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Linear Algebra : Matrix Calculus

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Example Questions

Example Question #31 : Matrix Calculus

Give the Hessian matrix of the function $f(x, y) = \frac{1}{x^{2}y^{3}}$

Possible Answers:

$\frac{1}{x^{4}y^{5}}\begin{bmatrix}6x^{2}&6xy\\ \\ 6xy&12y^{2}\end{bmatrix}$

$\frac{1}{x^{4}y^{5}}\begin{bmatrix}6x^{2}&-6xy\\ \\ -6xy&12y^{2}\end{bmatrix}$

$\frac{1}{x^{4}y^{5}}\begin{bmatrix}6y^{2}&-6xy\\ \\- 6xy&12x^{2}\end{bmatrix}$

$\frac{1}{x^{4}y^{5}}\begin{bmatrix}6x^{2}& 6xy\\ \\ -6xy&12y^{2}\end{bmatrix}$

$\frac{1}{x^{4}y^{5}}\begin{bmatrix}6y^{2}&6xy\\ \\ 6xy&12x^{2}\end{bmatrix}$

Correct answer:

$\frac{1}{x^{4}y^{5}}\begin{bmatrix}6y^{2}&6xy\\ \\ 6xy&12x^{2}\end{bmatrix}$

Explanation:

The Hessian matrix of a function f(x,y) is the matrix of partial second derivatives

$\begin{bmatrix} \frac{\partial^2 f}{\partial x^2} & \frac{\partial^2 f}{\partial x\partial y } \\ \\ \frac{\partial^2 f}{\partial y\partial x } & \frac{\partial^2 f}{ \partial y^2 } \end{bmatrix}$

First, rewrite

$f(x, y) = \frac{1}{x^{2}y^{3}}$

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

Find each partial second derivative separately:

$\frac{\partial f }{\partial x} =\frac{\partial }{\partial x} ( x^{-2}y^{-3})$

$=-2 x^{-3}y^{-3}$

$\frac{\partial^2 f}{\partial x^2} =\frac{\partial }{\partial x} \left ( -2 x^{-3}y^{-3} \right )$

$= -2 (-3)x^{-4}y^{-3} \right )$

$=6x^{-4}y^{-3}$

$=\frac{6}{x^{4}y^{3}}$

$\frac{\partial^2 f}{\partial y \partial x} = \frac{\partial^2 f}{\partial x \partial y} =\frac{\partial }{\partial y } \left ( -2 x^{-3}y^{-3} \right )$

$=-2 (-3)x^{-3}y^{-4}$

$=\frac{6}{x^{3}y^{4}}$

$\frac{\partial f }{\partial y} =\frac{\partial }{\partial y} ( x^{-2}y^{-3})$

$= -3 x^{-2}y^{-4}$

$\frac{\partial ^2 f}{\partial y^2} = \frac{\partial }{\partial y} \left ( -3 x^{-2}y^{-4} \right )$

$=-3 (-4) x^{-2}y^{-5}$

$=12x^{-2}y^{-5}$

$=\frac{12}{x^{2}y^{5}}$

The Hessian of is

$\begin{bmatrix}\frac{6}{x^{4}y^{3}} &\frac{6}{x^{3}y^{4}}\\ \\ \frac{6}{x^{3}y^{4}}&\frac{12}{x^{2}y^{5}}\end{bmatrix}$ ,

which can be rewritten as

$\frac{1}{x^{4}y^{5}}\begin{bmatrix}6y^{2}&6xy\\ \\ 6xy&12x^{2}\end{bmatrix}$ .

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Example Question #11 : The Hessian

Give the Hessian matrix of the function

$f(x,y) = \ln ( x^{2}+y)$ .

