Unit 4: Functions Involving Parameters, Vectors, and Matrices

Parameters and Families of Functions

A lot of precalculus is about learning how a function’s input controls its output. In this unit, you add one more layer: sometimes a function depends not only on an input variable, but also on a parameter.

A parameter is a value that acts like a constant within a specific situation but can change from one situation to another. A parameter does not “vary along the graph” the way the input does; instead, changing the parameter gives you a different graph or a different model.

For example, the family

f(x)=a(x2)2+1f(x)=a(x-2)^2+1

has input xx and parameter aa. Each fixed value of aa produces one specific parabola; varying aa produces a whole family of functions.

Why parameters matter

Parameters show up constantly in modeling. In physics, parameters might represent mass, drag, or initial velocity. In economics, parameters might represent interest rates or growth factors. In transformations, parameters control scaling or rotation amounts. AP-style questions often test whether you can interpret how a graph changes when a parameter changes.

How to think about a “family” of graphs

When you see a form like

f(x)=a(xh)2+kf(x)=a(x-h)^2+k

a helpful habit is to separate:

  • Within one graph: what changes as xx changes.
  • Across the family: what changes as the parameter changes.

A strong mental model is that parameters are knobs: turning the knob changes the entire function.

Common roles parameters play include vertical stretch/compression, horizontal stretch/compression, shifts, and deeper shape changes (curvature or asymptotes).

Example 1: Interpreting a parameter in a quadratic

Consider

f(x)=a(x3)24f(x)=a(x-3)^2-4

The vertex is always (3,4)\left(3,-4\right) regardless of aa. If a>0a>0 the parabola opens up; if a<0a<0 it opens down. If a>1|a|>1 the graph is narrower; if 0<a<10<|a|<1 it is wider.

A common mistake is to say “aa changes the vertex.” It does not here, because the vertex location is controlled by the shift values inside and outside the square.

Example 2: A parameter controlling an exponential model

Suppose

P(t)=P0(1+r)tP(t)=P_0(1+r)^t

Here tt is time, while P0P_0 and rr are parameters. For a fixed situation, P0P_0 and rr behave like constants, but different scenarios have different starting values and rates. Increasing rr increases the growth factor 1+r1+r, so the function grows faster.

Exam Focus
  • Typical question patterns:
    • Describe how changing a parameter affects key features (vertex, intercepts, asymptotes, end behavior).
    • Given a graph of one member of a family, determine parameter values.
    • Compare two parameter choices and decide which graph matches.
  • Common mistakes:
    • Treating a parameter like an input variable (mixing up what changes along the curve vs. what changes the curve).
    • Confusing horizontal and vertical effects (especially when a parameter appears inside the input).
    • Assuming a parameter affects every feature (for example, thinking a vertical stretch always moves an asymptote when it may not).

Parametric Functions and Parametric Equations

Usually you describe a curve by writing yy as a function of xx, but some curves are easier (or only possible) to describe by using a third variable (often time). A parametric representation gives both coordinates as functions of a parameter.

A parametric function can be viewed as a single object

f(t)=(x(t),y(t))f(t)=\left(x(t),y(t)\right)

where the coordinate functions might be written as

x=f(t)x=f(t)

y=g(t)y=g(t)

As tt varies over an interval, the point (x,y)\left(x,y\right) traces a curve.

Example: A basic parametric function

For instance,

f(t)=(t2,2t)f(t)=\left(t^2,2t\right)

means

x(t)=t2x(t)=t^2

y(t)=2ty(t)=2t

Here the two dependent variables xx and yy both depend on the single independent parameter tt.

Graphing and tabulating parametric functions

A standard way to visualize a parametric curve is to make a table of values. For selected parameter values, compute the corresponding points, then plot them in the plane. This makes the “trace” idea concrete and helps you see direction and which portion of a curve is actually produced.

Why parametric form matters

Parametric equations are powerful because they can:

  • Describe curves that fail the vertical line test (like circles).
  • Describe motion naturally (time as the parameter).
  • Encode direction (which way the curve is traced).
Orientation (direction) is part of the description

In parametric form, the direction comes from increasing tt. For example,

x=tx=t

y=t2y=t^2

traces the parabola from left to right as tt increases. If you replace tt with t-t (or reverse the interval), you trace the same curve in the opposite direction.

Eliminating the parameter (Cartesian equation)

Often you are asked to convert to a single relation between xx and yy by eliminating the parameter.

