## Matrices



My notes of the excellent lectures of “Denis Auroux. 18.02 Multivariable Calculus. Fall 2007. Massachusetts Institute of Technology: MIT OpenCourseWare, https://ocw.mit.edu. License: Creative Commons BY-NC-SA.”

Matrices can be used to express linear relations between variables. For example when we change coordinate systems from eg. $$(x_1,x_2,x_3)$$ to $$(u_1,u_2,u_3)$$ where \left\{ \begin{align} u_1 &= 2x_1+3x_2+3x_3 \nonumber \\ u_2 &= 2x_1+4x_2+5x_3 \nonumber \\ u_3 &= x_1+x_2+2x_3 \nonumber \end{align} \right. \label{eq:linear}

Expressed as matrix product \begin{align*} \underbrace{ \left[ \begin{matrix} 2 & 3 & 3 \\ 2 & 4 & 5 \\ 1 & 1 & 2 \end{matrix} \right] }_{A}\; \underbrace{ \left[ \begin{matrix} x_1 \\ x_2 \\ x_3 \end{matrix} \right] }_{X} &= \underbrace{ \left[ \begin{matrix} u_1 \\ u_2 \\ u_3 \end{matrix} \right] }_{U} \ A X &= U \end{align*} Here $$A$$ is a $$3\times 3$$ matrix, and $$X$$ is a vector or a $$3\times 1$$ matrix.

## Matrix Multiplication

Definition:

The entries in $$A X$$ are the dot-product between the rows in $$A$$ and the columns in $$X$$, as shown below

For example, the entries of $$AB$$ are $$\left[ \begin{matrix} 1 & 2 & 3 & 4 \\ \cdot & \cdot & \cdot & \cdot \\ \cdot & \cdot & \cdot & \cdot \\ \cdot & \cdot & \cdot & \cdot \\ \cdot & \cdot & \cdot & \cdot \end{matrix} \right]\; \left[ \begin{matrix} 0 & \cdot \\ 3 & \cdot \\ 0 & \cdot \\ 2 & \cdot \end{matrix} \right] = \left[ \begin{matrix} 14 & \cdot \\ \cdot & \cdot \\ \cdot & \cdot \end{matrix} \right]$$

Properties:

• The width of $$A$$ must equal the height of $$B$$.
• The product $$AB$$ has the same height as $$A$$ and the same width as $$B$$.
• Product $$AB$$ represents: do transformation $$B$$, then transformation $$A$$. Unfortunately, you multiply from right to left. Similar to $$f(g(x))$$, where you first apply $$g$$ and then $$f$$. The product $$BA$$ is not even be defined when the width of $$B$$ is not equal to the height of $$A$$. In other words $$AB\ne BA$$
• They are well behaved associative products: $$(AB)X =A(BX)$$
• $$BX$$ means we apply transformation $$B$$ to $$X$$.

## Identity matrix

Definition:

The identify matrix is a matrix that does no transformation: $$IX=X$$

The height of $$I$$ needs to match the width of $$X$$. $$I$$ has $$1$$’s on the diagonal, and $$0$$’s everywhere else. $$I_{n\times n} = \left[ \begin{matrix} 1 & & & \ldots & 0 \\ & 1 & & & \vdots \\ & & 1 & & \\ \vdots & & & \ddots & \\ 0 & \ldots & & & 1 \end{matrix} \right] \nonumber$$

For example: $$I_{3\times3} = \left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \right] \nonumber$$

### Rotation

Matrix $$R$$, gives a $$\frac{\pi}{2}$$ rotation. $$R = \left[ \begin{matrix} 0 & -1 \\ 1 & 0 \end{matrix} \right] \nonumber$$

In general $$R \left[ \begin{matrix} x \\ y \end{matrix} \right] = \left[ \begin{array}{r} -y \\ x \end{array} \right] \nonumber$$

