Precalculus

Semester 1

Angular Properties

Radians
A radian is a unit of angle measure, comparable to degrees (°).
The conversion from degrees to radians is: π = 180°, therefore 2π = 360°

Angles
Reference angles are the smallest angles that connects a ray to the x-axis. It is ALWAYS positive.
Coterminal angles are the same angles with different measures. A coterminal angle in degrees can be found through 360n° + θ (current angle meaure). A coterminal angle in radians can be found through 2πn + θ (current angle measure). In both cases, n is a positive or negative integer (meaning WHOLE NUMBER).

Example(s):
   The reference angle for 145° is 35°, because 180° − 145° = 35°
   The reference angle for -^π⁄₃ is ^π⁄₃, because 0 − (-^π⁄₃) = ^π⁄₃.
   Two coterminal angles (one negative, one positive) of 145° are 510°(360[n = 1]° + 145°) and -220°(-360[n = -1]° + 145°).
   Two coterminal angles (one negative, one positive) of -^π⁄₃ are ^5π⁄₃(2π[n = 1] + -^π⁄₃) and -^7π⁄₃(-2π[n = -1] + -^π⁄₃).

Trigonometric Ratios
There are six trigonometric ratios of a right triangle:
sinθ = ^opp⁄_hyp          cscθ = ^hyp⁄_opp
cosθ = ^adj⁄_hyp          secθ = ^hyp⁄_adj
tanθ = ^opp⁄_adj          cotθ = ^adj⁄_opp

cscθ, secθ, and cotθ are reciprocal functions becayse their ratios are reciprocals of the other three trig. ratios.
cscθ = ¹⁄_sinθ          secθ = ¹⁄_cosθ          cotθ = ¹⁄_tanθ

When given a set of coordinates to determine the trigonometric ratios, use the following formulae:
sinθ = ^y⁄_r          cscθ = ^r⁄_y
cosθ = ^x⁄_r          secθ = ^r⁄_x
tanθ = ^y⁄_x          cotθ = ^x⁄_y
In all of these, y is the y-coordinate, x is the x-coordinate, and r is √(x² + y²)



QUIZ

1. Determine the six trigonometric ratios for a right triangle with an opposite of 4 and hypotenuse of 5.
sinθ = ⁴⁄₅          cscθ = ⁵⁄₄
cosθ = ³⁄₅          secθ = ⁵⁄₃
tanθ = ⁴⁄₃          cotθ = ³⁄₄
2. Determine the reference angle for 163°
17°
3. The tanθ of a right triangle is ¹²⁄₁₃. Determine the other five trigonometric ratios.
sinθ = ⁵⁄₁₃          cscθ = ¹³⁄₅
cosθ = ¹²⁄₁₃        secθ = ¹³⁄₁₂
tanθ = ⁵⁄₁₂          cotθ = ¹²⁄₅
4. If a right triangle has a cscθ of ^-17⁄₈. What are the other ratios if cosθ > 0?
sinθ = ^-8⁄₁₇          cscθ = ^-17⁄₈
cosθ = ¹⁵⁄₁₇        secθ = ¹⁷⁄₁₅
tanθ = ^-8⁄₁₅          cotθ = ^-15⁄₈
5. Find the trigonometric ratios for an angle, θ, whose terminal side ends at (7, 24).
sinθ = ²⁴⁄₂₅        cscθ = ²⁵⁄₂₄
cosθ = ⁷⁄₂₅          secθ = ²⁵⁄₇
tanθ = ²⁴⁄₇          cotθ = ⁷⁄₂₄

Reveal Answers



Unit Circle & Trig Properties



The unit circle consists of various points and angle measurements. The first set of numbers is the measure of the angles in degrees (°), and the second set is the measure of the angles in radians. By using the unit circle, one can memorize any conversion between degrees and radians.
For all of the points on the unit circle, the hypotenuse (or r) is 1. Therefore, the six trigonometric ratios of the unit circle can be found through the following:
sinθ = y          cscθ = ¹⁄_y
cosθ = x          secθ = ¹⁄_x
tanθ = ^y⁄_x        cotθ = ^x⁄_y

Below is a table for the trig. ratios of the unit circle:


Sinusoidals & Other Trig Graphs


A sinusoidal is a graphed function that can be represented using the equation: f(x) = asin(bθ − h) + k, in which:
    • |a| is the amplitude
    • ^2π⁄_b is the period (distance from trough-trough or crest-crest)
    • ^h⁄_b is the phase shift (horizontal shift determined by period)
    • k is the vertical shift
Cosine graphs can also be represented through a sinusoidal. y = cosθ is the same as y = sin(θ + ^π⁄₂).

