If $$B$$ is a $$3 \times 3$$ matrix such that $$B^2 = 0$$, then $$\det[(I + B)^{50} - 50B]$$ is equal to:

Given that $$B$$ is a $$3 \times 3$$ matrix such that $$B^2 = 0$$, we need to find $$\det[(I + B)^{50} - 50B]$$.

Since $$B^2 = 0$$, $$B$$ is a nilpotent matrix of index 2. This means that any power of $$B$$ greater than or equal to 2 is the zero matrix. Specifically, $$B^k = 0$$ for all $$k \geq 2$$.

Now, consider $$(I + B)^{50}$$. The identity matrix $$I$$ commutes with every matrix, so $$I$$ commutes with $$B$$. Therefore, we can expand $$(I + B)^{50}$$ using the binomial theorem:

$$(I + B)^{50} = \sum_{k=0}^{50} \binom{50}{k} I^{50-k} B^k$$

Since $$I^m = I$$ for any positive integer $$m$$, this simplifies to:

$$(I + B)^{50} = \sum_{k=0}^{50} \binom{50}{k} B^k$$

Because $$B^k = 0$$ for $$k \geq 2$$, all terms where $$k \geq 2$$ vanish. Thus, only the terms for $$k = 0$$ and $$k = 1$$ remain:

• For $$k = 0$$: $$\binom{50}{0} B^0 = 1 \cdot I = I$$
• For $$k = 1$$: $$\binom{50}{1} B^1 = 50 B$$

So, $$(I + B)^{50} = I + 50B$$.

Now, substitute this into the expression:

$$(I + B)^{50} - 50B = (I + 50B) - 50B = I$$

Therefore, $$(I + B)^{50} - 50B = I$$, the identity matrix.

The determinant of the identity matrix of size $$3 \times 3$$ is 1:

$$\det(I) = 1$$

Hence, the correct answer is Option A.