Question

The given vectors are solutions of a system X’ = AX. Determine whether the vectors form a fundamental set on the interval $(-\infty ,\infty )$.

${\mathit{X}}_{\mathbf{1}}\mathbf{=}\mathbf{\left(}\begin{array}{c}\mathbf{1}\\ \mathbf{6}\\ \mathbf{-}\mathbf{13}\end{array}\mathbf{\right)}\mathbf{,}\mathbf{}\mathbf{}{\mathit{X}}_{\mathbf{2}}\mathbf{=}\mathbf{\left(}\begin{array}{c}\mathbf{1}\\ \mathbf{-}\mathbf{2}\\ \mathbf{-}\mathbf{1}\end{array}\mathbf{\right)}{\mathit{e}}^{\mathbf{-}\mathbf{4}\mathbf{t}}\mathbf{,}\mathbf{}\mathbf{}{\mathit{X}}_{\mathbf{3}}\mathbf{=}\mathbf{\left(}\begin{array}{c}\mathbf{2}\\ \mathbf{3}\\ \mathbf{-}\mathbf{2}\end{array}\mathbf{\right)}{\mathit{e}}^{\mathbf{3}\mathbf{t}}$

Short answer

The given vector is solution of system.

Step-by-step solution

Fundamental set of solution;

Any set ${\mathit{X}}_{\mathbf{1}}\mathbf{,}{\mathit{X}}_{\mathbf{2}}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}{\mathit{X}}_{\mathbf{n}}$ be a fundamental set of solution of the homogeneous system on an interval l is said to be a fundamental set of solution on interval.

Evaluating the given vector:

The given vector is

$X_1=\left(\begin{array}{c}1\\ 6\\ -13\end{array}\right),X_2=\left(\begin{array}{c}1\\ -2\\ -1\end{array}\right)e^{-4\mathrm{t}},X_3=\left(\begin{array}{c}2\\ 3\\ -2\end{array}\right)e^{3\mathrm{t}}$

Now use of determinant then,

$$\begin{gathered} W\left(X_1,X_2,X_3\right)=\left|\begin{array}{ccc}1& 6& -13\\ e^{-4t}& -e^{-4t}& -e^{-4t}\\ 2e^{3t}& 3e^{3t}& -2e^{3t}\end{array}\right| \\ W\left(X_1,X_2,X_3\right)=-84e^{-t} \\ \ne 0 \end{gathered}$$

For all real value of t. we conclude that $X_1,X_2,X_3$form a fundamental set of solution on $\left(\infty ,-\infty \right)$, thus the general solution of system on the interval is

$X=c_1{X}_1+c_2{X}_2+c_3{X}_3$

Hence, the given vector is solution of system.