Possible Answers:

$\frac{ - 1}{(x^{2}+y)^{2}} \begin{bmatrix} 2x^{2}-2y & 2 \\ 2 & 1 \end{bmatrix}$

$\frac{ - 1}{(x^{2}+y)^{2}} \begin{bmatrix} 2x^{2}+2y & -2x \\ -2x & 1 \end{bmatrix}$

$\frac{ - 1}{(x^{2}+y)^{2}} \begin{bmatrix} 2x^{2}-2y & -2x \\ -2x & 1 \end{bmatrix}$

$\frac{ - 1}{(x^{2}+y)^{2}} \begin{bmatrix} 2x^{2}+2y & 2x \\ 2x & 1 \end{bmatrix}$

$\frac{ - 1}{(x^{2}+y)^{2}} \begin{bmatrix} 2x^{2}-2y & 2x \\ 2x & 1 \end{bmatrix}$

Correct answer:

$\frac{ - 1}{(x^{2}+y)^{2}} \begin{bmatrix} 2x^{2}-2y & 2x \\ 2x & 1 \end{bmatrix}$

Explanation:

The Hessian matrix of a function f(x,y) is the matrix of partial second derivatives

$\begin{bmatrix} \frac{\partial^2 f}{\partial x^2} & \frac{\partial^2 f}{\partial x\partial y } \\ \\ \frac{\partial^2 f}{\partial y\partial x } & \frac{\partial^2 f}{ \partial y^2 } \end{bmatrix}$

Find each partial second derivative separately:

$\frac{\partial f }{\partial x} =\frac{\partial }{\partial x} \ln ( x^{2}+y)$

$=\frac{\frac{\partial }{\partial x} ( x^{2}+y)}{x^{2}+y}$

$= \frac{2x }{x^{2}+y}$

$\frac{\partial^2 f}{\partial x^2} =\frac{\partial }{\partial x} \left ( \frac{2x }{x^{2}+y} \right )$

$= \frac{(x^{2}+y) \frac{\partial }{\partial x}(2x)- 2x \frac{\partial }{\partial x}(x^{2}+y)}{(x^{2}+y)^{2}}$

$= \frac{(x^{2}+y) (2) - 2x (2x)}{(x^{2}+y)^{2}}$

$= \frac{(2x^{2}+2y) - 4x^{2}}{(x^{2}+y)^{2}}$

$= \frac{-2x^{2}+2y}{(x^{2}+y)^{2}}$

$\frac{\partial^2 f}{\partial x\partial y }= \frac{\partial^2 f}{\partial y\partial x } =\frac{\partial }{\partial y} \left ( \frac{2x }{x^{2}+y} \right )$

$=2x \frac{\partial }{\partial y} \left ( x^{2}+y \right )^{-1}$

$=2x (-1 )\left ( x^{2}+y \right )^{-2}$

$= \frac{ - 2x }{(x^{2}+y)^{2}}$

$\frac{\partial f }{\partial y} =\frac{\partial }{\partial y} \ln ( x^{2}+y)$

$=\frac{\frac{\partial }{\partial y} ( x^{2}+y)}{x^{2}+y}$

$=\frac{1 }{x^{2}+y}$

$\frac{\partial ^2 f}{\partial y^2} = \frac{\partial }{\partial y} \left ( \frac{1 }{x^{2}+y} \right )$

$= \frac{\partial }{\partial y} \left ( x^{2}+y \right )^{-1}$

$=-1\cdot \frac{\partial }{\partial y} \left ( x^{2}+y \right ) \left ( x^{2}+y \right )^{-2}$

$= (-1) \left ( x^{2}+y \right )^{-2}$

$= \frac{ - 1}{(x^{2}+y)^{2}}$

The Hessian of is

$\begin{bmatrix} \frac{-2x^{2}+2y}{(x^{2}+y)^{2}} & \frac{ - 2x }{(x^{2}+y)^{2}} \\ \frac{ - 2x }{(x^{2}+y)^{2}} & \frac{ - 1}{(x^{2}+y)^{2}} \end{bmatrix}$ ,

which can be rewritten as

$\frac{ - 1}{(x^{2}+y)^{2}} \begin{bmatrix} 2x^{2}-2y & 2x \\ 2x & 1 \end{bmatrix}$

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Example Question #33 : Matrix Calculus

Give the Hessian matrix for the function $f(x, y) = \sin (xy)+ \cos (xy)$ .

