Let's solve the system step by step: \[ \mathbf{X}' = \begin{pmatrix} 16 & 0 & 4 \\ 0 & 6 & 0 \\ -16 & 0 & 16 \end{pmatrix} \mathbf{X} \] Let \(\mathbf{X} = \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix}\). --- ## **Step 1: Find Eigenvalues** Set up the characteristic equation: \[ \det \left( A - \lambda I \right) = 0 \] \[ A - \lambda I = \begin{pmatrix} 16 - \lambda & 0 & 4 \\ 0 & 6 - \lambda & 0 \\ -16 & 0 & 16 - \lambda \end{pmatrix} \] Calculate the determinant: \[ \begin{vmatrix} 16-\lambda & 0 & 4 \\ 0 & 6-\lambda & 0 \\ -16 & 0 & 16-\lambda \end{vmatrix} = (6-\lambda) \begin{vmatrix} 16-\lambda & 4 \\ -16 & 16-\lambda \end{vmatrix} \] \[ = (6-\lambda)[(16-\lambda)^2 - (-16)(4)] \] \[ = (6-\lambda)[(16-\lambda)^2 + 64] \] \[ = (6-\lambda)[(256 - 32\lambda + \lambda^2) + 64] \] \[ = (6-\lambda)[\lambda^2 - 32\lambda + 320] \] Set to zero: \[ (6-\lambda)[\lambda^2 - 32\lambda + 320] = 0 \] So, eigenvalues: 1. \(\lambda_1 = 6\) 2. \(\lambda_{2,3}^2 - 32\lambda_{2,3} + 320 = 0\) Solve the quadratic: \[ \lambda^2 - 32\lambda + 320 = 0 \] \[ \lambda = \frac{32 \pm \sqrt{(-32)^2 - 4 \cdot 1 \cdot 320}}{2} \] \[ = \frac{32 \pm \sqrt{1024 - 1280}}{2} \] \[ = \frac{32 \pm \sqrt{-256}}{2} \] \[ = \frac{32 \pm 16i}{2} \] \[ = 16 \pm 8i \] So, eigenvalues are \(\boxed{6,\ 16+8i,\ 16-8i}\). --- ## **Step 2: Find Eigenvectors** ### For \(\lambda = 6\): \[ A - 6I = \begin{pmatrix} 10 & 0 & 4 \\ 0 & 0 & 0 \\ -16 & 0 & 10 \end{pmatrix} \] Let eigenvector \(\mathbf{v}_1 = \begin{pmatrix} a \\ b \\ c \end{pmatrix}\): First row: \(10a + 4c = 0\) Second row: \(0\) Third row: \(-16a + 10c = 0\) From third row: \(10c = 16a \implies c = \frac{8}{5}a\) From first row: \(10a + 4c = 0 \implies 10a + 4\left(\frac{8}{5}a\right) = 0 \implies 10a + \frac{32}{5}a = 0 \implies 50a + 32a = 0 \implies 82a = 0 \implies a = 0\) So \(a = 0\), then \(c = 0\) and \(b\) is free. So, eigenvector is \(\mathbf{v}_1 = \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}\). --- ### For \(\lambda = 16 + 8i\): \[ A - (16+8i)I = \begin{pmatrix} -8i & 0 & 4 \\ 0 & -10 - 8i & 0 \\ -16 & 0 & -8i \end{pmatrix} \] Second row: \(-10-8i = 0\) so \(b = 0\). First and third rows: \[ \begin{cases} -8i a + 4c = 0 \\ -16a - 8i c = 0 \end{cases} \] From first: \(4c = 8i a \implies c = 2i a\) From third: \(-16a - 8i c = 0 \implies -16a - 8i (2i a) = -16a - 16i^2 a = -16a - 16(-1)a = -16a + 16a = 0\) So, pick \(a=1\): \[ \mathbf{v}_2 = \begin{pmatrix} 1 \\ 0 \\ 2i \end{pmatrix} \] --- ### For \(\lambda = 16 - 8i\): Similarly, eigenvector is conjugate: \[ \mathbf{v}_3 = \begin{pmatrix} 1 \\ 0 \\ -2i \end{pmatrix} \] --- ## **Step 3: General Solution** The general solution is: \[ \mathbf{X}(t) = c_1 e^{6t} \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} + c_2 e^{(16+8i)t} \begin{pmatrix} 1 \\ 0 \\ 2i \end{pmatrix} + c_3 e^{(16-8i)t} \begin{pmatrix} 1 \\ 0 \\ -2i \end{pmatrix} \] Or, combine complex conjugate terms using Euler's formula: Let \(c_2, c_3\) be chosen so the sum is real: Let \(c_2 = \alpha + i \beta\), \(c_3 = \alpha - i \beta\). Or, write as: \[ \mathbf{X}(t) = c_1 e^{6t} \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} + e^{16t} \left[ A \cos(8t) + B \sin(8t) \right] \] where \(A, B\) are real vectors formed as real and imaginary parts of \(e^{8it} \begin{pmatrix} 1 \\ 0 \\ 2i \end{pmatrix}\): \[ e^{8it} \begin{pmatrix} 1 \\ 0 \\ 2i \end{pmatrix} = \begin{pmatrix} \cos 8t + i \sin 8t \\ 0 \\ 2i(\cos 8t + i \sin 8t) \end{pmatrix} = \begin{pmatrix} \cos 8t + i \sin 8t \\ 0 \\ 2i\cos 8t - 2\sin 8t \end{pmatrix} \] So, real part: \[ \text{Re} = \begin{pmatrix} \cos 8t \\ 0 \\ -2\sin 8t \end{pmatrix} \] Imaginary part: \[ \text{Im} = \begin{pmatrix} \sin 8t \\ 0 \\ 2\cos 8t \end{pmatrix} \] So the real general solution is: \[ \boxed{ \mathbf{X}(t) = c_1 e^{6t} \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} + c_2 e^{16t} \begin{pmatrix} \cos 8t \\ 0 \\ -2\sin 8t \end{pmatrix} + c_3 e^{16t} \begin{pmatrix} \sin 8t \\ 0 \\ 2\cos 8t \end{pmatrix} } \] Where \(c_1, c_2, c_3\) are arbitrary constants.