Try multiplying with unity vector $$\hat\imath$$, $$\hat\jmath$$, or take $$R$$ squared \begin{align*} R\; \hat\imath &= \left[ \begin{array}{rr} 0 & -1 \\ 1 & 0 \end{array} \right] \left[ \begin{matrix} 1 \\ 0 \end{matrix} \right] = \left[ \begin{matrix} 0 \\ 1 \end{matrix} \right] = \hat\jmath \\ R\;\hat\jmath &= \left[ \begin{array}{rr} 0 & -1 \\ 1 & 0 \end{array} \right] \left[ \begin{matrix} 0 \\ 1 \end{matrix} \right] = \left[ \begin{array}{r} -1 \\ 0 \end{array} \right] = -\hat\imath \\ R^2 &= \left[ \begin{array}{rr} 0 & -1 \\ 1 & 0 \end{array} \right] \left[ \begin{array}{rr} 0 & -1 \\ 1 & 0 \end{array} \right] = \left[ \begin{array}{rr} -1 & 0 \\ 0 & -1 \end{array} \right] = -I_{2\times 2} \end{align*} \nonumber

## Inverse Matrix

Definition:

The inverse of matrix $$A$$ is $$A^{-1}$$ such that \shaded{ \left\{ \begin{align*} A\;A^{-1} &= I \\ A^{-1}\;A &= I \end{align*} \right. } \nonumber
That implies that $$A$$ must be a square matrix ($$n \times n$$).

Referring to the system of equations $$\eqref{eq:linear}$$, to express variables $$u_i$$ in terms of $$x_i$$ values, we need to inverse the transformation. For instance: in $$AX=B$$; let matrix $$A$$ and $$B$$ be known what is $$X$$? \begin{align*} AX &= B \Rightarrow \\ A^{-1}(AX) &= A^{-1} B \Rightarrow \\ IX &= A^{-1} B \Rightarrow \\ X &= A^{-1} B \end{align*}

### Method

The inverse matrix is calculated using the adjoined matrix

$$A^{-1}=\frac{1}{\mathrm{det}(A)}\;\mathrm{adj}(A) \nonumber$$

For this $$3\times 3$$ example $$A=\left[ \begin{matrix} 2 & 3 & 3 \\ 2 & 4 & 5 \\ 1 & 1 & 2 \end{matrix} \right]$$

First, find the determinant of $$A$$ $$\det(A)= \left| \begin{array}{rrr} 2 & 3 & 3 \\ 2 & 4 & 5 \\ 1 & 1 & 2 \end{array} \right| = 3 \nonumber$$

Second, find the minors (matrix of determinants) of matrix $$A$$ $$\mathrm{minors} = \left[\begin{array}{rrr} \left|\begin{array}{rrr} 4 & 5 \\ 1 & 2 \end{array}\right| & \left|\begin{array}{rrr} 2 & 5 \\ 1 & 2 \end{array}\right| & \left|\begin{array}{rrr} 2 & 4 \\ 1 & 1 \end{array}\right| \\ \left|\begin{array}{rrr} 3 & 3 \\ 1 & 2 \end{array}\right| & \left|\begin{array}{rrr} 2 & 3 \\ 1 & 2 \end{array}\right| & \left|\begin{array}{rrr} 2 & 3 \\ 1 & 1 \end{array}\right| \\ \left|\begin{array}{rrr} 3 & 3 \\ 4 & 5 \end{array}\right| & \left|\begin{array}{rrr} 2 & 3 \\ 2 & 5 \end{array}\right| & \left|\begin{array}{rrr} 2 & 3 \\ 2 & 4 \end{array}\right| \end{array}\right] = \left[\begin{array}{rrr} 3 & -1 & -2 \\ 3 & 1 & -1 \\ 3 & 4 & 2 \end{array}\right] \nonumber$$

Third, find the cofactors. Flip the signs checker board $$\begin{array}{rrr} + & – & + \\ – & + & – \\ + & – & + \end{array} \nonumber$$ A ‘$$+$$’ means leave it alone. A ‘$$-$$’ means flip the sign. Apply the cofactors to the minors. $$\left[\begin{array}{rrr} 3 & 1 & -2 \\ -3 & 1 & 1 \\ 3 & -4 & 2 \end{array}\right] \nonumber$$

Fourth, transpose (switch rows and columns) to find the adjoined matrix $$\mathrm{adj}(A)$$. $$\mathrm{adj}(A) = \left[\begin{array}{rrr} 3 & -3 & 3 \\ 1 & 1 & -4 \\ -2 & 1 & 2 \end{array}\right] \nonumber$$

The inverse matrix $$A^{-1}$$ follows as $$A^{-1} = \frac{1}{\det(A)}\;\mathrm{adj}(A) = \frac{1}{3} \left[\begin{array}{rrr} 3 & -3 & 3 \\ 1 & 1 & -4 \\ -2 & 1 & 2 \end{array}\right] \nonumber$$

## Equations of planes

An equation of the form $$ax+by+cz=d$$, expresses the condition for the point $$(x,y,z)$$ to be in the plane. It defines a plane.

