- 3.3.1 Adjugate
- 3.3.2 Cramer's Rule
- 3.3.3 Homogeneous Coordinates
- 3.3.4 Area of a Parallelogram and Triangle
- 3.3.5 Collinearity in R² and R³
- 3.3.6 Equation of a Line in R²
- 3.3.7 Area and Volume in $\mathbb{R}^3$: Parallelepiped and Tetrahedron
- 3.3.8 Coplanarity of Four Points in $\mathbb{R}^3$
- 3.3.9 Equation of a Plane in $\mathbb{R}^3$
- 3.3.10 Exercises
- copyleft
Notation¶
Suppose we want to solve the $n\times n$ (square) linear system $A\mathbf{x}=\mathbf{b}$. Any solution $\mathbf{x}$ will describe the output vector $\mathbf{b}$ as a linear combination of the columns of matrix $A$,
$$\mathbf{b} = A\mathbf{x} = \sum_{k=1}^n x_k\mathbf{a}_k$$
In Theorem 3.2.1, part 1, we proved that $A\textrm{adj}(A) = |A|I_n$ by replacing a row of matrix $A$ before computing a determinant. In this section we replace a column of matrix $A$ with the output vector $\mathbf{b}$. There are couple of notations for this. The first notation explains everything. Replace the $i^{\text{th}}$ column of matrix $A$ with vector $\mathbf{b}$ is denoted
$$A\xleftarrow{i}\mathbf{b}.$$
Most textbooks use the shortest notation $A_i$, where
$$A_i = A_i(\mathbf{b}) = A\xleftarrow{i}\mathbf{b}.$$
The shortest notation is okay if you know that vector $\mathbf{b}$ is the same every time. However after you leave this class for Data Science, you will have $M = 1000$ samples $\mathbf{b}^{(j)}$, for $1\le j\le M$. Then keeping track of which $\mathbf{b}^{(j)}$ you are working on is important. I suggest the middle path. Denote replace the $i^{\text{th}}$ column of matrix $A$ with vector $\mathbf{b}^{(j)}$ by
$$A_i\left(\mathbf{b}^{(j)}\right).$$
Textbooks want to keep reading simple. However if you find yourself working in neural networks or statistical data analysis, and have $N\times M$ linear systems $A^{(k)}\mathbf{x}^{(k,j)} = \mathbf{b}^{(j)}$, then you will need to write down exactly what you mean:
$$A_i^{(k)}\left(\mathbf{b}^{(j)}\right)$$
This is where the arrow might come in handy:
$$A^{(k)}\xleftarrow{i}\mathbf{b}^{(j)}$$
In these types of statistical models the superscript enclosed in parenthesis denotes the $k^{\text{th}}$ of $N$ filters or models, and the $j^{\text{th}}$ output vector from several $1\le j\le M$ samples.
Lemma 1¶
If $A$ is an $n\times n$ matrix, then $A\,\textrm{adj}(A) = \textrm{adj}(A)\,A = |A|I_n$.
A lemma is a short theorem that will be used to provide bigger results. What makes this a lemma is that we are going to follow the proof of Theorem 3.2.1-1, almost line-by-line, changing only rows to columns. I have pedagogical reasons for repeating the proof of Theorem 3.2.1-1. However in a more advanced setting I might have simply stated, Refer to the proof of 3.2.1-1, mutatis mutandis. (Latin for "change where appropriate")
Proof of Lemma 1¶
We use the dual proof of Theorem 3.2.1-1.
Suppose that $A$ is an $n\times n$ matrix. In Theorem 3.2.1-1 we proved that $A\,\textrm{adj}(A) = |A|I_n$ expanding $\det(A)$ along row $i$ of matrix $A$. Now we will compute $\det(A)$ expanding along the $i^{\text{th}}$ column of matrix $A$.
$$ \begin{align*} |A| &= \sum_{j=1}^n a_{ji}\textrm{Cof}(a_{ji}) = \sum_{j=1}^n a_{ji}(-1)^{j+i}|A_{ji}| = \sum_{j=1}^n a_{ji}C_{ji} \\ \\ &= a_{1i}C_{1i} + a_{2i}C_{2i} + \cdots + a_{ni}C_{ni} \\ \\ &= \begin{bmatrix} C_{1i} & C_{2i} & \cdots & C_{ni} \end{bmatrix}\begin{bmatrix} a_{1i} \\ a_{2i} \\ \vdots \\ a_{ni} \end{bmatrix} \\ \\ &= \left( \textrm{adj}(A) \right)^i\mathbf{a}_i \end{align*} $$
The determinant of matrix $A$ is the product of the $i^{\text{th}}$ row of $\textrm{adj}(A)$ and the $i^{\text{th}}$ column of matrix $A$. What happens if we multiply the $k^{\text{th}}$ row of matrix $\textrm{adj}(A)$ by the $i^{\text{th}}$ column of matrix $A$? What is
$$\left(\textrm{adj}(A)\right)^k\mathbf{a}_i\ ?$$
As we have seen before, when one expands the determinant along the $i^{\text{th}}$ column of matrix $A$, the values of the $i^{\text{th}}$ column are never used. They are the factors, and not in the minor whose determinant is used to compute the cofactors. We can create a matrix $B$ whose columns are identical to matrix $A$, except for the $k^{\text{th}}$ column.
