Question

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

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

Short answer

The X₁ and X₂ are the fundamental solutions.

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

$$\begin{gathered} X_1=\left(\begin{array}{c}{\mathrm{e}}^{-2\mathrm{t}}\\ {\mathrm{e}}^{-2\mathrm{t}}\end{array}\right) \\ {X}_2=\left(\begin{array}{c}{\mathrm{e}}^{-6\mathrm{t}}\\ -{\mathrm{e}}^{-6\mathrm{t}}\end{array}\right) \end{gathered}$$

Now use of determinant then,

$$\begin{gathered} W\left(X_{1,}{X}_2\right)=\left|\begin{array}{cc}{e}^{-2t}& e^{-6t}\\ e^{-2t}& -e^{-6t}\end{array}\right| \\ =-e^{-8t}-e^{-8t} \\ =-2e^{-8t} \\ \ne 0 \end{gathered}$$

Hence, X₁ and X₂ are the fundamental solutions.