Tangent Functions:

Cosecant and Secant Functions:


The formulae for the graphs of tangent functions is similar to that of sine and cosine functions. The tangent function can be represented by the equation f(x) = a × ^sin⁄_cos(bθ − h) + k, where all previous rules apply except the period is ^π⁄_b. Unlike sine and cosine graphs, tangent functions contain undefined domain values, and thus their graphs have vertical asymptotes.

Cosecant and secant functions are the reciprocals of regular sine and cosine functions. Therefore, their graphs have reversed waves, where they are parabolas that face away from the central x-axis. Like tangent functions, they have vertical asymptotes. The difference between secant and cosecant graphs is where the asymptotes are positioned [ secθ is the same as csc(θ + ^π⁄₂) ]



QUIZ

1. Determine the phase shift, amplitude, and period for the following function: f(x) = -sin(θ⁄₂ − π) + 5
Period: 4π, Amplitude: 1, Phase Shift: right 2π
2. Determine five major points of following function within the domain [ -π, π ]: f(x) = 3cos(2θ + ^π⁄₂) − 1
(-^3π⁄₄, -4), (-^π⁄₂, -1), (-^π⁄₄, 2), (0, -1), (^π⁄₄, -4)
3. If a sinusoidal has a phase shift of "left ^π⁄₁₂", what would the trigonometric equation be if it the period is ^π⁄₂?
f(x) = cos(4θ + ^π⁄₃) or f(x) = sin(4θ + ^π⁄₃)
4. Determine the x-positions of the vertical asymptotes for the function f(x) = tan(θ⁄₃ + ^2π⁄₃)
x = ^3π⁄₂k (k ≠ 0, k ∈ Z)

Reveal Answers


Inverse Trig Functions
Each trigonometric function has what is known as an inverse function. Inverse functions, such as sin^-1x, return angular measures rather than values. For instance, sin^-1(^√2⁄₂) is ^π⁄₄. Thus, sin^-1x returns the θ from the equation x = sinθ (yes, it's the same x that's in the inverse).

There are other forms of inverse trig functions known as arc functions. Inverse functions have limited domains so that only one angle measure can be the result. Arc functions, however, have unlimited domains and has two results.

Example(s):
sin^-1(-^√2⁄₂) is -^π⁄₄, since the limited domain for inverse sin and tan functions is [ -^π⁄₂, ^π⁄₂ ].
cos^-1(-^√2⁄₂) is ^3π⁄₄, since the limited domain for inverse cos function [ 0, π ].
arcsin(-^√2⁄₂) is -^π⁄₄ + 2πk & ^5π⁄₄ + 2πk (k ∈ Z), which is determined by taking the original result and another angle that results in x (in this case, x = -^√2⁄₂) and adding 2πk (k ∈ Z) to include all coterminals for that angle. For arctan, πk (k ∈ Z) is added instead since the domain for tangent is π rather than 2π.

When give a problem that contains both a trig and inverse trig function, such as "find cos(sin^-1x)", construct a triangle and determine the side lengths.
For example, tan(cos^-1(³⁄₅)) = ⁴⁄₃.
Reasoning: The triangle has an adjacent of 3 and hypotenuse of 5 since the formula for cosine is ^adj⁄_hyp. If there is no denominator, use 1 in its place. The opposite can be found through √(hyp² − adj²), which in this case is equal to 4. The tangent formula is ^opp⁄_adj, so the final result is ⁴⁄₃.

Law of Sines and Cosines
Using trigonometry, how does one determine a side length or an angle measure in a triangle when given certain data? The answer is the Law of Sines and Cosines.

