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Linear Algebra : Matrix Calculus

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Example Questions

Example Question #1 : The Hessian

$\begin{align*}&\text{Determine the Hessian, }H\text{, of the function}\\&f(x,y,z)=6sin(6z)e^{(3x)}e^{(3y)}\end{align*}$

Possible Answers:

$\begin{bmatrix}108cos(6z)e^{(3x)}e^{(3y)}&54sin(6z)e^{(3x)}e^{(3y)}&54sin(6z)e^{(3x)}e^{(3y)}\\108cos(6z)e^{(3x)}e^{(3y)}&54sin(6z)e^{(3x)}e^{(3y)}&54sin(6z)e^{(3x)}e^{(3y)}\\-216sin(6z)e^{(3x)}e^{(3y)}&108cos(6z)e^{(3x)}e^{(3y)}&108cos(6z)e^{(3x)}e^{(3y)}\end{bmatrix}$

$\begin{bmatrix}54sin(6z)e^{(3x)}e^{(3y)}&54sin(6z)e^{(3x)}e^{(3y)}&108cos(6z)e^{(3x)}e^{(3y)}\\54sin(6z)e^{(3x)}e^{(3y)}&54sin(6z)e^{(3x)}e^{(3y)}&108cos(6z)e^{(3x)}e^{(3y)}\\108cos(6z)e^{(3x)}e^{(3y)}&108cos(6z)e^{(3x)}e^{(3y)}&-216sin(6z)e^{(3x)}e^{(3y)}\end{bmatrix}$

$\begin{bmatrix}108cos(6z)e^{(3x)}e^{(3y)}&108cos(6z)e^{(3x)}e^{(3y)}&-216sin(6z)e^{(3x)}e^{(3y)}\\54sin(6z)e^{(3x)}e^{(3y)}&54sin(6z)e^{(3x)}e^{(3y)}&108cos(6z)e^{(3x)}e^{(3y)}\\54sin(6z)e^{(3x)}e^{(3y)}&54sin(6z)e^{(3x)}e^{(3y)}&108cos(6z)e^{(3x)}e^{(3y)}\end{bmatrix}$

$\begin{bmatrix}54sin(6z)e^{(3x)}e^{(3y)}&54sin(6z)e^{(3x)}e^{(3y)}&108cos(6z)e^{(3x)}e^{(3y)}\\108cos(6z)e^{(3x)}e^{(3y)}&54sin(6z)e^{(3x)}e^{(3y)}&108cos(6z)e^{(3x)}e^{(3y)}\\108cos(6z)e^{(3x)}e^{(3y)}&108cos(6z)e^{(3x)}e^{(3y)}&-216sin(6z)e^{(3x)}e^{(3y)}\end{bmatrix}$

Correct answer:

Explanation:

$\begin{align*}&\text{The Hessian of a function is a matrix of second derivatives taken with}\\&\text{respect to its constituent variables. The form is as follows:}\\&\begin{bmatrix}\frac{d^2f}{d_{x_1}^2}&...&\frac{d^2f}{d_{x_1}d_{x_n}}\\...&...&...\\\frac{d^2f}{d_{x_n}d_{x_1}}&...&\frac{d^2f}{d_{x_n}^2}\end{bmatrix}\\&\text{Considering our function: }f(x,y,z)=6sin(6z)e^{(3x)}e^{(3y)}\\&\text{And utilizing derivative rules:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[e^u]=e^udu\\&d[sin(u)]=cos(u)du\\&Hf=\begin{bmatrix}54sin(6z)e^{(3x)}e^{(3y)}&54sin(6z)e^{(3x)}e^{(3y)}&108cos(6z)e^{(3x)}e^{(3y)}\\54sin(6z)e^{(3x)}e^{(3y)}&54sin(6z)e^{(3x)}e^{(3y)}&108cos(6z)e^{(3x)}e^{(3y)}\\108cos(6z)e^{(3x)}e^{(3y)}&108cos(6z)e^{(3x)}e^{(3y)}&-216sin(6z)e^{(3x)}e^{(3y)}\end{bmatrix}\end{align*}$

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Example Question #4 : The Hessian

