Worksheet 2-2

Worksheet 2-2#

Download: CMSE382-WS2_2.pdf

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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. \(A_4=\begin{bmatrix} -3 & 1 \\ 6 & -2 \end{bmatrix}\)

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 \(= \{-5,\; 0\}\)

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.)

Worksheet 2-2: Q4#

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

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