Possible Answers:

$\begin{bmatrix} y ^{2}( \sin(xy)+\cos (xy )) & \cos (xy ) - \sin(xy ) -xy (\sin(xy)+\cos(xy )) \\ \cos (xy ) - \sin(xy ) -xy (\sin(xy)+\cos(xy )) & x ^{2}( \sin(xy)+\cos (xy )) \end{bmatrix}$

$\begin{bmatrix}- y ^{2}( \sin(xy)+\cos (xy )) & \cos (xy ) - \sin(xy ) -xy (\sin(xy)+\cos(xy )) \\ \cos (xy ) - \sin(xy ) -xy (\sin(xy)+\cos(xy )) & -x ^{2}( \sin(xy)+\cos (xy )) \end{bmatrix}$

$\begin{bmatrix}- y ^{2}( \sin(xy)+\cos (xy )) & xy (\sin(xy)+\cos(xy )) \\ xy (\sin(xy)+\cos(xy )) & -x ^{2}( \sin(xy)+\cos (xy )) \end{bmatrix}$

$\begin{bmatrix}- y ^{2}( \sin(xy)+\cos (xy )) & \cos (xy ) - \sin(xy ) \\ \cos (xy ) - \sin(xy ) & -x ^{2}( \sin(xy)+\cos (xy )) \end{bmatrix}$

$\begin{bmatrix} y ^{2}( \sin(xy)+\cos (xy )) & xy (\sin(xy)+\cos(xy )) \\ xy (\sin(xy)+\cos(xy )) & x ^{2}( \sin(xy)+\cos (xy )) \end{bmatrix}$

Correct answer:

$\begin{bmatrix}- y ^{2}( \sin(xy)+\cos (xy )) & \cos (xy ) - \sin(xy ) -xy (\sin(xy)+\cos(xy )) \\ \cos (xy ) - \sin(xy ) -xy (\sin(xy)+\cos(xy )) & -x ^{2}( \sin(xy)+\cos (xy )) \end{bmatrix}$

Explanation:

The Hessian matrix of a function f(x,y) is the matrix of partial second derivatives

$\begin{bmatrix} \frac{\partial^2 f}{\partial x^2} & \frac{\partial^2 f}{\partial x\partial y } \\ \\ \frac{\partial^2 f}{\partial y\partial x } & \frac{\partial^2 f}{ \partial y^2 } \end{bmatrix}$

Find each of these derivatives as follows:

$\frac{\partial f}{\partial x} = \frac{\partial }{\partial x} ( \sin (xy)+\cos (xy ))$

$= \frac{\partial }{\partial x}(xy) \cdot ( \cos (xy)-\sin (xy ))$

$=y( \cos (xy)-\sin (xy ))$

$\frac{\partial^2 f}{\partial x^2}= \frac{\partial }{\partial x} (y( \cos (xy)-\sin (xy )))$

$= y \cdot \frac{\partial }{\partial x} ( \cos (xy)-\sin (xy ))$

$= y \cdot \frac{\partial }{\partial x}(xy ) \cdot ( -\sin(xy)-\cos (xy ))$

$= y \cdot y \cdot ( -\sin(xy)-\cos (xy ))$

$=- y ^{2}( \sin(xy)+\cos (xy ))$

$\frac{\partial f^{2}}{\partial y\partial x } = \frac{\partial f^{2}}{\partial x\partial y } =\frac{\partial }{\partial y} (y( \cos (xy)-\sin (xy )))$

$=y \cdot \frac{\partial }{\partial y} ( ( \cos (xy)-\sin (xy ))) + \frac{\partial }{\partial y} (y) ( \cos (xy)-\sin (xy ))$

$=y \cdot \frac{\partial }{\partial y}(xy)\cdot ( -\sin(xy)-\cos (xy )) + 1 ( \cos (xy)-\sin (xy ))$