$\begin{align*}&\text{Find the matrix }Hf \text{ for }f(x,y)=- 2\cdot 4^{(6y)} - 5y^{2}ln(2x)\end{align*}$

Possible Answers:

$\begin{bmatrix}\frac{(5y^{2})}{x^{2}}&-\frac{(10y)}{x}\\-\frac{(10y)}{x}&- 10ln(2x) - 72\cdot 4^{(6y)}ln(4)^{2}\end{bmatrix}$

$\begin{bmatrix}-\frac{(10y)}{x}&\frac{(5y^{2})}{x^{2}}\\- 10ln(2x) - 72\cdot 4^{(6y)}ln(4)^{2}&-\frac{(10y)}{x}\end{bmatrix}$

$\begin{bmatrix}-\frac{(10y)}{x}&- 10ln(2x) - 72\cdot 4^{(6y)}ln(4)^{2}\\\frac{(5y^{2})}{x^{2}}&-\frac{(10y)}{x}\end{bmatrix}$

$\begin{bmatrix}- 10ln(2x) - 72\cdot 4^{(6y)}ln(4)^{2}&-\frac{(10y)}{x}\\-\frac{(10y)}{x}&\frac{(5y^{2})}{x^{2}}\end{bmatrix}$

Correct answer:

$\begin{bmatrix}\frac{(5y^{2})}{x^{2}}&-\frac{(10y)}{x}\\-\frac{(10y)}{x}&- 10ln(2x) - 72\cdot 4^{(6y)}ln(4)^{2}\end{bmatrix}$

Explanation:

$\begin{align*}&\text{In asking for H, the question asks for the Hessian.}\\&\text{The Hessian of a function is a matrix of second derivatives taken with}\\&\text{respect to its constituent variables. The form is as follows:}\\&\begin{bmatrix}\frac{d^2f}{d_{x_1}^2}&...&\frac{d^2f}{d_{x_1}d_{x_n}}\\...&...&...\\\frac{d^2f}{d_{x_n}d_{x_1}}&...&\frac{d^2f}{d_{x_n}^2}\end{bmatrix}\\&\text{Considering our function: }f(x,y)=- 2\cdot 4^{(6y)} - 5y^{2}ln(2x)\\&\text{And utilizing derivative rules:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[ln(u)]=\frac{du}{u}\\&d[cos(u)]=-sin(u)du\\&d[tan(u)]=\frac{du}{cos^2(u)}=(tan^2(u)+1)du\\&d[a^u]=a^uduln(a)\\&Hf=\begin{bmatrix}\frac{(5y^{2})}{x^{2}}&-\frac{(10y)}{x}\\-\frac{(10y)}{x}&- 10ln(2x) - 72\cdot 4^{(6y)}ln(4)^{2}\end{bmatrix}\end{align*}$

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Example Question #5 : The Hessian

$\begin{align*}&\text{Determine the Hessian, }H\text{, of the function}\\&f(x,y,z)=3\cdot 4^yz^{2}e^{(6x)}\end{align*}$

Possible Answers:

$\begin{bmatrix}108\cdot 4^yz^{2}e^{(6x)}&18\cdot 4^yz^{2}e^{(6x)}ln(4)&36\cdot 4^yze^{(6x)}\\18\cdot 4^yz^{2}e^{(6x)}ln(4)&3\cdot 4^yz^{2}e^{(6x)}ln(4)^{2}&6\cdot 4^yze^{(6x)}ln(4)\\36\cdot 4^yze^{(6x)}&6\cdot 4^yze^{(6x)}ln(4)&6\cdot 4^ye^{(6x)}\end{bmatrix}$

$\begin{bmatrix}3\cdot 4^yz^{2}e^{(6x)}ln(4)^{2}&18\cdot 4^yz^{2}e^{(6x)}ln(4)&6\cdot 4^yze^{(6x)}ln(4)\\18\cdot 4^yz^{2}e^{(6x)}ln(4)&108\cdot 4^yz^{2}e^{(6x)}&36\cdot 4^yze^{(6x)}\\6\cdot 4^yze^{(6x)}ln(4)&36\cdot 4^yze^{(6x)}&6\cdot 4^ye^{(6x)}\end{bmatrix}$

