91影视

After calculating the determinant, we get: $$ (-4-\lambda)((-3-\lambda)(-2-\lambda)) - (-1)(-1)(-2-\lambda) = 0 $$ Now, we need to solve this equation for 位. Simplifying the equation, we obtain: $$ (\lambda^3 + 9\lambda^2 + 26\lambda + 24) - 2\lambda = 0 $$ $$ \lambda^3 + 9\lambda^2 + 24\lambda + 24 = 0 $$ By solving the cubic equation, we find the eigenvalues: 位鈧� = -6, 位鈧� = -3, and 位鈧� = 0. #tag_title#Step 2: Find the eigenvectors for each eigenvalue#tag_content# Now, we have to find the eigenvectors corresponding to each eigenvalue. Let's start with 位鈧� = -6. We need to solve the equation (A - 位鈧両)v = 0: $$ \left[\begin{array}{rrr} 2 & 0 & -1 \\ -1 & 3 & -1 \\ 1 & 0 & 4 \end{array}\right] \left[\begin{array}{r} x \\ y \\ z \end{array}\right] =0 $$ By solving this system of linear equations, we find the eigenvector corresponding to 位鈧� = -6: v鈧� = [1, 1, 2]. Next, let's find the eigenvector for 位鈧� = -3: $$ \left[\begin{array}{rrr} -1 & 0 & -1 \\ -1 & 0 & -1 \\ 1 & 0 & 1 \end{array}\right] \left[\begin{array}{r} x \\ y \\ z \end{array}\right] =0 $$ By solving this system of linear equations, we find the eigenvector corresponding to 位鈧� = -3: v鈧� = [1, 0, 1]. Finally, let's find the eigenvector for 位鈧� = 0: $$ \left[\begin{array}{rrr} -4 & 0 & -1 \\ -1 & -3 & -1 \\ 1 & 0 & -2 \end{array}\right] \left[\begin{array}{r} x \\ y \\ z \end{array}\right] =0 $$ By solving this system of linear equations, we find the eigenvector corresponding to 位鈧� = 0: v鈧� = [1, 1, -3]. #tag_title#Step 3: Write the general solution using eigenvalues and eigenvectors#tag_content# Now that we have eigenvalues and their corresponding eigenvectors, we can write the general solution to the given system of linear differential equations as follows: $$ \begin{aligned} X(t) = c_1e^{-6t}\begin{bmatrix}1\\1\\2\end{bmatrix} + c_2e^{-3t}\begin{bmatrix}1\\0\\1\end{bmatrix} + c_3e^{0t}\begin{bmatrix}1\\1\\-3\end{bmatrix}. \end{aligned} $$ Here, c鈧�, c鈧�, and c鈧� are arbitrary constants determined by the initial conditions.