$=y \cdot x\cdot ( -\sin(xy)-\cos (xy )) + ( \cos (xy)-\sin (xy ))$

$= \cos (xy ) - \sin(xy ) -xy (\sin(xy)+\cos(xy ))$

$\frac{\partial f}{\partial y} = \frac{\partial }{\partial y} ( \sin (xy)+\cos (xy ))$

$= \frac{\partial }{\partial y}(xy) \cdot ( \cos (xy)-\sin (xy ))$

$=x( \cos (xy)-\sin (xy ))$

$\frac{\partial^2 f}{\partial y^2}= \frac{\partial }{\partial y} (x( \cos (xy)-\sin (xy )))$

$= x \cdot \frac{\partial }{\partial y} ( \cos (xy)-\sin (xy ))$

$= x \cdot \frac{\partial }{\partial y}(xy ) \cdot ( -\sin(xy)-\cos (xy ))$

$= x \cdot x \cdot ( -\sin(xy)-\cos (xy ))$

$=- x ^{2}( \sin(xy)+\cos (xy ))$

The Hessian matrix is

$\begin{bmatrix}- y ^{2}( \sin(xy)+\cos (xy )) & \cos (xy ) - \sin(xy ) -xy (\sin(xy)+\cos(xy )) \\ \cos (xy ) - \sin(xy ) -xy (\sin(xy)+\cos(xy )) & -x ^{2}( \sin(xy)+\cos (xy )) \end{bmatrix}$

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Example Question #34 : Matrix Calculus

Give the Hessian matrix for the function $f(x,y) = xy^{2}e^{x}$ .

Possible Answers:

$e^{x}y^{2}\begin{bmatrix} x + 2 & 2 x + 1 \\ 2x+1 &2x \end{bmatrix}$

$e^{x}y^{2}\begin{bmatrix} x + 2 & 2 x -2 \\ 2x-2 &2x \end{bmatrix}$

$e^{x}\begin{bmatrix} x y^{2}- 2 y^{2} & 2 x y - 2y \\ 2xy-2y &2x \end{bmatrix}$

$e^{x}y \begin{bmatrix} x y - 2 y & 2 x - 2 \\ 2x -2 &2x \end{bmatrix}$

$e^{x}\begin{bmatrix} x y^{2}+ 2 y^{2} & 2 x y +2 y \\ 2xy+2y &2x \end{bmatrix}$

Correct answer:

$e^{x}\begin{bmatrix} x y^{2}+ 2 y^{2} & 2 x y +2 y \\ 2xy+2y &2x \end{bmatrix}$

Explanation:

The Hessian matrix of a function f(x,y) is the matrix of partial second derivatives

$\begin{bmatrix} \frac{\partial^2 f}{\partial x^2} & \frac{\partial^2 f}{\partial x\partial y } \\ \\ \frac{\partial^2 f}{\partial y\partial x } & \frac{\partial^2 f}{ \partial y^2 } \end{bmatrix}$

Find each of these derivatives as follows:

$\frac{\partial f}{\partial x} = \frac{\partial }{\partial x}(xy^{2}e^{x})$

$=y^{2} \frac{\partial }{\partial x}(xe^{x})$

$=y^{2}\left ( x \frac{\partial }{\partial x} e^{x}+ e^{x} \frac{\partial }{\partial x}x \right )$

$=y^{2}\left ( x \cdot e^{x}+ e^{x} \cdot 1 \right )$

$=y^{2}\left ( x e^{x}+ e^{x} \right )$

$\frac{ \partial^2 f}{\partial x^2}= \frac{\partial }{\partial x}(y^{2}\left ( x e^{x}+ e^{x} \right ))$

$=y^{2} \frac{\partial }{\partial x} ( x e^{x}+ e^{x} )$

$=y^{2} \left (\frac{\partial }{\partial x} ( x e^{x})+ \frac{\partial }{\partial x} (e^{x} ) \right )$