So for each $1\le k\le n$, consider matrix $B = A\xleftarrow{k}\mathbf{a}_i$. Note that $A = A\xleftarrow{i}\mathbf{a}_i$. For each $1\le k\le n$, the cofactors of the elements of column $k$ are the same for $B = A\xleftarrow{k}\mathbf{a}_i$ and $A$. For $1\le j\le n$,
$$\textrm{Cof}\left((A\xleftarrow{k}\mathbf{a}_i)_{jk}\right) = \textrm{Cof}\left(b_{jk}\right) = \textrm{Cof}(a_{jk}) = C_{jk}.$$
Since a row of the cofactor matrix is a column of the adjugate matrix,
$$\left(\textrm{adj}(A)\right)^k\mathbf{a}_i = \sum_{j=1}^n C_{jk}a_{ji} = \sum_{j=1}^n C_{jk}b_{ji} = \left(\textrm{adj}(B)\right)^k\mathbf{b}_i.$$
Furthermore, since columns $i$ and $k$ are the same in matrix $B = A\xleftarrow{k}\mathbf{a}_i$, we have for $i\neq k$,
$$\left(\textrm{adj}(A)\right)^k\mathbf{a}_i = \left(\textrm{adj}(B)\right)^k\mathbf{b}_i = \left(\textrm{adj}(B)\right)^k\mathbf{b}_k = |B| = 0.$$
Recall that $A\xleftarrow{i}\mathbf{a}_i = A$. Hence $\left(\textrm{adj}(A)\right)^i\mathbf{a}_i = \left| A\xleftarrow{i}\mathbf{a}_i \right| = |A|$. As in Theorem 3.2.1-1,
$$\left(\textrm{adj}(A)\right)^k\mathbf{a}_i = \left\{ \begin{array}{ccc} \det(A) & \text{ if } & i=k \\ 0 & \text{ if } & i \neq k \end{array} \right.$$
Therefore the matrix product,
$$\textrm{adj}(A)\,A = \left[ \left(\textrm{adj}(A) \right)^k\mathbf{a}_i \right] = \left[ \det(A)\delta_{ki} \right] = \det(A)I_n.\ \blacksquare$$
Application of Adjugate¶
Video Lecture 1: Cramer's Rule.¶
Cramer's RuleTheorem 2¶
Cramer's Rule
If $A$ is a non-singular $n\times n$ matrix, $\mathbf{b}$ is an $n\times 1$ vector, and we want to solve $A\mathbf{x}=\mathbf{b}$ for its unique solution, then for $1\le k\le n$,
$$x_k = \frac{\det(A\xleftarrow{k}\mathbf{b})}{\det(A)} = \frac{|A_k(\mathbf{b})|}{|A|}$$
Proof of Theorem 2¶
Suppose $A$ is an $n\times n$ matrix, $\mathbf{b}$ is an $n\times 1$ vector, and we want to solve $A\mathbf{x}=\mathbf{b}$. Multiplying both sides of the equation by $\textrm{adj}(A)$ yields
$$|A|\mathbf{x} = |A|I_n\mathbf{x} = \textrm{adj}(A)\,A\mathbf{x} = \textrm{adj}(A)\mathbf{b} = \textrm{adj}(A)\sum_{k=1}^n x_k\mathbf{a}_k.$$
Using Property 3, determinant is linear in one column as long as the other columns remain identical. Thus
$$ \begin{align*} \det(A\xleftarrow{k}\mathbf{b}) &= \det\left(A\xleftarrow{k}\left(\sum_{i=1}^n x_i\mathbf{a}_i\right)\right) \\ \\ &= \sum_{i=1}^n \det\left( A\xleftarrow{k} \left( x_i\mathbf{a}_i \right) \right)\qquad\text{by Property 3(b)} \\ \\ &= \sum_{i=1}^n x_i\det\left( A\xleftarrow{k} \mathbf{a}_i \right)\qquad\text{by Property 3(a)} \end{align*} $$
Remember in lemma 1, we showed that
$$\det(A\xleftarrow{k}\mathbf{a}_i) = \left\{ \begin{array}{ccc} |A| & \text{ if } & i=k \\ 0 & \text{ if } & i\neq k \end{array} \right.$$
Thus,
$$\det(A\xleftarrow{k}\mathbf{b}) = x_k\det(A).$$
As $A$ is a non-singular matrix we have for $1\le k\le n$,
$$x_k = \frac{\det(A\xleftarrow{k}\mathbf{b})}{\det(A)} = \frac{|A_k(\mathbf{b})|}{|A|}.\ \blacksquare$$
Example 1 - Cramer's Rule¶
Solve the linear system $\begin{bmatrix}\ 1\ &\ 1\ \\ \ 2\ &\ 4\ \end{bmatrix}\mathbf{x} = \begin{bmatrix}\ 4\ \\ \ 2\ \end{bmatrix}$
$$ \begin{align*} x_1 &= \dfrac{|A_1(\mathbf{b})|}{|A|} = \dfrac{\begin{vmatrix}\ 4\ &\ 1\ \\ \ 2\ &\ 4\ \end{vmatrix}}{\begin{vmatrix}\ 1\ &\ 1\ \\ \ 2\ &\ 4\ \end{vmatrix}} = \dfrac{14}{2} = 7 \\ \\ x_2 &= \dfrac{|A_2(\mathbf{b})|}{|A|} = \dfrac{\begin{vmatrix}\ 1\ &\ 4\ \\ \ 2\ &\ 2\ \end{vmatrix}}{\begin{vmatrix}\ 1\ &\ 1\ \\ \ 2\ &\ 4\ \end{vmatrix}} = \dfrac{-6}{2} = -3 \end{align*} $$
The solution $\mathbf{x} = \begin{bmatrix}\ \ 7\ \\ -3\ \end{bmatrix}$.
