It appears you haven't provided a specific question or problem to solve. Please provide the math problem or question you'd like help with, and I'll be happy to give a step-by-step solution with explanation and the final answer!

Let's carefully analyze the problem: you provided a matrix \(A\) as \([-7 \quad 4 \quad -1 \quad -3]\). However, this appears to be a 1x4 row vector, which cannot have eigenvalues or eigenvectors. **Possible clarification needed:** - If the matrix is intended to be a square matrix (e.g., 2x2, 3x3, 4x4), please specify the entire matrix. - If this is part of a larger matrix, please provide the complete matrix. - If you meant a different matrix, please clarify. --- ### Assumption: **Suppose you meant a 2x2 matrix, for example:** \[ A = \begin{bmatrix} -7 & 4 \\ -1 & -3 \end{bmatrix} \] **This is a common size for such problems.** --- ### Step 1: Find the eigenvalues Eigenvalues \(\lambda\) satisfy the characteristic equation: \[ \det(A - \lambda I) = 0 \] Where \(I\) is the identity matrix. \[ A - \lambda I = \begin{bmatrix} -7 - \lambda & 4 \\ -1 & -3 - \lambda \end{bmatrix} \] Calculate the determinant: \[ (-7 - \lambda)(-3 - \lambda) - (4)(-1) = 0 \] Expand: \[ (-7 - \lambda)(-3 - \lambda) + 4 = 0 \] Multiply out: \[ (-7)(-3) + (-7)(-\lambda) + (-\lambda)(-3) + (-\lambda)(-\lambda) + 4 = 0 \] which simplifies to: \[ 21 + 7\lambda + 3\lambda + \lambda^2 + 4 = 0 \] Combine like terms: \[ \lambda^2 + (7\lambda + 3\lambda) + (21 + 4) = 0 \] \[ \lambda^2 + 10\lambda + 25 = 0 \] --- ### Step 2: Solve the quadratic \[ \lambda^2 + 10\lambda + 25 = 0 \] Discriminant: \[ D = (10)^2 - 4 \times 1 \times 25 = 100 - 100 = 0 \] Since \(D=0\), there is one repeated eigenvalue: \[ \lambda = \frac{-10 \pm \sqrt{0}}{2} = \frac{-10}{2} = -5 \] **Eigenvalue:** \[ \boxed{\lambda = -5} \] --- ### Step 3: Find the eigenvector corresponding to \(\lambda = -5\) Solve: \[ (A - \lambda I) \mathbf{v} = 0 \] \[ A - (-5) I = A + 5 I = \begin{bmatrix} -7 + 5 & 4 \\ -1 & -3 + 5 \end{bmatrix} = \begin{bmatrix} -2 & 4 \\ -1 & 2 \end{bmatrix} \] Set up the system: \[ -2x + 4y = 0 \] \[ -1x + 2y = 0 \] Both equations are dependent; pick the first: \[ -2x + 4y = 0 \Rightarrow 4y = 2x \Rightarrow y = \frac{1}{2}x \] Let \(x = 1\), then: \[ y = \frac{1}{2} \times 1 = \frac{1}{2} \] **Eigenvector:** \[ \mathbf{v} = \begin{bmatrix} 1 \\ \frac{1}{2} \end{bmatrix} \] Or, scaled to clear fractions: \[ \mathbf{v} = \begin{bmatrix} 2 \\ 1 \end{bmatrix} \] --- ### **Final answer:** **Eigenvalues:** \[ \boxed{\lambda = -5 \quad (\text{double eigenvalue})} \] **Eigenvector corresponding to \(\lambda = -5\):** \[ \boxed{\mathbf{v} = \begin{bmatrix} 2 \\ 1 \end{bmatrix}} \] --- ### **Summary:** - The matrix \(A\) (assumed 2x2) has a single eigenvalue \(\lambda =