Solve the given initial-value problem.

$$\mathbf{X}^{\prime}=\left(\begin{array}{lll}{0} & {0} & {1} \\ {0} & {1} & {0} \\ {1} & {0} & {0}\end{array}\right) \mathbf{X}, \quad \mathbf{X}(0)=\left(\begin{array}{l}{1} \\ {2} \\ {5}\end{array}\right)$$

Find the general solution of the given system.

$$\mathbf{X}^{\prime}=\left(\begin{array}{lll}{4} & {1} & {0} \\ {0} & {4} & {1} \\ {0} & {0} & {4}\end{array}\right) \mathbf{X}$$

Consider the linear system

$$\mathbf{X}^{\prime}=\left(\begin{array}{rrr}{4} & {6} & {6} \\ {1} & {3} & {2} \\ {-1} & {-4} & {-3}\end{array}\right) \mathbf{X}$$

. Without attempting to solve the system, which one of the following vectors,

$$\mathbf{K}_{1}=\left(\begin{array}{l}{0} \\ {1} \\ {1}\end{array}\right), \quad \mathbf{K}_{2}=\left(\begin{array}{r}{1} \\ {1} \\ {-1}\end{array}\right), \quad \mathbf{K}_{3}=\left(\begin{array}{r}{3} \\ {1} \\ {-1}\end{array}\right), \quad \mathbf{K}_{4}=\left(\begin{array}{r}{6} \\ {2} \\ {-5}\end{array}\right)$$

is an eigenvector of the coefficient matrix? What is the solution of the system corresponding to this eigenvector?

Use diagonalization to solve the given system.

$$\mathbf{X}^{\prime}=\left(\begin{array}{ll}{1} & {\frac{1}{4}} \\ {1} & {1}\end{array}\right) \mathbf{X}$$

Write the given linear system in matrix form. $2 \frac{d x}{d t}=5 x-14 y+9 t$

Use (3) to compute $e^{\mathbf{A} t}$.

$$\mathbf{A}=\left(\begin{array}{rrr}{1} & {1} & {1} \\ {1} & {1} & {1} \\ {-2} & {-2} & {-2}\end{array}\right)$$

Find the general solution of the given system.

$$\begin{array}{l}{\frac{d x}{d t}=-4 x+2 y} \\ {\frac{d y}{d t}=-\frac{5}{2} x+2 y}\end{array}$$

Use the method of undetermined coefficients to solve the given system.

$$\mathbf{X}^{\prime}=\left(\begin{array}{ll}{1} & {3} \\ {3} & {1}\end{array}\right) \mathbf{X}+\left(\begin{array}{c}{-2 t^{2}} \\ {t+5}\end{array}\right)$$

Find the general solution of the given system.

$$\mathbf{X}^{\prime}=\left(\begin{array}{rrr}{1} & {0} & {0} \\ {2} & {2} & {-1} \\ {0} & {1} & {0}\end{array}\right) \mathbf{X}$$

Use variation of parameters to solve the given system.

$$\mathbf{X}^{\prime}=\left(\begin{array}{rr}{1} & {2} \\ {-\frac{1}{2}} & {1}\end{array}\right) \mathbf{X}+\left(\begin{array}{c}{\csc t} \\ {\sec t}\end{array}\right) e^{t}$$

Find the general solution of the given system.

$$\mathbf{X}^{\prime}=\left(\begin{array}{rrr}{1} & {0} & {0} \\ {0} & {3} & {1} \\ {0} & {-1} & {1}\end{array}\right) \mathbf{X}$$

Prove that the general solution of the nonhomogeneous linear system

$$\mathbf{X}^{\prime}=\left(\begin{array}{rr}{-1} & {-1} \\ {-1} & {1}\end{array}\right) \mathbf{X}+\left(\begin{array}{l}{1} \\ {1}\end{array}\right) t^{2}+\left(\begin{array}{r}{4} \\ {-6}\end{array}\right) t+\left(\begin{array}{r}{-1} \\ {5}\end{array}\right)$$

on the interval $(-\infty, \infty)$ is

$$\begin{aligned} \mathbf{X}=& c_{1}\left(\begin{array}{c}{1} \\ {-1-\sqrt{2}}\end{array}\right) e^{\sqrt{2} t}+c_{2}\left(\begin{array}{c}{1} \\ {-1+\sqrt{2}}\end{array}\right) e^{-\sqrt{2} t} \\ &+\left(\begin{array}{c}{1} \\ {0}\end{array}\right) t^{2}+\left(\begin{array}{c}{-2} \\ {4}\end{array}\right) t+\left(\begin{array}{c}{1} \\ {0}\end{array}\right) \end{aligned}$$

.

Use (1) to find the general solution of

$$\mathbf{X}^{\prime}=\left(\begin{array}{rrrr}{-4} & {0} & {6} & {0} \\ {0} & {-5} & {0} & {-4} \\ {-1} & {0} & {1} & {0} \\ {0} & {3} & {0} & {2}\end{array}\right) \mathbf{X}$$

. Use a CAS to find $e^{\mathbf{A} t}$.

Use variation of parameters to solve the given system.

$$\mathbf{X}^{\prime}=\left(\begin{array}{rr}{0} & {1} \\ {-1} & {0}\end{array}\right) \mathbf{X}+\left(\begin{array}{c}{1} \\ {\cot t}\end{array}\right)$$