Exercise 1 - Cramer's Rule¶
Using Cramer's Rule solve
$$ \begin{array}{rcrcrcl} x &-& 3y &+& 3z &=& -1 \\ -4x &+& 4y &-& 3z &=& 2 \\ -4x &-& 3y &+& 5z &=& 0 \end{array} $$
View Solution
$$ \begin{align*} |A| &= \begin{vmatrix}\ \ 1\ & -3\ &\ \ 3\ \\ -4\ &\ \ 4\ & -3\ \\ -4\ & -3\ &\ \ 5\ \end{vmatrix} = \begin{vmatrix}\ \ 1\ & -3\ &\ \ 3\ \\ -4\ &\ \ 4\ & -3\ \\ \ \ 0\ & -7\ &\ \ 8\ \end{vmatrix} = \begin{vmatrix}\ \ 1\ & -3\ &\ \ 3\ \\ \ \ 0\ & -8\ &\ \ 9\ \\ \ \ 0\ & -7\ &\ \ 8\ \end{vmatrix} = 1( -64 + 63 ) = -1 \\ \\ |A_1(\mathbf{b})| &= \begin{vmatrix} -1\ & -3\ &\ \ 3\ \\ \ \ 2\ &\ \ 4\ & -3\ \\ \ \ 0\ & -3\ &\ \ 5\ \end{vmatrix} = \begin{vmatrix} -1\ & -3\ &\ \ 3\ \\ \ \ 0\ & -2\ &\ \ 3\ \\ \ \ 0\ & -3\ &\ \ 5\ \end{vmatrix} = (-1)\begin{vmatrix} -2\ &\ \ 3\ \\ -3\ &\ \ 5\ \end{vmatrix} = (-1)(-10+9)= 1 \\ \\ |A_2(\mathbf{b})| &= \begin{vmatrix}\ \ 1\ & -1\ &\ \ 3\ \\ -4\ &\ \ 2\ & -3\ \\ -4\ &\ \ 0\ &\ \ 5\ \end{vmatrix} = \begin{vmatrix}\ \ 1\ & -1\ &\ \ 3\ \\ -2\ &\ \ 0\ &\ \ 3\ \\ -4\ &\ \ 0\ &\ \ 5\ \end{vmatrix} = -(-1)( -10 + 12 ) = 2 \\ \\ |A_3(\mathbf{b})| &= \begin{vmatrix}\ \ 1\ & -3\ & -1\ \\ -4\ &\ \ 4\ &\ \ 2\ \\ -4\ & -3\ &\ \ 0\ \end{vmatrix} = \begin{vmatrix}\ \ 1\ & -3\ & -1\ \\ -2\ & -2\ &\ \ 0\ \\ -4\ & -3\ &\ \ 0\ \end{vmatrix} = (-1)(6-8) = 2 \\ \\ x &= \frac{|A_1(\mathbf{b})|}{|A|} = \frac{1}{-1} = -1 \\ \\ y &= \frac{|A_2(\mathbf{b})|}{|A|} = \frac{2}{-1} = -2 \\ \\ z &= \frac{|A_3(\mathbf{b})|}{|A|} = \frac{2}{-1} = -2 \\ \\ \\ \mathbf{x} &= \begin{bmatrix} -1\ \\ -2\ \\ -2\ \end{bmatrix} \end{align*} $$\
Solving the Problem of Translations¶
In 
Section 2.1.3 - Linear Transformations, we learned that rotations, shear, dilations, and contractions are examples of linear transformations that can be represented by a matrix. However, translations are not. The only coordinate system we have yet to deal with are Euclidean coordinate systems in $\mathbb{R}^n$ because they are familiar. In order to represent translations with a matrix, one needs a new coordinate system.
In our familiar coordinate system each point in $2$-dimensional space (the plane) consists of two coordinates $(x,y)$ that give us all the information we need to locate the point relative to the origin on the coordinate plane. We represent vectors by their standard coordinates. That means that the tail of the vector is the origin of the coordinate system, and the tip of the vector lies on the point described by the coordinates $\begin{pmatrix} x \\ y \end{pmatrix}$. As we learn more about linear algebra, we will also learn more about rotations, shear, and other linear transformations in $\mathbb{R}^n$. We must also start grappling with new and different coordinate systems of vectors completely different from familiar $\mathbb{R}^n$.
Different coordinate systems will have different geometries and applications. For example, in computer graphics, your avatar must be allowed to move to new positions in the virtual environment. Translations are necessary to provide analogs of humans and objects moving through the environment. In addition to providing motion from one point in space to another, emulating perspective provides important elements of realism to virtual surroundings. As your avatar moves through a simulation, objects that get closer to the avatar's new position appear larger. Objects your avatar moves away from appear smaller.
Recall that an affine transformation in Euclidean coordinates takes the form
$$\begin{pmatrix} x \\ y \end{pmatrix} \mapsto A\begin{pmatrix} x \\ y \end{pmatrix} + \mathbf{b}.$$
Example 2 - Translation+Shear+Dilation¶
Consider $A = \begin{pmatrix} 2 & 1 \\ 0 & 2 \end{pmatrix}$ and $\mathbf{b} = \begin{pmatrix} 3 \\ 2 \end{pmatrix}$. Now let us re-write our matrix and vector in homogeneous coordinates.