In the Law of Sines, ^sinα⁄_a = ^sinβ⁄_b, where α and β are angle measures in a triangle, and a and b are the opposites of their respective angles. Law of Sines is best used when given an angle, its opposite, and another angle or side length (AAS/SAA, SAS, SSA/ASS).
In the Law of Cosines, a² = b² + c² − 2bccosα, where a, b, and c are side lengths and α is the angle that has a as its opposite. Law of Cosines is best used when given an angle and two side lengths, where neither side is the angle's opposite (SAS, ASA). Or, when given 3 side lengths, it can determine angle measures (SSS).

Regarding the Law of Sines, when given an SSA/ASS triangle, also known as an ambiguous case, there can be a certain number of possible solutions. You should know how to identify and solve ambiguous case triangles.
In these ambiguous cases, one must determine the value of h, or the height of the triangle. The formula is h = b × sinα. The following conditions determine the number of solutions:
    • If α is acute and a (opposite of α) < h:   NO SOLUTIONS
    • If α is acute and a = h:   ONE SOLUTION
    • If α is acute and a > b:   ONE SOLUTION
    • If α is acute and h < a < b:   TWO SOLUTIONS
    • If α is obtuse and a ≤ b:   NO SOLUTIONS
    • If α is obtuse and a > b:   ONE SOLUTION

Example(s):
Suppose a triangle ABC has the sides a (length of 10) and b (length of 13), as well as the angle α, which has a measure of 35°. This triangle has an h-value of approx. 7.46, which fulfills the 4^th condition (h < a < b). Therefore, ΔABC has two possible solutions.
After solving for your first solution, the next step is to take subtract the angle measure you had determined for β (angle measure of ∠B; has b as its opposite) from 180° (180° − β = β' [beta prime: the new measure for ∠B]). Then, solve for the new values of c (other side length) and γ (angle measure for ∠C; has c as its opposite).

There are two formulae for determining the area of a triangle:
    • A = ¹⁄₂bcsinα
    • A = √(s(s − a)(s − b)(s − c)) {s = ¹⁄₂(a + b + c)}

Trig Identities
There are several trigonometric identities that are used to simplify/prove trig expressions. Below is a list of several identities:
    • sin²θ + cos²θ = 1
    • tan²θ + 1 = sec²θ
    • cot²θ + 1 = csc²θ
    • sin(α ± β) = sinαcosβ ± cosαsinβ
    • cos(α ± β) = cosαcosβ ∓ sinαsinβ
    • tan(α ± β) = ^{tanα ± tanβ}⁄_{1 ∓ tanαtanβ}
    • tan2θ = ^2tanθ⁄_{1 − tan²(θ)}
    • sin2θ = 2sinθcosθ
    • cos2θ = cos²θ − sin²θ (OR) 1 − 2sin²θ (OR) 2cos²θ − 1

By using the above identities, one can verify and simplify trig expressions. BE CAREFUL, everything requires the proper procedure, so thoroughly analyze an expression before attempting to do anything! Practice makes perfect.

Solving Trigonometric Equations
There are multiple ways as to solve trigonometric equations. The first method is isolating trigonometric expressions. The following is an example:
    5cos x = 3cos x + √3
    → 2cos x = √3
    → cos x = ^√3⁄₂
    → x = arccos(^√3⁄₂)
    → x = ^π⁄₆ + 2nπ, ^11π⁄₆ + 2nπ [n ∈ Z]

Another method is taking the square root of both sides. The following is an example:
    3tan² x − 4 = -3
    → 3tan² x = 1
    → tan² x = ¹⁄₃
    → tan x = ± ¹⁄_√3
    → x = arctan(¹⁄_√3), arctan(- ¹⁄_√3)
    → x = ^π⁄₆ + nπ, ^5π⁄₆ + nπ [n ∈ Z]

The next method is solving by factoring. The following are examples:
    2sin x cos x = √3 sin x
    → 2sin x cos x − √3 sin x = 0
    → sin x (2cos x − √3) = 0
    → sin x = 0; cos x = ^√3⁄₂

    2sin² x + sin x − 1 = 0
    → (2sin x − 1)(sin x + 1) = 0
    → sin x = ¹⁄₂; sin x = -1

Basically, the same ways one can solve a nontrig equation can be applied to trigonometric equations as well. Furthermore, the use of identities is also an applicable solution.

