Question

Verify that the given vector X is a solution of the given homogenous linear equation;

$$\begin{gathered} \mathit{X}\mathbf{\text{'}}\mathbf{=}\mathbf{\left(}\begin{array}{cc}\mathbf{-}\mathbf{1}& \frac{\mathbf{1}}{\mathbf{4}}\\ \mathbf{1}& \mathbf{-}\mathbf{1}\end{array}\mathbf{\right)}\mathit{X} \\ \mathbf{}\mathit{X}\mathbf{=}\mathbf{\left(}\begin{array}{c}\mathbf{-}\mathbf{1}\\ \mathbf{2}\end{array}\mathbf{\right)}{\mathit{e}}^{\mathbf{-}\mathbf{3}\mathbf{t}\mathbf{/}\mathbf{2}} \end{gathered}$$

Short answer

Vector X is verified, it is the solution of given homogenous linear system:

$\therefore AX=X\text{'}$

Step-by-step solution

Homogeneous linear equation;

Let ${\mathit{X}}_{\mathbf{1}}\mathbf{,}{\mathit{X}}_{\mathbf{2}}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}{\mathit{X}}_{\mathbf{n}}$be a fundamental set of solutions of homogeneous system on an interval l. Then the general solution of the system on the intervals is

$\mathit{X}\mathbf{=}{\mathit{c}}_{\mathbf{1}}{\mathit{X}}_{\mathbf{1}}\mathbf{+}{\mathit{c}}_{\mathbf{2}}{\mathit{X}}_{\mathbf{2}}\mathbf{+}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{+}{\mathit{c}}_{\mathbf{n}}{\mathit{X}}_{\mathbf{n}}$

Where the${\mathit{c}}_{\mathbf{i}}\mathbf{,}\mathit{i}\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{2}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{,}\mathit{n}$ are arbitrary constant.

Verification of the vector:

Here we will verify that the given vector X is the solution of the given homogenous linear system;

$A=\left(\begin{array}{cc}-1& \frac{1}{4}\\ 1& -1\end{array}\right)$

The product of A and X is:

$$\begin{gathered} X=\left(\begin{array}{c}-1\\ 2\end{array}\right)e^{-3\mathrm{t}/2}=\left(\begin{array}{c}-{\mathrm{e}}^{-3\mathrm{t}/2}\\ 2{\mathrm{e}}^{-3\mathrm{t}/2}\end{array}\right) \\ X\text{'}=\left(\begin{array}{c}\frac{3}{2}{\mathrm{e}}^{-3\mathrm{t}/2}\\ -3{\mathrm{e}}^{-3\mathrm{t}/2}\end{array}\right) \\ X\text{'}=AX \\ \left(\begin{array}{c}\frac{{\displaystyle 3}}{{\displaystyle 2}}{\mathrm{e}}^{-3\mathrm{t}/2}\\ -3{\mathrm{e}}^{-3\mathrm{t}/2}\end{array}\right)=\left(\begin{array}{cc}-1& \frac{1}{4}\\ 1& -1\end{array}\right)\left(\begin{array}{c}-{\mathrm{e}}^{-3\mathrm{t}/2}\\ 2{\mathrm{e}}^{-3\mathrm{t}/2}\end{array}\right) \\ \left(\begin{array}{c}\frac{{\displaystyle 3}}{{\displaystyle 2}}{\mathrm{e}}^{-3\mathrm{t}/2}\\ -3{\mathrm{e}}^{-3\mathrm{t}/2}\end{array}\right)=\left(\begin{array}{c}\frac{{\displaystyle 3}}{{\displaystyle 2}}{\mathrm{e}}^{-3\mathrm{t}/2}\\ -3{\mathrm{e}}^{-3\mathrm{t}/2}\end{array}\right) \end{gathered}$$

$\therefore AX=X\text{'}$

Hence, the vector X is verified:

$\therefore AX=X\text{'}$