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Linear Algebra : Linear Algebra

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Example Questions

Example Question #1 : Matrix Calculus

What is the the gradient vector of the following function?

$\displaystyle f(x,y,z)=18xyz+50x^2z+15y^2$

Possible Answers:

$\nabla_x f(x)=\begin{bmatrix} 18xz+30y \\18yz+100xz\\ 18xy+50x^2 \end{bmatrix}$ $\displaystyle \nabla_x f(x)=\begin{bmatrix} 18xz+30y \\18yz+100xz\\ 18xy+50x^2 \end{bmatrix}$

$\nabla_x f(x)=\begin{bmatrix} 18yz+100xz\\ 18xz+30y \\ 18xy+50x^2 \end{bmatrix}$ $\displaystyle \nabla_x f(x)=\begin{bmatrix} 18yz+100xz\\ 18xz+30y \\ 18xy+50x^2 \end{bmatrix}$

$\nabla_x f(x)=\begin{bmatrix} 1 \\1 \\18yz+100xz\end{bmatrix}$ $\displaystyle \nabla_x f(x)=\begin{bmatrix} 1 \\1 \\18yz+100xz\end{bmatrix}$

$\nabla_x f(x)=\begin{bmatrix} 18xz+30y \\18xy+50x^2 \\18yz+100xz\end{bmatrix}$ $\displaystyle \nabla_x f(x)=\begin{bmatrix} 18xz+30y \\18xy+50x^2 \\18yz+100xz\end{bmatrix}$

$\nabla_x f(x)=\begin{bmatrix} 0 \\18xy+50x^2 \\0\end{bmatrix}$ $\displaystyle \nabla_x f(x)=\begin{bmatrix} 0 \\18xy+50x^2 \\0\end{bmatrix}$

Correct answer:

$\nabla_x f(x)=\begin{bmatrix} 18yz+100xz\\ 18xz+30y \\ 18xy+50x^2 \end{bmatrix}$ $\displaystyle \nabla_x f(x)=\begin{bmatrix} 18yz+100xz\\ 18xz+30y \\ 18xy+50x^2 \end{bmatrix}$

Explanation:

Recall that

$\nabla_x f(x)=\begin{bmatrix} \frac{\partial f}{\partial x_1}\\ \\ \frac{\partial f}{\partial x_2} \\ \\ \frac{\partial f}{\partial x_3} \\ \vdots \\ \frac{\partial f}{\partial x_n} \end{bmatrix}$ $\displaystyle \nabla_x f(x)=\begin{bmatrix} \frac{\partial f}{\partial x_1}\\ \\ \frac{\partial f}{\partial x_2} \\ \\ \frac{\partial f}{\partial x_3} \\ \vdots \\ \frac{\partial f}{\partial x_n} \end{bmatrix}$

All we need to do is calculate 3 partial derivatives, and put them into this form.

$\frac{\partial f}{\partial x}=18yz+100xz$ $\displaystyle \frac{\partial f}{\partial x}=18yz+100xz$

$\frac{\partial f}{\partial y}=18xz+30y$ $\displaystyle \frac{\partial f}{\partial y}=18xz+30y$

$\frac{\partial f}{\partial z}=18xy+50x^2$ $\displaystyle \frac{\partial f}{\partial z}=18xy+50x^2$

Put these into vector form to get

$\nabla_x f(x)=\begin{bmatrix} 18yz+100xz\\ 18xz+30y \\ 18xy+50x^2 \end{bmatrix}$ $\displaystyle \nabla_x f(x)=\begin{bmatrix} 18yz+100xz\\ 18xz+30y \\ 18xy+50x^2 \end{bmatrix}$

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Example Question #1 : Linear Algebra

Find the gradient vector of the following function.

$f(x,y)=x^2y^2+10\ln(xy)$ $\displaystyle f(x,y)=x^2y^2+10\ln(xy)$

Possible Answers:

$\begin{bmatrix} 1+\frac{10}{x}\\ \\ 1+\frac{10}{y} \end{bmatrix}$ $\displaystyle \begin{bmatrix} 1+\frac{10}{x}\\ \\ 1+\frac{10}{y} \end{bmatrix}$

$\begin{bmatrix} \frac{10}{x}\\ \\ \frac{10}{y} \end{bmatrix}$ $\displaystyle \begin{bmatrix} \frac{10}{x}\\ \\ \frac{10}{y} \end{bmatrix}$

$\begin{bmatrix} 2xy^2\\ \\ 2yx^2 \end{bmatrix}$ $\displaystyle \begin{bmatrix} 2xy^2\\ \\ 2yx^2 \end{bmatrix}$

$\begin{bmatrix} 2xy^2+\frac{10}{x}\\ \\ 2yx^2+\frac{10}{y} \end{bmatrix}$ $\displaystyle \begin{bmatrix} 2xy^2+\frac{10}{x}\\ \\ 2yx^2+\frac{10}{y} \end{bmatrix}$

Correct answer:

$\begin{bmatrix} 2xy^2+\frac{10}{x}\\ \\ 2yx^2+\frac{10}{y} \end{bmatrix}$ $\displaystyle \begin{bmatrix} 2xy^2+\frac{10}{x}\\ \\ 2yx^2+\frac{10}{y} \end{bmatrix}$

Explanation:

To find the gradient vector, we need to find the partial derivatives in respect to x and y.

