Matrices how to prove $A^{-1} = A$ Apologies mix up from earlier the wrong values where placed in $x_2$ and $x_3$.
Question 1
Proof that the following is true for matrix $A$,   $A^{-1}$ = $A^{T}$ = $A$
$A$=
$$
   1/7 \begin{pmatrix}
    2 & 3 & 6 \\
    3 & -6 & 2 \\
    6 & 2 & -3 \\
    \end{pmatrix}
$$
$A^T$=
$$
    1/7 \begin{pmatrix}
    2 & 3 & 6 \\
    3 & -6 & 2 \\
    6 & 2 & -3 \\
    \end{pmatrix}
$$
The determinant is $343$
The rule has already been applied to the matrix $(+ - +)
$
$A^{-1}$=
$$
   1/343 \begin{pmatrix}
    14 & 21 & 42 \\
    -14 & -21 & -42 \\
    42 & 14 & 21 \\
    \end{pmatrix}
$$
This is as far as I can go the identity rule is not producing $1$ in the diagonal how can I solve it from here?
$A^{-1}$
$x_1$ = $2/49$
$A$
$x_1$ = $2/7$

To check that $A^{-1}=A$, you don't need to "calculate" $A^{-1}$. If $A^{-1}=A$, then $A^2=A^{-1}A=I$; and viceversa, if $A^2=I$, then you know that $A^{-1}=A$. Here you can calculate directly that $A^2=I$. 
Now, in light of the above, your calculation of $A^{-1}$ is wrong. You don't say what computations you made, so I cannot comment on that. 
