SEHH1069 Last minutes
Linear System
Consistent: Have solution(s)
-Unique solution (Independent) -Infinitely many solution (Dependent)
-Homogeneous 0 solution = trivial sol non-0 solution = non-trivial sol
Inconsistent: -No solutions
Elementary row operations
- Interchange row 𝑅𝑖 ⟷ 𝑅𝑗
1
- Multiply and divide element of a row 𝑘𝑅𝑖 → 𝑅𝑖
𝑘
𝑅𝑖 → 𝑅𝑖 (𝑘 ≠ 0)
- Add and subtract row to another row 𝑘𝑅𝑖 + 𝑅𝑗 → 𝑅𝑗
To be:
row echelon form (REF)
(Gaussian Elimination)
1 ∗
(0 1
0 0
reduced row echelon form (RREF)
(Gauss-Jordan Elimination)
∗ ∗
∗| ∗)
0 0
1
(0
0
0 ∗ ∗
1 ∗| ∗ )
0 0 0
Requirement of row echelon form:
 Non-0 rows are above any 0 rows
 First non-0 entry(left->right) in row is 1 (Leading 1)
 All entries below leading 1 are 0
Requirement of reduced row echelon form:
 It is in REF
 Each leading 1 is the only non-0 entry in its column
Matrix Algebra
Row/Column = Horizontal/Vertical (𝑚[𝑟𝑜𝑤𝑠] × 𝑛[𝑐𝑜𝑙𝑢𝑚𝑛𝑠] 𝑚𝑎𝑡𝑟𝑖𝑥)
Row/Column matrix (1 × 𝑛 𝑚𝑎𝑡𝑟𝑖𝑥)/(𝑚 × 1 𝑚𝑎𝑡𝑟𝑖𝑥)
Null(0) matrix All element are 0
Square matrix (𝑚 × 𝑚 𝑚𝑎𝑡𝑟𝑖𝑥)
Diagonal matrix
A square matrix in which all the entries off the main diagonal are zero.
𝑎
e.g., 𝐴 = (0
0
0
𝑏
0
0
𝑎𝑛
𝑛
0),Note that 𝐴 = ( 0
𝑐
0
0
𝑏𝑛
0
0
0 ) , 𝑛 ∈ ℤ+ .
𝑐𝑛