$\frac{\partial f}{\partial x}=2xy^2+\frac{10}{x}$ $\displaystyle \frac{\partial f}{\partial x}=2xy^2+\frac{10}{x}$

$\frac{\partial f}{\partial y}=2yx^2+\frac{10}{y}$ $\displaystyle \frac{\partial f}{\partial y}=2yx^2+\frac{10}{y}$

Then our final answer looks like

$\begin{bmatrix} 2xy^2+\frac{10}{x}\\ \\ 2yx^2+\frac{10}{y} \end{bmatrix}$ $\displaystyle \begin{bmatrix} 2xy^2+\frac{10}{x}\\ \\ 2yx^2+\frac{10}{y} \end{bmatrix}$

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Example Question #1 : Matrix Calculus

$\begin{align*}&\text{Determine the gradient, }\nabla\text{, of the function}\\&f(x,y)=5xcos(y)\end{align*}$ $\displaystyle \begin{align*}&\text{Determine the gradient, }\nabla\text{, of the function}\\&f(x,y)=5xcos(y)\end{align*}$

Possible Answers:

$\begin{bmatrix}-5sin(y)\end{bmatrix}$ $\displaystyle \begin{bmatrix}-5sin(y)\end{bmatrix}$

$\begin{bmatrix}0&-5sin(y)\\-5sin(y)&-5xcos(y)\end{bmatrix}$ $\displaystyle \begin{bmatrix}0&-5sin(y)\\-5sin(y)&-5xcos(y)\end{bmatrix}$

$\begin{bmatrix}5cos(y)\\-5xsin(y)\end{bmatrix}$ $\displaystyle \begin{bmatrix}5cos(y)\\-5xsin(y)\end{bmatrix}$

$\begin{bmatrix}5cos(y) - 5xsin(y)\end{bmatrix}$ $\displaystyle \begin{bmatrix}5cos(y) - 5xsin(y)\end{bmatrix}$

Correct answer:

$\begin{bmatrix}5cos(y)\\-5xsin(y)\end{bmatrix}$ $\displaystyle \begin{bmatrix}5cos(y)\\-5xsin(y)\end{bmatrix}$

Explanation:

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Example Question #3 : Matrix Calculus

$\begin{align*}&\text{Determine the gradient, }\nabla\text{, of the function}\\&f(x,y)=15\cdot 4^{(2x)}ln(3y) - 15xy^{2}\end{align*}$ $\displaystyle \begin{align*}&\text{Determine the gradient, }\nabla\text{, of the function}\\&f(x,y)=15\cdot 4^{(2x)}ln(3y) - 15xy^{2}\end{align*}$

Possible Answers:

$\begin{bmatrix}30\cdot 4^{(2x)}ln(3y)ln(4) - 15y^{2}\\\frac{(15\cdot 4^{(2x)})}{y}- 30xy\end{bmatrix}$ $\displaystyle \begin{bmatrix}30\cdot 4^{(2x)}ln(3y)ln(4) - 15y^{2}\\\frac{(15\cdot 4^{(2x)})}{y}- 30xy\end{bmatrix}$

$\begin{bmatrix}\frac{(30\cdot 4^{(2x)}ln(4))}{y}- 30y\end{bmatrix}$ $\displaystyle \begin{bmatrix}\frac{(30\cdot 4^{(2x)}ln(4))}{y}- 30y\end{bmatrix}$

$\begin{bmatrix}60\cdot 4^{(2x)}ln(3y)ln(4)^{2}&\frac{(30\cdot 4^{(2x)}ln(4))}{y}- 30y\\\frac{(30\cdot 4^{(2x)}ln(4))}{y}- 30y&- 30x -\frac{ (15\cdot 4^{(2x)})}{y^{2}}\end{bmatrix}$ $\displaystyle \begin{bmatrix}60\cdot 4^{(2x)}ln(3y)ln(4)^{2}&\frac{(30\cdot 4^{(2x)}ln(4))}{y}- 30y\\\frac{(30\cdot 4^{(2x)}ln(4))}{y}- 30y&- 30x -\frac{ (15\cdot 4^{(2x)})}{y^{2}}\end{bmatrix}$

$\begin{bmatrix}\frac{(15\cdot 4^{(2x)})}{y}- 30xy - 15y^{2} + 30\cdot 4^{(2x)}ln(3y)ln(4)\end{bmatrix}$ $\displaystyle \begin{bmatrix}\frac{(15\cdot 4^{(2x)})}{y}- 30xy - 15y^{2} + 30\cdot 4^{(2x)}ln(3y)ln(4)\end{bmatrix}$

Correct answer:

Explanation:

$\begin{align*}&\text{The gradient of a function is a vector of first derivatives taken with}\\&\text{respect to its constituent variables. The form is as follows:}\\&\begin{bmatrix}\frac{d^f}{d_{x_1}}\\\frac{d^f}{d_{x_2}}\\...\\\frac{df}{d_{x_n}}\end{bmatrix}\\&\text{Considering our function: }f(x,y)=15\cdot 4^{(2x)}ln(3y) - 15xy^{2}\\&\text{And utilizing derivative rules:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[sin(u)]=cos(u)du\\&d[a^u]=a^uduln(a)\\&d[ln(u)]=\frac{du}{u}\\&d[e^u]=e^udu\\&\nabla f=\begin{bmatrix}30\cdot 4^{(2x)}ln(3y)ln(4) - 15y^{2}\\\frac{(15\cdot 4^{(2x)})}{y}- 30xy\end{bmatrix}\end{align*}$ $\displaystyle \begin{align*}&\text{The gradient of a function is a vector of first derivatives taken with}\\&\text{respect to its constituent variables. The form is as follows:}\\&\begin{bmatrix}\frac{d^f}{d_{x_1}}\\\frac{d^f}{d_{x_2}}\\...\\\frac{df}{d_{x_n}}\end{bmatrix}\\&\text{Considering our function: }f(x,y)=15\cdot 4^{(2x)}ln(3y) - 15xy^{2}\\&\text{And utilizing derivative rules:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[sin(u)]=cos(u)du\\&d[a^u]=a^uduln(a)\\&d[ln(u)]=\frac{du}{u}\\&d[e^u]=e^udu\\&\nabla f=\begin{bmatrix}30\cdot 4^{(2x)}ln(3y)ln(4) - 15y^{2}\\\frac{(15\cdot 4^{(2x)})}{y}- 30xy\end{bmatrix}\end{align*}$