$=y^{2} ( x e^{x}+ e^{x} + e^{x} )$

$=y^{2} ( x e^{x}+ 2 e^{x} )$

$=( x + 2 )e^{x} y^{2}$

$\frac{\partial^2 f}{\partial x\partial y }= \frac{\partial^2 f}{\partial y\partial x } = \frac{\partial }{\partial y}(y^{2}\left ( x e^{x}+ e^{x} \right ))$

$=\left ( x e^{x}+ e^{x} \right ) \frac{\partial }{\partial y}(y^{2})$

$= \left ( x e^{x}+ e^{x} \right ) \cdot 2y$

$=2 \left ( x + 1 \right )e^{x}y$

$\frac{\partial f}{\partial y} = \frac{\partial }{\partial y}(xy^{2}e^{x})$

$=xe^{x} \frac{\partial }{\partial y}y^{2}$

$=xe^{x} \cdot 2y$

$=2xe^{x} y$

$\frac{\partial f^{2}}{\partial^{2} y} = \frac{\partial }{\partial y}(2xe^{x} y)$

$=2xe^{x} \frac{\partial }{\partial y}y$

$=2xe^{x} \cdot 1$

$=2xe^{x }$

The Hessian matrix is

$\begin{bmatrix} ( x + 2 )e^{x} y^{2} & 2 \left ( x + 1 \right )e^{x}y \\ 2 \left ( x + 1 \right )e^{x}y &2xe^{x } \end{bmatrix}$ ,

which can be rewritten, after a little algebra, as

$e^{x}\begin{bmatrix} x y^{2}+ 2 y^{2} & 2 x y +2 y \\ 2xy+2y &2x \end{bmatrix}$ .

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Example Question #11 : The Hessian

Give the Hessian matrix of the function $f(x,y) = \cos x \sin y$ .

Possible Answers:

$\begin{bmatrix} \cos x \sin y & - \sin x \cos y \\ -\sin x \cos y & \cos x \sin y \end{bmatrix}$

$\begin{bmatrix} \cos x \sin y & 0 \\0 & \cos x \sin y \end{bmatrix}$

$\begin{bmatrix} -\cos x \sin y &0 \\0& -\cos x \sin y \end{bmatrix}$

$\begin{bmatrix} -\cos x \sin y & - \sin x \cos y \\ - \sin x \cos y & -\cos x \sin y \end{bmatrix}$

$\begin{bmatrix} -\cos x \sin y & \sin x \cos y \\ \sin x \cos y & -\cos x \sin y \end{bmatrix}$

Correct answer:

$\begin{bmatrix} -\cos x \sin y & - \sin x \cos y \\ - \sin x \cos y & -\cos x \sin y \end{bmatrix}$

Explanation:

The Hessian matrix of a function f(x,y ) is the matrix of partial second derivatives:

$\begin{bmatrix} \frac{\partial^2 f}{\partial x^2} & \frac{\partial^2 f}{\partial x \partial y} \\ \\ \frac{\partial^2 f}{\partial y \partial x } & \frac{\partial^2 f}{\partial y^2} \end{bmatrix}$ .

To get the entries, find these derivatives as follows:

$\frac{\partial^2 f}{\partial x^2} = \frac{\partial^2 }{\partial x^2} ( \cos x \sin y)$

$=\sin y \frac{\partial^2 }{\partial x^2} ( \cos x )$

$=\sin y ( - \cos x )$

$= -\cos x \sin y$

$\frac{\partial^2 f}{\partial x \partial y}= \frac{\partial^2 f}{\partial y \partial x } = \frac{\partial }{\partial x} \frac{\partial }{\partial y} (\cos x \sin y)$

$= \frac{\partial }{\partial x}(\cos x \frac{\partial }{\partial y} ( \sin y))$

$= \frac{\partial }{\partial x}(\cos x \cos y)$

$= \cos y \frac{\partial }{\partial x}(\cos x )$

$= \cos y (- \sin x )$

$= - \sin x \cos y$

$\frac{\partial^2 f}{\partial y^2} = \frac{\partial^2 }{\partial y^2} ( \cos x \sin y)$

$=\cos x \frac{\partial^2 }{\partial y^2} ( \sin y)$

$=\cos x \frac{\partial }{\partial y } ( \cos y)$

$=\cos x ( - \sin y)$

$= -\cos x \sin y$

The Hessian matrix is $\begin{bmatrix} -\cos x \sin y & - \sin x \cos y \\ - \sin x \cos y & -\cos x \sin y \end{bmatrix}$ .