Semester 2

Conic Sections

Parabola
The parabola is a graph structure that is normally produced by a quadratic equation, such as y = x². However, the conic formulae for parabolas are the following:
    • (x − h)² = 4p(y − k), if the parabola opens up or down.
    • (y − k)² = 4p(x − h), if the parabola opens left or right.

You're probably wondering what k, h, and p represent. k is the vertical shift, h is the horizontal shift, and p is the dilation/reflection factor.

When given a parabolic equation, one must determine the vertex, focus, axis of symmetry, and directrix. The following are the formulae for determining these through the given equation:
    • Vertex: (h, k)
    • Focus: (h, k + p), if x is squared; (h + p, k), if y is squared
    • Axis of Symmetry: x = h, if x is squared; y = k, if y is squared
    • Directrix: y = k − p, if x is squared; x = h − p, if y is squared

Therefore, if the parabolic equation were (x + 8)² = 8(y − 3), then these would be the results because p = 2:
    • Vertex: (-8, 3)
    • Focus: (-8, 5)
    • Axis of Symmetry: x = -8
    • Directrix: y = 1

Ellipse
An ellipse is a shape in which all of the edges are curves. A circle is a type of ellipse, much like how a square is a type of rectangle.
The conic formula for an ellipse (non-circle) is ^{(x − h)²}⁄_a² + ^{(y − k)²}⁄_b² = 1. If a is greater than b, then the orientation (or direction of the major/longer axis) is horizontal. If b is greater than a, then the orientation is vertical.

When given an elliptical equation, one must determine the vertices, the co-vertices, the major axis, the minor axis, and the foci. The following are the formulae with which one can find these results:
    • Vertices: (h + a, k), if horizontal; (h, k + b), if vertical
    • Co-Vertices: (h, k + b), if horizontal; (h + a, k), if vertical
    • Major Axis: y = k, len = 2a (if horizontal); x = h, len = 2b (if vertical)
    • Minor Axis: x = h, len = 2b (if horizontal); y = k, len = 2a (if vertical)
    • Foci: (h ± c, k), if horizontal; (h, k ± c), if vertical

The foci are fixed points such that the sum of the distances between each focus and any point on the ellipse is the same for all points on the ellipse. This sum is actually the length of the major axis because the relationship between the variables c (see above), a, and b is represented by the formula c = √( |a² − b²| ). This formula is derived from the fact that the distance between a focus and a co-vertex is equal to the larger a or b variable.

Another thing that is useful to find when given an ellipse is its eccentricity, which measures an ellipse's elongation. The formula for eccentricity is e = c⁄a if the orientation is horizontal, and e = c⁄b if the orientation is vertical.

Hyperbola
A hyperbola is the locus of all points in a plane such that the difference of their distances from two foci is constant. The conic formula for a hyperbola is ^{(y − k)²}⁄_a² − ^{(x − h)²}⁄_b² = 1.
However, it could also be ^{(x − h)²}⁄_b² − ^{(y − k)²}⁄_a² = 1.
There are two conic formulae for a hyperbola because either the x or y could be negative. If y is negative, the orientation of the transverse axis (see below) is horizontal. If x is negative, the orientation of the transervse axis is vertical.