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Example Question #3 : Matrix Calculus

$\begin{align*}&\text{Determine the gradient, }\nabla\text{, of the function}\\&f(x,y,z)=12ln(3y)ln(2z)tan(2x)\end{align*}$ $\displaystyle \begin{align*}&\text{Determine the gradient, }\nabla\text{, of the function}\\&f(x,y,z)=12ln(3y)ln(2z)tan(2x)\end{align*}$

Possible Answers:

$\begin{bmatrix}48ln(3y)ln(2z)tan(2x)\cdot (2tan(2x)^{2} + 2)&\frac{(12ln(2z)\cdot (2tan(2x)^{2} + 2))}{y}&\frac{(12ln(3y)\cdot (2tan(2x)^{2} + 2))}{z}\\\frac{(12ln(2z)\cdot (2tan(2x)^{2} + 2))}{y}&-\frac{(12ln(2z)tan(2x))}{y^{2}}&\frac{(12tan(2x))}{(yz)}\\\frac{(12ln(3y)\cdot (2tan(2x)^{2} + 2))}{z}&\frac{(12tan(2x))}{(yz)}&-\frac{(12ln(3y)tan(2x))}{z^{2}}\end{bmatrix}$ $\displaystyle \begin{bmatrix}48ln(3y)ln(2z)tan(2x)\cdot (2tan(2x)^{2} + 2)&\frac{(12ln(2z)\cdot (2tan(2x)^{2} + 2))}{y}&\frac{(12ln(3y)\cdot (2tan(2x)^{2} + 2))}{z}\\\frac{(12ln(2z)\cdot (2tan(2x)^{2} + 2))}{y}&-\frac{(12ln(2z)tan(2x))}{y^{2}}&\frac{(12tan(2x))}{(yz)}\\\frac{(12ln(3y)\cdot (2tan(2x)^{2} + 2))}{z}&\frac{(12tan(2x))}{(yz)}&-\frac{(12ln(3y)tan(2x))}{z^{2}}\end{bmatrix}$

$\begin{bmatrix}12ln(3y)ln(2z)\cdot (2tan(2x)^{2} + 2) +\frac{ (12ln(2z)tan(2x))}{y}+\frac{ (12ln(3y)tan(2x))}{z}\end{bmatrix}$ $\displaystyle \begin{bmatrix}12ln(3y)ln(2z)\cdot (2tan(2x)^{2} + 2) +\frac{ (12ln(2z)tan(2x))}{y}+\frac{ (12ln(3y)tan(2x))}{z}\end{bmatrix}$

$\begin{bmatrix}12ln(3y)ln(2z)\cdot (2tan(2x)^{2} + 2)\\\frac{(12ln(2z)tan(2x))}{y}\\\frac{(12ln(3y)tan(2x))}{z}\end{bmatrix}$ $\displaystyle \begin{bmatrix}12ln(3y)ln(2z)\cdot (2tan(2x)^{2} + 2)\\\frac{(12ln(2z)tan(2x))}{y}\\\frac{(12ln(3y)tan(2x))}{z}\end{bmatrix}$

$\begin{bmatrix}48ln(3y)ln(2z)tan(2x)\cdot (2tan(2x)^{2} + 2)\\-\frac{(12ln(2z)tan(2x))}{y^{2}}\\-\frac{(12ln(3y)tan(2x))}{z^{2}}\end{bmatrix}$ $\displaystyle \begin{bmatrix}48ln(3y)ln(2z)tan(2x)\cdot (2tan(2x)^{2} + 2)\\-\frac{(12ln(2z)tan(2x))}{y^{2}}\\-\frac{(12ln(3y)tan(2x))}{z^{2}}\end{bmatrix}$

Correct answer:

Explanation:

$\begin{align*}&\text{The gradient of a function is a vector of first derivatives taken with}\\&\text{respect to its constituent variables. The form is as follows:}\\&\begin{bmatrix}\frac{d^f}{d_{x_1}}\\\frac{d^f}{d_{x_2}}\\...\\\frac{df}{d_{x_n}}\end{bmatrix}\\&\text{Considering our function: }f(x,y,z)=12ln(3y)ln(2z)tan(2x)\\&\text{And utilizing derivative rules:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[tan(u)]=\frac{du}{cos^2(u)}=(tan^2(u)+1)du\\&d[ln(u)]=\frac{du}{u}\\&\nabla f=\begin{bmatrix}12ln(3y)ln(2z)\cdot (2tan(2x)^{2} + 2)\\\frac{(12ln(2z)tan(2x))}{y}\\\frac{(12ln(3y)tan(2x))}{z}\end{bmatrix}\end{align*}$ $\displaystyle \begin{align*}&\text{The gradient of a function is a vector of first derivatives taken with}\\&\text{respect to its constituent variables. The form is as follows:}\\&\begin{bmatrix}\frac{d^f}{d_{x_1}}\\\frac{d^f}{d_{x_2}}\\...\\\frac{df}{d_{x_n}}\end{bmatrix}\\&\text{Considering our function: }f(x,y,z)=12ln(3y)ln(2z)tan(2x)\\&\text{And utilizing derivative rules:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[tan(u)]=\frac{du}{cos^2(u)}=(tan^2(u)+1)du\\&d[ln(u)]=\frac{du}{u}\\&\nabla f=\begin{bmatrix}12ln(3y)ln(2z)\cdot (2tan(2x)^{2} + 2)\\\frac{(12ln(2z)tan(2x))}{y}\\\frac{(12ln(3y)tan(2x))}{z}\end{bmatrix}\end{align*}$