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Example Question #36 : Matrix Calculus

Give the Hessian matrix of the function $f(x,y) = \frac{x^{2}}{y^{2}}$ .

Possible Answers:

$\frac{1}{y^{4}}\begin{bmatrix}2y^{2} & - 4xy \\ - 4xy & 6x^{2} \end{bmatrix}$

$\frac{1}{y^{4}}\begin{bmatrix}2y^{2} & 4xy \\ 4xy & 6x^{2} \end{bmatrix}$

$\frac{1}{y^{4}}\begin{bmatrix}2y^{2} & 4xy \\ 4xy &- 6x^{2} \end{bmatrix}$

$\frac{1}{y^{4}}\begin{bmatrix}2y^{2} & - 4xy \\ - 4xy & - 6x^{2} \end{bmatrix}$

None of the other choices gives the correct response.

Correct answer:

$\frac{1}{y^{4}}\begin{bmatrix}2y^{2} & - 4xy \\ - 4xy & 6x^{2} \end{bmatrix}$

Explanation:

The Hessian matrix of a function f(x,y ) is the matrix of partial second derivatives:

$\begin{bmatrix} \frac{\partial^2 f}{\partial x^2} & \frac{\partial^2 f}{\partial x \partial y} \\ \\ \frac{\partial^2 f}{\partial y \partial x } & \frac{\partial^2 f}{\partial y^2} \end{bmatrix}$ .

Find the partial derivatives as follows:

$\frac{\partial^2 f}{\partial x^2} = \frac{\partial^2 }{\partial x^2} \left ( \frac{x^{2}}{y^{2}} \right )$

$\frac{\partial^2 f}{\partial x^2} =\frac{1}{y^{2}} \frac{\partial^2 }{\partial x^2} \left ( x^{2} \right )$

$= \frac{1}{y^{2}} \frac{\partial }{\partial x } \left (2x \right )$

$= \frac{1}{y^{2}} \cdot 2$

$= \frac{2 }{y^{2}}$

$\frac{\partial^2 f}{\partial x \partial y } = \frac{\partial^2 f}{\partial y \partial x } = \frac{\partial^2 f}{\partial y \partial x } \left ( \frac{x^{2}}{y^{2}} \right )$

$= \frac{\partial }{\partial y } \left (\frac{1}{y^{2}} \cdot \frac{\partial }{\partial x } \left (x^{2}\right ) \right )$

$= \frac{\partial }{\partial y } \left (\frac{1}{y^{2}} \cdot2x \right )$

$=2x \cdot \frac{\partial }{\partial y } \left (\frac{1}{y^{2}} \right )$

$=2x \cdot \frac{-2}{y^{3}}$

$= - \frac{4x}{y^{3}}$

$\frac{\partial^2 f}{\partial y^2} = \frac{\partial^2 }{\partial y^2} \left ( \frac{x^{2}}{y^{2}} \right )$

$= x^{2} \cdot \frac{\partial^2 }{\partial y^2} \left ( \frac{1}{y^{2}} \right )$

$= x^{2} \cdot \frac{\partial }{\partial y } \left ( \frac{-2}{y^{3}} \right )$

$= x^{2} \cdot \frac{6}{y^{4}}$

$= \frac{6x^{2} }{y^{4}}$

The Hessian matrix is

$\begin{bmatrix}\frac{2 }{y^{2}} & - \frac{4x}{y^{3}} \\ \\ - \frac{4x}{y^{3}}& \frac{6x^{2} }{y^{4}}\end{bmatrix}$ ,

$\frac{1}{y^{4}}\begin{bmatrix}2y^{2} & - 4xy \\ - 4xy & 6x^{2} \end{bmatrix}$ .

