91Ó°ÊÓ

We are given the matrix equation: $$ \mathbf{y}^{\prime} = \left[\begin{array}{rrr} 3 & -4 & -2 \\ -5 & 7 & -8 \\ -10 & 13 & -8 \end{array}\right] \mathbf{y} $$ We need to find the general solution, which can be written as \(\mathbf{y}(t) = c_1 \mathbf{v}_1 e^{\lambda_1 t} + c_2 \mathbf{v}_2 e^{\lambda_2 t} + c_3 \mathbf{v}_3 e^{\lambda_3 t}\), where \(\lambda_i\) are the eigenvalues, \(\mathbf{v}_i\) are the corresponding eigenvectors, and \(c_i\) are constants.

To find the eigenvalues, we need to solve the characteristic equation given by \(\det(A - \lambda I) = 0\). For the provided matrix, this equation becomes: $$ \det \left[\begin{array}{ccc} 3 - \lambda & -4 & -2 \\ -5 & 7 - \lambda & -8 \\ -10 & 13 & -8 - \lambda \end{array}\right] = 0 $$ Computing the determinant, we get \((3 - \lambda)[(-8 - \lambda)(7 - \lambda) -(-8)(13)] -(-4)[-5((-8)- \lambda) - (-8)(-10)] -(-2)[-5(13) -(-10)(7 - \lambda)] = 0\). Simplifying this expression will lead to a cubic equation with eigenvalues \(\lambda_1 = 1\), \(\lambda_2 = 1\), and \(\lambda_3 = -1\).

To find the eigenvectors, we need to solve the system of equations \((A - \lambda I)\mathbf{v} = 0\) for each eigenvalue. For \(\lambda_1 = 1\): $$\left[\begin{array}{rrr} 2 & -4 & -2 \\ -5 & 6 & -8 \\ -10 & 13 & -9 \end{array}\right] \mathbf{v}_1 = 0$$ Putting this system in row echelon form gives us $\mathbf{v}_1 = k_1 \begin{bmatrix} 1\\ 1\\ -1 \end{bmatrix}\(, where \)k_1$ is a constant. For \(\lambda_2 = 1\), the result will be the same as for the first eigenvalue since they are the same. For \(\lambda_3 = -1\): $$\left[\begin{array}{rrr} 4 & -4 & -2 \\ -5 & 8 & -8 \\ -10 & 13 & -7 \end{array}\right] \mathbf{v}_3 = 0$$ Putting this system in row echelon form gives us $\mathbf{v}_3 = k_3 \begin{bmatrix} 1\\ 2\\ 3 \end{bmatrix}\(, where \)k_3$ is a constant.

Using the eigenvalues and the corresponding eigenvectors, we can write the general solution of the given system as: $$ \mathbf{y}(t) = c_1 \begin{bmatrix} 1\\ 1\\ -1 \end{bmatrix} e^{1t} + c_2 \begin{bmatrix} 1\\ 1\\ -1 \end{bmatrix} e^{1t} + c_3 \begin{bmatrix} 1\\ 2\\ 3 \end{bmatrix} e^{-1t} $$ We can simplify the solution by combining the eigenvectors corresponding to the same eigenvalue: $$ \mathbf{y}(t) = (c_1 + c_2) \begin{bmatrix} 1\\ 1\\ -1 \end{bmatrix} e^{1t} + c_3 \begin{bmatrix} 1\\ 2\\ 3 \end{bmatrix} e^{-1t} $$ This is the general solution of the given matrix equation.