Step-by-step Solution:

Correct Answer: A ($AB+A+3B+3I$)

To find the inverse of the matrix C, we will first simplify the expression for C, then find the inverses of its factors using the given polynomial equations for A and B.

Step 1: Factor the Matrix C

The matrix C is given by $C = BA + 2B + 2A + 4I$. We can factor this expression by grouping:

$C = (BA + 2B) + (2A + 4I)$

$C = B(A + 2I) + 2(A + 2I)$

Factoring out the common term $(A+2I)$, we get:

$C = (B+2I)(A+2I)$

Step 2: Find the Inverse of C

Using the property for the inverse of a product of matrices, $(XY)^{-1} = Y^{-1}X^{-1}$, we can write the inverse of C as:

$C^{-1} = ((B+2I)(A+2I))^{-1} = (A+2I)^{-1}(B+2I)^{-1}$

Our next task is to find the inverses of $(A+2I)$ and $(B+2I)$.

Step 3: Find the Inverses of $(A+2I)$ and $(B+2I)$

Finding $(A+2I)^{-1}$:

We use the given equation for A: $A^2 + 5A + 5I = 0$. We manipulate this equation to find a matrix X such that $(A+2I)X = I$.

$A^2 + 5A + 5I = 0$

$A^2 + 5A + 6I = I$

$(A^2 + 2A) + (3A + 6I) = I$

$A(A+2I) + 3(A+2I) = I$

$(A+3I)(A+2I) = I$

From this, we can see that the inverse of $(A+2I)$ is $(A+3I)$.

Finding $(B+2I)^{-1}$:

We use the given equation for B: $B^2 + 3B + I = 0$. We follow a similar procedure.

$B^2 + 3B + I = 0$

$B^2 + 3B + 2I = I$

$(B^2 + 2B) + (B + 2I) = I$

$B(B+2I) + 1(B+2I) = I$

$(B+I)(B+2I) = I$

From this, we can see that the inverse of $(B+2I)$ is $(B+I)$.

Step 4: Calculate the Final Expression for $C^{-1}$

Now we substitute the individual inverses back into our expression for $C^{-1}$:

$C^{-1} = (A+2I)^{-1}(B+2I)^{-1} = (A+3I)(B+I)$

Finally, we expand this product:

$C^{-1} = A(B+I) + 3I(B+I)$

$C^{-1} = AB + AI + 3IB + 3I^2$

$C^{-1} = AB + A + 3B + 3I$