Identity Matrix(Ι) = a diagonal matrix with all diagonal element = 1.
Triangular matrix
-Lower triangular -Upper triangular
∗
∗ 0 0
𝐴 = (∗ ∗ 0) 𝐴 = (0
0
∗ ∗ ∗
∗ ∗
∗ ∗)
0 ∗
Transpose of a matrix
𝑎
(𝑑
𝑔
𝑏
𝑒
ℎ
𝑐 𝑇
𝑎
𝑓 ) = (𝑏
𝑖
𝑐
(𝐴𝑇 )𝑇 = 𝐴
𝑑
𝑒
𝑓
𝑔
ℎ)
𝑖
(𝐴 ± 𝐵)𝑇 = 𝐴𝑇 ± 𝐵 𝑇
(𝑘𝐴)𝑇 = 𝑘𝐴𝑇
(𝐴𝐵)𝑇 = 𝐵 𝑇 𝐴𝑇
Trace of a matrix (for square matrix)
𝑎
𝐴 = (𝑑
𝑔
𝑏
𝑒
ℎ
𝑐
𝑓 ) , 𝑡𝑟(𝐴) = 𝑎 + 𝑒 + 𝑖
𝑖
𝑡𝑟(𝐴 + 𝐵) = 𝑡𝑟(𝐴) + 𝑡𝑟(𝐵)
𝑡𝑟(𝑐𝐴) = 𝑐𝑡𝑟(𝐴) 𝑡𝑟(𝐴) = 𝑡𝑟(𝐴𝑇 ) 𝑡𝑟(𝐴𝐵) = 𝑡𝑟(𝐵𝐴) 𝑡𝑟(𝑃−1 𝐴𝑃) = 𝑡𝑟(𝐴)
Inverse of a matrix (for square matrix)
𝐼𝑓 𝐴𝐵 = 𝐵𝐴 = 𝐼, 𝐴−1 = 𝐵 𝑎𝑛𝑑 𝐵 −1 = 𝐴
Invertible = non-singular : det 𝐴 ≠ 0
Not invertible = singular : det 𝐴 = 0
𝑎
𝑐
𝐴=(
𝑏
)
𝑑
𝐴−1 =
1
𝑑
(
𝑎𝑑 − 𝑏𝑐 −𝑐
(𝐴𝐵)−1 = 𝐵 −1 𝐴−1 𝐴−𝑘 = (𝐴−1 )𝑘 = (𝐴𝑘 )−1 (𝐴𝑇 )−1 = (𝐴−1 )𝑇 (𝑘𝐴)−1 = 𝑘 −1 𝐴−1
Inversion Algorithm (𝐴|𝐼)~(𝐼|𝐵)
Adjoint Method 𝐴−1 = det(𝐴)−1 𝑎𝑑𝑗(𝐴)
Cramer’s rule
𝑎 𝑗 𝑐
𝑎
𝑗 𝑏 𝑐
|𝑑 𝑘 𝑓 |
|𝑑
|𝑘 𝑒 𝑓 |
𝑎 𝑏 𝑐 𝑥1
𝑗
𝑔
𝑙
𝑖
𝑔
(𝑑 𝑒 𝑓) (𝑥2 ) = (𝑘) 𝑥1 = 𝑙 ℎ 𝑖 𝑥2 =
𝑥3 =
𝑎 𝑏 𝑐
𝑎 𝑏 𝑐
𝑎
𝑥3
𝑔 ℎ 𝑖
𝑙
|𝑑 𝑒 𝑓 |
|𝑑 𝑒 𝑓 |
|𝑑
𝑔 ℎ 𝑖
𝑔 ℎ 𝑖
𝑔
Determinants
det(𝐴𝑇 ) = det(𝐴)
det(𝑘𝐴) = 𝑘 𝑛 det(𝐴) 𝑓𝑜𝑟 𝑛 × 𝑛 𝑚𝑎𝑡𝑟𝑖𝑥 𝐴
1
det(𝐴𝐵) = det(𝐴) det(𝐵) det(𝐴−1 ) = det(𝐴)
𝑎
𝐴 = |𝑑
𝑔
𝑑
|𝑎
𝑔
𝑒
𝑏
ℎ
𝑏
𝑒
ℎ
𝑐
𝑓|
𝑖
𝑘𝑎
|𝑑
𝑔
𝑘𝑏
𝑒
ℎ
𝑘𝑐
𝑓 | = 𝑘𝑑𝑒𝑡(𝐴)
𝑖
𝑎 + 𝑘𝑑
𝑓
𝑐 | = − det(𝐴) | 𝑑
𝑖
𝑔
𝑏 + 𝑘𝑒
𝑒
ℎ
𝑐 + 𝑘𝑓
