1. **Problem Statement:** Determine the geometric multiplicity and algebraic multiplicity of the eigenvalue $5$ for the matrix $$A = \begin{bmatrix}3 & 2 & 0 \\ 0 & 1 & 3 \\ -4 & -4 & -5\end{bmatrix}.$$\n\n2. **Recall definitions:**\n- The **algebraic multiplicity** of an eigenvalue is its multiplicity as a root of the characteristic polynomial $\det(A - \lambda I) = 0$.\n- The **geometric multiplicity** is the dimension of the eigenspace corresponding to that eigenvalue, i.e., $\dim(\ker(A - 5I))$.\n\n3. **Calculate $A - 5I$: **\n$$A - 5I = \begin{bmatrix}3-5 & 2 & 0 \\ 0 & 1-5 & 3 \\ -4 & -4 & -5-5\end{bmatrix} = \begin{bmatrix}-2 & 2 & 0 \\ 0 & -4 & 3 \\ -4 & -4 & -10\end{bmatrix}.$$\n\n4. **Find the null space of $A - 5I$: **\nSolve $(A - 5I)\mathbf{x} = \mathbf{0}$ for $\mathbf{x} = \begin{bmatrix}x_1 \\ x_2 \\ x_3\end{bmatrix}$.\nThe system is:\n$$\begin{cases} -2x_1 + 2x_2 = 0 \\ -4x_2 + 3x_3 = 0 \\ -4x_1 -4x_2 -10x_3 = 0 \end{cases}.$$\n\n5. **Solve the system:**\nFrom the first equation: $-2x_1 + 2x_2 = 0 \implies x_1 = x_2$.\nFrom the second: $-4x_2 + 3x_3 = 0 \implies 3x_3 = 4x_2 \implies x_3 = \frac{4}{3}x_2$.\nSubstitute $x_1 = x_2$ and $x_3 = \frac{4}{3}x_2$ into the third equation:\n$$-4x_1 -4x_2 -10x_3 = -4x_2 -4x_2 -10 \cdot \frac{4}{3}x_2 = -8x_2 - \frac{40}{3}x_2 = -\frac{24}{3}x_2 - \frac{40}{3}x_2 = -\frac{64}{3}x_2 = 0.$$\nThis implies $x_2 = 0$, and thus $x_1 = 0$, $x_3 = 0$.\n\n6. **Conclusion on geometric multiplicity:**\nThe only solution is the trivial one, so the eigenspace has dimension $0$. But since eigenvalues must have at least one eigenvector, this suggests a miscalculation. Let's re-check the third equation substitution carefully.\n\nRe-checking the third equation with substitutions:\n$$-4x_1 -4x_2 -10x_3 = -4x_2 -4x_2 -10 \cdot \frac{4}{3}x_2 = -8x_2 - \frac{40}{3}x_2 = -\frac{24}{3}x_2 - \frac{40}{3}x_2 = -\frac{64}{3}x_2.$$\nFor this to be zero, $x_2$ must be zero. So the only solution is the trivial vector.\n\n7. **Algebraic multiplicity:**\nCalculate the characteristic polynomial $p(\lambda) = \det(A - \lambda I)$.\n$$\det\begin{bmatrix}3-\lambda & 2 & 0 \\ 0 & 1-\lambda & 3 \\ -4 & -4 & -5-\lambda\end{bmatrix}.$$\nExpanding along the first row:\n$$ (3-\lambda) \det \begin{bmatrix}1-\lambda & 3 \\ -4 & -5-\lambda \end{bmatrix} - 2 \det \begin{bmatrix}0 & 3 \\ -4 & -5-\lambda \end{bmatrix} + 0.$$\nCalculate minors:\n$$\det \begin{bmatrix}1-\lambda & 3 \\ -4 & -5-\lambda \end{bmatrix} = (1-\lambda)(-5-\lambda) - (-4)(3) = -(5+\lambda)(1-\lambda) + 12.$$\nSimplify:\n$-(5+\lambda)(1-\lambda) + 12 = -[(5)(1) - 5\lambda + \lambda - \lambda^2] + 12 = -[5 - 4\lambda - \lambda^2] + 12 = -5 + 4\lambda + \lambda^2 + 12 = \lambda^2 + 4\lambda + 7.$\n\nNext minor:\n$$\det \begin{bmatrix}0 & 3 \\ -4 & -5-\lambda \end{bmatrix} = 0 \cdot (-5-\lambda) - (-4)(3) = 12.$$\n\nSo characteristic polynomial is:\n$$p(\lambda) = (3-\lambda)(\lambda^2 + 4\lambda + 7) - 2(12) = (3-\lambda)(\lambda^2 + 4\lambda + 7) - 24.$$\nExpand:\n$$(3)(\lambda^2 + 4\lambda + 7) - \lambda(\lambda^2 + 4\lambda + 7) - 24 = 3\lambda^2 + 12\lambda + 21 - (\lambda^3 + 4\lambda^2 + 7\lambda) - 24.$$\nSimplify:\n$$-\lambda^3 + (3\lambda^2 - 4\lambda^2) + (12\lambda - 7\lambda) + (21 - 24) = -\lambda^3 - \lambda^2 + 5\lambda - 3.$$\n\n8. **Check if $5$ is a root:**\n$$p(5) = -(5)^3 - (5)^2 + 5(5) - 3 = -125 - 25 + 25 - 3 = -128 \neq 0.$$\n\nSince $5$ is not a root of the characteristic polynomial, it is not an eigenvalue of $A$.\n\n**Therefore, the eigenvalue $5$ does not exist for matrix $A$, so algebraic multiplicity is $0$ and geometric multiplicity is $0$.**