Question

find a fundamental matrix for the given system of equations. In each case also find the fundamental matrix \(\mathbf{\Phi}(t)\) satisfying \(\Phi(0)=\mathbf{1}\) $$ \mathbf{x}^{\prime}=\left(\begin{array}{rrr}{1} & {1} & {1} \\ {2} & {1} & {-1} \\ {-8} & {-5} & {-3}\end{array}\right) \mathbf{x} $$

Short answer

Based on the given information, determine the fundamental matrix \(\mathbf{\Phi}(t)\), where \(\mathbf{\Phi}(0) = \mathbf{1}\). Solution: The fundamental matrix \(\mathbf{\Phi}(t)\) is given by: $$\mathbf{\Phi}(t)=\frac{1}{2}\left(\begin{array}{ccc}{1} & {1} & {1} \\\ {1} & {-1} & {1} \\\ {-2} & {1} & {-1}\end{array}\right)\left(\begin{array}{ccc}{e^{-t}} & {0} & {0} \\\ {0} & {e^t} & {0} \\\ {0} & {0} & {e^{-3t}}\end{array}\right) \left(\begin{array}{ccc}{1} & {0} & {0} \\\ {0} & {1} & {0} \\\ {0} & {0} & {1}\end{array}\right)$$

Step-by-step solution

Determine the eigenvalues and eigenvectors of the given matrix

To find eigenvalues and their respective eigenvectors, we need to solve the following equation: $$\text{det} \ |A - \lambda I| = 0$$ where \(A\) is the given matrix and \(\lambda\) represents eigenvalues. The given matrix is: $$A=\left(\begin{array}{rrr}{1} & {1} & {1} \\\ {2} & {1} & {-1} \\\ {-8} & {-5} & {-3}\end{array}\right)$$ The equation becomes: $$\text{det} \ \left(\begin{array}{ccc}{1-\lambda} & {1} & {1} \\\ {2} & {1-\lambda} & {-1} \\\ {-8} & {-5} & {-3-\lambda}\end{array}\right) = 0$$ Solve the determinant to find the eigenvalues. After solving the determinant, we get the eigenvalues: \(\lambda_1 = -1\), \(\lambda_2 = 1\), and \(\lambda_3 = -3\). Now, find the eigenvectors corresponding to each eigenvalue. For \(\lambda_1 = -1\): $$\left(\begin{array}{ccc}{2} & {1} & {1} \\\ {2} & {2} & {-1} \\\ {-8} & {-5} & {-2}\end{array}\right)\mathbf{x} = \mathbf{0}$$ The eigenvector is: \(\mathbf{x}_1 = (1,1,-2)^T\) For \(\lambda_2 = 1\): $$\left(\begin{array}{ccc}{0} & {1} & {1} \\\ {2} & {0} & {-1} \\\ {-8} & {-5} & {-4}\end{array}\right)\mathbf{x} = \mathbf{0}$$ The eigenvector is: \(\mathbf{x}_2 = (1,-1,1)^T\) For \(\lambda_3 = -3\): $$\left(\begin{array}{ccc}{4} & {1} & {1} \\\ {2} & {4} & {-1} \\\ {-8} & {-5} & {0}\end{array}\right)\mathbf{x} = \mathbf{0}$$ The eigenvector is: \(\mathbf{x}_3 = (1,1,-1)^T\)

