1. **State the problem:** We are given a matrix $$A = \begin{bmatrix} 5 & -2 & -1 \\ 5 & -1 & 1 \\ 7 & 4 & \lambda \end{bmatrix}$$ and asked to find the characteristic polynomial by computing $$\det(A - \lambda I)$$ and then find the Jordan canonical form of matrix $$A$$. 2. **Form the matrix $$A - \lambda I$$:** Subtract $$\lambda$$ from each diagonal element of $$A$$: $$ A - \lambda I = \begin{bmatrix} 5 - \lambda & -2 & -1 \\ 5 & -1 - \lambda & 1 \\ 7 & 4 & \lambda - \lambda \end{bmatrix} = \begin{bmatrix} 5 - \lambda & -2 & -1 \\ 5 & -1 - \lambda & 1 \\ 7 & 4 & 0 \end{bmatrix} $$ 3. **Calculate the determinant $$\det(A - \lambda I)$$:** Using cofactor expansion along the third row: $$ \det = 7 \cdot \det \begin{bmatrix} -2 & -1 \\ -1 - \lambda & 1 \end{bmatrix} - 4 \cdot \det \begin{bmatrix} 5 - \lambda & -1 \\ 5 & 1 \end{bmatrix} + 0 \cdot (...) $$ Calculate each minor: $$ \det \begin{bmatrix} -2 & -1 \\ -1 - \lambda & 1 \end{bmatrix} = (-2)(1) - (-1)(-1 - \lambda) = -2 - (1 + \lambda) = -3 - \lambda $$ $$ \det \begin{bmatrix} 5 - \lambda & -1 \\ 5 & 1 \end{bmatrix} = (5 - \lambda)(1) - (-1)(5) = 5 - \lambda + 5 = 10 - \lambda $$ So, $$ \det(A - \lambda I) = 7(-3 - \lambda) - 4(10 - \lambda) = -21 - 7\lambda - 40 + 4\lambda = -61 - 3\lambda $$ 4. **Find eigenvalues:** Set $$\det(A - \lambda I) = 0$$: $$ -61 - 3\lambda = 0 \implies 3\lambda = -61 \implies \lambda = -\frac{61}{3} $$ 5. **Jordan canonical form:** Since the characteristic polynomial is linear, there is only one eigenvalue $$\lambda = -\frac{61}{3}$$ with algebraic multiplicity 1. For a 3x3 matrix, this implies the Jordan form is a diagonal matrix with this eigenvalue repeated 3 times only if the geometric multiplicity equals 3. But here, the characteristic polynomial is degree 1, so the matrix is not diagonalizable and the Jordan form will have one Jordan block of size 3 with eigenvalue $$-\frac{61}{3}$$. **Final answer:** The Jordan canonical form of $$A$$ is $$ J = \begin{bmatrix} -\frac{61}{3} & 1 & 0 \\ 0 & -\frac{61}{3} & 1 \\ 0 & 0 & -\frac{61}{3} \end{bmatrix} $$