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Example Question #4 : Matrix Calculus

$\begin{align*}&\text{Find the vector }\nabla f \text{ for }f(x,y)=9x^{2}tan(2y)\end{align*}$ $\displaystyle \begin{align*}&\text{Find the vector }\nabla f \text{ for }f(x,y)=9x^{2}tan(2y)\end{align*}$

Possible Answers:

$\begin{bmatrix}18xtan(2y)\\9x^{2}\cdot (2tan(2y)^{2} + 2)\end{bmatrix}$ $\displaystyle \begin{bmatrix}18xtan(2y)\\9x^{2}\cdot (2tan(2y)^{2} + 2)\end{bmatrix}$

$\begin{bmatrix}18xtan(2y) + 9x^{2}\cdot (2tan(2y)^{2} + 2)\end{bmatrix}$ $\displaystyle \begin{bmatrix}18xtan(2y) + 9x^{2}\cdot (2tan(2y)^{2} + 2)\end{bmatrix}$

$\begin{bmatrix}18tan(2y)\\36x^{2}tan(2y)\cdot (2tan(2y)^{2} + 2)\end{bmatrix}$ $\displaystyle \begin{bmatrix}18tan(2y)\\36x^{2}tan(2y)\cdot (2tan(2y)^{2} + 2)\end{bmatrix}$

$\begin{bmatrix}18tan(2y)&18x\cdot (2tan(2y)^{2} + 2)\\18x\cdot (2tan(2y)^{2} + 2)&36x^{2}tan(2y)\cdot (2tan(2y)^{2} + 2)\end{bmatrix}$ $\displaystyle \begin{bmatrix}18tan(2y)&18x\cdot (2tan(2y)^{2} + 2)\\18x\cdot (2tan(2y)^{2} + 2)&36x^{2}tan(2y)\cdot (2tan(2y)^{2} + 2)\end{bmatrix}$

Correct answer:

$\begin{bmatrix}18xtan(2y)\\9x^{2}\cdot (2tan(2y)^{2} + 2)\end{bmatrix}$ $\displaystyle \begin{bmatrix}18xtan(2y)\\9x^{2}\cdot (2tan(2y)^{2} + 2)\end{bmatrix}$

Explanation:

$\begin{align*}&\text{In asking for }\nabla\text{, the question asks for the gradient.}\\&\text{The gradient of a function is a vector of first derivatives taken with}\\&\text{respect to its constituent variables. The form is as follows:}\\&\begin{bmatrix}\frac{d^f}{d_{x_1}}\\\frac{d^f}{d_{x_2}}\\...\\\frac{df}{d_{x_n}}\end{bmatrix}\\&\text{Considering our function: }f(x,y)=9x^{2}tan(2y)\\&\text{And utilizing derivative rules:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[tan(u)]=\frac{du}{cos^2(u)}=(tan^2(u)+1)du\\&\nabla f=\begin{bmatrix}18xtan(2y)\\9x^{2}\cdot (2tan(2y)^{2} + 2)\end{bmatrix}\end{align*}$ $\displaystyle \begin{align*}&\text{In asking for }\nabla\text{, the question asks for the gradient.}\\&\text{The gradient of a function is a vector of first derivatives taken with}\\&\text{respect to its constituent variables. The form is as follows:}\\&\begin{bmatrix}\frac{d^f}{d_{x_1}}\\\frac{d^f}{d_{x_2}}\\...\\\frac{df}{d_{x_n}}\end{bmatrix}\\&\text{Considering our function: }f(x,y)=9x^{2}tan(2y)\\&\text{And utilizing derivative rules:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[tan(u)]=\frac{du}{cos^2(u)}=(tan^2(u)+1)du\\&\nabla f=\begin{bmatrix}18xtan(2y)\\9x^{2}\cdot (2tan(2y)^{2} + 2)\end{bmatrix}\end{align*}$

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Example Question #1 : Linear Algebra

$\begin{align*}&\text{Determine the gradient, }\nabla\text{, of the function}\\&f(x,y,z)=- xcos(3z)e^{(5y)} - 20\cdot 2^{(3z)}y^{2}ln(4x)\end{align*}$ $\displaystyle \begin{align*}&\text{Determine the gradient, }\nabla\text{, of the function}\\&f(x,y,z)=- xcos(3z)e^{(5y)} - 20\cdot 2^{(3z)}y^{2}ln(4x)\end{align*}$

Possible Answers:

$\begin{bmatrix}\frac{(20\cdot 2^{(3z)}y^{2})}{x^{2}}\\- 40\cdot 2^{(3z)}ln(4x) - 25xcos(3z)e^{(5y)}\\9xcos(3z)e^{(5y)} - 180\cdot 2^{(3z)}y^{2}ln(4x)ln(2)^{2}\end{bmatrix}$ $\displaystyle \begin{bmatrix}\frac{(20\cdot 2^{(3z)}y^{2})}{x^{2}}\\- 40\cdot 2^{(3z)}ln(4x) - 25xcos(3z)e^{(5y)}\\9xcos(3z)e^{(5y)} - 180\cdot 2^{(3z)}y^{2}ln(4x)ln(2)^{2}\end{bmatrix}$