$$ \begin{align*} \mathbf{b} = \begin{pmatrix} 3 \\ 2 \end{pmatrix} &\mapsto \begin{bmatrix} 3 \\ 2 \\ 1 \end{bmatrix} = \mathbf{b}', \\ \\ \begin{pmatrix} x \\ y \end{pmatrix} &\mapsto \begin{bmatrix} x \\ y \\ 1 \end{bmatrix} \\ \\ A = \begin{pmatrix} 2 & 1 \\ 0 & 2 \end{pmatrix} &\mapsto \begin{bmatrix} A & \mathbf{b} \\ O & 1 \end{bmatrix} = \begin{bmatrix} 2 & 1 & 3 \\ 0 & 2 & 2 \\ 0 & 0 & 1 \end{bmatrix} = A' \end{align*} $$
Now
$$A'\begin{bmatrix} x \\ y \\ 1 \end{bmatrix} = \begin{bmatrix} A & \mathbf{b} \\ 0 & 1 \end{bmatrix}\begin{bmatrix} x \\ y \\ 1 \end{bmatrix} = \begin{bmatrix} A\begin{pmatrix} x \\ y \end{pmatrix} + \mathbf{b} \\ 1 \end{bmatrix}$$
We can now perform a translation in homogeneous coordinates using matrix multiplications. We will learn in chapter 6, this makes the translation a linear transformation in homogeneous coordinates. We can still perform any other linear transformation by letting $\mathbf{b}=0$. This is why the science of computer graphics uses homogeneous coordinates instead of Euclidean coordinates.
Definition¶
Homogeneous Coordinates
The equation of a line passing through the origin of the Euclidean plane $(0,0)$ and the point $(x,y)$ may be written $yu - xv = 0$. This is the equation of the line in the $uv$-plane (The variables are $u$ and $v$), where $x$ and $y$ are not both zero. In parametric form, the equation of the line is given by
$$ \begin{align*} u &= xt \\ v &= yt. \end{align*} $$
Define $w = 1/t$. Then the coordinates of any point on this line is given by $\left(\,\dfrac{x}{w},\ \dfrac{y}{w}\,\right)$. The homogeneous coordinates for this point is a triple,
$$\left( x,\ y,\ \frac{1}{t} \right) \sim \left( u,\ v,\ 1 \right) .$$
When the third coordinate is $1$, the homogeneous coordinate is said to be normalized. Setting $t=1$ projects the homogeneous coordinates onto our familiar 2-dimensional coordinate system with coordinates $(x,y)$.
A point in projective space is a line passing through the origin of the plane. Since all lines in the plane are parallel to one passing through the origin, all lines with the same slope are the same point in projective space. This is just like our notion that all vectors with the same magnitude and direction, but different initial points (tail) are the same vector.
Moreover, if $t\neq0$ is any real number then the triple $(tx, ty, t)$ are all Euclidean points on the line $yu - xv = 0$. Hence they represent the same line. In a sense, they all represent the same line, so they are valid coordinates for the same point in projective space.
Mastering new coordinate systems is pivotal to this course¶
Homogeneous coordinates are used by computer graphics,
computer vision,
elliptic-curve cryptography, and
robotics
Exercise 2 - Find the 2D Coordinate¶
Find the 2D coordinate of the projective point $(7,8,2)$.
View Solution
Setting $t = \dfrac{1}{2}$, the homogeneous coordinate $(7,8,2) \sim (7t,8t,2t) = \left(\frac{7}{2},\,4,\,1\right) \mapsto \left(\frac{7}{2},\,4\right).$Theorem 3¶
Area of a Parallelogram
The area of the parallelogram spanned by $\mathbf{u}=(u_1,u_2)$ and $\mathbf{v}=(v_1,v_2)$ is
$$\text{Area} = \textrm{abs} \left( \det \left( \begin{bmatrix} u_1 & u_2 \\ v_1 & v_2 \end{bmatrix} \right) \right) = \textrm{abs} \left( \begin{vmatrix} u_1 & u_2 \\ v_1 & v_2 \end{vmatrix} \right).$$
(Remember that determinate is signed, however areas are always positive.)
Proof of Theorem 3¶
We can create a linear transformation such that $\ihat\mapsto\mathbf{u}$ and $\jhat\mapsto\mathbf{v}$. This transformation $\mathbf{y} = A\mathbf{x}$,
$$ \mathbf{y} = \begin{bmatrix} y_1 \\ y_2 \end{bmatrix} = \begin{bmatrix} u_1 & v_1 \\ u_2 & v_2 \end{bmatrix}\begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = A\mathbf{x}. $$
The vectors $\mathbf{u}$ and $\mathbf{v}$ in standard position are adjacent sides of a parallelogram with vertices $(0,0)$, $(u_1, u_2)$, $(v_1,v_2)$, and $(u_1+v_1, u_2+v_2)$. The signed area of this parallelogram divided by the area of the unit square (1) is the determinant of $A$. Hence
$$\text{Area} = \textrm{abs} \left( \begin{vmatrix} u_1 & v_1 \\ u_2 & v_2 \end{vmatrix} \right).$$
Using property 10 of determinants, we have
$$\text{Area} = \textrm{abs} \left( \begin{vmatrix} u_1 & u_2 \\ v_1 & v_2 \end{vmatrix} \right). \blacksquare$$
Theorem 4¶
Area of a Triangle and Parallelogram using Homogeneous Coordinates
The triangle with vertices $P_1=(x_1,y_1)$, $P_2=(x_2,y_2)$, $P_3=(x_3,y_3)$ has area
$$\text{Area of Triangle} = \frac{1}{2} \textrm{abs} \left( \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} \right).$$
The area of the parallelogram whose fourth vertex must be $P_4=(x_2+x_3-x_1,y_2+y_3-y_1)$ has area
$$\text{Area of Parallelogram} = \textrm{abs} \left( \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} \right).$$
Proof of Theorem 4¶
Suppose that a triangle has vertices $P_1=(x_1,y_1)$, $P_2=(x_2,y_2)$, and $P_3=(x_3,y_3)$. Furthermore consider the parallelogram with vertices $P_1$, $P_2$, $P_3$, and $P_4=(x_2+x_3-x_1,y_2+y_3-y_1)$.