$\begin{bmatrix}36\cdot 4^yze^{(6x)}&6\cdot 4^yze^{(6x)}ln(4)&6\cdot 4^ye^{(6x)}\\18\cdot 4^yz^{2}e^{(6x)}ln(4)&3\cdot 4^yz^{2}e^{(6x)}ln(4)^{2}&6\cdot 4^yze^{(6x)}ln(4)\\108\cdot 4^yz^{2}e^{(6x)}&18\cdot 4^yz^{2}e^{(6x)}ln(4)&36\cdot 4^yze^{(6x)}\end{bmatrix}$

$\begin{bmatrix}36\cdot 4^yze^{(6x)}&18\cdot 4^yz^{2}e^{(6x)}ln(4)&108\cdot 4^yz^{2}e^{(6x)}\\6\cdot 4^yze^{(6x)}ln(4)&3\cdot 4^yz^{2}e^{(6x)}ln(4)^{2}&18\cdot 4^yz^{2}e^{(6x)}ln(4)\\6\cdot 4^ye^{(6x)}&6\cdot 4^yze^{(6x)}ln(4)&36\cdot 4^yze^{(6x)}\end{bmatrix}$

Correct answer:

Explanation:

$\begin{align*}&\text{The Hessian of a function is a matrix of second derivatives taken with}\\&\text{respect to its constituent variables. The form is as follows:}\\&\begin{bmatrix}\frac{d^2f}{d_{x_1}^2}&...&\frac{d^2f}{d_{x_1}d_{x_n}}\\...&...&...\\\frac{d^2f}{d_{x_n}d_{x_1}}&...&\frac{d^2f}{d_{x_n}^2}\end{bmatrix}\\&\text{Considering our function: }f(x,y,z)=3\cdot 4^yz^{2}e^{(6x)}\\&\text{And utilizing derivative rules:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[e^u]=e^udu\\&d[a^u]=a^uduln(a)\\&Hf=\begin{bmatrix}108\cdot 4^yz^{2}e^{(6x)}&18\cdot 4^yz^{2}e^{(6x)}ln(4)&36\cdot 4^yze^{(6x)}\\18\cdot 4^yz^{2}e^{(6x)}ln(4)&3\cdot 4^yz^{2}e^{(6x)}ln(4)^{2}&6\cdot 4^yze^{(6x)}ln(4)\\36\cdot 4^yze^{(6x)}&6\cdot 4^yze^{(6x)}ln(4)&6\cdot 4^ye^{(6x)}\end{bmatrix}\end{align*}$

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Example Question #1 : The Hessian

$\begin{align*}&\text{Determine the Hessian, }H\text{, of the function}\\&f(x,y)=20xy^{2}\end{align*}$

Possible Answers:

$\begin{bmatrix}40x&40y\\40y&0\end{bmatrix}$

$\begin{bmatrix}0&40y\\40y&40x\end{bmatrix}$

$\begin{bmatrix}40y&40x\\0&40y\end{bmatrix}$

$\begin{bmatrix}40y&0\\40x&40y\end{bmatrix}$

Correct answer:

$\begin{bmatrix}0&40y\\40y&40x\end{bmatrix}$

Explanation:

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Example Question #1 : The Hessian

$\begin{align*}&\text{Find the matrix }Hf \text{ for }f(x,y,z)=11\cdot 3^{(4y)}\cdot 4^{(3x)}z\end{align*}$

Possible Answers:

$\begin{bmatrix}33\cdot 3^{(4y)}\cdot 4^{(3x)}ln(4)&44\cdot 3^{(4y)}\cdot 4^{(3x)}ln(3)&0\\132\cdot 3^{(4y)}\cdot 4^{(3x)}zln(3)ln(4)&176\cdot 3^{(4y)}\cdot 4^{(3x)}zln(3)^{2}&44\cdot 3^{(4y)}\cdot 4^{(3x)}ln(3)\\99\cdot 3^{(4y)}\cdot 4^{(3x)}zln(4)^{2}&132\cdot 3^{(4y)}\cdot 4^{(3x)}zln(3)ln(4)&33\cdot 3^{(4y)}\cdot 4^{(3x)}ln(4)\end{bmatrix}$