𝑓 | = det(𝐴)
𝑖
𝑏
𝑒
ℎ
𝑏
𝑒
ℎ
𝑗
𝑘|
𝑙
𝑐
𝑓|
𝑖
−𝑏
)
𝑎
Adjoint Matrix
𝑎
𝑐
2×2𝐴 =(
𝑎
𝑑
3×3𝐴 =(
𝑔
𝐴−1 =
𝑏
)
𝑑
𝑏
𝑒
ℎ
𝑑
−𝑐
𝑎𝑑𝑗(𝐴) = (
−𝑏
)
𝑎
𝑒
|
ℎ
𝑐
𝑓 ) 𝑎𝑑𝑗(𝐴) = − |𝑏
ℎ
𝑖
𝑏
|
( 𝑒
1
𝑎𝑑𝑗(𝐴)
det(𝐴)
adj(𝐴) = det(𝐴) 𝐴−1
det(𝑎𝑑𝑗(𝐴)) = det(det(𝐴) 𝐴−1 )
𝑑𝑒𝑡(𝑎𝑑𝑗(𝐴)) = det(𝐴)𝑛 × det(𝐴)−1
𝑑𝑒𝑡(𝑎𝑑𝑗(𝐴)) = det(𝐴)𝑛−1
adj(𝐴) = det(𝐴) 𝐴−1
(𝑎𝑑𝑗(𝐴))
(𝑎𝑑𝑗(𝐴))
−1
−1
= (det(A) 𝐴−1 )−1 =
=
1
det(𝑎𝑑𝑗(𝐴))
𝐴
det(𝐴)
𝑎𝑑𝑗(𝑎𝑑𝑗(𝐴))
𝑎𝑑𝑗(𝑎𝑑𝑗(𝐴)) = det(𝑎𝑑𝑗(𝐴))
𝐴
det(𝐴)
𝑎𝑑𝑗(𝑎𝑑𝑗(𝐴)) = det(𝐴)𝑛−2 𝐴
Symmetric Matrix 𝐴 = 𝐴𝑇
Skew symmetric Matrix 𝐴 = −𝐴𝑇
𝑇
𝑑 𝑓
𝑑 𝑒
𝑓
| −|
| |
|
𝑔 ℎ
𝑔 𝑖
𝑖
𝑎
𝑐
𝑎 𝑏
𝑐
|
|𝑔 𝑖 | − |
|
𝑔
ℎ
𝑖
𝑎 𝑐
𝑐
𝑎 𝑏
| − |𝑑 𝑓 | |
|
𝑓
𝑑 𝑒 )
Function
[open intervals] including border (closed intervals) excluding border
Inverse function y = 𝑓(𝑥) ~ 𝑥 = 𝑓 −1 (𝑦) ~ 𝑦 = 𝑓 −1 (𝑥)
Limit
- Simplification
- Rationalization (sqrt)
-
0
±∞
𝑓 ′ (𝑥)
𝑓(𝑥)
L'Hospital's Rule (0 \ ±∞) lim 𝑔(𝑥) = lim 𝑔′ (𝑥)
𝑥→𝑐
𝑥→𝑐
Continuity
lim 𝑓(𝑥) = 𝑓(𝑎)
𝑥→𝑎
Differentiation
Differentiable
Continuity and no “corner” at 𝑥 → 𝑎
Chain rule
𝑑𝑦
𝑑𝑥
=
𝑑𝑦 𝑑𝑢
∙
𝑑𝑢 𝑑𝑥
Implicit Differentiation
Inverse function theorem
(𝑓 −1 )′ (𝑥 ) =
1
𝑓 ′ (𝑓−1 (𝑥))
Logarithmic Differentiation
𝑦 = 𝑥𝑟
ln 𝑦 = 𝑟 ln 𝑥
1 𝑑𝑦
𝑑 ln 𝑥
∙
=𝑟
𝑦 𝑑𝑥
𝑑𝑦
Increasing/Decreasing 𝑓 ′ (𝑥) > 0 / 𝑓′(𝑥) < 0
