Question

classify the critical point \\((0, 0)\\) of the given linear system by computing the trace \\(\tau\\) and determinant \\(\delta\\) and using the figure.

\\
\

$$\begin{aligned} x &= -5x + 6y \\\\ y &= 3x + 4y \\end{aligned}$$

\\

unstable spiral
degenerate unstable node
center
saddle
stable node
stable spiral
degenerate stable node
unstable node

Explanation:

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<reasoning>

Set up the coefficient matrix

\[
A =

$$\begin{pmatrix} -5 & 6 \\ 3 & 4 \end{pmatrix}$$

\]

Calculate the trace and determinant

\[

$$\begin{aligned} \tau &= \operatorname{tr}(A) = -5 + 4 = -1 \\ \Delta &= \det(A) = (-5)(4) - (6)(3) = -20 - 18 = -38 \end{aligned}$$

\]

Classify the critical point

Since \(\Delta = -38 < 0\), the critical point \((0,0)\) is a saddle.
</reasoning>

<answer>
<mcq-option>unstable spiral</mcq-option>
<mcq-option>degenerate unstable node</mcq-option>
<mcq-option>center</mcq-option>
<mcq-correct>saddle</mcq-correct>
<mcq-option>stable node</mcq-option>
<mcq-option>stable spiral</mcq-option>
<mcq-option>degenerate stable node</mcq-option>
<mcq-option>unstable node</mcq-option>
</answer>

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Answer:

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<reasoning>

Set up the coefficient matrix

\[
A =

$$\begin{pmatrix} -5 & 6 \\ 3 & 4 \end{pmatrix}$$

\]

Calculate the trace and determinant

\[

$$\begin{aligned} \tau &= \operatorname{tr}(A) = -5 + 4 = -1 \\ \Delta &= \det(A) = (-5)(4) - (6)(3) = -20 - 18 = -38 \end{aligned}$$

\]

Classify the critical point

Since \(\Delta = -38 < 0\), the critical point \((0,0)\) is a saddle.
</reasoning>

<answer>
<mcq-option>unstable spiral</mcq-option>
<mcq-option>degenerate unstable node</mcq-option>
<mcq-option>center</mcq-option>
<mcq-correct>saddle</mcq-correct>
<mcq-option>stable node</mcq-option>
<mcq-option>stable spiral</mcq-option>
<mcq-option>degenerate stable node</mcq-option>
<mcq-option>unstable node</mcq-option>
</answer>

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