When given a hyperbolic equation, one must determine the vertices, foci, transverse axis, conjugate axis, and asymptotes. The following are the formulae with which one can find these results:




QUIZ

1. Rewrite in standard form, determine the conic, and analyze the data: -x² + 10x − 8y − 9 = 0
    • Standard Form: (x − 5)² = -8(y − 2)
    • Conic: Parabola
    • Vertex: (5, 2)
    • Focus: (5, 0)
    • Axis of Symmetry: x = 5
    • Directrix: y = 4
2. Use the following data to write the equation of a hyperbola in standard form:
    • Foci: (5 + 11√2, -9), (5 − 11√2, -9)
    • Asymptotes: y = x − 14, y = -x − 4
    • Standard Form: (x − 5)² − (y + 9)² = 121
3. Use the following data to write the equation of a parabola in standard form:
    • Focus: (2, -5√3)
    • Directrix: x = -1
    • Standard Form: (y + 5√3)² = 6(x − ½)
4. Rewrite in standard form, determine the conic, and analyze the data: 4x² + 9y² − 16x + 90y + 205 = 0
    • Standard Form: 4(x − 2)² + 9(y + 5)² = 36
    • Conic: Ellipse
    • Vertices: (5, -5), (-1, -5)
    • Co-Vertices: (2, -3), (2, -7)
    • Foci: (2 ± √5, -5)
    • Major Axis: y = -5, len = 6
    • Minor Axis: x = 2, len = 4
    • Eccentricity: 0.745

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Vectors

Below are some images depicting references for data regarding vectors.

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A vector is a type of line that has a specified direction and data value.
This direction is expressed through an angle, θ. This angle can be found with the component form of the vector (〈a, b〉) through the formula θ = arctan(b⁄a) in the restricted domain of [0, 2π].
The data value is known as the vector's magnitude, which can be calculated via the formula above (second image).

One can write the component form of a vector when given its angle of direction and magnitude. The following is the formula:
    → v = ‖v‖〈cosθ, sinθ〉

Aside from the component form, there is also linear combination. Linear combination is represented through scalar multiplication of variable, where i =〈1, 0〉 and j =〈0, 1〉
For instance, the linear combination of the vector〈7, -3〉would be 7i − 3j. In other words, for a vector, v, with component form〈a, b〉the linear combination would be ai + bj.

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Here is some more data that can be acquired from vectors:





The other value that is used to represent the vector u as a sum of two vectors including the projection of u onto v is found via the following:
    → w₂ = u − projvu

In order to determine the angle, ψ, between two vectors within the domain [0, 2π], one must use the following formula when given vectors u and v:
    → cosψ = ^{(u • v)}⁄_{(‖v‖ × ‖u‖)}

Parametric Equations

A parametric equation is a set of two functions of the variable t that represent x and y values.
x = f(t)     y = g(t)

The variable t is known as the parameter. For each value of t, there is a set of (x, y) coordinates. Below is a table and graph that serve as a provided example:

t	x = t² − 1	y = t⁄4 + 2
-3	8	1.25
-2	3	1.5
-1	0	1.75
0	-1	2
1	0	2.25
2	3	2.5
3	8	2.75

Parametric to Rectangular
In calculus, the term rectangular form is used very commonly. The recntangular form of a equation is to derirve that equation in terms of x and y.
An example of converting parametric form to rectangular form is below:
    • y = 2t
    • x = t² + 2
    → t = y⁄₂
    → x = ( y⁄₂ )² + 2
    → ( y⁄₂ )² = x − 2
    → y⁄₂ = √(x − 2)
    → y = 2√(x − 2)
For the conversion, the first step is to solve for t using one of the functions. Then, substitute the parameter in the other function with the result acquired from the first step. Finally, simplify until the result is a rectangular equation in terms of x and y.

Sometimes, the parameter is an angle, θ. In this case, the goal is to put both parameters in terms of the trigonometric functions being used. Here are a few examples:
1. y = 5sinθ     x = 3cosθ
→ sinθ = y⁄₅     cosθ = x⁄₃
→ ( y⁄₅ )² + ( x⁄₃ )² = 1    [Identity: sin²θ + cos²θ = 1]
→ ^y²⁄₂₅ + ^x²⁄₉ = 1

2. y = ^√2⁄_4sinθ         x = ^cotθ⁄₆
→ cscθ = 2y√2     cotθ = 6x
→ (2y√2)² = (6x)² + 1    [Identity: cot²θ + 1 = csc²θ]
→ 8y² − 36x² = 1