$\begin{bmatrix}176\cdot 3^{(4y)}\cdot 4^{(3x)}zln(3)^{2}&132\cdot 3^{(4y)}\cdot 4^{(3x)}zln(3)ln(4)&44\cdot 3^{(4y)}\cdot 4^{(3x)}ln(3)\\132\cdot 3^{(4y)}\cdot 4^{(3x)}zln(3)ln(4)&99\cdot 3^{(4y)}\cdot 4^{(3x)}zln(4)^{2}&33\cdot 3^{(4y)}\cdot 4^{(3x)}ln(4)\\44\cdot 3^{(4y)}\cdot 4^{(3x)}ln(3)&33\cdot 3^{(4y)}\cdot 4^{(3x)}ln(4)&0\end{bmatrix}$

$\begin{bmatrix}33\cdot 3^{(4y)}\cdot 4^{(3x)}ln(4)&132\cdot 3^{(4y)}\cdot 4^{(3x)}zln(3)ln(4)&99\cdot 3^{(4y)}\cdot 4^{(3x)}zln(4)^{2}\\44\cdot 3^{(4y)}\cdot 4^{(3x)}ln(3)&176\cdot 3^{(4y)}\cdot 4^{(3x)}zln(3)^{2}&132\cdot 3^{(4y)}\cdot 4^{(3x)}zln(3)ln(4)\\0&44\cdot 3^{(4y)}\cdot 4^{(3x)}ln(3)&33\cdot 3^{(4y)}\cdot 4^{(3x)}ln(4)\end{bmatrix}$

$\begin{bmatrix}99\cdot 3^{(4y)}\cdot 4^{(3x)}zln(4)^{2}&132\cdot 3^{(4y)}\cdot 4^{(3x)}zln(3)ln(4)&33\cdot 3^{(4y)}\cdot 4^{(3x)}ln(4)\\132\cdot 3^{(4y)}\cdot 4^{(3x)}zln(3)ln(4)&176\cdot 3^{(4y)}\cdot 4^{(3x)}zln(3)^{2}&44\cdot 3^{(4y)}\cdot 4^{(3x)}ln(3)\\33\cdot 3^{(4y)}\cdot 4^{(3x)}ln(4)&44\cdot 3^{(4y)}\cdot 4^{(3x)}ln(3)&0\end{bmatrix}$

Correct answer:

Explanation:

$\begin{align*}&\text{In asking for H, the question asks for the Hessian.}\\&\text{The Hessian of a function is a matrix of second derivatives taken with}\\&\text{respect to its constituent variables. The form is as follows:}\\&\begin{bmatrix}\frac{d^2f}{d_{x_1}^2}&...&\frac{d^2f}{d_{x_1}d_{x_n}}\\...&...&...\\\frac{d^2f}{d_{x_n}d_{x_1}}&...&\frac{d^2f}{d_{x_n}^2}\end{bmatrix}\\&\text{Considering our function: }f(x,y,z)=11\cdot 3^{(4y)}\cdot 4^{(3x)}z\\&\text{And utilizing derivative rules:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[a^u]=a^uduln(a)\\&Hf=\begin{bmatrix}99\cdot 3^{(4y)}\cdot 4^{(3x)}zln(4)^{2}&132\cdot 3^{(4y)}\cdot 4^{(3x)}zln(3)ln(4)&33\cdot 3^{(4y)}\cdot 4^{(3x)}ln(4)\\132\cdot 3^{(4y)}\cdot 4^{(3x)}zln(3)ln(4)&176\cdot 3^{(4y)}\cdot 4^{(3x)}zln(3)^{2}&44\cdot 3^{(4y)}\cdot 4^{(3x)}ln(3)\\33\cdot 3^{(4y)}\cdot 4^{(3x)}ln(4)&44\cdot 3^{(4y)}\cdot 4^{(3x)}ln(3)&0\end{bmatrix}\end{align*}$

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Example Question #7 : The Hessian

$\begin{align*}&\text{Calculate the Hessian matrix of the function}\\&f(x,y,z)=14e^{(5x)}e^{(y)}e^{(z)}\end{align*}$

Possible Answers:

$\begin{bmatrix}14e^{(5x)}e^{(y)}e^{(z)}&70e^{(5x)}e^{(y)}e^{(z)}&14e^{(5x)}e^{(y)}e^{(z)}\\70e^{(5x)}e^{(y)}e^{(z)}&350e^{(5x)}e^{(y)}e^{(z)}&70e^{(5x)}e^{(y)}e^{(z)}\\14e^{(5x)}e^{(y)}e^{(z)}&70e^{(5x)}e^{(y)}e^{(z)}&14e^{(5x)}e^{(y)}e^{(z)}\end{bmatrix}$

$\begin{bmatrix}70e^{(5x)}e^{(y)}e^{(z)}&70e^{(5x)}e^{(y)}e^{(z)}&350e^{(5x)}e^{(y)}e^{(z)}\\14e^{(5x)}e^{(y)}e^{(z)}&14e^{(5x)}e^{(y)}e^{(z)}&70e^{(5x)}e^{(y)}e^{(z)}\\14e^{(5x)}e^{(y)}e^{(z)}&14e^{(5x)}e^{(y)}e^{(z)}&70e^{(5x)}e^{(y)}e^{(z)}\end{bmatrix}$

$\begin{bmatrix}70e^{(5x)}e^{(y)}e^{(z)}&14e^{(5x)}e^{(y)}e^{(z)}&14e^{(5x)}e^{(y)}e^{(z)}\\70e^{(5x)}e^{(y)}e^{(z)}&14e^{(5x)}e^{(y)}e^{(z)}&14e^{(5x)}e^{(y)}e^{(z)}\\350e^{(5x)}e^{(y)}e^{(z)}&70e^{(5x)}e^{(y)}e^{(z)}&70e^{(5x)}e^{(y)}e^{(z)}\end{bmatrix}$

$\begin{bmatrix}350e^{(5x)}e^{(y)}e^{(z)}&70e^{(5x)}e^{(y)}e^{(z)}&70e^{(5x)}e^{(y)}e^{(z)}\\70e^{(5x)}e^{(y)}e^{(z)}&14e^{(5x)}e^{(y)}e^{(z)}&14e^{(5x)}e^{(y)}e^{(z)}\\70e^{(5x)}e^{(y)}e^{(z)}&14e^{(5x)}e^{(y)}e^{(z)}&14e^{(5x)}e^{(y)}e^{(z)}\end{bmatrix}$

Correct answer:

Explanation:

$\begin{align*}&\text{The Hessian of a function is a matrix of second derivatives taken with}\\&\text{respect to its constituent variables. The form is as follows:}\\&\begin{bmatrix}\frac{d^2f}{d_{x_1}^2}&...&\frac{d^2f}{d_{x_1}d_{x_n}}\\...&...&...\\\frac{d^2f}{d_{x_n}d_{x_1}}&...&\frac{d^2f}{d_{x_n}^2}\end{bmatrix}\\&\text{Considering our function: }f(x,y,z)=14e^{(5x)}e^{(y)}e^{(z)}\\&\text{And utilizing derivative rules:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[e^u]=e^udu\\&Hf=\begin{bmatrix}350e^{(5x)}e^{(y)}e^{(z)}&70e^{(5x)}e^{(y)}e^{(z)}&70e^{(5x)}e^{(y)}e^{(z)}\\70e^{(5x)}e^{(y)}e^{(z)}&14e^{(5x)}e^{(y)}e^{(z)}&14e^{(5x)}e^{(y)}e^{(z)}\\70e^{(5x)}e^{(y)}e^{(z)}&14e^{(5x)}e^{(y)}e^{(z)}&14e^{(5x)}e^{(y)}e^{(z)}\end{bmatrix}\end{align*}$

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Example Question #21 : Linear Algebra

$\begin{align*}&\text{Determine the Hessian, }H\text{, of the function}\\&f(x,y,z)=14cos(6z)tan(6x)sin(y)\end{align*}$

Possible Answers:

$\begin{bmatrix}-84sin(6z)sin(y)\cdot (6tan(6x)^{2} + 6)&14cos(6z)cos(y)\cdot (6tan(6x)^{2} + 6)&168cos(6z)tan(6x)sin(y)\cdot (6tan(6x)^{2} + 6)\\-84tan(6x)sin(6z)cos(y)&-14cos(6z)tan(6x)sin(y)&14cos(6z)cos(y)\cdot (6tan(6x)^{2} + 6)\\-504cos(6z)tan(6x)sin(y)&-84tan(6x)sin(6z)cos(y)&-84sin(6z)sin(y)\cdot (6tan(6x)^{2} + 6)\end{bmatrix}$

$\begin{bmatrix}-84sin(6z)sin(y)\cdot (6tan(6x)^{2} + 6)&-84tan(6x)sin(6z)cos(y)&-504cos(6z)tan(6x)sin(y)\\14cos(6z)cos(y)\cdot (6tan(6x)^{2} + 6)&-14cos(6z)tan(6x)sin(y)&-84tan(6x)sin(6z)cos(y)\\168cos(6z)tan(6x)sin(y)\cdot (6tan(6x)^{2} + 6)&14cos(6z)cos(y)\cdot (6tan(6x)^{2} + 6)&-84sin(6z)sin(y)\cdot (6tan(6x)^{2} + 6)\end{bmatrix}$

$\begin{bmatrix}-14cos(6z)tan(6x)sin(y)&14cos(6z)cos(y)\cdot (6tan(6x)^{2} + 6)&-84tan(6x)sin(6z)cos(y)\\14cos(6z)cos(y)\cdot (6tan(6x)^{2} + 6)&168cos(6z)tan(6x)sin(y)\cdot (6tan(6x)^{2} + 6)&-84sin(6z)sin(y)\cdot (6tan(6x)^{2} + 6)\\-84tan(6x)sin(6z)cos(y)&-84sin(6z)sin(y)\cdot (6tan(6x)^{2} + 6)&-504cos(6z)tan(6x)sin(y)\end{bmatrix}$

$\begin{bmatrix}168cos(6z)tan(6x)sin(y)\cdot (6tan(6x)^{2} + 6)&14cos(6z)cos(y)\cdot (6tan(6x)^{2} + 6)&-84sin(6z)sin(y)\cdot (6tan(6x)^{2} + 6)\\14cos(6z)cos(y)\cdot (6tan(6x)^{2} + 6)&-14cos(6z)tan(6x)sin(y)&-84tan(6x)sin(6z)cos(y)\\-84sin(6z)sin(y)\cdot (6tan(6x)^{2} + 6)&-84tan(6x)sin(6z)cos(y)&-504cos(6z)tan(6x)sin(y)\end{bmatrix}$

Correct answer:

Explanation:

$\begin{align*}&\text{The Hessian of a function is a matrix of second derivatives taken with}\\&\text{respect to its constituent variables. The form is as follows:}\\&\begin{bmatrix}\frac{d^2f}{d_{x_1}^2}&...&\frac{d^2f}{d_{x_1}d_{x_n}}\\...&...&...\\\frac{d^2f}{d_{x_n}d_{x_1}}&...&\frac{d^2f}{d_{x_n}^2}\end{bmatrix}\\&\text{Considering our function: }f(x,y,z)=14cos(6z)tan(6x)sin(y)\\&\text{And utilizing derivative rules:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[tan(u)]=\frac{du}{cos^2(u)}=(tan^2(u)+1)du\\&d[sin(u)]=cos(u)du\\&d[cos(u)]=-sin(u)du\\&Hf=\begin{bmatrix}168cos(6z)tan(6x)sin(y)\cdot (6tan(6x)^{2} + 6)&14cos(6z)cos(y)\cdot (6tan(6x)^{2} + 6)&-84sin(6z)sin(y)\cdot (6tan(6x)^{2} + 6)\\14cos(6z)cos(y)\cdot (6tan(6x)^{2} + 6)&-14cos(6z)tan(6x)sin(y)&-84tan(6x)sin(6z)cos(y)\\-84sin(6z)sin(y)\cdot (6tan(6x)^{2} + 6)&-84tan(6x)sin(6z)cos(y)&-504cos(6z)tan(6x)sin(y)\end{bmatrix}\end{align*}$

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Example Question #21 : Matrix Calculus

$\begin{align*}&\text{Calculate the Hessian matrix of the function}\\&f(x,y)=10x^{2}ln(6y)\end{align*}$