### Examples

#### Plane through the origin

Find the equation of the plane through the origin with normal vector $$\vec N = \left\langle 1, 5, 10 \right\rangle$$.

Point $$P=(x,y,z)$$ is in the plane when $$\vec{OP}\perp\vec{N}$$. Therefore, their dot-product must equal zero (see vectors). \begin{align*} \overrightarrow{OP}\cdot\vec{N} = 0 \\ \Leftrightarrow \left\langle x, y, z \right\rangle \cdot \left\langle 1, 5, 10 \right\rangle = 0 \\ \Leftrightarrow x + 5y + 10z = 0 \end{align*}

#### Plane not through the origin

Find the equation of the plane through $$P_0=(2,1,-1)$$ with normal vector $$\vec N = \left\langle 1, 5, 10 \right\rangle$$.

The normal vector is the same as in the first example, therefore it will be the same plane, but shifted so that it passes through $$P_0$$.

Point $$P=(x,y,z)$$ is in the plane when $$\overrightarrow{P_0P}\perp\overrightarrow{N}$$. Therefore, their dot-product must equal zero (see vectors). This vector $$\overrightarrow{P_0P}$$ equals $$P-P_0$$. \begin{align*} \left\langle x-2, y-1, z+1 \right\rangle \cdot \left\langle 1, 5, 10 \right\rangle &= 0 \\ \Leftrightarrow (x-2)+5(y-1)+10(z+1) &= 0 \\ \Leftrightarrow \underline{1}x+\underline{5}y+\underline{10}z &= -3 \end{align*}

In the equation $$ax+by+cz=d$$, the coefficients $$\left\langle a,b,c\right\rangle$$ is the normal vector $$\vec{N}$$. Constant $$d$$ indicates how far the plane is from the origin.

##### How could we have found the $$-3$$ more quickly?

The first part of the equation is based on the normal vector $$x + 5y + 10z = d \label{eq:planeequations2a}$$

We know $$P_0$$ is in the plane. Substituting $$\left\langle x,y,z\right\rangle=P_0$$ in $$\eqref{eq:planeequations2a}$$ \begin{align*} 1(2)+5(1)+10(-1) &= d \\ \Leftrightarrow d &= -3 \end{align*} \nonumber

#### Parallel or perpendicular?

Are vector $$\vec{v}=\left\langle 1,2,-1 \right\rangle$$ and plane $$x+y+3z=5$$ parallel, perpendicular or neither?

Vector $$\vec{v}$$ is perpendicular to the plane when $$\vec{v}$$=$$s\;\vec{N}$$, where $$s$$ is a scalar. The normal vector follows from the coefficients of the plane equation $$\vec{N} = \left\langle 1,1,3 \right\rangle \nonumber$$ Therefore $$\vec{V}$$ is not perpendicular to the plane.

If $$\vec{v}$$ is perpendicular to $$\vec{N}$$, it is parallel to the plane. $$\vec{v}\perp\vec{N}$$ when the dot-product equals zero. (see vectors) \begin{align*} \vec{v}\cdot\vec{N} &= \left\langle 2, 1, -1 \right\rangle \cdot \left\langle 1, 1, 3 \right\rangle \\ &= 1+2-3 = 0 \end{align*} Therefore, $$\vec{v}$$ is parallel to the plane.

## Solving systems of equations

To solve a system of equations, you try to find a point that is on several planes at the same time.

### Example

Find the $$x,y,z$$ that satisfies the conditions of the $$3\times 3$$ linear system: \left\{ \begin{align*} x+ z = 1 \\ x + y = 2 \\ x + 2y + 3z = 3 \end{align*} \right.

The first 2 equations represent two planes that intersect in line $$P_1\cap P_2$$. The third plane intersects that line at the point $$P(x,y,z)$$, the solution to the linear system.

Unless:

• if the line $$P_1\cap P_2$$ is contained in $$P_3$$, there are infinite many solutions. (Any point on the line is a solution.)
• if the line $$P_1\cap P_2$$ is parallel to $$P_3$$, then there are no solutions.