Critical/Maximum/Minimum Point 𝑓 ′ (𝑥) = 0
Convex/Concave 𝑓 ′′ (𝑥) > 0 / 𝑓′′(𝑥) < 0
Point of inflexion 𝑓 ′ ′(𝑥) = 0
SAsymtotes
o Vertical Asymptote
lim 𝑓(𝑥) = ±∞ 𝑜𝑟 lim 𝑓(𝑥) = ±∞
𝑥→𝑎+
𝑥→𝑎−
o Horizontal Asymptote
lim 𝑓(𝑥) = 𝐿 𝑜𝑟 lim 𝑓(𝑥) = 𝐿
𝑥→+∞
𝑥→−∞𝑆
Integrals
- Partial Fractions
-
Substitution
Part
∫ 𝑓(𝑥)𝑔′ (𝑥)𝑑𝑥 = 𝑓(𝑥)𝑔(𝑥) − ∫ 𝑔(𝑥)𝑓′(𝑥) 𝑑𝑥
-
FTC
o FTC I
𝑥
𝐹(𝑥) = ∫ 𝑓(𝑡)𝑑𝑡 𝑓𝑜𝑟 𝑎 ≤ 𝑥 ≤ 𝑏
𝑎
𝑥
𝑑
[∫ 𝑓(𝑡)𝑑𝑡] = 𝑓(𝑥) 𝑓𝑜𝑟 𝑎 ≤ 𝑥 ≤ 𝑏
𝑑𝑥 𝑎
o FTC II
𝑏
∫ 𝑓(𝑥) 𝑑𝑥 = ∫ 𝑓(𝑏)𝑑𝑥 − ∫ 𝑓(𝑎) 𝑑𝑥
𝑎
-
Area/Volume
𝑏
o Area ∫𝑎 [𝑔(𝑥) − 𝑓(𝑥)]𝑑𝑥
Formulas and Identities(Provided/Learnt)
sin θ
1
1
1
sec 𝜃 =
csc 𝜃 =
cot 𝜃 =
cos θ
cos 𝜃
sin 𝜃
tan 𝜃
2
2
2
2
2
sin 𝜃 + cos 𝜃 = 1
1 + tan 𝜃 = sec 𝜃
1 + cot 𝜃 = csc 2 𝜃
sin(𝑎 ± 𝑏) = sin 𝑎 cos 𝑏 ± cos 𝑎 sin 𝑏
cos(𝑎 ± 𝑏) = cos 𝑎 cos 𝑏 ∓ sin 𝑎 sin 𝑏
tan θ =
tan(𝑎 ± 𝑏) =
tan 𝑎 ± cos 𝑏
1 ∓ tan 𝑎 tan 𝑏
sin 𝑥 cos 𝑦 =
sin(𝑥 − 𝑦) + sin(𝑥 + 𝑦)
2
cos 𝑥 cos 𝑦 =
cos(𝑥 − 𝑦) + cos(𝑥 + 𝑦)
cos(2𝑥) = 2 cos 2 𝑥 − 1 = 1 − 2 sin2 𝑥
2
sin 𝑥 sin 𝑦 =
cos(𝑥 − 𝑦) − cos(𝑥 + 𝑦)
2
sin(2𝑥) = 2 sin 𝑥 cos 𝑥
1
1
sin 𝛼 ± sin 𝛽 = 2 sin (𝛼 ± 𝛽) cos (𝛼 ∓ 𝛽)
2
2
1
1
cos 𝛼 + cos 𝛽 = 2 cos (𝛼 + 𝛽) cos (𝛼 − 𝛽)
2
2
1
1
cos 𝛼 − cos 𝛽 = −2 sin (𝛼 + 𝛽) sin (𝛼 − 𝛽)
2
2
1 n
lim (1 + ) = e
n→∞
n
𝑓(x)
c
x
r
1
lim (1 + 𝑛)𝑛 = 𝑒
𝑛→0
𝑓 ′ (x)
0
