Find a vector orthogonal to the vectors $2e_0+e_1+e_2$ and $e_0+e_1+e_2$ in $\R_1^3$ and show that the matrix
$$
A\colon=\left(\begin{array}{ccc}
\sqrt2&1&0\\
1/\sqrt2&1&-1/\sqrt2\\
1/\sqrt2&1&1/\sqrt2
\end{array}\right)
$$
defines a Lorentz transformation in $\R_1^3$. Is it a boost, a rotation? Is it diagonalizable? Finally compute the inverse matrix on your own!
By the formula given in exam a vector orthogonal to both $2e_0+e_1+e_2$ and $e_0+e_1+e_2$ is $-e_1+e_2$. The last three vectors are pairwise orthogonal and the columns of $A$ are just the components of normalized versions of these vectors, i.e. $A$ is the matrix of an isometry. Finally $\det A=+1$ and $a_{00} > 0$.
2. To find the eigenvalues and eigenvectors we employ sage:
sage: a=sqrt(2)
sage: A=matrix([[a,1,0],[1/a,1,-1/a],[1/a,1,1/a]])
sage: A.eigenvectors_right()
The eigenvalues of $A$ are $1,\sqrt2$ and $1/\sqrt2$, i.e. $A$ is diagonalizable and the corresponding eigenvectors are
$$
b_1\colon=e_0+(1-\sqrt2)e_1+e_2,
b_2\colon=e_0+e_2,
b_3\colon=e_0-1/\sqrt2\,e_1+1/\sqrt2\,e_2
$$
$A$ has exactly one eigenvalue $1$ and the corresponding eigenvector $b_1$ is space-like. Hence $A$ is not a rotation. Since $\la b_2,b_1\ra=\la b_3,b_1\ra=0$ the subspace $\lhull{b_2,b_3}=b_1^\perp$ is Lorentz, i.e. $A$ is a boost and as $b_2$ and $b_3$ are light-like, the direction of the boost is either $(b_2+b_3)/\norm{b_2+b_3}$ or $(b_2-b_3)/\norm{b_2-b_3}$. Now
$$
\frac{b_2+b_3}{\norm{b_2+b_3}}
=\frac{2e_0-1/\sqrt2 e_1+(1+1/\sqrt2)e_2}{\sqrt{2-\sqrt2}}
$$
is time-like and therefore the direction of the boost must be
$$
\frac{b_2-b_3}{\norm{b_2-b_3}}
=\frac{1/\sqrt2e_1+(1-1/\sqrt2)e_2}{\sqrt{2-\sqrt2}}
\quad\mbox{and}\quad
\cosh(\vp)=\frac3{\sqrt8}~.
$$
3. Since $A$ is the matrix of an isometry with respect to the ONB $e_0,e_1,e_2$ the inverse is the transposed with the first row multiplied by $-1=\la e_0,e_0\ra$ and then the first column multiplied by $-1$, i.e.
$$
\left(\begin{array}{ccc}
\sqrt2&-1/\sqrt2&-1/\sqrt2\\
-1&1&1\\
0&-1/\sqrt2&1/\sqrt2
\end{array}\right)
$$