Step 1: Understand the task.
Matrix $A$ has an unknown $a$ and its inverse $A^{-1}$ has an unknown $c$. We find both using the fact that $A\,A^{-1} = I$, the identity matrix.
Step 2: Pick the easy positions.
Instead of multiplying all nine entries, we use a few positions of $I$ that are zero. The $(1,3)$ entry of $I$ is 0.
Step 3: Use position (1,3) for $c$.
Row 1 of $A$ is $(0,1,2)$; column 3 of $A^{-1}$ is $(1, 2c, 1)$ (times $\tfrac12$):
\[ \frac{1}{2}\big[(0)(1) + (1)(2c) + (2)(1)\big] = 0 \;\Rightarrow\; 2c + 2 = 0 \;\Rightarrow\; c = -1 \]
Step 4: Check with position (2,3).
Row 2 is $(1,2,3)$; with $c = -1$:
\[ \frac{1}{2}\big[1 + 2(2)(-1) + 3\big] = \frac{1}{2}(1 - 4 + 3) = 0 \]
This is 0 as it should be, so $c = -1$ is confirmed.
Step 5: Use position (3,1) for $a$.
Row 3 of $A$ is $(3, a, 1)$; column 1 of $A^{-1}$ is $(1, -8, 5)$:
\[ \frac{1}{2}\big[(3)(1) + (a)(-8) + (1)(5)\big] = 0 \;\Rightarrow\; 3 - 8a + 5 = 0 \]
Step 6: Solve for $a$.
\[ 8 - 8a = 0 \;\Rightarrow\; a = 1 \]
So $a = 1$ and $c = -1$.
\[ \boxed{a = 1,\ c = -1} \]