linear algebra
Find the area of the pentagon of the five vertices $(1,2), (4,1), (5,3), (3,7), (2,6)$ . Please, use the way of using determinant.
My idea is to cut the pentagon into some triangles, then calculate each triangle, then sum them. I wonder if there is some other way to directly calculate it using a bigger matrix calculating its determinant?
The method at that site ignores the sign of the determinant, and is phrased in a way that makes the generalization to higher dimensions a bit less clear than I'd like. I'll try here to provide an altered version of the approach up to four dimensions. I'll use bold for vectors, subscripts for components, and double vertical lines for length, so that $\left\Vert \mathbf{a}\right\Vert =\sqrt{a_{1}^{2}+a_{2}^{2}+\cdots}$.
As was mentioned by LutzL in a comment, this method is very closely connected to using the $QR$-decomposition to find the absolute value of the determinant, as described on Wikipedia here.
Let's calculate $\det\left(\mathbf{a}\right)$ where $\mathbf{a}$ has one nonzero component. It's $\left\Vert \mathbf{a}\right\Vert$ if $\mathbf{a}$ has a positive component and $-\left\Vert \mathbf{a}\right\Vert$ if $\mathbf{a}$ has a negative component.
Let's calculate $\det\left(\mathbf{a},\mathbf{b}\right)$ where $\mathbf{a}$ and $\mathbf{b}$ are not collinear.
Let's ignore $\mathbf{a}$ for now. The first step is to find a vector $\mathbf{n}$ that's orthogonal to $\mathbf{b}$. We set $\mathbf{n}\bullet\mathbf{b}$ equal to $\boldsymbol{0}$. That's two unknowns and only one equation. In a typical case, the component $n_{1}$ of $\mathbf{n}$ is not forced to be $0$, so it can be whatever we want that's nonzero (e.g. $1$). (In a special case, $n_{1}$ might be forced to be $0$, but then $n_{2}$ can be chosen freely.)
Now scale $\mathbf{n}$ to get a new vector $\mathbf{n}'$ so that $\left\Vert \mathbf{n}'\right\Vert =\left\Vert \mathbf{b}\right\Vert$ . By some geometry, the area of the parallelogram formed by $\mathbf{a}$ and $\mathbf{b}$ is then $\left|\mathbf{n}'\bullet\mathbf{a}\right|$. The determinant is $\pm\mathbf{n}'\bullet\mathbf{a}$ where the $\pm$ sign here (which may not be the sign of the determinant) is positive exactly when the rotation to get from $\mathbf{a}$ to $\mathbf{b}$ is in the same direction (clockwise or counterclockwise) as the rotation to get from $\mathbf{b}$ to $\mathbf{n}'$. Unfortunately, that can't be determined by a dot-product calculation.
Let's calculate $\det\left(\mathbf{a},\mathbf{b},\mathbf{c}\right)$ where $\mathbf{a},\mathbf{b},\mathbf{c}$ are not coplanar.
Let's ignore $\mathbf{a}$ for now. The first step is to find a vector $\mathbf{n}$ that's orthogonal to both $\mathbf{b}$ and $\mathbf{c}$. We set $\mathbf{n}\bullet\mathbf{b},\mathbf{n}\bullet\mathbf{c}$ equal to $\boldsymbol{0}$. That's three unknowns and only two equations. In a typical case, the component $n_{1}$ of $\mathbf{n}$ is not forced to be $0$, so it can be whatever we want that's nonzero (e.g. $1$). (In a special case, $n_{1}$ might be forced to be $0$, but then $n_{2}$ or $n_{3}$ can be chosen freely.)
The second step is to find a vector $\mathbf{o}$ that's orthogonal to $\mathbf{c}$ (this choice differs from the original author), but lies in the same plane as $\mathbf{b}$ and $\mathbf{c}$. To keep it in that plane, we need $\mathbf{o}$ to be orthogonal to $\mathbf{n}$. So we have $\mathbf{o}\bullet\mathbf{n}=\boldsymbol{0}$ as well as $\mathbf{o}\bullet\mathbf{c}=\boldsymbol{0}$. Again that's three unknowns and two equations, so we have a degree of freedom and could choose a particular value for some component.
Now scale $\mathbf{o}$ to get a new vector $\mathbf{o}'$ so that $\left\Vert \mathbf{o}'\right\Vert =\left\Vert \mathbf{c}\right\Vert$ . By some geometry, the area of the parallelogram formed by $\mathbf{b}$ and $\mathbf{c}$ is then $\left|\mathbf{o}'\bullet\mathbf{b}\right|$. Now scale $\mathbf{n}$ to get a new vector $\mathbf{n}'$ so that $\left\Vert \mathbf{n}'\right\Vert =\left|\mathbf{o}'\bullet\mathbf{b}\right|$. By some geometry, the volume of the parallelepiped formed by $\mathbf{a}, \mathbf{b}, \mathbf{c}$ is then $\left|\mathbf{n}'\bullet\mathbf{a}\right|$. The determinant is then $\pm\mathbf{n}'\bullet\mathbf{a}$ where I'm pretty sure the $\pm$ sign is positive when the handedness (right or left) of $\mathbf{a},\mathbf{b},\mathbf{c}$ is the same as that of $\mathbf{c},\mathbf{n}',\mathbf{o}'$.
