0&0&\beta \gamma \\
0 & 1 & 0 & 0\\
0 & 0 & 1 & 0\\
\beta \gamma & 0 & 0 & \gamma 
\end{bmatrix}
$$
Soo i tried to derive inverse matrix $\Lambda^{-1}$ using adjugate formula:
$$\Lambda^{-1} = \frac{1}{|\Lambda|} \textrm{adj}(\Lambda)$$
I did it quite metodically by first calculating the determinant of $\Lambda$, then matrix of minors, matrix of cofactors, 
adjugate matrix and in the end using the above formula to find the inverse (here is my whole derivation). Long story short, at the end I end up with this:
$$
\Lambda^{-1} = \frac{1}{|\Lambda|} \textrm{adj}(\Lambda) = \frac{1}{\gamma^2 (1 - \beta^2)}
\begin{bmatrix}
\gamma & 0 & 0 &\beta \gamma\\
0 & \gamma^2(1-\beta^2) & 0 & 0\\
0 & 0 & \gamma^2(1-\beta^2) & 0\\
\beta \gamma & 0 & 0 & \gamma\\
\end{bmatrix}
$$
The result is not what my professor says i should get. In my adjugate matrix parts with $\gamma$ and $\beta \gamma$ seem wrong. 
Is it possible my professor wrote down wrong inverse matrix?
Could anyone point me to the right direction? 
I am kind of lost here, but i am sure i have done a lot of work and am near the solution.
 A: In special relativity, the quantity $\gamma$ is defined as
$$\gamma^2=\frac{1}{1-\beta^2},$$
so that your result does, in fact, agree with your professor's.
To be completely explicit, $\beta=v/c$ is a one-dimensional velocity in units of the speed of light, and $\gamma=1/\sqrt{1-(v/c)^2}$ is the factor which shows up in all computations in special relativity (see Lorentz Factor)
A: It depends on what exactly you want to prove. If you're given two formulas for matrices and want to prove them inverses of each other, just multiplying out would usualy be much simpler them plugging one into the adjugate formula and checking that it gives the other. (That's computing $16$ scalar products versus $16$ order-$3$ determinants and one order-$4$ determinant as bonus.)
A: If what you need to know is if your computation was ok, then the question has been answered. However, it seems important to notice that you don't actually have to invert this matrix. If one knows the effect $\Lambda$ has $-$rotating in the $z$-$ct$ plane$-$, and
$$
\Lambda=\begin{bmatrix}
\gamma&0&0&-\beta \gamma \\
0 & 1 & 0 & 0\\
0 & 0 & 1 & 0\\
-\beta \gamma & 0 & 0 & \gamma 
\end{bmatrix}
=\begin{bmatrix}
\cosh \phi&0&0&-\sinh\phi \\
0 & 1 & 0 & 0\\
0 & 0 & 1 & 0\\
-\sinh \phi & 0 & 0 & \cosh \phi
\end{bmatrix}
=\exp(-\phi K_3),$$
where
$$
K_3=\begin{bmatrix}
0&0&0&1\\
0&0&0&0\\
0&0&0&0\\
1&0&0&0
\end{bmatrix}
$$
The last equality follows from expanding the exponential. Notice that $K_3^2=$diag$(1,0,0,1)$.
Then the inverse of this rotation is $\Lambda^{-1}=\exp(+\phi K_3)$, that is
$$
\Lambda^{-1}
=\begin{bmatrix}
\cosh \phi&0&0&-\sinh\phi \\
0 & 1 & 0 & 0\\
0 & 0 & 1 & 0\\
-\sinh \phi & 0 & 0 & \cosh \phi
\end{bmatrix}=
\begin{bmatrix}
\gamma&0&0&+\beta \gamma \\
0 & 1 & 0 & 0\\
0 & 0 & 1 & 0\\
+\beta \gamma & 0 & 0 & \gamma 
\end{bmatrix}.
$$
In physical terms, the velocity $\beta$ parametrizes the rotation, and in this matrix it is the only parameter. It shifts a reference system to another which moves along the $z$ axis with velocity $c\beta$. 
Hence by setting the parameter to $-\beta$ one finds the inverse. 