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Example Question #21 : The Hessian

Give the Hessian matrix of the function $f(x) =( x+y)^{5}$ .

Possible Answers:

$\begin{bmatrix} 20 ( x+y)^{3} &20 ( x+y)^{3} \\ 30( x+y)^{3} &30( x+y)^{3} \end{bmatrix}$

$\begin{bmatrix} 30 ( x+y)^{3} &20( x+y)^{3} \\ 20 ( x+y)^{3} &30( x+y)^{3} \end{bmatrix}$

$\begin{bmatrix} 20 ( x+y)^{3} &20 ( x+y)^{3} \\ 20 ( x+y)^{3} &20 ( x+y)^{3} \end{bmatrix}$

$\begin{bmatrix} 60 ( x+y)^{3} &20( x+y)^{3} \\ 20 ( x+y)^{3} &60( x+y)^{3} \end{bmatrix}$

$\begin{bmatrix} 20 ( x+y)^{3} &10( x+y)^{3} \\ 10( x+y)^{3} &20 ( x+y)^{3} \end{bmatrix}$

Correct answer:

$\begin{bmatrix} 20 ( x+y)^{3} &20 ( x+y)^{3} \\ 20 ( x+y)^{3} &20 ( x+y)^{3} \end{bmatrix}$

Explanation:

The Hessian matrix of a function f(x,y ) is the matrix of partial second derivatives:

$\begin{bmatrix} \frac{\partial^2 f}{\partial x^2} & \frac{\partial^2 f}{\partial x \partial y} \\ \\ \frac{\partial^2 f}{\partial y \partial x } & \frac{\partial^2 f}{\partial y^2} \end{bmatrix}$ .

To get the entries in the Hessian matrix, find these derivatives as follows:

$\frac{\partial^2 f}{\partial x^2} =\frac{\partial^2 }{\partial x^2} \left [( x+y)^{5} \right ]$

$=\frac{\partial }{\partial x } \left [5 ( x+y)^{4} \right ]$

$= 5 \cdot 4 ( x+y)^{3}$

$= 20 ( x+y)^{3}$

By symmetry,

$\frac{\partial^2 f}{\partial y^2} =20 ( x+y)^{3}$

$\frac{\partial^2 f}{\partial x \partial y}= \frac{\partial^2 f}{\partial y \partial x } =\frac{\partial^2 }{\partial y \partial x } \left [( x+y)^{5} \right ]$

$=\frac{\partial }{\partial y } \left [5 ( x+y)^{4} \right ]$

$= 5 \cdot 4 ( x+y)^{3}$

$= 20 ( x+y)^{3}$

The Hessian matrix is

$\begin{bmatrix} 20 ( x+y)^{3} &20 ( x+y)^{3} \\ 20 ( x+y)^{3} &20 ( x+y)^{3} \end{bmatrix}$ .

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Example Question #22 : The Hessian

f(x,y) is a continuous function such that $f_{x} (0,0) = f_{y}(0,0)= 0$ .

The Hessian matrix for , evaluated at (0,0) , is

$H(f) = \begin{bmatrix} a & 3 \\ 3 & 4 \end{bmatrix}$

From the set $\left \{ 1,2,3,4,5\right \}$ , which value(s) can be assigned to so that the graph of has a saddle point at (0,0) ?

Possible Answers:

$\left \{ 1,2 \right \}$

$\left \{ 3,4,5\right \}$

$\left \{ 4,5\right \}$

$\left \{ 1,2,3,4,5\right \}$

$\left \{ 1,2,3\right \}$

Correct answer:

$\left \{ 1,2 \right \}$

Explanation:

The graph of has a saddle point at (0,0) if and only

$\det H(f) < 0$

when evaluated at this point.

Calculate the determinant of the Hessian at this point in terms of by subtracting the upper-right to lower-left product by from the upper-left to lower-right product; set this less than 0 and solve for .

