# Worksheet 2-2

Download: [CMSE382-WS2_2.pdf](CMSE382-WS2_2.pdf)

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This is an AI-generated transcript of the worksheet and may contain errors or inaccuracies. Please refer to the original course materials for authoritative content.
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## Worksheet 2-2: Q1

For the following symmetric matrices, cross out the classifications it **CANNOT** be due to the diagonal entries (don't calculate the eigenvalues).

For each matrix, circle those that remain possible: *pos def*   *pos semidef*   *indefinite*   *neg def*   *neg semidef*

1. $A_1=\begin{bmatrix} 0 & 1 \\ 1 & 5 \end{bmatrix}$

2. $A_2=\begin{bmatrix} 2 & -1 \\ -1 & 1 \end{bmatrix}$

3. $A_3=\begin{bmatrix} -1 & 2 \\ 2 & 1 \end{bmatrix}$

4. $A_4=\begin{bmatrix} -3 & 1 \\ 6 & -2 \end{bmatrix}$

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## Worksheet 2-2: Q2

For the following matrices with their given eigenvalues, what is the correct classification and why?

1. $A_1=\begin{bmatrix} 0 & 1 \\ 1 & 5 \end{bmatrix}$; eigenvalues $= \{-0.193,\; 5.193\}$

2. $A_2=\begin{bmatrix} 2 & -1 \\ -1 & 1 \end{bmatrix}$; eigenvalues $= \{0.382,\; 2.618\}$

3. $A_3=\begin{bmatrix} -1 & 2 \\ 2 & 1 \end{bmatrix}$; eigenvalues $= \{-2.236,\; 2.236\}$

4. $A_4=\begin{bmatrix} -3 & 1 \\ 6 & -2 \end{bmatrix}$; eigenvalues $= \{-5,\; 0\}$

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## Worksheet 2-2: Q3

On the last worksheet, we found the stationary points of

$$f(x,y)=6x^{2}y-3x^{3}+2y^{3}-150y$$

were at $p_1 = (0,5)$, $p_2 = (0,-5)$, $p_3 = (4,3)$, $p_4 = (-4,-3)$.

Use the second optimality condition to classify each stationary point as a local minimum, local maximum, or saddle point. You may use a computational tool to compute eigenvalues. *(Visual check: [desmos.com/3d/xa4komuwmb](https://www.desmos.com/3d/xa4komuwmb).)*

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## Worksheet 2-2: Q4

Find and classify the stationary points of $f(x,y)=2x^3 + 3y^2 + 3x^2 y - 24y$.