Rectangular to Parametric
The first step of converting a rectangular equation to parametric is to find the value of x or y in terms of t or θ. Then, substitution and simplification. The following are examples:
1.  t = x − 1     y = x² + 2
→ x = t + 1     y = (t + 1)² + 2
→ x = t + 1     y = t² + 2t + 3

2.  t = 4y²     y = √x
→ y = ^√t⁄₂     ^√t⁄₂ = √x
→ y = ^√t⁄₂     x = t⁄₄

3.  sinθ = ^{(x − 1)}⁄₂     ^{(x − 1)²}⁄₄ + ^y²⁄₉ = 1
→ x − 1 = 2sinθ     ^4sin²θ⁄₄ + ^y²⁄₉ = 1
→ x = 2sinθ + 1     1 − sin²θ = ^y²⁄₉
→ x = 2sinθ + 1     cos²θ = ^y²⁄₉
→ x = 2sinθ + 1     y⁄₃ = cosθ
→ x = 1 + 2sinθ     y = 3cosθ

Projectile Motion
When an object is thrown, it curves upward and then falls back down to the ground. The movement of the object is known as projectile motion. This is caused by gravitational forces represented by a constant, g ( g = 32 ^ft⁄_s² or 9.8 ^m⁄_s² ).


The following data can be derived from projectile motion:
    1. The time at which the object hits the ground ( t_f ).
        → Solve for t when y = 0.
    2. The highest point the object reaches during its travel.
        → Solve for y when t = ½ × t_f
    3. The total distance travelled by the object.
        → Solve for x when t = t_f



QUIZ

1. What is the direction and magnitude of a vector that has the initial point (-1, 5) and the terminal point (2, 3).
    • v = 3i − 2j
    • θ = 326.31°
    • ‖v‖ = √13
2. Convert the following parametric equation to rectangular:
    • x = t² − 1
    • y = ³⁄t
          → y = ³⁄_{√(x + 1)}
3. Give the component form of a vector with a magnitude of 13√2 and direction of ^5π⁄₄
    →〈-13, -13〉
4. Convert the following equation to parametric form: ^{(x + 3)²}⁄₂₅ + ^{(y − 5)²}⁄₁₆₉ = 1
    • x = 5sinθ − 3
    • y = 13cosθ + 5
5. Find the measure of the angle between the vectors〈7, 0〉and〈-3, -4〉
ψ = 126.87°
6. Write vector u as a sum of two vectors, including the projection of u onto v:
    u =〈-5, 10〉
    v =〈1, 2〉
          → u =〈3, 6〉+〈-8, 4〉
7. A soccer ball is kicked across a field at an angle of 60° with an initial speed of 18 m/s.
    a) Write a set of parametric equations for the motion of the soccer ball.
    b) Determine when the soccer ball hit the ground.
    c) Determine how far the soccer ball travelled.
    d) Determine the soccer ball's maximum height.

a) x = 9t, y = 9t√3 − 4.9t²
b) 3.18 seconds
c) 28.63 m
d) 12.4 m

Reveal Answers

Polar and Rectangular Systems

Polar Coordinates
In mathematics, thepolar coordinate system is a two-dimensional coordinate system in which each point on a plane is determined by a distance from a reference point and an angle from a reference direction.
A polar coordinate is a point on the coordinate system. It is expressed by P_pol = (r, θ), where r is the distance from the reference point, and θ is the angle of direction.
A rectangular coordinate is one that can be found on the standard coordinate plane.
    • To convert a polar coordinate to rectangular: P_rect = (rcosθ, rsinθ)

Converting from rectangular to polar is not as simple. Consider the rectangular coordinate, P. When given P_rect = (x, y), one must determine the θ and r values. This can be determined thusly:
    → r = √(x² + y²)
    → θ = arctan(y⁄x)

The distance, r, is usually greater than 1. However, when it is less than one, there is a way to rewrite the polar coordinate with a positive r value. The following is that method:
    • (-r, θ) → (r, θ ± 180°)

Polar Equations
A polar equation is a function that exists on the polar coordinate system. It is written in terms of r. The following are rectangular to polar conversions:
   • x = rcosθ
   • y = rsinθ
    • r² = x² + y²
    • tanθ = y⁄x

Example(s):
• Rect: x² + y² = 9  →  Pol: r = 3
• Pol: r = ⁶⁄_{(2cosθ − 3sinθ)}  →  Rect: 2x − 3y = 6




Complex Numbers in Polar and Rectangular Systems

A complex number is one that contains both an imaginary and real number. In reality, all numbers can be expressed as a complex number by appending 0i.
A complex number is normally represented by the variable z. In a rectangular system, complex numbers are expressed by z = a + bi.