Define vectors $\mathbf{u} = \overrightarrow{P_1P_2}$ and $\mathbf{v} = \overrightarrow{P_1P_3}$. Consider the determinant of the $3\times 3$ matrix
$$\begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix}.$$
Sutracting the first row from the second and third row yields
$$\begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} = \begin{vmatrix} x_1 & y_1 & 1 \\ x_2-x_1 & y_2-y_1 & 0 \\ x_3-x_1 & y_3-y_1 & 0 \end{vmatrix} = \begin{vmatrix} x_2-x_1 & y_2-y_1 \\ x_3-x_1 & y_3-y_1 \end{vmatrix} = \begin{vmatrix} u_1 & u_2 \\ v_1 & v_2 \end{vmatrix}.$$
By Theorem 3, the absolute value of the last determinant is the area of the parallelogram. The area of the triangle subtended by $P_1$, $P_2$, and $P_3$ is half of the area of the parallelogram. $\blacksquare$
Example 3 - Find Area¶
Find the Area of the triangle with vertices $(2,1)$, $(8,3)$, and $(5,7)$:
$$\frac12\,\textrm{abs}\left( \begin{vmatrix} 2 & 1 & 1 \\ 8 & 3 & 1 \\ 5 & 7 & 1 \end{vmatrix} \right) = \frac12\,\textrm{abs}\left( \begin{vmatrix} 2 & 1 & 1 \\ 6 & 2 & 0 \\ 3 & 6 & 0 \end{vmatrix} \right) = \frac12\,\textrm{abs}\left( \begin{vmatrix} 6 & 2 \\ 3 & 6 \end{vmatrix} - 0 + 0 \right) = \frac12\,|36 - 6| = 15.$$
Exercise 3 - Find Area¶
Find the area of the parallelogram with vertices $P_1=(1,8)$, $P_2=(8,3)$ and $P_3=(6,5)$.
View Solution
$$\textrm{abs}\left( \begin{vmatrix} 1 & 8 & 1 \\ 8 & 3 & 1 \\ 6 & 5 & 1 \end{vmatrix} \right) = \textrm{abs} \left( \begin{vmatrix} 1 & 8 & 1 \\ 7 & -5 & 0 \\ 5 & -3 & 0 \end{vmatrix} \right) = \textrm{abs} \left( \begin{vmatrix} 7 & -5 \\ 5 & -3 \end{vmatrix} \right) = | -21 + 25 | = 4 $$Theorem 5¶
Determine Collinearity on the Plane
Three points $P_1=(x_1,y_1)$, $P_2=(x_2,y_2)$, and $P_3=(x_3,y_3)$ in $\mathbb{R}^2$ are collinear if and only if the triangle they form has zero area:
$$\begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} = 0.$$
Proof of Theorem 5¶
It should be clear that the area of the triangle has area zero if and only if the height of the triangle is zero; not a triangle at all. $\blacksquare$
Example 4 - Determine Collinearity¶
Determine if points $P_1=(1,1,1)$, $P_2=(2,3,4)$, and $P_3=(4,7,10)$ are collinear in $\mathbb{R}^3$.
Define $\mathbf{u} = \overrightarrow{P_1P_2} = \langle 1,2,3 \rangle$ and $\mathbf{v} = \overrightarrow{P_1P_3} = \langle 3,6,9 \rangle$. $\mathbf{v} = 3\mathbf{u}$, so $\mathbf{v}$ is on the line spanned by $\mathbf{u}$. Therefore, they are collinear.
Exercise 4 - Determine Collinearity¶
Determine if the points $P_1=(8,8)$, $P_2=(2,4)$, and $P_3=(8,7)$ are collinear.
View Solution
$$\textrm{abs} \left( \begin{vmatrix} 8 & 8 & 1 \\ 2 & 4 & 1 \\ 8 & 7 & 1 \end{vmatrix} \right) = \textrm{abs} \left( \begin{vmatrix} 8 & 8 & 1 \\ -6 & -4 & 0 \\ 0 & -1 & 0 \end{vmatrix} \right) = \textrm{abs} \left( \begin{vmatrix} -6 & -4 \\ 0 & -1 \end{vmatrix} \right) = |6 - 0| = 6 \neq 0 $$The points are not collinear because the area of the triangle is positive $3$.
Theorem 6¶
Equation of a Line in $\mathbb{R}^2$
The equation of the line that passes through points $P_1=(x_1,y_1)$ and $P_2=(x_2,y_2)$ in $\mathbb{R}^2$ is given by the determinant,
$$\begin{vmatrix} x & y & 1 \\ x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \end{vmatrix} = 0.$$
Proof of Theorem 6¶
The collinearity test is the equation of the line once you let one point be variable. A point $(x,y)$ lies on the line through $P_1,P_2$ exactly when $(x,y),P_1,P_2$ are collinear. $\blacksquare$
Example 5 - Determine the Equation of the Line¶
Determine the equation of the line through $(-1,2)$ and $(3,-4)$:
$$ \begin{align*} \begin{vmatrix} x & y & 1 \\ -1 & 2 & 1 \\ 3 & -4 & 1 \end{vmatrix} &= 0 \\ \\ x\begin{vmatrix} 2 & 1 \\ -4 & 1 \end{vmatrix} - y\begin{vmatrix} -1 & 1 \\ 3 & 1 \end{vmatrix} + 1\begin{vmatrix} -1 & 2 \\ 3 & -4 \end{vmatrix} &= 0 \\ \\ x(2+4) - y(-1-3) + (4 - 6) &= 0 \\ \\ 6x + 4y - 2 &= 0 \\ \\ \Longrightarrow\, 3x + 2y &= 1. \end{align*} $$
Exercise 5 - Determine the Equation of the Line¶
Determine the Equation of the line that passes through the points $P_1=(4,8)$ and $P_2=(8,6)$.