Core strategy:

  1. Solve one equation for tt (if possible).
  2. Substitute into the other.
  3. Simplify.
Worked Example 1: A parabola in parametric form

Given

x=t1x=t-1

y=t22y=t^2-2

Solve for tt from the first equation:

t=x+1t=x+1

Substitute:

y=(x+1)22y=(x+1)^2-2

A key warning: eliminating the parameter captures the shape of the curve but not necessarily the tracing behavior or which part of the curve is produced, which depends on the allowed values of tt.

Worked Example 2: A circle (not a function of xx)

A standard parametric form for a circle of radius rr is

x=rcos(t)x=r\cos(t)

y=rsin(t)y=r\sin(t)

Use the identity

cos2(t)+sin2(t)=1\cos^2(t)+\sin^2(t)=1

Square and add:

x2+y2=r2x^2+y^2=r^2

This represents a shape that cannot be written as a single function y=f(x)y=f(x).

Special case: parametrizing a function and its inverse

A function written as

y=f(x)y=f(x)

can be parametrized by letting

x(t)=tx(t)=t

y(t)=f(t)y(t)=f(t)

If ff has an inverse, the inverse relation can be parametrized by swapping roles:

x(t)=f(t)x(t)=f(t)

y(t)=ty(t)=t

Exam Focus
  • Typical question patterns:
    • Convert between parametric and Cartesian forms (eliminate the parameter).
    • Identify a curve (line, parabola, circle) from parametric equations.
    • Determine direction of motion as tt increases.
    • Use a table of values (or create one) to graph a parametric curve.
  • Common mistakes:
    • Ignoring restrictions on tt and claiming the curve is “the entire” Cartesian graph.
    • Losing orientation information (direction is not captured by the Cartesian equation).
    • Algebra errors when solving for tt and substituting.

Modeling Planar Motion with Parametric Functions

One of the most natural uses of parametric equations is to model motion. Interpreting tt as time, the functions

x(t)x(t)

and

y(t)y(t)

describe the position of an object at time tt.

Analyzing particle motion by components

To understand the motion, analyze the coordinate functions separately. The horizontal component comes from x(t)x(t) and the vertical component comes from y(t)y(t). The parametric graph is the particle’s path; each point on the path corresponds to a position at a specific time.

Position, displacement, and average velocity (in coordinates)

If position is

(x(t),y(t))\left(x(t),y(t)\right)

then displacement from t=at=a to t=bt=b is

(x(b)x(a),y(b)y(a))\left(x(b)-x(a),y(b)-y(a)\right)

The average velocity vector over that interval is

(x(b)x(a)ba,y(b)y(a)ba)\left(\frac{x(b)-x(a)}{b-a},\frac{y(b)-y(a)}{b-a}\right)

This coordinate-by-coordinate approach is exactly why parametric models are effective.

Average rate of change in each direction

The average rate of change in the horizontal direction over [a,b][a,b] is

x(b)x(a)ba\frac{x(b)-x(a)}{b-a}

and in the vertical direction over [a,b][a,b] is

y(b)y(a)ba\frac{y(b)-y(a)}{b-a}

These tell you how the particle is changing in each direction during that time window.

Direction of motion

Direction is determined by whether x(t)x(t) and y(t)y(t) are increasing or decreasing as tt increases. For example, if x(t)x(t) increases while y(t)y(t) decreases, the motion is to the right and downward.

Horizontal and vertical extrema

The horizontal extrema of the motion correspond to the maximum and minimum values of x(t)x(t) on the given time interval. The **vertical extrema** correspond to the maximum and minimum values of y(t)y(t). Identifying these helps you locate the “furthest left/right” and “highest/lowest” points on the path.

Example 1: Constant velocity in a straight line

Suppose

x(t)=2t+1x(t)=2t+1

y(t)=3t+4y(t)=-3t+4

To find the line, eliminate tt. From the first equation,

t=x12t=\frac{x-1}{2}

Substitute:

y=3(x12)+4y=-3\left(\frac{x-1}{2}\right)+4

y=32x+112y=-\frac{3}{2}x+\frac{11}{2}

This is constant velocity because each coordinate changes linearly with time.

Example 2: A non-linear path from mixing function types

Consider

x(t)=tx(t)=t

y(t)=t2y(t)=t^2

This is the parabola y=x2y=x^2, but the _timing_ is not uniform: the vertical change accelerates as tt grows.