Possible Answers:

$\begin{bmatrix}\frac{(20x)}{y}&-\frac{(10x^{2})}{y^{2}}\\20ln(6y)&\frac{(20x)}{y}\end{bmatrix}$

$\begin{bmatrix}-\frac{(10x^{2})}{y^{2}}&\frac{(20x)}{y}\\\frac{(20x)}{y}&20ln(6y)\end{bmatrix}$

$\begin{bmatrix}20ln(6y)&\frac{(20x)}{y}\\\frac{(20x)}{y}&-\frac{(10x^{2})}{y^{2}}\end{bmatrix}$

$\begin{bmatrix}\frac{(20x)}{y}&20ln(6y)\\-\frac{(10x^{2})}{y^{2}}&\frac{(20x)}{y}\end{bmatrix}$

Correct answer:

$\begin{bmatrix}20ln(6y)&\frac{(20x)}{y}\\\frac{(20x)}{y}&-\frac{(10x^{2})}{y^{2}}\end{bmatrix}$

Explanation:

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Example Question #11 : The Hessian

$\begin{align*}&\text{Calculate the Hessian matrix of the function}\\&f(x,y)=10ysin(6x)\end{align*}$

Possible Answers:

$\begin{bmatrix}60cos(6x)&-360ysin(6x)\\0&60cos(6x)\end{bmatrix}$

$\begin{bmatrix}-360ysin(6x)&60cos(6x)\\60cos(6x)&0\end{bmatrix}$

$\begin{bmatrix}60cos(6x)&0\\-360ysin(6x)&60cos(6x)\end{bmatrix}$

Correct answer:

$\begin{bmatrix}-360ysin(6x)&60cos(6x)\\60cos(6x)&0\end{bmatrix}$

Explanation:

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Example Question #21 : Linear Algebra

$\begin{align*}&\text{Find the matrix }Hf \text{ for }f(x,y)=-7cos(3y)ln(4x)\end{align*}$

Possible Answers:

$\begin{bmatrix}\frac{(21sin(3y))}{x}&\frac{(7cos(3y))}{x^{2}}\\63cos(3y)ln(4x)&\frac{(21sin(3y))}{x}\end{bmatrix}$

$\begin{bmatrix}63cos(3y)ln(4x)&\frac{(21sin(3y))}{x}\\\frac{(21sin(3y))}{x}&\frac{(7cos(3y))}{x^{2}}\end{bmatrix}$

$\begin{bmatrix}\frac{(7cos(3y))}{x^{2}}&\frac{(21sin(3y))}{x}\\\frac{(21sin(3y))}{x}&63cos(3y)ln(4x)\end{bmatrix}$

$\begin{bmatrix}\frac{(21sin(3y))}{x}&63cos(3y)ln(4x)\\\frac{(7cos(3y))}{x^{2}}&\frac{(21sin(3y))}{x}\end{bmatrix}$

Correct answer:

$\begin{bmatrix}\frac{(7cos(3y))}{x^{2}}&\frac{(21sin(3y))}{x}\\\frac{(21sin(3y))}{x}&63cos(3y)ln(4x)\end{bmatrix}$

Explanation:

$\begin{align*}&\text{In asking for H, the question asks for the Hessian.}\\&\text{The Hessian of a function is a matrix of second derivatives taken with}\\&\text{respect to its constituent variables. The form is as follows:}\\&\begin{bmatrix}\frac{d^2f}{d_{x_1}^2}&...&\frac{d^2f}{d_{x_1}d_{x_n}}\\...&...&...\\\frac{d^2f}{d_{x_n}d_{x_1}}&...&\frac{d^2f}{d_{x_n}^2}\end{bmatrix}\\&\text{Considering our function: }f(x,y)=-7cos(3y)ln(4x)\\&\text{And utilizing derivative rules:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[ln(u)]=\frac{du}{u}\\&d[cos(u)]=-sin(u)du\\&Hf=\begin{bmatrix}\frac{(7cos(3y))}{x^{2}}&\frac{(21sin(3y))}{x}\\\frac{(21sin(3y))}{x}&63cos(3y)ln(4x)\end{bmatrix}\end{align*}$

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Linear Algebra : Matrix Calculus

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