View Solution
$$ \begin{align*} \begin{vmatrix} x & y & 1 \\ 4 & 8 & 1 \\ 8 & 6 & 1 \end{vmatrix} = 0 \\ \\ x\begin{vmatrix} 8 & 1 \\ 6 & 1 \end{vmatrix} - y\begin{vmatrix} 4 & 1 \\ 8 & 1 \end{vmatrix} + 1\begin{vmatrix} 4 & 8 \\ 8 & 6 \end{vmatrix} &= 0 \\ \\ x(8 - 6) - y(4 - 8) + (24 - 64) &= 0 \\ \\ 2x + 4y - 40 &= 0 \\ \\ \Longrightarrow\, x + 2y &= 20 \end{align*} $$Theorem 7¶
Volume of a Parallelepiped
The volume of the parallelepiped with sides $\mathbf{u}$, $\mathbf{v}$, and $\mathbf{w}$ in $\mathbb{R}^3$ equals the absolute value of the $3\times 3$ determinant:
$$V = \textrm{abs} \left( \begin{vmatrix} u_1 & u_2 & u_3 \\ v_1 & v_2 & v_3 \\ w_1 & w_2 & w_3 \end{vmatrix} \right).$$
Proof of Theorem 7¶
This proof will follow the steps of the proof of Theorem 3, mutatis mutandis. We construct a linear transformation such that $\ihat\mapsto\mathbf{u}$, $\jhat\mapsto\mathbf{v}$, and $\khat\mapsto\mathbf{w}$. The matrix for this transformation is given by $\mathbf{y}=A\mathbf{x}$, where
$$\mathbf{y} = \begin{bmatrix} y_1 \\ y_2 \\ y_3 \end{bmatrix} = \begin{bmatrix} u_1 & v_1 & w_1 \\ u_2 & v_2 & w_2 \\ u_3 & v_3 & w_3 \end{bmatrix}\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = A\mathbf{x}.$$
The determinant of this matrix is the signed ratio of the volume of the parallelepiped divided by the volume of the unit cube (1). Hence,
$$\text{Volume of parallelepiped} = \textrm{abs}\left( \begin{vmatrix}u_1 & v_1 & w_1 \\ u_2 & v_2 & w_2 \\ u_3 & v_3 & w_3 \end{vmatrix} \right).$$
By property 10 of determinants,
$$\text{Volume of parallelepiped} = \textrm{abs}\left( \begin{vmatrix} u_1 & u_2 & u_3 \\ v_1 & v_2 & v_3 \\ w_1 & w_2 & w_3 \end{vmatrix} \right).\ \blacksquare$$
Theorem 8¶
Volume of a Tetrahedron
The tetrahedron with vertices $P_1=(x_1,y_1,z_1)$, $P_2=(x_2,y_2,z_2)$, $P_3=(x_3,y_3,z_3)$, and $P_4=(x_4,y_4,z_4)$ in $\mathbb{R}^3$ has volume
$$V = \frac{1}{6} \textrm{abs} \left( \begin{vmatrix} x_1 & y_1 & z_1 & 1 \\ x_2 & y_2 & z_2 & 1 \\ x_3 & y_3 & z_3 & 1 \\ x_4 & y_4 & z_4 & 1 \end{vmatrix} \right).$$
Proof of Theorem 8¶
This proof follows the same steps as the proof of Threorem 4, mutatis mutandis.
Define $\mathbf{u}=P_2-P_1$, $\mathbf{v}=P_3-P_1$, and $\mathbf{w}=P_4-P_1$. The same row reduction as in Theorem 4, subtract row 1 from the others and expand the determinant along the last column, reduces the $4\times4$ form to the $3\times3$ edge determinant. Since the volume of the tetrahedron is one-sixth of the volume of the parallelepiped with the same three edges, we have our formula. $\blacksquare$
Example 6 - Compute the Volume¶
Determine volumes of the parallelepiped and the tetrahedron with edges $\mathbf{u}=\langle 2,0,0 \rangle$, $\mathbf{v}=\langle 1,3,0 \rangle$, and $\mathbf{w}=\langle 1,1,4\rangle$.
$$\text{parallelepiped volume} =\textrm{abs}\left( \begin{vmatrix} 2 & 0 & 0 \\ 1 & 3 & 0 \\ 1 & 1 & 4 \end{vmatrix} \right) = 2(3)(4) = 24,$$
tetrahedron volume $=24/6=4$.
Exercise 6 - Compute the Volume¶
Find the volume of the parallelepiped with edge vectors $\langle 6, -5, 4 \rangle$, $\langle 0, -1, 6 \rangle$, and $\langle 4, 5, 2 \rangle$.