$\begin{bmatrix}3xsin(3z)e^{(5y)} -\frac{ (20\cdot 2^{(3z)}y^{2})}{x}- 5xcos(3z)e^{(5y)} - 40\cdot 2^{(3z)}yln(4x) - cos(3z)e^{(5y)} - 60\cdot 2^{(3z)}y^{2}ln(4x)ln(2)\end{bmatrix}$ $\displaystyle \begin{bmatrix}3xsin(3z)e^{(5y)} -\frac{ (20\cdot 2^{(3z)}y^{2})}{x}- 5xcos(3z)e^{(5y)} - 40\cdot 2^{(3z)}yln(4x) - cos(3z)e^{(5y)} - 60\cdot 2^{(3z)}y^{2}ln(4x)ln(2)\end{bmatrix}$

$\begin{bmatrix}\frac{(20\cdot 2^{(3z)}y^{2})}{x^{2}}&- 5cos(3z)e^{(5y)} -\frac{ (40\cdot 2^{(3z)}y)}{x}&3sin(3z)e^{(5y)} -\frac{ (60\cdot 2^{(3z)}y^{2}ln(2))}{x}\\- 5cos(3z)e^{(5y)} -\frac{ (40\cdot 2^{(3z)}y)}{x}&- 40\cdot 2^{(3z)}ln(4x) - 25xcos(3z)e^{(5y)}&15xsin(3z)e^{(5y)} - 120\cdot 2^{(3z)}yln(4x)ln(2)\\3sin(3z)e^{(5y)} -\frac{ (60\cdot 2^{(3z)}y^{2}ln(2))}{x}&15xsin(3z)e^{(5y)} - 120\cdot 2^{(3z)}yln(4x)ln(2)&9xcos(3z)e^{(5y)} - 180\cdot 2^{(3z)}y^{2}ln(4x)ln(2)^{2}\end{bmatrix}$ $\displaystyle \begin{bmatrix}\frac{(20\cdot 2^{(3z)}y^{2})}{x^{2}}&- 5cos(3z)e^{(5y)} -\frac{ (40\cdot 2^{(3z)}y)}{x}&3sin(3z)e^{(5y)} -\frac{ (60\cdot 2^{(3z)}y^{2}ln(2))}{x}\\- 5cos(3z)e^{(5y)} -\frac{ (40\cdot 2^{(3z)}y)}{x}&- 40\cdot 2^{(3z)}ln(4x) - 25xcos(3z)e^{(5y)}&15xsin(3z)e^{(5y)} - 120\cdot 2^{(3z)}yln(4x)ln(2)\\3sin(3z)e^{(5y)} -\frac{ (60\cdot 2^{(3z)}y^{2}ln(2))}{x}&15xsin(3z)e^{(5y)} - 120\cdot 2^{(3z)}yln(4x)ln(2)&9xcos(3z)e^{(5y)} - 180\cdot 2^{(3z)}y^{2}ln(4x)ln(2)^{2}\end{bmatrix}$

$\begin{bmatrix}- cos(3z)e^{(5y)} -\frac{ (20\cdot 2^{(3z)}y^{2})}{x}\\- 5xcos(3z)e^{(5y)} - 40\cdot 2^{(3z)}yln(4x)\\3xsin(3z)e^{(5y)} - 60\cdot 2^{(3z)}y^{2}ln(4x)ln(2)\end{bmatrix}$ $\displaystyle \begin{bmatrix}- cos(3z)e^{(5y)} -\frac{ (20\cdot 2^{(3z)}y^{2})}{x}\\- 5xcos(3z)e^{(5y)} - 40\cdot 2^{(3z)}yln(4x)\\3xsin(3z)e^{(5y)} - 60\cdot 2^{(3z)}y^{2}ln(4x)ln(2)\end{bmatrix}$

Correct answer:

Explanation:

$\begin{align*}&\text{The gradient of a function is a vector of first derivatives taken with}\\&\text{respect to its constituent variables. The form is as follows:}\\&\begin{bmatrix}\frac{d^f}{d_{x_1}}\\\frac{d^f}{d_{x_2}}\\...\\\frac{df}{d_{x_n}}\end{bmatrix}\\&\text{Considering our function: }f(x,y,z)=- xcos(3z)e^{(5y)} - 20\cdot 2^{(3z)}y^{2}ln(4x)\\&\text{And utilizing derivative rules:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[e^u]=e^udu\\&d[cos(u)]=-sin(u)du\\&d[ln(u)]=\frac{du}{u}\\&d[a^u]=a^uduln(a)\\&\nabla f=\begin{bmatrix}- cos(3z)e^{(5y)} -\frac{ (20\cdot 2^{(3z)}y^{2})}{x}\\- 5xcos(3z)e^{(5y)} - 40\cdot 2^{(3z)}yln(4x)\\3xsin(3z)e^{(5y)} - 60\cdot 2^{(3z)}y^{2}ln(4x)ln(2)\end{bmatrix}\end{align*}$ $\displaystyle \begin{align*}&\text{The gradient of a function is a vector of first derivatives taken with}\\&\text{respect to its constituent variables. The form is as follows:}\\&\begin{bmatrix}\frac{d^f}{d_{x_1}}\\\frac{d^f}{d_{x_2}}\\...\\\frac{df}{d_{x_n}}\end{bmatrix}\\&\text{Considering our function: }f(x,y,z)=- xcos(3z)e^{(5y)} - 20\cdot 2^{(3z)}y^{2}ln(4x)\\&\text{And utilizing derivative rules:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[e^u]=e^udu\\&d[cos(u)]=-sin(u)du\\&d[ln(u)]=\frac{du}{u}\\&d[a^u]=a^uduln(a)\\&\nabla f=\begin{bmatrix}- cos(3z)e^{(5y)} -\frac{ (20\cdot 2^{(3z)}y^{2})}{x}\\- 5xcos(3z)e^{(5y)} - 40\cdot 2^{(3z)}yln(4x)\\3xsin(3z)e^{(5y)} - 60\cdot 2^{(3z)}y^{2}ln(4x)ln(2)\end{bmatrix}\end{align*}$