Let's calculate $\det\left(\mathbf{a},\mathbf{b},\mathbf{c},\mathbf{d}\right)$ where $\mathbf{a},\mathbf{b},\mathbf{c},\mathbf{d}$ are not in the same $3$-dimensional hyperplane.
Let's ignore $\mathbf{a}$ for now. The first step is to find a vector $\mathbf{n}$ that's orthogonal to all three of $\mathbf{b},\mathbf{c},\mathbf{d}$. We set $\mathbf{n}\bullet\mathbf{b},\mathbf{n}\bullet\mathbf{c},\mathbf{n}\bullet\mathbf{c}$ all to $\boldsymbol{0}$. That's four unknowns and only three equations. In a typical case the component $n_{1}$ of $\mathbf{n}$ is not forced to be $0$, so it can be whatever we want that's nonzero (e.g. $1$). (In a special case, $n_{1}$ might be forced to be $0$, but there will be at least one component we could choose freely.)
The second step is to find a vector $\mathbf{o}$ that's orthogonal to $\mathbf{c}$ and $\mathbf{d}$, but lies in the same $3$-dimensional hyperplane as $\mathbf{b},\mathbf{c},\mathbf{d}$. To keep it in that hyperplane, we need $\mathbf{o}$ to be orthogonal to $\mathbf{n}$. So we have $\mathbf{o}\bullet\mathbf{n}=\boldsymbol{0}$ as well as $\mathbf{o}\bullet\mathbf{c},\mathbf{o}\bullet\mathbf{d}=\boldsymbol{0}$. Again that's four unknowns and three equations, so we have a degree of freedom and could choose a particular value for some component.
The third step is to find a vector $\mathbf{p}$ that's orthogonal to $\mathbf{d}$, but lies in the same $2$-dimensional plane as $\mathbf{c},\mathbf{d}$. To keep it in that plane, it should be orthogonal to $\mathbf{n}$ as well as $\mathbf{o}$. So we have $\mathbf{p}\bullet\mathbf{n},\mathbf{p}\bullet\mathbf{o}=\boldsymbol{0}$ as well as $\mathbf{p}\bullet\mathbf{d}=\boldsymbol{0}$. Again that's four unknowns and three equations, so we have a degree of freedom and could choose a particular value for some component.
Now scale $\mathbf{p}$ to get a new vector $\mathbf{p}'$ so that $\left\Vert \mathbf{p}'\right\Vert =\left\Vert \mathbf{d}\right\Vert$ . By some geometry, the area of the parallelogram formed by $\mathbf{c}$ and $\mathbf{d}$ is then $\left|\mathbf{p}'\bullet\mathbf{c}\right|$. Now scale $\mathbf{o}$ to get a new vector $\mathbf{o}'$ so that $\left\Vert \mathbf{o}'\right\Vert =\left|\mathbf{p}'\bullet\mathbf{c}\right|$. By some geometry, the volume of the parallelepiped formed by $\mathbf{b}, \mathbf{c}, \mathbf{d}$ is then $\left|\mathbf{o}'\bullet\mathbf{b}\right|$. Now scale $\mathbf{n}$ to get a new vector $\mathbf{n}'$ so that $\left\Vert \mathbf{n}'\right\Vert =\left|\mathbf{o}'\bullet\mathbf{b}\right|$. By some geometry, the hypervolume of the hyperparallelepiped formed by $\mathbf{a},\mathbf{b}, \mathbf{c}, \mathbf{d}$ is then $\left|\mathbf{n}'\bullet\mathbf{a}\right|$. The determinant is then $\pm\mathbf{n}'\bullet\mathbf{a}$ where I think the $\pm$ sign is positive when the orientation of $\mathbf{a},\mathbf{b},\mathbf{c},\mathbf{d}$ is the same as that of $\mathbf{d},\mathbf{n}',\mathbf{o}',\mathbf{p}'$.
You can think of triangle area as sum of 2 trapezoids, then remove the base.
$$Area = \left|{y_1 + y_0 \over 2}(x_1 - x_0) + {y_2 + y_1 \over 2}(x_2 - x_1) - {y_2 + y_0 \over 2}(x_2 - x_0) \right|$$
The middle term can be rewrite as:
$${(y_2+y_1)(x_2-x_0) - (y_2+y_1)(x_1-x_0) \over 2}$$
$$Area = {1\over2}\left|(y_0-y_2)(x_1 - x_0) - (y_0 - y_1)(x_2 - x_0) \right|$$
Flip the sign inside absolute function, we get your ½ Det expression.
Best Answer
If you label your points such that they form a clockwise rotation about a central point you can use the following shoelace formula:
$$A =\frac12\bigg| \sum_{i=1}^{n} x_iy_{i+1} - x_{i+1}y_i \bigg| = \frac12\bigg|\sum_{i=1}^n\det\bigg(\begin{matrix} x_i & x_{i+1} \\ y_i & y_{i+1} \end{matrix}\bigg)\bigg|$$
Where $x_0 = x_n,\ x_{n+1} = x_1$ and similarly, $y_0 = y_n,\ y_{n+1} = y_1$