Motion around a circle and transformations

On the unit circle centered at the origin, each point can be written as

(cos(θ),sin(θ))\left(\cos(\theta),\sin(\theta)\right)

where θ\theta is the angle from the positive xx-axis. This is a core model for circular motion. Using transformations, you can adjust circular motion models:

  • Scaling changes the radius in one or both directions.
  • Translating shifts the center (done by adding constants to the coordinate functions).
  • Reflecting flips the motion across an axis.

A strong reference tool here is the unit circle with special angles labeled, since common angles (like those with known sine and cosine values) help you interpret positions and directions quickly.

Domain of the parameter and what part of the path you get

If tt is restricted, only part of the curve is traced. For example, with

x=tx=t

y=t2y=t^2

and 0t20\le t\le 2, only the segment from x=0x=0 to x=2x=2 appears.

Parametrizing a linear path in context

Parametric equations are also ideal for linear motion. For example, a car moving from point A to point B at constant speed can be modeled by choosing tt to represent time (or “progress”), and building coordinate functions that start at A when t=0t=0 and reach B at the ending time.

Exam Focus
  • Typical question patterns:
    • Interpret parametric equations as motion and compute position/displacement over an interval.
    • Compute average rates of change in each coordinate.
    • Identify direction of motion by checking where x(t)x(t) and y(t)y(t) increase or decrease.
    • Find horizontal/vertical extrema using the max/min of x(t)x(t) and y(t)y(t) on an interval.
    • Use parameter restrictions to identify only a portion of a curve.
  • Common mistakes:
    • Confusing the curve’s equation with the timing along the curve.
    • Forgetting to apply the given interval for tt.
    • Treating coordinate changes as unrelated, instead of tied to the same time parameter.

Implicitly Defined Functions

An implicitly defined relation is an equation involving two variables that does not explicitly solve for one variable in terms of the other. Unlike explicit functions, an implicit equation can represent one function, multiple functions, disjoint pieces, or even infinitely many curves, all within a single equation.

Variation in implicit graphs

To graph an implicitly defined relation, you look for points (x,y)\left(x,y\right) that satisfy the equation. Depending on the equation, the result might be a single connected curve, multiple disconnected curves, or a more complicated collection of points.

Exam Focus
  • Typical question patterns:
    • Decide whether a relation is explicit or implicit.
    • Describe (qualitatively) why an implicit relation might produce more than one curve or fail the vertical line test.
  • Common mistakes:
    • Assuming every equation in two variables defines exactly one function.
    • Treating an implicit curve as if it must pass the vertical line test.

Parametric Representations of Conic Sections

Curves can be represented explicitly (with yy as a function of xx) or parametrically. Parametric form is especially flexible for conic sections.

Basics of expressing a curve parametrically

In a parametric representation, you define

x=x(t)x=x(t)

y=y(t)y=y(t)

As tt varies, the resulting points trace the curve. One way to validate a parametrization is to substitute x(t)x(t) and y(t)y(t) into the original equation; if the equation is true for all allowed tt, the parametrization is correct.

Parabolas

A typical approach to parametrize a parabola is to solve for one variable and then set the other equal to tt. For a function form

y=f(x)y=f(x)

a direct parametrization is

x(t)=tx(t)=t

y(t)=f(t)y(t)=f(t)

or, if convenient, swap which variable you treat as the parameter.

Ellipses

Ellipses are commonly parametrized using sine and cosine. A standard form is

x=acos(t)x=a\cos(t)

y=bsin(t)y=b\sin(t)

which traces an ellipse centered at the origin with horizontal radius aa and vertical radius bb.

Hyperbolas

Hyperbolas can be parametrized using trigonometric functions such as secant and tangent. A common right/left-opening form is

x=asec(t)x=a\sec(t)

y=btan(t)y=b\tan(t)

with appropriate restrictions on tt to avoid where secant is undefined.

Exam Focus
  • Typical question patterns:
    • Recognize that different parameter intervals trace different portions of the same conic.
    • Verify a parametrization by substitution.
    • Match a parametrization to the type of conic suggested by the functions used (polynomial vs. trig).
  • Common mistakes:
    • Forgetting domain restrictions on trig-based parametrizations.
    • Concluding “the whole curve” is traced without checking the parameter interval.

Vectors in the Plane

A vector is a quantity with both magnitude and direction. Vectors can be drawn as directed line segments (with a tail and head) or represented algebraically by components.

A common component form is

v=a,b\vec{v}=\langle a,b\rangle

meaning move aa units horizontally and bb units vertically.

Why vectors matter in this unit

Vectors bridge several unit ideas:

  • Parametric motion tracks position.
  • Vectors describe displacement and direction.
  • Matrices transform vectors, transforming entire objects.
Vector components and “free” vectors

A key idea is that a vector is not tied to a specific location. The displacement

3,1\langle 3,1\rangle

means the same move anywhere in the plane.

