Let's solve this step by step: ### Step 1: **Find the Eigenvectors of B (which is already in Jordan form/diagonalizable)** Given: \[ A = \begin{bmatrix} -27 & 18 \\ -50 & 33 \\ \end{bmatrix}, \quad B = \begin{bmatrix} 3 & -2 \\ 0 & 3 \\ \end{bmatrix} \] Notice \( B \) is already in its **Jordan canonical form** (it’s almost diagonal, but has a single off-diagonal entry). Let's find the eigenvector(s) for \( B \) for eigenvalue \( \lambda = 3 \): \[ B - 3I = \begin{bmatrix} 0 & -2 \\ 0 & 0 \\ \end{bmatrix} \] Set \( (B-3I)\vec{v} = 0 \): \[ \begin{bmatrix} 0 & -2 \\ 0 & 0 \\ \end{bmatrix} \begin{bmatrix} v_1 \\ v_2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \] This yields \( -2v_2 = 0 \implies v_2 = 0 \). So, eigenvector: \( \begin{bmatrix} 1 \\ 0 \end{bmatrix} \). Now, find a generalized eigenvector by solving: \[ (B-3I)\vec{w} = \vec{v}_1 = \begin{bmatrix} 1 \\ 0 \end{bmatrix} \] So, \[ \begin{bmatrix} 0 & -2 \\ 0 & 0 \\ \end{bmatrix} \begin{bmatrix} w_1 \\ w_2 \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \end{bmatrix} \] So \( -2w_2 = 1 \implies w_2 = -\frac{1}{2} \), \( w_1 \) can be anything (let \( w_1 = 0 \) for simplicity). Generalized eigenvector: \( \begin{bmatrix} 0 \\ -\frac{1}{2} \end{bmatrix} \). **Explanation:** We found the eigenvector and generalized eigenvector for \( B \) corresponding to eigenvalue 3. --- ### Step 2: **Construct P using A and the eigenvectors of B** Let’s use the eigenvector and generalized eigenvector to construct \( P \). Recall, if \( A = P^{-1}BP \), then columns of \( P \) are mapped as: \( AP = PB \implies A\vec{p}_j = B\vec{p}_j \). Alternatively, since \( B \) is in Jordan form, **to find P, we find the corresponding vectors for A**: Let \( \vec{v}_1 \) and \( \vec{v}_2 \) (eigenvector and generalized eigenvector) for \( B \), and their images under \( P \) be \( \vec{w}_1 \) and \( \vec{w}_2 \) for \( A \): #### Find eigenvector of A for \( \lambda = 3 \): \[ A - 3I = \begin{bmatrix} -30 & 18 \\ -50 & 30 \end{bmatrix} \] Row 1 gives: \( -30x + 18y = 0 \implies 5x = 3y \implies y = \frac{5}{3}x \). So, eigenvector: \( \begin{bmatrix} 1 \\ \frac{5}{3} \end{bmatrix} \). #### Find generalized eigenvector \( (A - 3I)\vec{w}_2 = \vec{w}_1 \): \[ (A - 3I)\vec{w}_2 = \vec{w}_1 = \begin{bmatrix} 1 \\ \frac{5}{3} \end{bmatrix} \] Let \( \vec{w}_2 = \begin{bmatrix} a \\ b \end{bmatrix} \). \[ \begin{bmatrix} -30 & 18 \\ -50 & 30 \end{bmatrix} \begin{bmatrix} a \\ b \end{bmatrix} = \begin{bmatrix} 1 \\ \frac{5}{3} \end{bmatrix} \] This gives two equations: - \( -30a + 18b = 1 \) - \( -50a + 30b = \frac{5}{3} \) Solve the first: \( -30a + 18b = 1 \implies 18b = 30a + 1 \implies b = \frac{30a + 1}{18} \). Plug into the second: \[ -50a + 30b = \frac{5}{3} \] \[ -50a + 30\left(\frac{30a + 1}{18}\right) = \frac{5}{3} \] \[ -50a + \frac{900a + 30}{18} = \frac{5}{3} \] \[ -50a + 50a + \frac{30}{18} = \frac{5}{3} \] \[ 0a + \frac{5}{3} = \frac{5}{3} \] True for any \( a \). Let \( a = 0 \), \( b = \frac{1}{18} \). So, **generalized eigenvector**: \( \begin{bmatrix} 0 \\ \frac{1}{18} \end{bmatrix} \). Let’s scale for simplicity (multiply by 18): \( \begin{bmatrix} 0 \\ 1 \end{bmatrix} \). So, the two vectors: - \( \vec{w}_1 = \begin{bmatrix} 1 \\ \frac{5}{3} \end{bmatrix} \) - \( \vec{w}_2 = \begin{bmatrix} 0 \\ 1 \end{bmatrix} \) So, \[ P = \begin{bmatrix} 1 & 0 \\ \frac{5}{3} & 1 \end{bmatrix} \] Calculate \( P^{-1} \): \[ P^{-1} = \frac{1}{\det P} \begin{bmatrix} 1 & 0 \\ -\frac{5}{3} & 1 \end{bmatrix} \] \[ \det P = 1 \cdot 1 - 0 \cdot \frac{5}{3} = 1 \] So, \[ P^{-1} = \begin{bmatrix} 1 & 0 \\ -\frac{5}{3} & 1 \end{bmatrix} \] --- ## **Final Answer** \[ P = \boxed{ \begin{bmatrix} 1 & 0 \\ \frac{5}{3} & 1 \end{bmatrix} } \qquad P^{-1} = \boxed{ \begin{bmatrix} 1 & 0 \\ -\frac{5}{3} & 1 \end{bmatrix} } \] **Explanation:** We found the eigenvector and generalized eigenvector for both \( A \) and \( B \), constructed \( P \) from these vectors, and calculated \( P^{-1} \). Thus, \( A \) and \( B \) are similar, with \( A = P^{-1}BP \).