Vectors and Matrices II

Module 2 was arithmetic. This one is about what vectors and matrices are actually for: proving geometrical facts, and undoing operations.

Vector geometry

Set up two base vectors — usually $\overrightarrow{OA} = \mathbf{a}$ and $\overrightarrow{OB} = \mathbf{b}$ — and write every other vector in terms of them.

The rule for travelling between points never changes:

$$\overrightarrow{AB} = \mathbf{b} - \mathbf{a}$$

Midpoints. If $M$ is the midpoint of $AB$, go to $A$ first, then half way along $AB$:

$$\overrightarrow{OM} = \mathbf{a} + \tfrac12(\mathbf{b} - \mathbf{a}) = \mathbf{a} + \tfrac12\mathbf{b} - \tfrac12\mathbf{a} = \tfrac12(\mathbf{a} + \mathbf{b})$$

A tidy result worth knowing: the position vector of a midpoint is the average of the two position vectors.

Ratios. If $P$ divides $AB$ in the ratio $1:2$, then $P$ is one third of the way along:

$$\overrightarrow{OP} = \mathbf{a} + \tfrac13(\mathbf{b} - \mathbf{a}) = \tfrac23\mathbf{a} + \tfrac13\mathbf{b}$$

Parallel lines. Two vectors are parallel when one is a scalar multiple of the other. If $\overrightarrow{PQ} = 3\mathbf{a} + 6\mathbf{b}$ and $\overrightarrow{RS} = \mathbf{a} + 2\mathbf{b}$, then $\overrightarrow{PQ} = 3\overrightarrow{RS}$, so $PQ$ is parallel to $RS$ and three times as long.

Collinear points

Three points are collinear when they lie on one straight line. To prove it, show two vectors joining them are parallel and share a common point.

Show that $A(1,2)$, $B(3,5)$ and $C(7,11)$ are collinear.

$$\overrightarrow{AB} = \begin{pmatrix} 3 \\ 5 \end{pmatrix} - \begin{pmatrix} 1 \\ 2 \end{pmatrix} = \begin{pmatrix} 2 \\ 3 \end{pmatrix}$$

$$\overrightarrow{BC} = \begin{pmatrix} 7 \\ 11 \end{pmatrix} - \begin{pmatrix} 3 \\ 5 \end{pmatrix} = \begin{pmatrix} 4 \\ 6 \end{pmatrix} = 2\begin{pmatrix} 2 \\ 3 \end{pmatrix} = 2\overrightarrow{AB}$$

So $\overrightarrow{BC}$ is parallel to $\overrightarrow{AB}$, and both pass through $B$. Therefore $A$, $B$ and $C$ lie on the same straight line.

The determinant

For a $2 \times 2$ matrix,

$$A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \qquad \det A = ad - bc$$

Multiply the leading diagonal, multiply the other diagonal, subtract.

$$\det\begin{pmatrix} 3 & 1 \\ 5 & 2 \end{pmatrix} = (3)(2) - (1)(5) = 6 - 5 = 1$$

$$\det\begin{pmatrix} 4 & 2 \\ 3 & 5 \end{pmatrix} = 20 - 6 = 14$$

When the determinant is zero the matrix is singular and has no inverse:

$$\det\begin{pmatrix} 2 & 4 \\ 1 & 2 \end{pmatrix} = 4 - 4 = 0$$

The inverse of a 2 × 2 matrix

$$A^{-1} = \frac{1}{ad-bc}\begin{pmatrix} d & -b \\ -c & a \end{pmatrix}$$

Swap the leading diagonal, change the sign of the other two, divide by the determinant.

Find the inverse of $\begin{pmatrix} 3 & 1 \\ 5 & 2 \end{pmatrix}$.

The determinant is 1, so

$$A^{-1} = \begin{pmatrix} 2 & -1 \\ -5 & 3 \end{pmatrix}$$

Check by multiplying:

$$\begin{pmatrix} 3 & 1 \\ 5 & 2 \end{pmatrix}\begin{pmatrix} 2 & -1 \\ -5 & 3 \end{pmatrix} = \begin{pmatrix} 6-5 & -3+3 \\ 10-10 & -5+6 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \ \checkmark$$

That last matrix is the identity, $I$ — the matrix equivalent of the number 1, and the definition of an inverse is $AA^{-1} = I$.

Another: $\begin{pmatrix} 4 & 2 \\ 3 & 5 \end{pmatrix}$, with determinant 14:

$$A^{-1} = \frac{1}{14}\begin{pmatrix} 5 & -2 \\ -3 & 4 \end{pmatrix}$$

Solving equations with matrices

Any pair of simultaneous linear equations can be written as one matrix equation.

$$3x + y = 11 \qquad 5x + 2y = 18$$

becomes

$$\begin{pmatrix} 3 & 1 \\ 5 & 2 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 11 \\ 18 \end{pmatrix}$$

In the form $A\mathbf{x} = \mathbf{b}$, multiply both sides by $A^{-1}$ on the left:

$$\begin{pmatrix} x \\ y \end{pmatrix} = A^{-1}\mathbf{b} = \begin{pmatrix} 2 & -1 \\ -5 & 3 \end{pmatrix}\begin{pmatrix} 11 \\ 18 \end{pmatrix}$$

$$= \begin{pmatrix} 22 - 18 \\ -55 + 54 \end{pmatrix} = \begin{pmatrix} 4 \\ -1 \end{pmatrix}$$

So $x = 4$ and $y = -1$. Check: $3(4) + (-1) = 11$ ✓ and $5(4) + 2(-1) = 18$ ✓

Matrix transformations

A $2 \times 2$ matrix acting on a position vector transforms the point. These are the ones to know, all centred on the origin:

Rotate $(2,3)$ through $90°$ anticlockwise about the origin.

$$\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}\begin{pmatrix} 2 \\ 3 \end{pmatrix} = \begin{pmatrix} (0)(2) + (-1)(3) \\ (1)(2) + (0)(3) \end{pmatrix} = \begin{pmatrix} -3 \\ 2 \end{pmatrix}$$

Matching the rule from Module 2: $(x,y) \mapsto (-y, x)$ ✓