Vector components from points

Given two points

P(x1,y1)P\left(x_1,y_1\right)

and

Q(x2,y2)Q\left(x_2,y_2\right)

a vector from PP to QQ is

PQ=x2x1,y2y1\overrightarrow{PQ}=\langle x_2-x_1,y_2-y_1\rangle

Magnitude (length) and direction

For

v=a,b\vec{v}=\langle a,b\rangle

the magnitude is

v=a2+b2|\vec{v}|=\sqrt{a^2+b^2}

Direction can be viewed as the direction of the line segment from the origin to the point a,b\langle a,b\rangle, and trigonometry can be used to connect angles to components.

Vector operations

Vector addition and subtraction are done component-wise:

a,b+c,d=a+c,b+d\langle a,b\rangle+\langle c,d\rangle=\langle a+c,b+d\rangle

a,bc,d=ac,bd\langle a,b\rangle-\langle c,d\rangle=\langle a-c,b-d\rangle

Scalar multiplication stretches/shrinks and possibly reverses direction:

ka,b=ka,kbk\langle a,b\rangle=\langle ka,kb\rangle

A frequent error is adding magnitudes instead of components. Magnitudes only add directly when vectors point in the same direction.

Unit vectors

A unit vector has magnitude 1 and represents direction. For nonzero v\vec{v},

v^=1vv\hat{v}=\frac{1}{|\vec{v}|}\vec{v}

Dot product, angles, and perpendicularity

For

u=a,b\vec{u}=\langle a,b\rangle

and

v=c,d\vec{v}=\langle c,d\rangle

the dot product is

uv=ac+bd\vec{u}\cdot\vec{v}=ac+bd

A powerful geometric test:

  • If uv=0\vec{u}\cdot\vec{v}=0 and both vectors are nonzero, then the vectors are perpendicular.

The dot product also connects to the angle θ\theta between vectors:

uv=uvcos(θ)\vec{u}\cdot\vec{v}=|\vec{u}||\vec{v}|\cos(\theta)

so

cos(θ)=uvuv\cos(\theta)=\frac{\vec{u}\cdot\vec{v}}{|\vec{u}||\vec{v}|}

Law of Cosines and Law of Sines (vector-formed triangles)

When vectors form the sides of a triangle, classical triangle relationships can be applied.

Law of Cosines (conceptually connecting side lengths and the included angle):

c2=a2+b22abcos(C)c^2=a^2+b^2-2ab\cos(C)

Law of Sines (connecting side lengths to opposite angles):

sin(A)a=sin(B)b=sin(C)c\frac{\sin(A)}{a}=\frac{\sin(B)}{b}=\frac{\sin(C)}{c}

Worked Example 1: Vector magnitude and scaling

Let

v=6,8\vec{v}=\langle 6,-8\rangle

Then

v=62+(8)2=10|\vec{v}|=\sqrt{6^2+(-8)^2}=10

Scaling by 12\frac{1}{2} gives

12v=3,4\frac{1}{2}\vec{v}=\langle 3,-4\rangle

with magnitude 5, same direction.

Worked Example 2: Perpendicular vectors via dot product

Check

2,5\langle 2,5\rangle

and

5,2\langle -5,2\rangle

Compute

2,55,2=2(5)+5(2)=0\langle 2,5\rangle\cdot\langle -5,2\rangle=2(-5)+5(2)=0

So they are perpendicular.

Exam Focus
  • Typical question patterns:
    • Compute magnitude, unit vectors, or results of vector addition/subtraction.
    • Interpret vectors as displacements between points.
    • Use the dot product to test perpendicularity and to find angles.
  • Common mistakes:
    • Mixing up point coordinates and vector components (a point (a,b)\left(a,b\right) is not automatically a vector unless interpreted from the origin).
    • Adding magnitudes instead of adding components.
    • Forgetting that scalar multiplication by a negative reverses direction.

Vector-Valued Functions (Position in the Plane)

A particle’s position can be described by a position vector: a vector that starts at the origin and ends at the particle’s location.

A common representation is

p(t)=x(t),y(t)\mathbf{p}(t)=\langle x(t),y(t)\rangle

Another notation uses unit direction vectors i\mathbf{i} and j\mathbf{j}:

p(t)=x(t)i+y(t)j\mathbf{p}(t)=x(t)\mathbf{i}+y(t)\mathbf{j}

The magnitude p(t)|\mathbf{p}(t)| is the particle’s distance from the origin.