$\begin{vmatrix} a & 3 \\ 3 & 4 \end{vmatrix} < 0$

a(4) - 3(3) < 0

4a- 9 < 0

4a< 9

$a< \frac{9}{4}$

Therefore, the graph of has a saddle point at (0,0) if $a< \frac{9}{4}$ . The correct choice is therefore $\left \{ 1,2 \right \}$ .

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Example Question #31 : Matrix Calculus

Define $f(x,y)= x^{2} + \sin y$ .

Use the Hessian matrix H(f) , if applicable, to answer this question:

Does the graph of have a local maximum, a local minimum, or a saddle point at (0,0) ?

Possible Answers:

The graph of has a critical point at (0,0) , but the Hessian matrix test is inconclusive.

The graph of has a local maximum at (0,0) .

The graph of does not have a critical point at (0,0) .

The graph of has a local minimum at (0,0) .

The graph of has a saddle point at (0,0) .

Correct answer:

The graph of does not have a critical point at (0,0) .

Explanation:

First, it must be established that the graph of has a critical point at (0,0) ; this holds if $f_{x}(0,0) = f_{y}(0,0) = 0$ , so the first partial derivatives of must be evaluated at (0,0) :

$f(x,y)= x^{2} + \sin y$

$f_{x}(x,y)= 2x$

$f_{x}(0,0)= 2(0) = 0$

$f_{y}(x,y)= \cos y$

$f_{y}(0,0)= \cos 0 = 1$

Since $f_{y}(0,0) \ne 0$ , the graph of does not have a critical point at (0,0) .

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Example Question #21 : The Hessian

Define $f(x,y)= x^{2} + \cos y$ .

Use the Hessian matrix H(f) , if applicable, to answer this question:

Does the graph of have a local maximum, a local minimum, or a saddle point at (0,0) ?

Possible Answers:

The graph of has a critical point at (0,0) , but the Hessian matrix test is inconclusive.

The graph of has a local minimum at (0,0) .

The graph of has a saddle point at (0,0) .

The graph of does not have a critical point at (0,0) .

Correct answer:

The graph of has a saddle point at (0,0) .

Explanation:

First, it must be established that the graph of has a critical point at (0,0) ; this holds if $f_{x}(0,0) = f_{y}(0,0) = 0$ , so the first partial derivatives of must be evaluated at (0,0) :

$f(x,y)= x^{2} + \cos y$

$f_{x}(x,y)= 2x$

$f_{x}(0,0)= 2(0) = 0$

$f_{y}(x,y)= - \sin y$

$f_{y}(0,0)= -\sin 0 = 0$

The graph of has a critical point at (0,0) , so the Hessian matrix test applies.

The Hessian matrix H(f) is the matrix of partial second derivatives

$\begin{bmatrix} f_{xx} & f_{xy} \\ f_{yx} & f_{yy} \end{bmatrix}$ ,

the determinant of which can be used to determine whether a critical point of is a local maximum, a local minimum, or a saddle point. Find the partial second derivatives of :

$f_{x}(x,y)= 2x$

$f_{xx}(x,y)= 2$

$f_{xy}(x,y)=0$

$f_{y}(x,y)= - \sin y$

$f_{yx}(x,y)= 0$

$f_{yy}(x,y)= - \cos y$

$f_{xx}, f_{xy}, f_{yx}$ all are constant functions.

$f_{yy}(x,y)= - \cos y$ ,

$f_{yy}(0,0)= - \cos 0 = -1$

The Hessian matrix, evaluated at (0,0) , is

$\begin{bmatrix} 2 & 0 \\ 0 & -1 \end{bmatrix}$ .

Its determinant is the upper-left to lower-right product minus the upper-right to lower-left product;

$\begin{vmatrix} 2 & 0 \\ 0 & -1 \end{vmatrix} = 2(-1) - 0(0) = -2$

The determinant of the Hessian is negative, so the graph of has a saddle point at (0,0) .

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