Question

Consider the linear system $\mathit{X}\mathbf{\text{'}}\mathbf{}\mathbf{=}\left(\begin{array}{ccc}4& 6& 6\\ 1& 3& 2\\ -1& -4& -3\end{array}\right)\mathit{X}$.

Without attempting to solve the system, determine which one of the vectors

$K_1=\left[\begin{array}{c}0\\ 1\\ 1\end{array}\right],K_2=\left[\begin{array}{c}1\\ 1\\ -1\end{array}\right],K_3=\begin{array}{c}\left[\begin{array}{c}3\\ 1\\ -1\end{array}\right]\end{array},K_4=\begin{array}{c}\left[\begin{array}{c}6\\ 2\\ -5\end{array}\right]\end{array}$

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

Short answer

K₃is the eigenvector of the coefficient matrix. The solution of the system corresponding to this eigenvector is $X_3=\begin{array}{c}\left[\begin{array}{c}3\\ 1\\ -1\end{array}\right]\end{array}{e}^{4t}$.

Step-by-step solution

Define matrix exponential.

Consider a square matrix A of size n. This matrix can contain either complex numbers or real numbers. The matrix can be calculated as:

where I is the unit matrix of order n.

Therefore, the infinite matrix power series is .

Now, the matrix exponential is defined as the sum of the infinite matrix power series.

It is denoted by the expression e^At. It is given by the formula,.

Find the solution of the system corresponding to the eigenvector.

It is given that, $X\text{'}=\left(\begin{array}{ccc}4& 6& 6\\ 1& 3& 2\\ -1& -4& -3\end{array}\right)X$where, A = $\begin{array}{c}\begin{array}{c}\left(\begin{array}{ccc}4& 6& 6\\ 1& 3& 2\\ -1& -4& -3\end{array}\right)\end{array}\end{array}$.

Now for$K_1=\left[\begin{array}{c}0\\ 1\\ 1\end{array}\right]$the solution is,

$$\begin{gathered} A{K}_1=\begin{array}{c}\begin{array}{c}\left(\begin{array}{ccc}4& 6& 6\\ 1& 3& 2\\ -1& -4& -3\end{array}\right)\end{array}\end{array}\left(\begin{array}{c}0\\ 1\\ 1\end{array}\right) \\ =\left(\begin{array}{c}12\\ 5\\ -7\end{array}\right) \\ \ne \lambda \left(\begin{array}{c}0\\ 1\\ 1\end{array}\right) \end{gathered}$$

For $K_2=\left[\begin{array}{c}1\\ 1\\ -1\end{array}\right]$, the solution is,

$$\begin{gathered} A{K}_2=\begin{array}{c}\begin{array}{c}\left(\begin{array}{ccc}4& 6& 6\\ 1& 3& 2\\ -1& -4& -3\end{array}\right)\end{array}\end{array}\left(\begin{array}{c}1\\ 1\\ -1\end{array}\right) \\ =\left(\begin{array}{c}4\\ 2\\ -2\end{array}\right) \\ \ne \lambda \left(\begin{array}{c}0\\ 1\\ 1\end{array}\right) \end{gathered}$$

For $K_3=\begin{array}{c}\left[\begin{array}{c}3\\ 1\\ -1\end{array}\right]\end{array}$, the solution is

$$\begin{gathered} A{K}_3=\begin{array}{c}\begin{array}{c}\left(\begin{array}{ccc}4& 6& 6\\ 1& 3& 2\\ -1& -4& -3\end{array}\right)\end{array}\end{array}\left(\begin{array}{c}3\\ 1\\ -1\end{array}\right) \\ =\left(\begin{array}{c}12\\ 4\\ -4\end{array}\right) \\ =4\left(\begin{array}{c}3\\ 1\\ -1\end{array}\right) \end{gathered}$$

Hence, the solution for the system corresponding to this eigenvector is$X_3=\begin{array}{c}\left[\begin{array}{c}3\\ 1\\ -1\end{array}\right]\end{array}{e}^{4t}$.