rx r−1
ax
ax ln 𝑎
sin 𝑥
cos 𝑥
cos 𝑥
− sin 𝑥
tan 𝑥
sec 2 𝑥
cot 𝑥
− csc 2 𝑥
sec 𝑥
sec 𝑥 tan 𝑥
csc 𝑥
− csc 𝑥 cot 𝑥
𝑎
−1
sin (𝑎𝑥)
cos−1 (𝑎𝑥)
tan−1(𝑎𝑥)
√1 − (𝑎𝑥)2
𝑎
−
√1 − (𝑎𝑥)2
𝑎
1 + (𝑎𝑥)2
𝑓(x)
∫ 𝑓(x)𝑑𝑥
1
𝑥
ln|𝑥|
𝑎𝑥
(𝑎 > 0 𝑎𝑛𝑑 𝑎 ≠ 1)
𝑎𝑥
ln 𝑎
𝑥𝑟
sin 𝑥
𝑥 𝑟+1
𝑟+1
− cos 𝑥
cos 𝑥
sin 𝑥
2
sec 𝑥
tan 𝑥
csc 2 𝑥
− cot 𝑥
sec 𝑥 tan 𝑥
sec 𝑥
csc 𝑥 cot 𝑥
1
− csc 𝑥
√1 −
(𝑎𝑥)2
1
1 + (𝑎𝑥)2
1 −1
sin (𝑎𝑥)
𝑎
1
tan−1(𝑎𝑥)
𝑎
∫ ln 𝑥 𝑑𝑥
∫ tan2 𝑥 𝑑𝑥
1
= ln(𝑥) 𝑥 − ∫ 𝑥 ( ) 𝑑𝑥
𝑥
= ∫ sec 2 𝑥 + 1 𝑑𝑥
= 𝑥 ln(𝑥) − ∫ 𝑑𝑥
= 𝑥 ln(𝑥) − 𝑥 + 𝐶
sin 𝑥
𝑑𝑥
cos 𝑥
= −∫
1
𝑑 cos 𝑥
cos 𝑥
= − ln|cos 𝑥| + 𝐶
= ln|sec 𝑥| + 𝐶
∫ sec 𝑥 𝑑𝑥
= ∫ sec 𝑥 ×
∫ sec 2 𝑥 𝑑𝑥
= 𝑡𝑎𝑛 𝑥 + 𝐶
∫ tan 𝑥 𝑑𝑥
=∫
= tan 𝑥 + 𝑥 + 𝐶
sec 𝑥 + tan 𝑥
𝑑𝑥
sec 𝑥 + tan 𝑥
(sec 2 𝑥 + sec 𝑥 tan 𝑥)
𝑑𝑥
sec 𝑥 + tan 𝑥
1
=∫
𝑑(sec 𝑥 + tan 𝑥)
sec 𝑥 + tan 𝑥
∫ sin−1 𝑥 𝑑𝑥
= ∫ 1 × sin−1 𝑥 𝑑𝑥
= 𝑥 sin−1 𝑥 − ∫ 𝑥 ×
1
𝑑𝑥
√1 − 𝑥 2
1
1
= 𝑥 sin−1 𝑥 − ∫
𝑑(1 − 𝑥 2 )
2 √1 − 𝑥 2
1
1
= 𝑥 sin−1 𝑥 − × 2 × (1 − 𝑥 2 )2 + 𝐶
2
= 𝑥 sin−1 𝑥 − √1 − 𝑥 2 + 𝐶
=∫
= ln|sec 𝑥 + tan 𝑥| + 𝐶
∫ tan−1 𝑥 𝑑𝑥
= ∫ 1 × tan−1 𝑥 𝑑𝑥
= 𝑥 tan−1 𝑥 − ∫ 𝑥 ×
∫ sin2 𝑥 𝑑𝑥
1 cos 2𝑥
=∫ −
dx
2
2
=
x 1
− ∫ cos 2𝑥 𝑑𝑥
2 2
=
x 1 1
− × ∫ cos 2𝑥 𝑑2𝑥
2 2 2
=
x sin 2𝑥
−
+C
2
4
1
𝑑𝑥
1 + 𝑥2
1
1
= 𝑥 tan−1 𝑥 − ∫
𝑑(1 + 𝑥 2 )
2 1 + 𝑥2
1
= 𝑥 tan−1 𝑥 − ln(1 + 𝑥 2 ) + 𝐶
2