Example: Writing a position vector

If

x(t)=2t+1x(t)=2t+1

y(t)=3t2y(t)=3t-2

then

p(t)=(2t+1)i+(3t2)j\mathbf{p}(t)=(2t+1)\mathbf{i}+(3t-2)\mathbf{j}

Velocity idea (rates of change)

In many math courses, a velocity vector describes how position changes. In this unit’s non-calculus setting, the most reliable tool is average velocity over an interval, computed from displacement divided by elapsed time (as in the motion section). When calculus is used, instantaneous velocity is modeled by derivatives, but average velocity already captures direction (signs of components) and speed over the interval via magnitude.

Exam Focus
  • Typical question patterns:
    • Write a position vector from parametric coordinate functions.
    • Use displacement and average velocity vectors to interpret motion.
    • Use magnitude to interpret distance from the origin (for position) or average speed over an interval (for average velocity).
  • Common mistakes:
    • Confusing the position vector with displacement (displacement depends on two times; position uses one time).
    • Treating vector magnitude as a signed quantity (magnitudes are nonnegative).

Vector Forms of Lines and Connections to Parametrics

Vectors give a clean way to describe lines and constant-direction motion.

Parametric equation of a line

If a line passes through (x0,y0)\left(x_0,y_0\right) with direction vector a,b\langle a,b\rangle, then a parametric form is

x=x0+atx=x_0+at

y=y0+bty=y_0+bt

This works because at t=0t=0 you are at the starting point, and changing tt moves you along multiples of the direction vector.

A common misconception is that aa and bb must be the slope and intercept. They are not; they are the components of a direction vector.

Example 1: Line through two points

Find a parametric equation for the line through

A(1,2)A\left(1,-2\right)

and

B(5,4)B\left(5,4\right)

Direction vector:

AB=4,6\overrightarrow{AB}=\langle 4,6\rangle

So one parametrization is

x=1+4tx=1+4t

y=2+6ty=-2+6t

Any nonzero scalar multiple of 4,6\langle 4,6\rangle works (for example 2,3\langle 2,3\rangle).

Example 2: Converting parametric line form to slope-intercept form

Given

x=32tx=3-2t

y=1+5ty=1+5t

Solve for tt:

t=3x2t=\frac{3-x}{2}

Substitute:

y=17252xy=\frac{17}{2}-\frac{5}{2}x

Interpreting parametric line motion

If tt is time, then

x=x0+atx=x_0+at

y=y0+bty=y_0+bt

models constant velocity with coordinate velocities aa and bb.

Exam Focus
  • Typical question patterns:
    • Write parametric equations for a line given two points or a point and a direction.
    • Convert a parametric line to an equation in xx and yy.
    • Interpret aa and bb as rates of change in each coordinate.
  • Common mistakes:
    • Treating tt like an xx value (it is a separate parameter, not the horizontal coordinate).
    • Using slope alone as the “direction vector” (slope must be turned into a 2D direction like 1,m\langle 1,m\rangle).
    • Forgetting that different direction vectors can describe the same line (scalar multiples).

Matrices: Structure, Dimensions, and Basic Operations

A matrix is a rectangular array of numbers that can represent data, systems, transformations, and dynamic models.

Matrix dimensions and types

A matrix with nn rows and mm columns has dimension n×mn\times m.

Common classifications:

  • Square matrix: n×nn\times n
  • Row matrix: 1×n1\times n
  • Column matrix: n×1n\times 1

Entries are often denoted by amna_{mn} for the entry in the mm-th row and nn-th column.

A general 2×2 matrix is

A=[abcd]A=\left[\begin{array}{cc}a&b\cr c&d\end{array}\right]

Matrix equality

Two matrices are equal only if they have the same dimensions and every corresponding entry is equal.

Matrix multiplication (when it is defined)

Matrices can be multiplied when the number of columns in the first equals the number of rows in the second. If an m×km\times k matrix multiplies a k×nk\times n matrix, the result is an m×nm\times n matrix.

Matrix multiplication is not element-by-element. Each entry is the dot-product-like combination of a row and a column.

In particular, if

A=[abcd]A=\left[\begin{array}{cc}a&b\cr c&d\end{array}\right]

and

x=[xy]\vec{x}=\left[\begin{array}{c}x\cr y\end{array}\right]

then

Ax=[ax+bycx+dy]A\vec{x}=\left[\begin{array}{c}ax+by\cr cx+dy\end{array}\right]

A very common mistake is to multiply corresponding entries only (missing the cross terms).