In matrix notation $$\underbrace{ \left[\begin{array}{rrr} 1 & 0 & 1 \\ 1 & 1 & 0 \\ 1 & 2 & 3 \end{array}\right] }_{A}\; \underbrace{ \left[\begin{array}{ccc} x \\ y \\ z \end{array}\right] }_{X} = \underbrace{ \left[\begin{array}{rrr} 1 \\ 2 \\ 3 \end{array}\right] }_{B} \nonumber$$

The solution to $$AX=B$$ is given by (see Inverse matrix) $$X = A^{-1}B \nonumber$$

Recall

$$A^{-1}=\frac{1}{\det (A)}\mathrm{adj}(A) \nonumber$$

This implies that matrix $$A$$ is only invertible when $$\shaded{ \det (A)\ne 0 } \nonumber$$

### Theory

#### Homogeneous case

Homogeneous means that equations are invariant under scaling. In matrix notation: $$AX=0$$.

For example: \left\{ \begin{align*} x + z = 0 \\ x + y = 0 \\ x + 2y + 3z = 0 \end{align*} \right.

There is always the trivial solution: $$(0,0,0)$$.

Depending on the $$\det(A)$$:

• If the $$\det (A)\ne 0$$: $$A$$ can be inverted. $$AX=0 \Leftrightarrow X=A^{-1}.0=0$$. No other solutions.
• If the $$\det (A)= 0$$: the determinant of $$\vec{N_1},\vec{N_2},\vec{N_3}$$ equals $$0$$. This implies that the plane’s normal vectors $$\vec{N_1}$$, $$\vec{N_2}$$ and $$\vec{N_3}$$ are coplanar. A line through origin, perpendicular to plane of $$\vec{N_1}, \vec{N_2}, \vec{N_3}$$ is parallel to all 3 planes and contained in them. Therefore there are infinite many solutions. To find the solutions, one can take the cross-product of two of the normals. It’s a nontrivial solution.

### General case

The system $$AX=B \nonumber$$

Depending on the $$\det(A)$$

• if the $$\det {A}\ne 0$$: there is an unique solution $$X=A^{-1}B$$
• if the $$\det {A}=0$$: either no solution, or infinitely many solutions. If you would solve it by hand and end up with $$0=0$$, there are infinite solutions; if you end up with 1=2, there are no solutions.

## Vectors



My notes of the excellent lectures of “Denis Auroux. 18.02 Multivariable Calculus. Fall 2007. Massachusetts Institute of Technology: MIT OpenCourseWare, https://ocw.mit.edu. License: Creative Commons BY-NC-SA.”

Description will use a plane $$\mathbb{R}^2$$, or space $$\mathbb{R}^3$$, but the same principles apply to higher dimensions.

Vectors are commonly displayed on the $$xyz$$-axis, with unit vectors $$\hat\imath\, \hat\jmath, \hat k$$. $$x,y,z$$-axis and $$\hat\imath,\hat\jmath,\hat k$$-unit vectors

Vectors do not have a start point, but do have a magnitude (length) and direction. They are described in terms of the unit vectors $$\hat\imath, \hat\jmath, \hat k$$, or using angle brackets notation. $$\vec{A} = \hat\imath\;a_1 + \hat\jmath\;a_2 + \hat\;k a_3 = \left\langle \;a_1,\;a_2,\;a_3\; \right\rangle$$

You can find the length of a vector $$|\vec{A}|$$, by applying the Pythagorean theorem twice. $$\shaded{ |\vec{A}| = \sqrt{(a_1)^2 + (a_2)^2 + (a_3)^2} } \nonumber$$

## Rotation $$\vec{A}$$ rotated over $$\tfrac{\pi}{2}$$

Let $$\vec{A}=\left\langle a_1, a_2\right\rangle$$, and let $$\vec{A}’$$ be $$\vec{A}$$ rotated over $$\frac{\pi}{2}$$. Then $$\shaded{ \vec{A}’=\left\langle -a_2, a_1\right\rangle } \label{eq:rotation}$$

Let $$\vec{A}=\left\langle a_1, a_2, a_3\right\rangle$$, and $$\vec{B}=\left\langle b_1, b_2, b_3\right\rangle$$. Then $$\vec{A}$$ plus $$\vec{B}$$ is defined as $$\shaded{ \vec{A}+\vec{B} = \left\langle a_1+b_1, a_2+b_2, a_3+b_3 \right\rangle } \nonumber$$

Geometric, the sum is the vector to the corner of the parallelogram. $$\vec{A} + \vec{B}$$

