# What do the column of 1's signify when calculating the area of an image using the determinant? [duplicate]

When calculating the area of an image such as that given by the three coordinates (4,3), (4,10), and (-4,-3), why when writing it out as a matrix is there an extra column of 1's?

$$\begin{matrix} 4 & 3 & 1 \\ 4 & 10 & 1 \\ -4 & -3 & 1 \\ \end{matrix}$$ I understand that it is not possible to find the determinant of a non-square matrix, so my questions are:

1) Are 1's just used to make the matrix a square?
2) If so, is there any reason why 1's and not 0's are used?

• This is a tangential comment, but it would be better to be more careful when stating what you are taking the area of. The title states that you are computing the "area of a matrix", which does not make sense because a matrix does not have an area. The question body states that you have to find the area of the three given coordinates, which also does not make sense. I believe your goal is to compute the area of the triangle whose vertices are the three given points. – littleO Aug 14 '17 at 20:08
• You mean the area of the triangle determined by the three points ? – Peter Aug 14 '17 at 20:09
• – levap Aug 14 '17 at 20:11
• Thanks for the link, but I'm more confused as to why 1's are used in the column rather than how to solve it – Kyzen Aug 14 '17 at 20:15
• Perhaps a more intuitive answer is the result at the end of that link, that the area of the triangle is half of that determinant. Then the rest of the answer, read backwards, can be thought of as manipulating that result into a nicer form. – Ian Aug 14 '17 at 20:33

Algebraically, we are using the properties of determinant to express the area of a triangle $T$ with vertices at $A : (x_1,y_1)$, $B : (x_2,y_2)$, $C : (x_3,y_3)$ in a simpler form:
$$\verb/Area/(T) = \frac12 \left| \begin{matrix} x_2 - x_1 & y_2 - y_1\\ x_3 - x_1 & y_3 - y_1\\ \end{matrix}\right| = \frac12 \left| \begin{matrix} 0 & 0 & 1\\ x_2 - x_1 & y_2 - y_1 & 1\\ x_3 - x_1 & y_3 - y_1 & 1\\ \end{matrix}\right| = \frac12 \left| \begin{matrix} x_1 & y_1 & 1\\ x_2 & y_2 & 1\\ x_3 & y_3 & 1\\ \end{matrix}\right|$$
Geometrically, we can embed $\mathbb{R}^2$ as the plane $z = 1$ in $\mathbb{R}^3$. The vertices of $T$ becomes the points $A' : (x_1,y_1,1)$, $B' : (x_1,y_2,1)$, $C' : (x_3,y_3,1)$ on $\mathbb{R}^3$. Let $T'$ be the tetrahedron spanned by $A', B', C'$ and origin $O : (0,0,0)$. The area of $T$ is 3 times the volume of tetrahedron $T'$. The volume of $T'$ is $\frac16$ of the volume of the parallelepiped $P$ with one vertex at $O$ and spanned by the 3 vectors $A', B', C'$. Since the volume of a parallelepiped can be expressed as a cross product which equals to corresponding determinant, we have:
$$\verb/Area/(T) = 3\verb/Volume/(T') = \frac{3}{6}\verb/Volume/(P) = \frac12 \left| \begin{matrix} x_1 & y_1 & 1\\ x_2 & y_2 & 1\\ x_3 & y_3 & 1\\ \end{matrix}\right|$$ In certain sense, the introduction of ones here reflect the possibility to express geometric relations for objects living on the plane as geometric relations on $\mathbb{R}^3$. It allows one to look at plane geometry problem from a completely different angle and offer us new insight how to solve a problem.