Identity matrix

The 2×2 identity matrix is

I=[1001]I=\left[\begin{array}{cc}1&0\cr 0&1\end{array}\right]

and it satisfies

Ix=xI\vec{x}=\vec{x}

Determinant (2×2)

For

A=[abcd]A=\left[\begin{array}{cc}a&b\cr c&d\end{array}\right]

the determinant is

det(A)=adbc\det(A)=ad-bc

In the plane, det(A)|\det(A)| is the factor by which areas scale under the transformation represented by AA. If det(A)=0\det(A)=0, the transformation collapses the plane (not invertible).

Inverse of a 2×2 matrix

If det(A)0\det(A)\ne 0, then

A1=1adbc[dbca]A^{-1}=\frac{1}{ad-bc}\left[\begin{array}{cc}d&-b\cr -c&a\end{array}\right]

Worked Example 1: Multiply a matrix by a vector

Let

A=[2134]A=\left[\begin{array}{cc}2&-1\cr 3&4\end{array}\right]

and

x=[52]\vec{x}=\left[\begin{array}{c}5\cr 2\end{array}\right]

Then

Ax=[823]A\vec{x}=\left[\begin{array}{c}8\cr 23\end{array}\right]

So (5,2)\left(5,2\right) transforms to (8,23)\left(8,23\right).

Worked Example 2: Invertibility via the determinant

Let

A=[1236]A=\left[\begin{array}{cc}1&2\cr 3&6\end{array}\right]

Then

det(A)=0\det(A)=0

So AA has no inverse; the transformation is not one-to-one.

Exam Focus
  • Typical question patterns:
    • Identify matrix dimensions and decide whether a product is defined.
    • Multiply matrices and vectors using row-by-column multiplication.
    • Use determinants to decide whether a matrix is invertible.
    • Compute the inverse of a 2×2 matrix (when it exists).
  • Common mistakes:
    • Doing element-wise multiplication instead of row-by-column multiplication.
    • Forgetting that matrix multiplication is not generally commutative.
    • Computing an inverse without checking adbc0ad-bc\ne 0.

Linear Transformations and Matrices in R2\mathbb{R}^2

A major theme of this unit is that a 2×2 matrix can represent a linear transformation of the plane.

What makes a transformation linear

A transformation LL from R2\mathbb{R}^2 to R2\mathbb{R}^2 is linear if it preserves vector addition and scalar multiplication:

L(u+v)=L(u)+L(v)L(\vec{u}+\vec{v})=L(\vec{u})+L(\vec{v})

L(cv)=cL(v)L(c\vec{v})=cL(\vec{v})

A consequence is that linear transformations send the zero vector to the zero vector.

Every linear transformation corresponds to a unique 2×2 matrix

For a linear transformation LL on R2\mathbb{R}^2, there exists a unique 2×2 matrix AA such that

L(v)=AvL(\vec{v})=A\vec{v}

Conversely, any 2×2 matrix defines a linear transformation by the rule above.

The columns tell you where the basis vectors go

If

A=[abcd]A=\left[\begin{array}{cc}a&b\cr c&d\end{array}\right]

then the first column is where (1,0)\left(1,0\right) goes and the second column is where (0,1)\left(0,1\right) goes:

A[10]=[ac]A\left[\begin{array}{c}1\cr 0\end{array}\right]=\left[\begin{array}{c}a\cr c\end{array}\right]

A[01]=[bd]A\left[\begin{array}{c}0\cr 1\end{array}\right]=\left[\begin{array}{c}b\cr d\end{array}\right]

This is one of the fastest ways to interpret a matrix geometrically.

Common transformation matrices

Horizontal scaling by factor kk:

[k001]\left[\begin{array}{cc}k&0\cr 0&1\end{array}\right]

Vertical scaling by factor kk:

[100k]\left[\begin{array}{cc}1&0\cr 0&k\end{array}\right]

Reflection across the x-axis:

[1001]\left[\begin{array}{cc}1&0\cr 0&-1\end{array}\right]

Reflection across the y-axis:

[1001]\left[\begin{array}{cc}-1&0\cr 0&1\end{array}\right]

Swap coordinates (reflection across the line y=xy=x):

[0110]\left[\begin{array}{cc}0&1\cr 1&0\end{array}\right]

Rotation counterclockwise by angle θ\theta:

[cos(θ)sin(θ)sin(θ)cos(θ)]\left[\begin{array}{cc}\cos(\theta)&-\sin(\theta)\cr \sin(\theta)&\cos(\theta)\end{array}\right]

These are linear transformations: they keep the origin fixed and map lines through the origin to lines through the origin.

