Let A be an m × n matrix and Ban n × m matrix. It is given that determinant ($$I$$m + AB) = determinant ($$I$$n + BA), where $$I$$k is the k × k identity matrix. Using the above property, the determinant of the matrix given below is \[\left[ {\begin{array}{*{20}{c}} 2&1&1&1 \\ 1&2&1&1 \\ 1&1&2&1 \\ 1&1&1&2 \end{array}} \right]\] A. 2 B. 5 C. 8 D. 16

The correct answer is $\boxed{\text{D}}$.

The determinant of a 4×4 matrix can be computed using the formula below:

$$\det(A) = \sum_{cyc} (-1)^{r+s+t} a_{1r} a_{2s} a_{3t} a_{4(r+s+t-1)}$$

where $r$, $s$, and $t$ are distinct integers from 1 to 4, and $(-1)^{r+s+t}$ is the sign of the permutation $(r,s,t)$.

In this case, the matrix $A$ is given by

$$A = \left[ {\begin{array}{*{20}{c}} 2&1&1&1 \ 1&2&1&1 \ 1&1&2&1 \ 1&1&1&2 \end{array}} \right]$$

We can compute the determinant of $A$ as follows:

$$\det(A) = \sum_{cyc} (-1)^{r+s+t} a_{1r} a_{2s} a_{3t} a_{4(r+s+t-1)} = (-1)^{1+2+3} 2 \cdot 2 \cdot 1 \cdot 1 + (-1)^{2+1+3} 1 \cdot 1 \cdot 2 \cdot 1 + (-1)^{3+1+2} 1 \cdot 2 \cdot 1 \cdot 1 + (-1)^{1+3+2} 1 \cdot 1 \cdot 2 \cdot 2 = 16$$

Therefore, the determinant of the matrix given below is $\boxed{16}$.

Here is a brief explanation of each option:

• Option A: The determinant of the matrix is $2$. This is incorrect because the determinant of the matrix is $16$.
• Option B: The determinant of the matrix is $5$. This is incorrect because the determinant of the matrix is $16$.
• Option C: The determinant of the matrix is $8$. This is incorrect because the determinant of the matrix is $16$.
• Option D: The determinant of the matrix is $16$. This is correct because the determinant of the matrix is $16$.