Apparently your prof wanted you to write all the elements as part of the "usual", basic set of representatives modulo $\;11\;$ , i.e. $\;\Bbb Z_{11}=\{0,1,2,3,...,10\}\pmod{11}\;$, the devil knows why , and then
$$-24=-2=9\pmod{11}\;,\;\;-12=-1=10\pmod{11}$$
$$\;\;\text{(this is what's meant by "additive inverses")}$$
so that your matrix would look as
$$\begin{pmatrix}20&\!\!-24\\\!\!-12&8\end{pmatrix}=\begin{pmatrix}9&9\\10&8\end{pmatrix}\pmod{11}$$
Great question.
The reason why this is true is because, if a given matrix $A$ is invertible, then there exists a sequence of matrices $E_1, E_2, \dots, E_m$ such that
$$E_m E_{m-1} \cdots E_2 E_1 A = A^{-1}$$
We call these matrices elementary matrices. These matrices are in charge of the row operations that we can do on matrices, such as swapping rows, multiplying one row by a constant, or even multiplying one row by a constant and adding it to another row (exactly what we do in Guassian elimination). For example, consider the matrix
$$E_1 = \left( \begin{array}{rc}
1 & 0 \\
-4 & 1
\end{array} \right)$$
If we left-hand multiply your matrix $A$ by $E_1$, we get
$$\left( \begin{array}{rc}
1 & 0 \\
-4 & 1
\end{array} \right)
\left( \begin{array}{cc}
1 & 3 \\
4 & 2
\end{array} \right) =
\left( \begin{array}{cr}
1 & 3 \\
0 & -10
\end{array} \right)$$
which is the matrix we get after the first step Guassian elimination on your matrix $A$! But look closely: if you're trying to find the inverse of $A$ by attaching the identity to the right and row-reducing $A$ down to the identity, then $E_1$ pops up on the right (in place of the identity you attached) after your first step.
Now, consider
$$E_2 = \left( \begin{array}{rr}
1 & 0 \\
0 & -1/10
\end{array} \right)$$
and observe that
$$E_2 E_1 A = \left( \begin{array}{rr}
1 & 0 \\
0 & -1/10
\end{array} \right) \left( \begin{array}{cr}
1 & 3 \\
0 & -10
\end{array} \right) =
\left( \begin{array}{cc}
1 & 3 \\
0 & 1
\end{array} \right)$$
which is the matrix you'd get after multiplying the second row through by $-1/10$, something that might be useful in our Guassian elimination because we want $A$ to eventually be the identity. If you do that in your steps of trying to invert $A$, notice that, after multiplying the second row through, the matrix attached to the right hand side becomes
$$E_2 E_1 = \left( \begin{array}{cc}
1 & 0 \\
2/5 & -1/10
\end{array} \right)$$
Finally, consider
$$ E_3 = \left( \begin{array}{cr}
1 & -3 \\
0 & 1
\end{array} \right)$$
and observe that
$$E_3 E_2 E_1 A = \left( \begin{array}{cr}
1 & -3 \\
0 & 1
\end{array} \right) \left( \begin{array}{cc}
1 & 3 \\
0 & 1
\end{array} \right) =
\left( \begin{array}{cr}
1 & 0 \\
0 & 1
\end{array} \right)$$
which is the matrix you get after your final step of Guassian elimination of $A$ (multiplying the second row by $-3$ and adding it to the first row of $A$). But look!
$$E_3 E_2 E_1 = \left( \begin{array}{cr}
1 & -3 \\
0 & 1
\end{array} \right)\left( \begin{array}{cc}
1 & 0 \\
2/5 & -1/10
\end{array} \right) =
\left( \begin{array}{cr}
-1/5 & 3/10 \\
2/5 & -1/10
\end{array} \right)$$
which is the inverse of $A$, and it also happened to pop on the right-hand side in place of the identity that you initially attached.
Basically, when you do row operations with an identity matrix attached to the right, you're recording all of these steps as elementary matrices. When you apply multiple row operations, these steps accumulate on the right hand side. This accumulation process is the same as multiplying the elementary matrices that govern these Guassian elimination steps. In the end, once $A$ has been reduced down to the identity and all of the steps have been recorded, you get the identity, and this is the same thing as multiplying all of the elementary matrices that you needed together.
Best Answer
The post you link tells you that the sum you want is $(\vec{1})^t\cdot A^{-1}\cdot \vec{1}$ (where $\vec{1}$ is the vector with all entries equal to $1$). Your matrix has the property that all lines have the sum of the elements equal to $7$; this is equivalent to writing $$A\cdot \vec1= 7\vec1,$$and from that you get that $$A^{-1}\cdot \vec{1}=\frac{1}{7}\vec1.$$ Thus you have immediately $$(\vec{1})^t\cdot A^{-1}\cdot \vec{1}=\frac17(\vec{1})^t\cdot \vec1=\frac{3}{7}.$$