Vectors as matrices

A single vector in R2\mathbb{R}^2 can be represented as a 2×1 column matrix. A set of nn vectors can be represented as a 2×nn matrix; multiplying by a 2×2 transformation matrix produces a 2×nn matrix of transformed output vectors.

Exam Focus
  • Typical question patterns:
    • Decide whether a rule is linear by checking additivity and scalar-multiplication properties.
    • Build a matrix by tracking where basis vectors map.
    • Identify transformations (scales, reflections, swaps, rotations) from a matrix.
  • Common mistakes:
    • Assuming translations are possible with 2×2 matrices alone (translations require adding a vector or using a different setup).
    • Confusing “any transformation” with “linear transformation” (linear ones must keep the origin fixed).

Matrices as Transformations of the Plane (Applying Them to Shapes)

Once you treat points as vectors, applying a matrix transformation becomes systematic.

If

A=[abcd]A=\left[\begin{array}{cc}a&b\cr c&d\end{array}\right]

then the transformation sends

[xy]\left[\begin{array}{c}x\cr y\end{array}\right]

to

[ax+bycx+dy]\left[\begin{array}{c}ax+by\cr cx+dy\end{array}\right]

Example 1: Transforming a triangle by a scaling matrix

Triangle vertices:

P(1,1)P\left(1,1\right)

Q(3,1)Q\left(3,1\right)

R(2,4)R\left(2,4\right)

Apply horizontal scaling by factor 2:

A=[2001]A=\left[\begin{array}{cc}2&0\cr 0&1\end{array}\right]

Transform:

A[11]=[21]A\left[\begin{array}{c}1\cr 1\end{array}\right]=\left[\begin{array}{c}2\cr 1\end{array}\right]

A[31]=[61]A\left[\begin{array}{c}3\cr 1\end{array}\right]=\left[\begin{array}{c}6\cr 1\end{array}\right]

A[24]=[44]A\left[\begin{array}{c}2\cr 4\end{array}\right]=\left[\begin{array}{c}4\cr 4\end{array}\right]

New vertices: (2,1)\left(2,1\right), (6,1)\left(6,1\right), (4,4)\left(4,4\right).

Example 2: Area scaling via the determinant

For

A=[2001]A=\left[\begin{array}{cc}2&0\cr 0&1\end{array}\right]

det(A)=2\det(A)=2

So areas should double, which is a useful check.

Determinant sign and orientation

The absolute value det(A)|\det(A)| gives the area scale factor. A negative determinant indicates the transformation flips orientation (a “flip” effect).

Exam Focus
  • Typical question patterns:
    • Apply a matrix to points or vertices of a shape.
    • Use det(A)\det(A) to reason about area scaling and invertibility.
    • Use the sign of det(A)\det(A) to identify whether orientation flips.
  • Common mistakes:
    • Applying a matrix to only some vertices (inconsistent transformation).
    • Expecting distances/angles to be preserved under general matrices (only certain transformations preserve them).
    • Forgetting that linear transformations always send the origin to the origin.

Composing Transformations and Why Order Matters

If transformation T1T_1 is represented by matrix AA and transformation T2T_2 by matrix BB, then applying T1T_1 followed by T2T_2 corresponds to

B(Ax)=(BA)xB(A\vec{x})=(BA)\vec{x}

So the combined matrix is BABA.

Why order matters

Matrix multiplication is not generally commutative:

ABBAAB\ne BA

Geometrically, “stretch then rotate” usually differs from “rotate then stretch.”

Example 1: Two transformations in different orders (a special case)

Let

A=[1001]A=\left[\begin{array}{cc}-1&0\cr 0&1\end{array}\right]

and

B=[2001]B=\left[\begin{array}{cc}2&0\cr 0&1\end{array}\right]

Then

BA=[2001]BA=\left[\begin{array}{cc}-2&0\cr 0&1\end{array}\right]

and

AB=[2001]AB=\left[\begin{array}{cc}-2&0\cr 0&1\end{array}\right]

They match here because both are diagonal and act independently on the axes.

Example 2: A case where order changes the outcome

Let

A=[0110]A=\left[\begin{array}{cc}0&1\cr 1&0\end{array}\right]

and

B=[2001]B=\left[\begin{array}{cc}2&0\cr 0&1\end{array}\right]

Then

BA=[0210]BA=\left[\begin{array}{cc}0&2\cr 1&0\end{array}\right]

and

AB=[0120]AB=\left[\begin{array}{cc}0&1\cr 2&0\end{array}\right]

Different results, so order matters.