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Example Question #2 : Matrix Calculus

$\begin{align*}&\text{Determine the gradient, }\nabla\text{, of the function}\\&f(x,y)=-18e^{(5x)}tan(y)\end{align*}$ $\displaystyle \begin{align*}&\text{Determine the gradient, }\nabla\text{, of the function}\\&f(x,y)=-18e^{(5x)}tan(y)\end{align*}$

Possible Answers:

$\begin{bmatrix}-450e^{(5x)}tan(y)&-90e^{(5x)}\cdot (tan(y)^{2} + 1)\\-90e^{(5x)}\cdot (tan(y)^{2} + 1)&-36e^{(5x)}tan(y)\cdot (tan(y)^{2} + 1)\end{bmatrix}$ $\displaystyle \begin{bmatrix}-450e^{(5x)}tan(y)&-90e^{(5x)}\cdot (tan(y)^{2} + 1)\\-90e^{(5x)}\cdot (tan(y)^{2} + 1)&-36e^{(5x)}tan(y)\cdot (tan(y)^{2} + 1)\end{bmatrix}$

$\begin{bmatrix}-450e^{(5x)}tan(y)\\-36e^{(5x)}tan(y)\cdot (tan(y)^{2} + 1)\end{bmatrix}$ $\displaystyle \begin{bmatrix}-450e^{(5x)}tan(y)\\-36e^{(5x)}tan(y)\cdot (tan(y)^{2} + 1)\end{bmatrix}$

$\begin{bmatrix}- 18e^{(5x)}\cdot (tan(y)^{2} + 1) - 90e^{(5x)}tan(y)\end{bmatrix}$ $\displaystyle \begin{bmatrix}- 18e^{(5x)}\cdot (tan(y)^{2} + 1) - 90e^{(5x)}tan(y)\end{bmatrix}$

$\begin{bmatrix}-90e^{(5x)}tan(y)\\-18e^{(5x)}\cdot (tan(y)^{2} + 1)\end{bmatrix}$ $\displaystyle \begin{bmatrix}-90e^{(5x)}tan(y)\\-18e^{(5x)}\cdot (tan(y)^{2} + 1)\end{bmatrix}$

Correct answer:

$\begin{bmatrix}-90e^{(5x)}tan(y)\\-18e^{(5x)}\cdot (tan(y)^{2} + 1)\end{bmatrix}$ $\displaystyle \begin{bmatrix}-90e^{(5x)}tan(y)\\-18e^{(5x)}\cdot (tan(y)^{2} + 1)\end{bmatrix}$

Explanation:

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Example Question #2 : The Gradient

$\begin{align*}&\text{Calculate the gradient of the function}\\&f(x,y,z)=14tan(5y)e^{(4z)}tan(x) - 13\cdot 5^{(4x)}ln(2z)tan(2y)\end{align*}$ $\displaystyle \begin{align*}&\text{Calculate the gradient of the function}\\&f(x,y,z)=14tan(5y)e^{(4z)}tan(x) - 13\cdot 5^{(4x)}ln(2z)tan(2y)\end{align*}$

Possible Answers:

$\begin{bmatrix}14tan(5y)e^{(4z)}\cdot (tan(x)^{2} + 1) - 13\cdot 5^{(4x)}ln(2z)\cdot (2tan(2y)^{2} + 2) -\frac{ (13\cdot 5^{(4x)}tan(2y))}{z}+ 56tan(5y)e^{(4z)}tan(x) + 14e^{(4z)}tan(x)\cdot (5tan(5y)^{2} + 5) - 52\cdot 5^{(4x)}ln(2z)tan(2y)ln(5)\end{bmatrix}$ $\displaystyle \begin{bmatrix}14tan(5y)e^{(4z)}\cdot (tan(x)^{2} + 1) - 13\cdot 5^{(4x)}ln(2z)\cdot (2tan(2y)^{2} + 2) -\frac{ (13\cdot 5^{(4x)}tan(2y))}{z}+ 56tan(5y)e^{(4z)}tan(x) + 14e^{(4z)}tan(x)\cdot (5tan(5y)^{2} + 5) - 52\cdot 5^{(4x)}ln(2z)tan(2y)ln(5)\end{bmatrix}$

$\begin{bmatrix}28tan(5y)e^{(4z)}tan(x)\cdot (tan(x)^{2} + 1) - 208\cdot 5^{(4x)}ln(2z)tan(2y)ln(5)^{2}\\140tan(5y)e^{(4z)}tan(x)\cdot (5tan(5y)^{2} + 5) - 52\cdot 5^{(4x)}ln(2z)tan(2y)\cdot (2tan(2y)^{2} + 2)\\\frac{(13\cdot 5^{(4x)}tan(2y))}{z^{2}}+ 224tan(5y)e^{(4z)}tan(x)\end{bmatrix}$ $\displaystyle \begin{bmatrix}28tan(5y)e^{(4z)}tan(x)\cdot (tan(x)^{2} + 1) - 208\cdot 5^{(4x)}ln(2z)tan(2y)ln(5)^{2}\\140tan(5y)e^{(4z)}tan(x)\cdot (5tan(5y)^{2} + 5) - 52\cdot 5^{(4x)}ln(2z)tan(2y)\cdot (2tan(2y)^{2} + 2)\\\frac{(13\cdot 5^{(4x)}tan(2y))}{z^{2}}+ 224tan(5y)e^{(4z)}tan(x)\end{bmatrix}$