## Scalar product

Let $$s$$ be a scalar, and $$\vec{A}=\left\langle a_1, a_2, a_3\right\rangle$$. Then the scalar product of $$s$$ and $$\vec{A}$$ is defined as $$\shaded{ s\;\vec{A} = \left\langle s\;a_1, s\;a_2, s\;a_3\right\rangle } \nonumber$$

Geometrically, it makes the vector longer or shorter. $$s\;\vec{A}$$

## Dot-product

Let $$\vec{A}=\left\langle a_1, a_2, a_3\right\rangle$$, and $$\vec{B}=\left\langle b_1, b_2, b_3\right\rangle$$. The dot-product of $$\vec{A}$$ and $$\vec{B}$$ is defined as the scalar $$\shaded{ \vec{A} \cdot \vec{B} = \sum_i a_i\,b_i = a_1 b_1 + a_2 b_2 + a_3 b_3 } \nonumber$$ $$\vec{A}\cdot\vec{B}$$
For a geometric interpretation, start with the dot-product of $$\vec{A}$$ with itself $$\vec{A}\cdot\vec{A} = |\vec{A}|^2 \cos 0 = |\vec{A}|^2 \label{eq:vecsquare}$$

Let $$\vec{C}=\vec{A}-\vec{B}$$, and expand $$|\vec{C}|^2$$ by applying $$\eqref{eq:vecsquare}$$ \begin{align} |\vec{C}|^2 &= \vec{C} \vec{C} = \left(\vec{A} – \vec{B} \right) \cdot \left(\vec{A} – \vec{B} \right) \nonumber \\ &= \vec{A}\cdot\vec{A} – \vec{A}\cdot\vec{B} – \vec{B}\cdot\vec{A} + \vec{B}\cdot\vec{B} \nonumber \\ &= |\vec{A}|^2 + |\vec{B}|^2 – 2 \vec{A}\cdot\vec{B} \label{eq:expanded} \end{align}

Recall, the law of cosines from geometry.

$$c^2 = a^2 b^2 – 2 a b\cos\theta \nonumber$$

Apply the law of cosines to $$|\vec{A}|$$, $$|\vec{B}|$$ and $$|\vec{C}|$$ $$|\vec{C}|^2 = |\vec{A}|^2 + |\vec{B}|^2 – 2 |\vec{A}| |\vec{B}|\cos\theta \label{eq:lawofcos}$$

Combining equations $$\eqref{eq:expanded}$$ and $$\eqref{eq:lawofcos}$$ gives the geometric equation $$\shaded{ \vec{A}\cdot\vec{B} = |\vec{A}|\,|\vec{B}|\, \cos\theta } \nonumber$$

The dot-product can be used to compute length and angles in $$\mathbb{R}^3$$, or find components of $$\vec{A}$$ along unit vector $$\hat u$$ $$\shaded{ \vec{A}\cdot \hat u } \nonumber$$

## Determinant

### In 2 dimensions

Let $$\vec{A}=\left\langle a_1, a_2\right\rangle$$ and $$\vec{B}=\left\langle b_1, b_2\right\rangle$$. The $$\mathbb{R}^2$$-determinant is defined as \shaded{ \begin{align*} \mathrm{det}(\vec{A}, \vec{B}) &= \left|\begin{matrix} a_1 & a_2 \\ b_1 & b_2 \\ \end{matrix}\right| \\ &= a_1b_2-a_2b_1 \end{align*} } \nonumber

### In 3 dimensions

Let $$\vec{A}=\left\langle a_1, a_2, a_3\right\rangle$$, $$\vec{B}=\left\langle b_1, b_2, b_3\right\rangle$$ and $$\vec{C}=\left\langle c_1, c_2, c_3\right\rangle$$. The $$\mathbb{R}^3$$-determinant is defined as \shaded{ \begin{align*} \mathrm{det}(\vec{A}, \vec{B}, \vec{C}) &= \left|\begin{matrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{matrix}\right| \\ &= a_1 \left|\begin{matrix} b_2 & b_3 \\ c_2 & c_3 \end{matrix}\right| – a_2 \left|\begin{matrix} b_1 & b_3 \\ c_1 & c_3 \end{matrix}\right| + a_3 \left|\begin{matrix} b_1 & b_2 \\ c_1 & c_2 \end{matrix}\right| \end{align*} } \nonumber

### Area of a parallelogram

Let $$\vec{A}=\left\langle a_1, a_2\right\rangle$$, and $$\vec{B}=\left\langle b_1, b_2\right\rangle$$.