View Solution
$$ \begin{align*} \text{parallelepiped volume} &= \textrm{abs}\left( \begin{vmatrix}\ \ 6\ & -5\ &\ \ 4\ \\ \ \ 0\ & -1\ &\ \ 6\ \\ \ \ 4\ &\ \ 5\ &\ \ 2\ \end{vmatrix} \right) \\ \\ &= \textrm{abs}\left( \begin{vmatrix}\ \ 2\ & -10\ &\ \ 2\ \\ \ \ 0\ & -1\ &\ \ 6\ \\ \ \ 4\ &\ \ 5\ &\ \ 2\ \end{vmatrix} \right) \\ \\ &= \textrm{abs}\left( \begin{vmatrix}\ \ 2\ & -10\ &\ \ 2\ \\ \ \ 0\ & -1\ &\ \ 6\ \\ \ \ 0\ &\ 25\ & -2\ \end{vmatrix} \right) \\ \\ &= \textrm{abs}\left( 2\begin{vmatrix} -1\ &\ \ 6\ \\ \ 25\ & -2\ \end{vmatrix} - 0 + 0 \right) \\ \\ &= \textrm{abs}\left( 2(2 - 150) \right) = \textrm{abs}\left( 2(-148) \right) = \textrm{abs}(-296) = 296 \end{align*} $$Theorem 9¶
Coplanarity
Four points $P_1=(x_1,y_1,z_1)$, $P_2=(x_2,y_2,z_2)$, $P_3=(x_3,y_3,z_3)$, and $P_4=(x_4,y_4,z_4)$ in $\mathbb{R}^3$ are coplanar if and only if the tetrahedron they span has zero volume,
$$\begin{vmatrix} x_1 & y_1 & z_1 & 1 \\ x_2 & y_2 & z_2 & 1 \\ x_3 & y_3 & z_3 & 1 \\ x_4 & y_4 & z_4 & 1 \end{vmatrix} = 0.$$
This is the exact 3-dimensional analogue of Theorem 5, collinear $\Leftrightarrow$ zero area, and coplanar $\Leftrightarrow$ zero volume.
Example 7 - Determine whether points are Coplanar¶
Determine if the points $(0,0,0)$, $(1,2,3)$, $(2,1,0)$, and $(3,3,3)$ are coplanar.
$$\begin{vmatrix} 0 & 0 & 0 & 1 \\ 1 & 2 & 3 & 1 \\ 2 & 1 & 0 & 1 \\ 3 & 3 & 3 & 1 \end{vmatrix} = 0 - 0 + 0 - 1\begin{vmatrix} 1 & 2 & 3 \\ 2 & 1 & 0 \\ 3 & 3 & 3 \end{vmatrix} $$
Since row 3 of the last $3\times 3$ matrix is the sum of rows 1 and 2, the determinant is zero. $\Longrightarrow$ The points are coplanar. That is to say that the matrix of homogeneous points is singular (degenerate) in $\mathbb{R}^4$.
Exercise 7 - Determine whether points are Coplanar¶
Determine if the points $(-6,3,-4)$, $(5,3,3)$, $(6,-1,-6)$, and $(1,-4,-9)$ are coplanar.
View Solution
Consider the $4\times4$ determinant $$ \begin{align*} \begin{vmatrix} -6\ &\ \ 3\ & -4\ &\ \ 1\ \\ \ \ 5\ &\ \ 3\ &\ \ 3\ &\ \ 1\ \\ \ \ 6\ & -1\ & -6\ &\ \ 1\ \\ \ \ 1\ & -4\ & -9\ &\ \ 1\ \end{vmatrix} &= \begin{vmatrix} -7\ &\ \ 7\ &\ \ 5\ &\ \ 0\ \\ \ \ 4\ &\ \ 7\ &\ 12\ &\ \ 0\ \\ \ \ 5\ &\ \ 3\ &\ \ 3\ &\ \ 0\ \\ \ \ 1\ & -4\ & -9\ &\ \ 1\ \end{vmatrix} \\ \\ &= -0 + 0 - 0 + 1\begin{vmatrix} -7\ &\ \ 7\ &\ \ 5\ \\ \ \ 4\ &\ \ 7\ &\ 12\ \\ \ \ 5\ &\ \ 3\ &\ \ 3\ \end{vmatrix} \\ \\ &= \begin{vmatrix}\ \ 0\ &\ \ 7\ & -2\ \\ \ 11\ &\ \ 7\ &\ \ 5\ \\ \ \ 8\ &\ \ 3\ &\ \ 0\ \end{vmatrix} = \begin{vmatrix}\ \ 0\ &\ \ 7\ & -2\ \\ \ 11\ &\ 21\ &\ \ 1\ \\ \ \ 8\ &\ \ 3\ &\ \ 0\ \end{vmatrix} \\ \\ &= \begin{vmatrix}\ 22\ &\ 49\ &\ \ 0\ \\ \ 11\ &\ 21\ &\ \ 1\ \\ \ \ 8\ &\ \ 3\ &\ \ 0\ \end{vmatrix} = 0 - 1\begin{vmatrix}\ 22\ &\ 49\ \\ \ \ 8\ &\ \ 3\ \end{vmatrix} - 0 \\ \\ &= -1\begin{vmatrix} -2\ &\ 40\ \\ \ \ 8\ &\ \ 3\ \end{vmatrix} = -1\begin{vmatrix} -2\ &\ 40 \\ \ \ 0\ &\ 163\ \end{vmatrix} \\ \\ &= -1(-326) = 326 \\ \\ &\Longrightarrow \text{NOT coplanar} \end{align*} $$Theorem 10¶
Equation of a Plane
The equation of the plane that passes through points $P_1=(x_1,y_1,z_1)$, $P_2=(x_2,y_2,z_2)$, and $P_3=(x_3,y_3,z_3)$ in $\mathbb{R}^3$ is given by the determinant,
$$\begin{vmatrix} x & y & z & 1 \\ x_1 & y_1 & z_1 & 1 \\ x_2 & y_2 & z_2 & 1 \\ x_3 & y_3 & z_3 & 1 \end{vmatrix} = 0.$$
Just as in the proof of theorem 6, let one point be variable. A point $(x,y,z)$ lies on the plane through $P_1$, $P_2$, and $P_3$ exactly when all four points are coplanar.