$\begin{bmatrix}28tan(5y)e^{(4z)}tan(x)\cdot (tan(x)^{2} + 1) - 208\cdot 5^{(4x)}ln(2z)tan(2y)ln(5)^{2}&14e^{(4z)}\cdot (5tan(5y)^{2} + 5)\cdot (tan(x)^{2} + 1) - 52\cdot 5^{(4x)}ln(2z)ln(5)\cdot (2tan(2y)^{2} + 2)&56tan(5y)e^{(4z)}\cdot (tan(x)^{2} + 1) -\frac{ (52\cdot 5^{(4x)}tan(2y)ln(5))}{z}\\14e^{(4z)}\cdot (5tan(5y)^{2} + 5)\cdot (tan(x)^{2} + 1) - 52\cdot 5^{(4x)}ln(2z)ln(5)\cdot (2tan(2y)^{2} + 2)&140tan(5y)e^{(4z)}tan(x)\cdot (5tan(5y)^{2} + 5) - 52\cdot 5^{(4x)}ln(2z)tan(2y)\cdot (2tan(2y)^{2} + 2)&56e^{(4z)}tan(x)\cdot (5tan(5y)^{2} + 5) -\frac{ (13\cdot 5^{(4x)}\cdot (2tan(2y)^{2} + 2))}{z}\\56tan(5y)e^{(4z)}\cdot (tan(x)^{2} + 1) -\frac{ (52\cdot 5^{(4x)}tan(2y)ln(5))}{z}&56e^{(4z)}tan(x)\cdot (5tan(5y)^{2} + 5) -\frac{ (13\cdot 5^{(4x)}\cdot (2tan(2y)^{2} + 2))}{z}&\frac{(13\cdot 5^{(4x)}tan(2y))}{z^{2}}+ 224tan(5y)e^{(4z)}tan(x)\end{bmatrix}$ $\displaystyle \begin{bmatrix}28tan(5y)e^{(4z)}tan(x)\cdot (tan(x)^{2} + 1) - 208\cdot 5^{(4x)}ln(2z)tan(2y)ln(5)^{2}&14e^{(4z)}\cdot (5tan(5y)^{2} + 5)\cdot (tan(x)^{2} + 1) - 52\cdot 5^{(4x)}ln(2z)ln(5)\cdot (2tan(2y)^{2} + 2)&56tan(5y)e^{(4z)}\cdot (tan(x)^{2} + 1) -\frac{ (52\cdot 5^{(4x)}tan(2y)ln(5))}{z}\\14e^{(4z)}\cdot (5tan(5y)^{2} + 5)\cdot (tan(x)^{2} + 1) - 52\cdot 5^{(4x)}ln(2z)ln(5)\cdot (2tan(2y)^{2} + 2)&140tan(5y)e^{(4z)}tan(x)\cdot (5tan(5y)^{2} + 5) - 52\cdot 5^{(4x)}ln(2z)tan(2y)\cdot (2tan(2y)^{2} + 2)&56e^{(4z)}tan(x)\cdot (5tan(5y)^{2} + 5) -\frac{ (13\cdot 5^{(4x)}\cdot (2tan(2y)^{2} + 2))}{z}\\56tan(5y)e^{(4z)}\cdot (tan(x)^{2} + 1) -\frac{ (52\cdot 5^{(4x)}tan(2y)ln(5))}{z}&56e^{(4z)}tan(x)\cdot (5tan(5y)^{2} + 5) -\frac{ (13\cdot 5^{(4x)}\cdot (2tan(2y)^{2} + 2))}{z}&\frac{(13\cdot 5^{(4x)}tan(2y))}{z^{2}}+ 224tan(5y)e^{(4z)}tan(x)\end{bmatrix}$

$\begin{bmatrix}14tan(5y)e^{(4z)}\cdot (tan(x)^{2} + 1) - 52\cdot 5^{(4x)}ln(2z)tan(2y)ln(5)\\14e^{(4z)}tan(x)\cdot (5tan(5y)^{2} + 5) - 13\cdot 5^{(4x)}ln(2z)\cdot (2tan(2y)^{2} + 2)\\56tan(5y)e^{(4z)}tan(x) -\frac{ (13\cdot 5^{(4x)}tan(2y))}{z}\end{bmatrix}$ $\displaystyle \begin{bmatrix}14tan(5y)e^{(4z)}\cdot (tan(x)^{2} + 1) - 52\cdot 5^{(4x)}ln(2z)tan(2y)ln(5)\\14e^{(4z)}tan(x)\cdot (5tan(5y)^{2} + 5) - 13\cdot 5^{(4x)}ln(2z)\cdot (2tan(2y)^{2} + 2)\\56tan(5y)e^{(4z)}tan(x) -\frac{ (13\cdot 5^{(4x)}tan(2y))}{z}\end{bmatrix}$

Correct answer:

Explanation:

$\begin{align*}&\text{The gradient of a function is a vector of first derivatives taken with}\\&\text{respect to its constituent variables. The form is as follows:}\\&\begin{bmatrix}\frac{d^f}{d_{x_1}}\\\frac{d^f}{d_{x_2}}\\...\\\frac{df}{d_{x_n}}\end{bmatrix}\\&\text{Considering our function: }f(x,y,z)=14tan(5y)e^{(4z)}tan(x) - 13\cdot 5^{(4x)}ln(2z)tan(2y)\\&\text{And utilizing derivative rules:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[tan(u)]=\frac{du}{cos^2(u)}=(tan^2(u)+1)du\\&d[e^u]=e^udu\\&d[a^u]=a^uduln(a)\\&d[ln(u)]=\frac{du}{u}\\&\nabla f=\begin{bmatrix}14tan(5y)e^{(4z)}\cdot (tan(x)^{2} + 1) - 52\cdot 5^{(4x)}ln(2z)tan(2y)ln(5)\\14e^{(4z)}tan(x)\cdot (5tan(5y)^{2} + 5) - 13\cdot 5^{(4x)}ln(2z)\cdot (2tan(2y)^{2} + 2)\\56tan(5y)e^{(4z)}tan(x) -\frac{ (13\cdot 5^{(4x)}tan(2y))}{z}\end{bmatrix}\end{align*}$ $\displaystyle \begin{align*}&\text{The gradient of a function is a vector of first derivatives taken with}\\&\text{respect to its constituent variables. The form is as follows:}\\&\begin{bmatrix}\frac{d^f}{d_{x_1}}\\\frac{d^f}{d_{x_2}}\\...\\\frac{df}{d_{x_n}}\end{bmatrix}\\&\text{Considering our function: }f(x,y,z)=14tan(5y)e^{(4z)}tan(x) - 13\cdot 5^{(4x)}ln(2z)tan(2y)\\&\text{And utilizing derivative rules:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[tan(u)]=\frac{du}{cos^2(u)}=(tan^2(u)+1)du\\&d[e^u]=e^udu\\&d[a^u]=a^uduln(a)\\&d[ln(u)]=\frac{du}{u}\\&\nabla f=\begin{bmatrix}14tan(5y)e^{(4z)}\cdot (tan(x)^{2} + 1) - 52\cdot 5^{(4x)}ln(2z)tan(2y)ln(5)\\14e^{(4z)}tan(x)\cdot (5tan(5y)^{2} + 5) - 13\cdot 5^{(4x)}ln(2z)\cdot (2tan(2y)^{2} + 2)\\56tan(5y)e^{(4z)}tan(x) -\frac{ (13\cdot 5^{(4x)}tan(2y))}{z}\end{bmatrix}\end{align*}$

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Example Question #2 : The Gradient

$\begin{align*}&\text{Determine the gradient, }\nabla\text{, of the function}\\&f(x,y)=11ysin(5x) + 11x^{2}tan(y)\end{align*}$ $\displaystyle \begin{align*}&\text{Determine the gradient, }\nabla\text{, of the function}\\&f(x,y)=11ysin(5x) + 11x^{2}tan(y)\end{align*}$

Possible Answers:

$\begin{bmatrix}22tan(y) - 275ysin(5x)\\22x^{2}tan(y)\cdot (tan(y)^{2} + 1)\end{bmatrix}$ $\displaystyle \begin{bmatrix}22tan(y) - 275ysin(5x)\\22x^{2}tan(y)\cdot (tan(y)^{2} + 1)\end{bmatrix}$

$\begin{bmatrix}11sin(5x) + 55ycos(5x) + 22xtan(y) + 11x^{2}\cdot (tan(y)^{2} + 1)\end{bmatrix}$ $\displaystyle \begin{bmatrix}11sin(5x) + 55ycos(5x) + 22xtan(y) + 11x^{2}\cdot (tan(y)^{2} + 1)\end{bmatrix}$

$\begin{bmatrix}22tan(y) - 275ysin(5x)&55cos(5x) + 22x\cdot (tan(y)^{2} + 1)\\55cos(5x) + 22x\cdot (tan(y)^{2} + 1)&22x^{2}tan(y)\cdot (tan(y)^{2} + 1)\end{bmatrix}$ $\displaystyle \begin{bmatrix}22tan(y) - 275ysin(5x)&55cos(5x) + 22x\cdot (tan(y)^{2} + 1)\\55cos(5x) + 22x\cdot (tan(y)^{2} + 1)&22x^{2}tan(y)\cdot (tan(y)^{2} + 1)\end{bmatrix}$

$\begin{bmatrix}55ycos(5x) + 22xtan(y)\\11sin(5x) + 11x^{2}\cdot (tan(y)^{2} + 1)\end{bmatrix}$ $\displaystyle \begin{bmatrix}55ycos(5x) + 22xtan(y)\\11sin(5x) + 11x^{2}\cdot (tan(y)^{2} + 1)\end{bmatrix}$

Correct answer:

Explanation:

$\begin{align*}&\text{The gradient of a function is a vector of first derivatives taken with}\\&\text{respect to its constituent variables. The form is as follows:}\\&\begin{bmatrix}\frac{d^f}{d_{x_1}}\\\frac{d^f}{d_{x_2}}\\...\\\frac{df}{d_{x_n}}\end{bmatrix}\\&\text{Considering our function: }f(x,y)=11ysin(5x) + 11x^{2}tan(y)\\&\text{And utilizing derivative rules:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[tan(u)]=\frac{du}{cos^2(u)}=(tan^2(u)+1)du\\&d[sin(u)]=cos(u)du\\&\nabla f=\begin{bmatrix}55ycos(5x) + 22xtan(y)\\11sin(5x) + 11x^{2}\cdot (tan(y)^{2} + 1)\end{bmatrix}\end{align*}$ $\displaystyle \begin{align*}&\text{The gradient of a function is a vector of first derivatives taken with}\\&\text{respect to its constituent variables. The form is as follows:}\\&\begin{bmatrix}\frac{d^f}{d_{x_1}}\\\frac{d^f}{d_{x_2}}\\...\\\frac{df}{d_{x_n}}\end{bmatrix}\\&\text{Considering our function: }f(x,y)=11ysin(5x) + 11x^{2}tan(y)\\&\text{And utilizing derivative rules:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[tan(u)]=\frac{du}{cos^2(u)}=(tan^2(u)+1)du\\&d[sin(u)]=cos(u)du\\&\nabla f=\begin{bmatrix}55ycos(5x) + 22xtan(y)\\11sin(5x) + 11x^{2}\cdot (tan(y)^{2} + 1)\end{bmatrix}\end{align*}$

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