The area of the parallelogram shown above is calculated as width $$\times$$ height. $$\mathrm{area}_\triangle = |\vec{A}| |\vec{B}| \sin\theta \label{eq:triangle}$$

Change from $$\sin\theta$$ to $$\cos\theta$$ so it fits the dot-product. $$\vec{A}’\cdot\vec{B}$$

Obtain $$\vec{A}’$$ by rotating $$\vec{A}$$ over $$\frac{\pi}{2}$$, see equation $$\eqref{eq:rotation}$$. Apply $$sin\;\theta = \cos(\tfrac{\pi}{2}-\theta)$$ $$\left. \begin{array}{l} \theta ‘ = \tfrac{\pi}{2} – \theta \\ \cos(\tfrac{\pi}{2}-\theta) = \sin\theta \end{array} \right\} \Rightarrow \cos(\theta’) = sin(\theta) \label{eq:sincos}$$

Substitute $$\eqref{eq:sincos}$$ in $$\eqref{eq:triangle}$$ $$\mathrm{area} = |\vec{A}’| \cdot |\vec{B}| \cos\theta = \tfrac{1}{2}\vec{A}’\cdot \vec{B}$$

Expand the dot-product between $$\vec{A}’$$ and $$\vec{B}$$, and find the determinant \begin{align*} \mathrm{area} &= \left\langle -a_2, a_1 \right\rangle \cdot \left\langle b_1, b_2 \right\rangle \\ &= \left( a_1 b_2 – a_2 b_1 \right) \\ &= \left|\begin{array}{cc} a_1 & a_2 \\ b_1 & b_2 \end{array}\right| \end{align*}

The area of a parallelogram follows $$\shaded{ \mathrm{area} = \mathrm{det}\left(\vec{A},\vec{B}\right) } \label{eq:area}$$

## Cross-product

Let $$\vec{A}=\left\langle a_1, a_2, a_3\right\rangle$$, and $$\vec{B}=\left\langle b_1, b_2, b_3\right\rangle$$. The cross product of $$\vec{A}$$ and $$\vec{B}$$ in $$\mathbb{R}^3$$ is defined as the pseudo determinant vector \shaded{ \begin{align*} \vec{A}\times\vec{B} &= \left| \begin{array}{ccc} \hat\imath & \hat\jmath & \hat k \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{array} \right| \\ &= \hat\imath \left| \begin{array}{cc} a_2 & a_3 \\ b_2 & b_3 \end{array} \right| – \hat\jmath \left| \begin{array}{cc} a_1 & a_3 \\ b_1 & b_3 \end{array} \right| + \hat k \left| \begin{array}{cc} a_1 & a_2 \\ b_1 & b_2 \end{array} \right| \end{align*} } \nonumber

Theorem:

• the area of the parallelogram from the vectors $$\vec{A}$$ and $$\vec{B}$$ is $$|\vec{A}\times\vec{B}|$$
• the direction of $$\vec{A}\times\vec{B}$$ is perpendicular to the plane of the parallelogram. $$\vec{A}\times\vec{B}$$

The direction of the vector $$|\vec{A}\times\vec{B}|$$ is determined by the right-hand rule

For example: $$\hat\imath\times\hat\jmath=\hat k$$ \begin{align*} \hat\imath\times\hat\jmath &= \left| \begin{array}{ccc} \hat\imath & \hat\jmath & \hat k \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{array} \right| \\ &= \hat\imath \left| \begin{array}{cc} 0 & 0 \\ 1 & 0 \end{array} \right| – \hat\jmath \left| \begin{array}{cc} 1 & 0 \\ 0 & 0 \end{array} \right| + \hat z \left| \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right| \\ &= \hat z \end{align*}

Some properties

The right-hand rule shows that $$\shaded{ \vec{A}\times\vec{B}=-\vec{B}\times\vec{A} }$$

The parallelogram of $$\vec{A}\times\vec{A}$$ has area zero. $$\vec{A}\times\vec{A}=\vec{0}$$

### Volume in space

Let $$\vec{A}, \vec{B}, \vec{C}$$ in space $$\mathbb{R}^3$$.

