Question

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

$$\begin{gathered} {\mathit{X}}_{\mathbf{1}}\mathbf{=}\mathbf{\left(}\begin{array}{c}\mathbf{1}\\ \mathbf{-}\mathbf{1}\end{array}\mathbf{\right)}{\mathit{e}}^{\mathbf{t}} \\ {\mathit{X}}_{\mathbf{2}}\mathbf{=}\mathbf{\left(}\begin{array}{c}\mathbf{2}\\ \mathbf{6}\end{array}\mathbf{\right)}{\mathit{e}}^{\mathbf{t}}\mathbf{+}\mathbf{\left(}\begin{array}{c}\mathbf{8}\\ \mathbf{-}\mathbf{8}\end{array}\mathbf{\right)}\mathit{t}{\mathit{e}}^{\mathbf{t}} \end{gathered}$$

Short answer

The X₁ and X₂ are linearly dependent on interval (∞,-∞).

∴ $W(X_1,X_2)=8t{e}^{2t}\ne 0$

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

$$\begin{gathered} X_1=\left(\begin{array}{c}1\\ -1\end{array}\right)e^{\mathrm{t}} \\ {X}_2=\left(\begin{array}{c}2\\ 6\end{array}\right)e^{\mathrm{t}}+\left(\begin{array}{c}8\\ -8\end{array}\right)t{e}^{\mathrm{t}} \end{gathered}$$

Now use of determinant then,

role="math" localid="1663885793341" $$\begin{gathered} X_1=\left(\begin{array}{c}{e}^t\\ -e^t\end{array}\right) \\ {X}_2=\left(\begin{array}{c}2e^t+8t{e}^t\\ 6e^t-8t{e}^t\end{array}\right) \end{gathered}$$

$$\begin{gathered} W(X_1,X_2)=\left|\begin{array}{cc}{e}^t& 2e^t+8t{e}^t\\ -e^t& 6e^t-8t{e}^t\end{array}\right| \\ =e^t(6e^t-8t{e}^t)-(-e^t)(2e^t+8t{e}^t) \\ =6e^{2t}-8t{e}^{2t}+2e^{2t}+8e^{2t} \\ =8t{e}^{2t} \\ \ne 0 \end{gathered}$$

Hence, X₁ and X₂ are linearly dependent on interval (∞,-∞).

$W(X_1,X_2)=8t{e}^{2t}\ne 0$