We have a given a tensor $(2,0)$ and its components are, $M^{\alpha \beta} = \begin{pmatrix} 0 & 1 & 0 & 0 \\ 1 & -1 & 0 & 2 \\ 2 & 0 & 0 & 1 \\ 1 & 0 & -2 & 0 \end{pmatrix}$
The questions ask us to find the components of
(i) $M^{\alpha}_{\beta}$
(ii) $M^{\beta}_{\alpha}$
(iii) $M_{\alpha \beta}$
Answers
(i)
I used the $M^{\alpha}_{\beta} = \eta_{\beta \mu}M^{\alpha \mu}$ for $\eta_{\beta \mu} = diag(-1,1,1,1)$
If my calculations are corret I find $M^{\alpha}_{\beta} = \begin{pmatrix} 0 & -1 & -2 & -1 \\ 1 & -1 & 0 & 0 \\ 0 & 0 & 0 & -2 \\ 0 & -2 & -1 & 0 \end{pmatrix}$
(ii)
Similarly I thought I can write
$M^{\beta}_{\alpha} = \eta_{\alpha \mu}M^{\beta \mu}$
(iii)
$M_{\alpha \beta} = \eta_{\alpha \gamma}\eta_{\beta \mu}M^{\gamma \mu}$
These can be calculated by using algebra but my question is, can we take a shortcut ?
Questions
Q1) Can we say that the $M^{\beta}_{\alpha} = (M^{\alpha}_{\beta})^T$ ?
Q2) Can we say that $M_{\alpha \beta} M^{\alpha \beta} = I$ Where I is the identity matrix?
Q3) For a given $(m,n)$ tensor can we always say that, the components of the $(m,n)$ tensor times the components of the $(n,m)$ tensor gives an identity matrix (i.e $T^{\alpha \beta}_{\gamma}T^{\gamma} _{\alpha \beta} = I$)?
Q4) Similarly for a given $(n,n)$ matrix (i.e $T^{\beta \gamma}_{\alpha \mu}$) can we say that the transpose of the components will always gives the components of the $T^{\alpha \mu}_{\beta \gamma}$
To answer your questions:
$1)$ No, since
$$M_\beta^\alpha = \begin{pmatrix} 0 & 1 & 0 & 0 \\ -1 & -1 & 0 & 2 \\ -2 & 0 & 0 & 1 \\ -1 & 0 & -2 & 0 \end{pmatrix}$$
$$M_\alpha^\beta = \begin{pmatrix} 0 & -1 & 0 & 0 \\ 1 & -1 & 0 & 2 \\ 2 & 0 & 0 & 1 \\ 1 & 0 & -2 & 0 \end{pmatrix}$$
These don't look like transposes to me.
$2)$ No, since the quantity you describe contracts all of the indices, there is no way it could be anything other than a scalar.
$3)$ No, for the same reason as $2$
$4)$ No, for the same reason as $1$