The polar form of a complex number is expressed as z = r(cosθ, + i sinθ). A shorter method of writing this is z = r cis θ.





Matrices

In mathematics, a matrix (plural: matrices) is a rectangular array of numbers, symbols, or expressions, arranged in rows and columns. For example, the dimensions of the matrix below are 2 × 3, because there are two rows and three columns:


Basic Operations
The basic operations of matrices are addition and subtraction. Matrices can only do this if they share the same dimensions. The following is an example:


Matrix Multplication
Matrix multiplication can only be done if the number of columns in matrix A and the number of rows in matrix B are the same. Furthermore, matrix multiplication is not cummutative, meaning AB ≠ BA. An example of two matrices not commuting with each other is:

whereas

If the dimensions of matrix A are m × n, and the dimensions in matrix B are n × p, the dimensions of the resultant matrix ( AB ) are m × p.


Inverse Matrices
It is impossible to divide by a matrix. However, you can multiply the inverse of the matrix, represented as A^-1. Only square matrices with the same number of rows and columns can have inverses. For pre-calculus, you will only ever need to determine the inverse of a 2 × 2 matrix by hand. However, you can find the inverse of other square matrices using a graphing calculator.

What even is an inverse matrix? Well, it is a matrix that makes A^-1A = In. In is a square identity matrix with dimensions n × n. An identity matrix is a matrix that, when multiplied with a matrix with dimensions n × p, the result is that same matrix. Basically: AI_n = A.
An identity matrix contains all zeros, but the main diagonal is consisted of ones.

When finding the inverse of a 2 × 2 matrix, one must find its determinant. The determinant can be found as follows:

The determinant of a 3 × 3 matrix can be found thusly:

The inverse of a 2 × 2 matrix is calculated via the following:


Partial Fractions
Do you remember what a rational function is? In mathematics, a rational function is any function which can be defined by a rational fraction, i.e. an algebraic fraction such that both the numerator and the denominator are polynomials.
Partial fractions make use of these rational fractions to represent a rational function.

Partial Fraction Decomposition of Proper Rational Expressions
The number of items in the partial fraction decomposition is the same as the number of factors in the denominator of the rational expression. A rational expression is considered proper if the degree of the leading term of the denominator is greater than that of the numerator; furthermore, the denominator can be factored into two or more prime factors. The following is an example:

      ex: ^{(3x − 7)}⁄_{(x² − 7x + 12)}
    → ^{(3x − 7)}⁄_{(x − 4)(x − 3)} = A⁄_{(x − 4)} + B⁄_{(x − 3)}
    → 3x − 7 = A(x − 3) + B(x − 4)
      • A = 5,   B = -2
    → ⁵⁄_{(x − 4)} − ²⁄_{(x − 3)}

Partial Fraction Decomposition of Improper Rational Expressions
If the rational expression is improper, the polynomials must be divided to acquire a simplified proper expression.
The decomposition occurs with the remainder ( r(x) ) of the quotient ( q(x) ). An example is below:

      ex: ^{(3x² + 5x + 2)} ⁄ _{(x² + x)}
    → q(x) = 3 + ^{(2 − x)}⁄_{(x² + 2x)}
    → r(x) = ^{(2 − x)}⁄_{x(x + 2)} = A⁄x + B⁄_{(x + 2)}
    → r(x) = 2 − 2 = A(x + 2) + B(x)
      • A = 1,   B = -2
    → r(x) = ¹⁄x − ²⁄_{(x + 2)}
    → q(x) = 3 + r(x) = 3 + ¹⁄x − ²⁄_{(x + 2)}