Example 8 - Find the Equation of the Plane¶
Find the equation of the plane that passes through the points $(1,2,3)$, $(0,1,1)$, and $(2,0,1)$.
$$ \begin{align*} \begin{vmatrix} x & y & z & 1 \\ 1 & 2 & 3 & 1 \\ 0 & 1 & 1 & 1 \\ 2 & 0 & 1 & 1 \end{vmatrix} &= x \begin{vmatrix} 2 & 3 & 1 \\ 1 & 1 & 1 \\ 0 & 1 & 1 \end{vmatrix} - y\begin{vmatrix} 1 & 3 & 1 \\ 0 & 1 & 1 \\ 2 & 1 & 1 \end{vmatrix} + z\begin{vmatrix} 1 & 2 & 1 \\ 0 & 1 & 1 \\ 2 & 0 & 1 \end{vmatrix} - 1\begin{vmatrix} 1 & 2 & 3 \\ 0 & 1 & 1 \\ 2 & 0 & 1 \end{vmatrix} \\ \\ &= x\begin{vmatrix} 2 & 3 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 1 \end{vmatrix} - y\begin{vmatrix} 1 & 3 & 1 \\ 0 & 1 & 1 \\ 2 & 0 & 0 \end{vmatrix} + z\begin{vmatrix} 1 & 2 & 1 \\ 0 & 1 & 1 \\ 0 & -4 & -1 \end{vmatrix} - 1\begin{vmatrix} 1 & 2 & 3 \\ 0 & 1 & 1 \\ 0 & -4 & -5 \end{vmatrix} \\ \\ &= x\left(-1\begin{vmatrix} 3 & 1 \\ 1 & 1 \end{vmatrix} + 0 - 0\right) - y\left(2\begin{vmatrix} 3 & 1 \\ 1 & 1 \end{vmatrix} - 0 + 0 \right) + z\left(-1\begin{vmatrix} 1 & 1 \\ 4 & 1 \end{vmatrix} - 0 + 0 \right) - 1\left(-1\begin{vmatrix} 1 & 1 \\ 4 & 5 \end{vmatrix}\right) \\ \\ &= x(-2) - y(4) + z(3) - 1(-1) \\ \\ &= -2x - 4y + 3z + 1 = 0 \\ \\ &\Longrightarrow 2x + 4y - 3z = 1 \end{align*} $$
Exercise 8 - Find the Equation of the Plane¶
Find the equation of the plane that passes through the points $(2, 0, 2)$, $(1, 0, 1)$, and $(-3, 4, -2)$.
View Solution
$$ \begin{align*} \begin{vmatrix}\ \ x\ &\ \ y\ &\ \ z\ &\ \ 1\ \\ \ \ 2\ &\ \ 0\ &\ \ 2\ &\ \ 1\ \\ \ \ 1\ &\ \ 0\ &\ \ 1\ &\ \ 1\ \\ -3\ &\ \ 4\ & -2\ &\ \ 1\ \end{vmatrix} &= x\begin{vmatrix}\ \ 0\ &\ \ 2\ &\ \ 1\ \\ \ \ 0\ &\ \ 1\ &\ \ 1\ \\ \ \ 4\ & -2\ &\ \ 1\ \end{vmatrix} - y\begin{vmatrix}\ \ 2\ &\ \ 2\ &\ \ 1\ \\ \ \ 1\ &\ \ 1\ &\ \ 1\ \\ -3\ & -2\ &\ \ 1\ \end{vmatrix} + z\begin{vmatrix}\ \ 2\ &\ \ 0\ &\ \ 1\ \\ \ \ 1\ &\ \ 0\ &\ \ 1\ \\ -3\ &\ \ 4\ &\ \ 1\ \end{vmatrix} - 1\begin{vmatrix}\ \ 2\ &\ \ 0\ &\ \ 2\ \\ \ \ 1\ &\ \ 0\ &\ \ 1\ \\ -3\ & \ \ 4\ & -2\ \end{vmatrix} \\ \\ &= 4x\begin{vmatrix}\ 2\ &\ 1\ \\ \ 1\ &\ 1\ \end{vmatrix} - y\begin{vmatrix}\ \ 1\ &\ \ 1\ &\ \ 0\ \\ \ \ 1\ &\ \ 1\ &\ \ 1\ \\ -4\ & -3\ &\ \ 0\ \end{vmatrix} - 4z\begin{vmatrix}\ 2\ &\ 1\ \\ \ 1\ &\ 1\ \end{vmatrix} + 4\begin{vmatrix}\ 2\ &\ 2\ \\ \ 1\ &\ 1\ \end{vmatrix} \\ \\ &= 4x(2 - 1) + y\begin{vmatrix}\ \ 1\ &\ \ 1\ \\ -4\ & -3\ \end{vmatrix} - 4z(2 - 1) + 4(0) \\ \\ &= 4x + y(-3+4) - 4z = 0 \\ \\ &\Longrightarrow 4x + y - 4z = 0 \end{align*} $$Exercise 9 - Find the Equation of the Circle¶
Find the circle through $(0,0)$, $(2,0)$, $(0,2)$, then give its center and radius.
One may include the quadratic terms using homogeneous coordinates similar to Theorem 10. For three non-collinear points, compute
$$\begin{vmatrix} x^2+y^2 & x & y & 1 \\ x_1^2+y_1^2 & x_1 & y_1 & 1 \\ x_2^2+y_2^2 & x_2 & y_2 & 1 \\ x_3^2+y_3^2 & x_3 & y_3 & 1 \end{vmatrix} = 0$$
