Let's proceed with part (g): **Is matrix \(A\) similar to matrix \(B\)? If yes, find an invertible matrix \(P\) such that \(P^{-1} A P = B\). (Hint: First check if \(B\) is diagonalizable.)** --- ## **Step 1: Recall matrix \(B\)** From the problem: \[ B = \begin{bmatrix} 0 & 0 & 2 \\ -1 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \] --- ## **Step 2: Check if \(B\) is diagonalizable** ### **Find eigenvalues of \(B\):** Calculate \(\det(B - \lambda I)\): \[ B - \lambda I = \begin{bmatrix} -\lambda & 0 & 2 \\ -1 & 1 - \lambda & 0 \\ 0 & 0 & 1 - \lambda \end{bmatrix} \] Determinant: \[ \det(B - \lambda I) = \det \begin{bmatrix} -\lambda & 0 & 2 \\ -1 & 1 - \lambda & 0 \\ 0 & 0 & 1 - \lambda \end{bmatrix} \] Because it's upper block, expand along third row: \[ = (1 - \lambda) \cdot \det \begin{bmatrix} -\lambda & 0 \\ -1 & 1 - \lambda \end{bmatrix} \] Compute the minor: \[ (-\lambda)(1 - \lambda) - (0)(-1) = -\lambda (1 - \lambda) = -\lambda + \lambda^2 \] So, \[ \det(B - \lambda I) = (1 - \lambda)(-\lambda + \lambda^2) = (1 - \lambda)(\lambda^2 - \lambda) \] Factor: \[ = (1 - \lambda) \lambda (\lambda - 1) = - (1 - \lambda)^2 \lambda \] Eigenvalues are roots of: \[ - (1 - \lambda)^2 \lambda = 0 \] which gives: \[ \boxed{\lambda = 0, \quad \lambda = 1} \] with multiplicities: - \(\lambda=0\): multiplicity 1 - \(\lambda=1\): multiplicity 2 --- ## **Step 3: Check if \(B\) is diagonalizable** ### **Eigenvalue \(\lambda=0\):** Solve \((B - 0 I) \mathbf{x} = 0\): \[ B \mathbf{x} = 0 \] \[ \begin{bmatrix} 0 & 0 & 2 \\ -1 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \] From row 3: \[ x_3 = 0 \] From row 2: \[ - x_1 + x_2 = 0 \Rightarrow x_2 = x_1 \] From row 1: \[ 0 \cdot x_1 + 0 \cdot x_2 + 2 x_3 = 0 \Rightarrow 0 = 0 \] So the eigenvector for \(\lambda=0\): \[ \mathbf{v}_0 = \begin{bmatrix} x_1 \\ x_1 \\ 0 \end{bmatrix} = x_1 \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} \] Dimension of eigenspace: 1 --- ### **Eigenvalue \(\lambda=1\):** Solve \((B - I) \mathbf{x} = 0\): \[ B - I = \begin{bmatrix} -1 & 0 & 2 \\ -1 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \] Equation system: - From row 2: \[ - x_1 + 0 \cdot x_2 + 0 \cdot x_3 = 0 \Rightarrow x_1 = 0 \] - From row 1: \[ - x_1 + 0 \cdot x_2 + 2 x_3 = 0 \Rightarrow 0 + 2 x_3 = 0 \Rightarrow x_3=0 \] - \(x_2\) is free (since row 3 is all zeros). Eigenvectors: \[ \mathbf{v}_1 = \begin{bmatrix} 0 \\ x_2 \\ 0 \end{bmatrix} = x_2 \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} \] Dimension of eigenspace: 1 --- ## **Step 4: Check for diagonalizability** - For \(\lambda=0\): geometric multiplicity 1 (matches algebraic multiplicity 1). - For \(\lambda=1\): geometric multiplicity 1 (less than algebraic multiplicity 2). Since the algebraic multiplicity of \(\lambda=1\) is 2 but its geometric multiplicity is only 1, **\(B\) is not diagonalizable**. --- ## **Conclusion:** - Since \(A\) is diagonalizable but \(B\) is not, **\(A\) is not similar to \(B\)**. --- ## **Final answer:** \[ \boxed{ \text{Matrix } A \text{ is NOT similar to matrix } B. } \] **Because** \(B\) is not diagonalizable, whereas \(A\) is, they cannot be similar matrices. --- **Summary:** - \(A\) is diagonalizable with eigenvalues 2 and 3. - \(B\) has eigenvalues 0 and 1, with algebraic multiplicities 1 and 2, but is not diagonalizable. - Therefore, \(A\) and \(B\) are **not similar**. --- Would you like any further clarification or proceed to another question?