The volume equals the area of the base times the height. The area base follows from equation $$\eqref{eq:area}$$. The height is the component of $$\vec{A}$$ that is perpendicular to the base. Call the direction perpendicular to the base unit vector $$\hat n$$. $$\mathrm{volume} = |\vec{B}\times\vec{C}|\;(\vec{A}\cdot\hat n) \label{eq:volume1}$$

The unit vector $$\hat n$$ can be derived from the cross-product of $$\vec{B}$$ and $$\vec{C}$$. To make it a unit vector, we divide by its length. $$\hat n = \frac{\vec{B}\times\vec{C}}{|\vec{B}\times\vec{C}|} \nonumber$$

Substitute this back in $$\eqref{eq:volume1}$$ \begin{align*} \mathrm{volume} &= \bcancel{|\vec{B}\times\vec{C}|}\;\left(\vec{A}\cdot \frac{\left(\vec{B}\times\vec{C}\right)}{\bcancel{|\vec{B}\times\vec{C}|}}\right) \\ &= \vec{A}\ \cdot\ \left(\vec{B}\times\vec{C}\right) \end{align*}

This equals the determinant of $$\vec{A}, \vec{B}, \vec{C}$$, the so called “triple product” rule $$\shaded{ \mathrm{det}\left(\vec{A},\vec{B},\vec{C}\right) =\vec{A}\ \cdot\ \left(\vec{B}\times\vec{C}\right) } \label{eq:tripleproduct}$$

Because $$a_1 \left| \begin{matrix} b_2 & b_3 \\ c_2 & c_3 \end{matrix} \right| – a_2 \left| \begin{matrix} b_1 & b_3 \\ c_1 & c_3 \end{matrix} \right| + a_3 \left| \begin{matrix} b_1 & b_2 \\ c_1 & c_2 \end{matrix} \right| \ = \left\langle a_1, a_2,a_3 \right\rangle \cdot \ \left( \hat\imath \left| \begin{array}{cc} b_2 & b_3 \\ c_2 & c_3 \end{array} \right| – \hat\jmath \left| \begin{array}{cc} b_1 & b_3 \\ c_1 & c_3 \end{array} \right| + \hat k \left| \begin{array}{cc} b_1 & b_2 \\ c_1 & c_2 \end{array} \right| \right) \nonumber$$

The volume in space described by $$\vec{A}$$, $$\vec{B}$$ and $$\vec{C}$$ follows as $$\shaded{ \mathrm{volume } = \mathrm{det}\left(\vec{A},\vec{B},\vec{C}\right) } \nonumber$$

### Equation of a plane from points

Find the plane that contains the points $$p$$, $$q$$ and $$r$$. Point $$p, q, r, s$$ in space

#### Solution

Consider $$\overrightarrow{qr}$$, $$\overrightarrow{qs}$$ and $$\overrightarrow{qp}$$ that form a parallelepiped. if these vectors are in the same plane, the parallelepiped will be flat. In other words, it will have no volume.

If $$p$$ is in the $$qrs$$-plane, the determinant should be $$0$$. $$\shaded{ \mathrm{det}\left( \overrightarrow{qp}, \overrightarrow{qr}, \overrightarrow{qs} \right) = 0 } \nonumber$$ with $$q$$, $$r$$ and $$s$$ known, and $$p$$ unknown, this equation will give the expression in $$x,y,z$$ for the plane.

#### A more intuitive solution Point $$p, q, r, s$$ and \vec{n} in space

Let a “normal vector” $$\overrightarrow n$$ be a vector perpendicular to the plane. Then $$p$$ is the plane when $$\overrightarrow{qp} \perp \overrightarrow n$$. Therefore the dot-product $$\overrightarrow{qp}\cdot \overrightarrow{n} = 0 \label{eq:moreintuitive}$$

$$\overrightarrow{n}$$ equals $$\overrightarrow{pr} \times \overrightarrow{qs}$$. Substituting this in equation $$\eqref{eq:moreintuitive}$$ $$\overrightarrow{qp} \cdot \left( \overrightarrow{pr} \times \overrightarrow{qs} \right) = 0 \nonumber$$

Applying the triple product equation $$\eqref{eq:tripleproduct}$$ gives the condition $$\shaded{ \mathrm{det}\left( \overrightarrow{qp}, \overrightarrow{pr}, \overrightarrow{qs} \right) = 0 }$$ with $$q$$, $$r$$ and $$s$$ known, and $$p$$ unknown, this equation will give the expression in $$x,y,z$$ for the plane.