1. Find eigenvalues and eigenvectors of the following matrices: \[ A=\left(\begin{array}{rr}-1 & 0 \\ 1 & -1\end{array}\right), \quad E=\left(\begin{array}{lll}2 & 0 & 1 \\ 0 & 3 & 4 \\ 0 & 0 & 1\end{array}\right) \]

To find the eigenvalues and eigenvectors of the given matrices, we first determine the eigenvalues by solving the characteristic polynomial \(\det(A - \lambda I) = 0\). **For Matrix A:** \[ A = \begin{pmatrix} -1 & 0 \\ 1 & -1 \end{pmatrix} \] The characteristic polynomial is: \[ \det\left(\begin{pmatrix} -1 - \lambda & 0 \\ 1 & -1 - \lambda \end{pmatrix}\right) = (-1 - \lambda)(-1 - \lambda) - (0)(1) = (\lambda + 1)^2 \] Setting this equal to zero gives: \[ (\lambda + 1)^2 = 0 \implies \lambda = -1 \quad (\text{with algebraic multiplicity 2}) \] Now, we find the eigenvectors by substituting \(\lambda = -1\) into \((A - \lambda I)\mathbf{v} = 0\): \[ \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} = 0 \implies 1 \cdot x_1 = 0 \implies x_1 = 0 \] Thus, we can choose \(x_2 = t\), where \(t\) is any scalar. The eigenvector corresponding to \(\lambda = -1\) is: \[ \mathbf{v} = \begin{pmatrix} 0 \\ t \end{pmatrix} \] where \(t\) is a free variable. **For Matrix E:** \[ E = \begin{pmatrix} 2 & 0 & 1 \\ 0 & 3 & 4 \\ 0 & 0 & 1 \end{pmatrix} \] To find the eigenvalues, calculate: \[ \det\left(\begin{pmatrix} 2 - \lambda & 0 & 1 \\ 0 & 3 - \lambda & 4 \\ 0 & 0 & 1 - \lambda \end{pmatrix}\right) = (2 - \lambda)(3 - \lambda)(1 - \lambda) \] Setting this product to zero gives the eigenvalues: \[ \lambda_1 = 2, \lambda_2 = 3, \lambda_3 = 1 \] Next, we find the eigenvectors for each eigenvalue. 1. **For \(\lambda = 2\)**: \[ (E - 2I) \mathbf{v} = \begin{pmatrix} 0 & 0 & 1 \\ 0 & 1 & 4 \\ 0 & 0 & -1 \end{pmatrix}\begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = 0 \implies x_1 + x_3 = 0, \quad x_2 + 4x_3 = 0 \] Choosing \(x_3 = t\), we have \(x_1 = -t\) and \(x_2 = -4t\), giving the eigenvector: \[ \mathbf{v_1} = \begin{pmatrix} -t \\ -4t \\ t \end{pmatrix} \] 2. **For \(\lambda = 3\)**: \[ (E - 3I) \mathbf{v} = \begin{pmatrix} -1 & 0 & 1 \\ 0 & 0 & 4 \\ 0 & 0 & -2 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = 0 \implies -x_1 + x_3 = 0, \quad 4x_3 = 0 \] This implies \(x_3 = 0\) and consequently \(x_1 = 0\). Therefore \(x_2\) is free, leading to: \[ \mathbf{v_2} = \begin{pmatrix} 0 \\ t \\ 0 \end{pmatrix} \] 3. **For \(\lambda = 1\