Let $$A=\begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 6 & -11 & 6\end{pmatrix}$$. What is the trace of $$A^5$$?

To find the trace of the matrix raised to the fifth power, we should first determine the eigenvalues of the original matrix $$A$$.

The characteristic equation of matrix $$A$$ is given by $$\vert{}A - \lambda I\vert{} = 0$$.

Let us compute this determinant.

$$\begin{vmatrix} -\lambda & 1 & 0 \\ 0 & -\lambda & 1 \\ 6 & -11 & 6-\lambda \end{vmatrix} = 0$$

Expanding this determinant along the first row yields the following equation.

$$-\lambda \cdot (-\lambda \cdot (6 - \lambda) + 11) - 1 \cdot (0 - 6) = 0$$

$$-\lambda \cdot (\lambda^2 - 6\lambda + 11) + 6 = 0$$

$$-\lambda^3 + 6\lambda^2 - 11\lambda + 6 = 0$$

Multiplying by negative one gives the standard characteristic polynomial.

$$\lambda^3 - 6\lambda^2 + 11\lambda - 6 = 0$$

By simple inspection, we can see that $$\lambda = 1$$ is a root since $$1 - 6 + 11 - 6 = 0$$.

Dividing the cubic polynomial by $$\lambda - 1$$ gives a quadratic quotient of $$\lambda^2 - 5\lambda + 6 = 0$$.

Factoring the quadratic equation yields roots of $$2$$ and $$3$$.

Thus, the eigenvalues of matrix $$A$$ are $$1, 2,$$ and $$3$$.

A standard property of matrices states that if a matrix has eigenvalues $$\lambda_1, \lambda_2,$$ and $$\lambda_3$$, then the matrix raised to the power of $$k$$ will have eigenvalues $$\lambda_1^k, \lambda_2^k,$$ and $$\lambda_3^k$$.

Therefore, the eigenvalues of $$A^5$$ are $$1^5, 2^5,$$ and $$3^5$$.

The trace of any square matrix is always equal to the sum of its eigenvalues.

$$\text{Trace}(A^5) = 1^5 + 2^5 + 3^5$$
$$\text{Trace}(A^5) = 1 + 32 + 243$$
$$\text{Trace}(A^5) = 276$$

The correct value is 276.