Find the matrix exponential of the given matrix

Using the eigenvectors, we can form a matrix \(X=[\mathbf{x}_1,\mathbf{x}_2,\mathbf{x}_3]\) which contains eigenvectors as its columns. We then find a diagonal matrix \(\Lambda\) with the eigenvalues as its elements along the diagonal, that is, $$X=\left(\begin{array}{ccc}1 & 1 & 1 \\\ 1 & -1 & 1 \\\ -2 & 1 & -1\end{array}\right), \quad \Lambda=\left(\begin{array}{ccc}-1 & 0 & 0 \\\ 0 & 1 & 0 \\\ 0 & 0 & -3\end{array}\right)$$ Now, we can compute the matrix exponential of \(A\) as follows, $$\Phi(t) = e^{A t} = X e^{\Lambda t}X^{-1}$$ where the exponential of a diagonal matrix is calculated by taking the exponential of all the diagonal elements separately. Thus we get, $$e^{\Lambda t} = \left(\begin{array}{ccc}{e^{-t}} & {0} & {0} \\\ {0} & {e^t} & {0} \\\ {0} & {0} & {e^{-3t}}\end{array}\right)$$ Now, we need to calculate \(X^{-1}\), which is the inverse of matrix \(X\). $$X^{-1} = \frac{1}{2}\left(\begin{array}{rrr}{-1} & {1} & {1} \\\ {2} & {0} & {1} \\\ {-1} & {-1} & {0}\end{array}\right)$$ Finally, calculate \(\Phi(t)\): $$\Phi(t) = Xe^{\Lambda t}X^{-1} = \left(\begin{array}{ccc}1 & 1 & 1 \\\ 1 & -1 & 1 \\\ -2 & 1 & -1\end{array}\right)\left(\begin{array}{ccc}{e^{-t}} & {0} & {0} \\\ {0} & {e^t} & {0} \\\ {0} & {0} & {e^{-3t}}\end{array}\right) \frac{1}{2}\left(\begin{array}{rrr}{-1} & {1} & {1} \\\ {2} & {0} & {1} \\\ {-1} & {-1} & {0}\end{array}\right)$$

Find the desired fundamental matrix satisfying \(\Phi(0)=\mathbf{1}\)

The fundamental matrix fulfilling this condition can be obtained by evaluating \(\Phi(t)\) at \(t=0\) and solving for the constant matrix \(C\). Then multiply \(\Phi(t)\) by the constant matrix \(C\). First, calculate \(\Phi(0)\): $$\Phi(0) = \left(\begin{array}{ccc}1 & 1 & 1 \\\ 1 & -1 & 1 \\\ -2 & 1 & -1\end{array}\right)\left(\begin{array}{ccc}{1} & {0} & {0} \\\ {0} & {1} & {0} \\\ {0} & {0} & {1}\end{array}\right) \frac{1}{2}\left(\begin{array}{rrr}{-1} & {1} & {1} \\\ {2} & {0} & {1} \\\ {-1} & {-1} & {0}\end{array}\right)$$ $$\Phi(0) = \left(\begin{array}{ccc}{1} & {1} & {1} \\\ {1} & {-1} & {1} \\\ {-2} & {1} & {-1}\end{array}\right)\frac{1}{2}\left(\begin{array}{rrr}{-1} & {1} & {1} \\\ {2} & {0} & {1} \\\ {-1} & {-1} & {0}\end{array}\right)$$ $$\Phi(0) = \left(\begin{array}{ccc}{0} & {0} & {2} \\\ {0} & {2} & {0} \\\ {0} & {0} & {2}\end{array}\right)$$ Now, to find the constant matrix \(C\), we need \(\Phi(0)\) to be equal to the identity matrix: $$\Phi(0)C = \mathbf{1}$$ $$\left(\begin{array}{ccc}{0} & {0} & {2} \\\ {0} & {2} & {0} \\\ {0} & {0} & {2}\end{array}\right)C = \left(\begin{array}{ccc}{1} & {0} & {0} \\\ {0} & {1} & {0} \\\ {0} & {0} & {1}\end{array}\right)$$ Solve for \(C\): $$C = \frac{1}{2}\left(\begin{array}{ccc}{1} & {0} & {0} \\\ {0} & {1} & {0} \\\ {0} & {0} & {1}\end{array}\right)$$ Finally, compute the desired fundamental matrix as \(\Phi(t)C\): $$\mathbf{\Phi}(t)=\Phi(t)C=\frac{1}{2}\left(\begin{array}{ccc}{1} & {1} & {1} \\\ {1} & {-1} & {1} \\\ {-2} & {1} & {-1}\end{array}\right)\left(\begin{array}{ccc}{e^{-t}} & {0} & {0} \\\ {0} & {e^t} & {0} \\\ {0} & {0} & {e^{-3t}}\end{array}\right) \left(\begin{array}{ccc}{1} & {0} & {0} \\\ {0} & {1} & {0} \\\ {0} & {0} & {1}\end{array}\right)$$