Linear Algebra

Biochemistry

### Solution #### a) Determine if the matrix is diagonalizable. To determine if the matrix \( A \) is diagonalizable, we need to find its eigenvalues and check if the algebraic multiplicity of each eigenvalue equals its geometric multiplicity. **Step 1: Find the eigenvalues of \( A \).** The eigenvalues are solutions to the characteristic equation \(\det(A - \lambda I) = 0\), where \( I \) is the identity matrix and \( \lambda \) represents the eigenvalues. Given: \[ A = \begin{pmatrix} -1 & 3 & 0 \\ 3 & -1 & 0 \\ -2 & -2 & 6 \end{pmatrix} \] Calculate \( A - \lambda I \): \[ A - \lambda I = \begin{pmatrix} -1-\lambda & 3 & 0 \\ 3 & -1-\lambda & 0 \\ -2 & -2 & 6-\lambda \end{pmatrix} \] Compute the determinant: \[ \det(A - \lambda I) = \begin{vmatrix} -1-\lambda & 3 & 0 \\ 3 & -1-\lambda & 0 \\ -2 & -2 & 6-\lambda \end{vmatrix} \] Expanding along the third column: \[ \det(A - \lambda I) = (6-\lambda)\begin{vmatrix} -1-\lambda & 3 \\ 3 & -1-\lambda \end{vmatrix} \] \[ = (6-\lambda)((-1-\lambda)(-1-\lambda) - 3 \cdot 3) \] \[ = (6-\lambda)(\lambda^2 + 2\lambda - 8) \] \[ = (6-\lambda)(\lambda^2 + 2\lambda - 8) \] \[ = (6-\lambda)(\lambda + 4)(\lambda - 2) \] The eigenvalues are \(\lambda = 6, -4, 2\). **Step 2: Check the geometric multiplicity of each eigenvalue.** For each eigenvalue \(\lambda\), solve \( (A - \lambda I)v = 0 \) for non-zero vectors \( v \). The dimension of the solution space (nullity) for each \(\lambda\) must equal the algebraic multiplicity found in the characteristic polynomial. - For \(\lambda = 6\): \[ A - 6I = \begin{pmatrix} -7 & 3 & 0 \\ 3 & -7 & 0 \\ -2 & -2 & 0 \end{pmatrix} \] Row reducing, we find the rank is 2, so the nullity (geometric multiplicity) is 1. - For \(\lambda = -4\): \[ A + 4I = \begin{pmatrix} 3 & 3 & 0 \\ 3 & 3 & 0 \\ -2 & -2 & 10 \end{pmatrix} \] Row reducing, we find the rank is 2, so the nullity is 1. - For \(\lambda = 2\): \[ A - 2I = \begin{pmatrix} -3 & 3 & 0 \\ 3 & -3 & 0 \\ -2 & -2 & 4 \end{pmatrix} \] Row reducing, we find the rank is 2, so the nullity is 1. Since the algebraic and geometric multiplicities match for all eigenvalues, \( A \) is diagonalizable. #### b) Diagonalize \( A \). **Step 1: Find eigenvectors for each eigenvalue.** - For \(\lambda = 6\), solve \( (A - 6I)v = 0 \). An eigenvector is \( v_1 = (1, 1, 2) \). - For \(\lambda = -4\), solve \( (A + 4I)v = 0 \). An eigenvector is \( v_2 = (1, 1, 0) \). - For \(\lambda = 2\), solve \( (A - 2I)v = 0 \). An eigenvector is \( v_3 = (1, -1, 0) \). **Step 2: Form the diagonalizing matrix \( P \) and the diagonal matrix \( D \).** \[ P = \begin{pmatrix} 1 & 1 & 1 \\ 1 & 1 & -1 \\ 2 & 0 & 0 \end{pmatrix}, \quad D = \begin{pmatrix} 6 & 0 & 0 \\ 0 & -4 & 0 \\ 0 & 0 & 2 \end{pmatrix} \] #### c) Compute \( A^{2024} \). Using the diagonalization, \( A = PDP^{-1} \), we have: \[ A^{2024} = PD^{2024}P^{-1} \] \[ D^{2024} = \begin{pmatrix} 6^{2024} & 0 & 0 \\ 0 & (-4)^{2024} & 0 \\ 0 & 0 & 2^{2024} \end{pmatrix} \] Compute \( P^{-1} \) and multiply to find \( A^{2024} \). #### d) Determine if \( A \) is invertible. If so, find \( A^{-1} \). Since the determinant of \( A \) (product of eigenvalues) is non-zero, \( A \) is invertible. Compute \( A^{-1} \) using: \[ A^{-1} = PD^{-1}P^{-1} \] \[ D^{-1} = \begin{pmatrix} \frac{1}{6} & 0 & 0 \\ 0 & -\frac{1}{4} & 0 \\ 0 & 0 & \frac{1}{2} \end{pmatrix} \] Multiply \( P \), \( D^{-1} \), and \( P^{-1} \) to find \( A^{-1} \).

Math

Given the matrix \[ A = \begin{pmatrix} -1 & 3 & 0 \\ 3 & -1 & 0 \\ -2 & -2 & 6 \end{pmatrix} \] a) Determine if the matrix is diagonalizable. b) Diagonalize A. c) Compute \( A^{2024} \). d) Determine if A is invertible. If so, find \( A^{-1} \).

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