Inverses as “undo” transformations

If AA is invertible, then

A1(Ax)=xA^{-1}(A\vec{x})=\vec{x}

This is the standard tool when you are given an output point and asked to find the original point.

Exam Focus
  • Typical question patterns:
    • Find a combined matrix for “first do AA, then do BB.”
    • Decide whether two transformations commute.
    • Use inverses to reverse a transformation.
  • Common mistakes:
    • Reversing multiplication order.
    • Assuming all transformations commute.
    • Forgetting that an inverse exists only when the determinant is nonzero.

Matrices Modeling Contexts (Transition Matrices)

Matrices are not only geometric tools; they also model systems that transition between states over time.

Transitions and state changes

A system can be described by a state vector (counts, proportions, or amounts in each category). Changes from one time step to the next can be encoded by percentage rates. A transition matrix (often 2×2 in simple models) records how much of each state moves to each other state each step.

Conceptually:

  • A transition represents a shift from one state to another.
  • The matrix encodes the rates of transition.
  • Repeated multiplication models the system evolving over discrete time intervals.
Exam Focus
  • Typical question patterns:
    • Interpret entries of a transition matrix as percentages/rates moving between states.
    • Use matrix multiplication to update a state vector over one or more time steps.
  • Common mistakes:
    • Mixing up “from” and “to” when interpreting matrix entries.
    • Using the wrong multiplication order for state updates.

Bringing It All Together: Parameters, Vectors, Parametrics, and Matrices

This unit is about moving fluently between representations:

  • Functions with parameters describe families of behaviors.
  • Parametric equations describe curves and motion using a parameter.
  • Vectors describe direction and displacement.
  • Matrices transform vectors and therefore transform geometric objects and graphs.
Parameters inside matrices

Matrices can contain parameters, creating families of transformations. For example,

A=[k001]A=\left[\begin{array}{cc}k&0\cr 0&1\end{array}\right]

describes a horizontal scaling by factor kk.

Transforming a parametric curve with a matrix

If a curve is

r(t)=[f(t)g(t)]\vec{r}(t)=\left[\begin{array}{c}f(t)\cr g(t)\end{array}\right]

then applying transformation matrix AA gives

R(t)=Ar(t)\vec{R}(t)=A\vec{r}(t)

which produces new parametric equations automatically.

Worked Example: Transforming a circle into an ellipse

Start with a circle of radius 1:

x=cos(t)x=\cos(t)

y=sin(t)y=\sin(t)

Apply horizontal scaling by factor 3:

A=[3001]A=\left[\begin{array}{cc}3&0\cr 0&1\end{array}\right]

Then

[X(t)Y(t)]=[3001][cos(t)sin(t)]=[3cos(t)sin(t)]\left[\begin{array}{c}X(t)\cr Y(t)\end{array}\right]=\left[\begin{array}{cc}3&0\cr 0&1\end{array}\right]\left[\begin{array}{c}\cos(t)\cr \sin(t)\end{array}\right]=\left[\begin{array}{c}3\cos(t)\cr \sin(t)\end{array}\right]

So

X(t)=3cos(t)X(t)=3\cos(t)

Y(t)=sin(t)Y(t)=\sin(t)

Eliminate the parameter:

(X3)2+Y2=1\left(\frac{X}{3}\right)^2+Y^2=1

This shows a complete chain: parametric curve, matrix transformation, new parametric equations, and a Cartesian equation.

What changes and what stays the same under transformations

A linear transformation keeps straight lines straight and keeps the origin fixed. It may change lengths and angles unless it is a distance-preserving transformation (like a pure rotation or reflection). The determinant tells you area scaling, invertibility, and whether orientation flips.

Technology and representation fluency

Graphing tools can help visualize parametric curves and transformations, but exam questions typically still require algebraic justification (elimination steps, matrix multiplication, interpreting parameters).

Exam Focus
  • Typical question patterns:
    • Given a parametric curve and a transformation matrix, find the transformed parametric equations.
    • Use a parameter in a model to match a graph or meet a condition.
    • Translate between representations: parametric form, vector form, and matrix form.
  • Common mistakes:
    • Treating parameter changes as if they change the input variable’s domain rather than changing the whole model.
    • Applying a matrix transformation inconsistently (transforming some points but not others).
    • Forgetting that order matters when combining transformations.