4x4 lorentz matrix and finding its inverse

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I have been struggling to find an inverse to a Lorentz matrix [itex]\Lambda[/itex] using formula: [itex]\Lambda^{-1}= \frac{1}{| \Lambda| }\textrm{adj}(\Lambda)[/itex] from linear algebra.

[itex] \Lambda =<br /> \begin{bmatrix}<br /> \gamma&0&0&-\beta \gamma \\<br /> 0 & 1 & 0 & 0\\<br /> 0 & 0 & 1 & 0\\<br /> -\beta \gamma & 0 & 0 & \gamma <br /> \end{bmatrix}[/itex]

My professor says that inverse to the matrix above is:

[itex] \Lambda^{-1} =<br /> \begin{bmatrix}<br /> \gamma&0&0&\beta \gamma \\<br /> 0 & 1 & 0 & 0\\<br /> 0 & 0 & 1 & 0\\<br /> \beta \gamma & 0 & 0 & \gamma <br /> \end{bmatrix}[/itex]

I tried to derive inverse matrix metodically by first calculating the determinant of [itex]\Lambda[/itex], then matrix of minors, matrix of cofactors,
adjugate matrix and in the end using the above formula to find the inverse. And i end up with this:

[itex] \Lambda^{-1} = \frac{1}{|\Lambda|} \textrm{adj}(\Lambda) = \frac{1}{\gamma^2 (1 - \beta^2)}<br /> \begin{bmatrix}<br /> \gamma & 0 & 0 &\beta \gamma\\<br /> 0 & \gamma^2(1-\beta^2) & 0 & 0\\<br /> 0 & 0 & \gamma^2(1-\beta^2) & 0\\<br /> \beta \gamma & 0 & 0 & \gamma\\<br /> \end{bmatrix}[/itex]

Well my result is not what my professor says i should get. In my adjugate matrix parts with [itex]\gamma[/itex] and [itex]\beta \gamma[/itex] seem wrong.
Is it possible my professor wrote down wrong inverse matrix?

I have tripple checked all the calculations, but i simply can't get the right result. Here is my whole derivation (please click the lower "DOWNLOAD" button).
Could anyone point me to the right direcction?
I am kind of lost here, but i am sure i have done a lot of work and am near the solution.
 
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Erm, what do you think [itex]\gamma^2 (1-\beta^2)[/itex] is?
 
Muphrid said:
Erm, what do you think [itex]\gamma^2 (1-\beta^2)[/itex] is?

OMG it is 1! I AM SUCH A NOOB :)