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UNENE Math Refresher Course
Algebra
1
Algebra — Exponential Functions © Wei-Chau Xie
Exponential Functions
y = ax is an exponential function.
Properties
1. a0 = 1, a 6=0
2. am · an = am+n
3.1
an= a−n,
am
an= am−n, a 6=0
4. (am)n = am ·n
5. (a b)n = an · bn
6.(a
b
)n= an
bn
2
Algebra — Exponential Functions © Wei-Chau Xie
For the special case y = ex, e = 2.718281828 . . .
exe−x
e−x→0, when x→∞
x0
5
10
15
20
−1−2−3 2 31
3
Algebra — Logarithmic Functions © Wei-Chau Xie
Logarithmic Functions
If x = a y, a>0, a 6= 1 =⇒ y = loga x, where a is called the base.
Properties
1. a = a1 =⇒ loga a = 1
2. 1 = a0 =⇒ loga 1 = 0
3. loga bx = x loga b
4. loga x + loga y = loga (x y)
5. loga x − loga y = loga
( x
y
)
6. aloga x = x
7. loga x = logb x
logb a
4
Algebra — Logarithmic Functions © Wei-Chau Xie
In mathematics and engineering, the most frequently used bases: a = 10 and e.
a = 10 : log10 x = lg x
a = e : loge x = ln x
−5
−4
−3
−2
−1
0
1
2
3
1 2 4 6 8 10
x
lnx
5
Algebra — Exponential and Logarithmic Functions © Wei-Chau Xie
Example
Simplify y =ln x2 − ln
1x
ln 3√
x.
y = ln x2 − ln x−1
ln x13
= 2 ln x − (− ln x)
13 ln x
= 3 ln x13 ln x
= 9
6
Algebra — Exponential and Logarithmic Functions © Wei-Chau Xie
Example
Solve for x in equation 20 = 500
(
1 − 4
4 + e−0.002x
)
.
Divide the equation by 500:1
25= 1 − 4
4 + e−0.002x
Simplify:4
4 + e−0.002x= 24
25=⇒ 4 + e−0.002x = 25
6
e−0.002x = 1
6
Take ln of both sides: ln e−0.002x = ln1
6=⇒ − 0.002 x = − ln 6
∴ x = 1
0.002ln 6 = 895.8797
7
Algebra — Exponential and Logarithmic Functions © Wei-Chau Xie
Example
• Given that y = y0 e−axb, x>0, solve for x.
• If y = 0.1, y0 = 10.2, a = 0.5, b = 2.1, evaluate x.
Taking ln of both sides of the equation
ln y = ln y0 + ln e−axb
= ln y0 − axb
∴ xb = ln y0 − ln y
a=⇒ x =
( ln y0 − ln y
a
) 1b
For y = 0.1, y0 = 10.2, a = 0.5, b = 2.1
x =( ln 10.2 − ln 0.1
0.5
) 12.1 = 9.2499
12.1 = 2.88
8
Algebra — Graphs © Wei-Chau Xie
Graphs of Functions
The following figure shows a plot of functions y = 2x, 5x, 10x, ex2.
It is obvious that we cannot see any detail of 2x for all values of x and not much
details of 5x, 10x, ex2for x<2.
9
Algebra — Graphs © Wei-Chau Xie
500
0 1 2 3 4
1000
1500
2000
2500
3000
3500
4000
4500
5000
5500
6000
6500
7000
7500
8000
8500
9000
9500
10000
x
y
2x
5x
10xex2
10
Algebra — Semi-Log Graphs © Wei-Chau Xie
Semi-Log Graphs
The same functions plotted using a semi-logarithmic scale. Details are revealed
for all values of x.
For semi-log plots, the numbers along the horizontal x-axis are (linearly)
evenly spaced, while along the vertical y-axis, powers of 10 are evenly spaced.
For exponential function y = ax, a>0, taking logarithm of both sides gives
log10 y = x · log10 a
The function appears as a straight line when plotted on semi-log paper.
11
Algebra — Semi-Log Graphs © Wei-Chau Xie
x
y
10
2x
5x
10xex2
10
2
0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 3 43.1 3.2 3.3 3.4 3.5 3.6 3.7 3.8 3.9
3
4
5
6789
100
20
30
40
50
60708090
1000
200
300
400
500
600700800900
10000
2000
3000
4000
5000
6000700080009000
12
Algebra — Log-Log Graphs © Wei-Chau Xie
Log-Log Graphs
The following figure shows a plot of functions y = x2, x3, x4, ex2, e−x2
plotted
using the logarithmic scales (both axes using log scales), in which powers of 10
are evenly spaced.
For function y = xa, x>0, taking logarithm of both sides gives
log10 y = a · log10 x
The function appears as a straight line when plotted on log-log paper.
13
Algebra — Log-Log Graphs © Wei-Chau Xie
x
x2
x3
x4
y
10
1000
200
300
400
500600700800900
100
20
30
40
5060708090
10
2
3
4
56789
1
0.2
0.3
0.4
0.50.60.70.80.9
0.1
0.02
0.03
0.04
0.050.060.070.080.09
0.01
0.002
0.003
0.004
0.0050.0060.0070.0080.009
0.001
0.0001
0.1
0.0002
0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 2 3 4 5 6 7 8 9
0.0003
0.0004
0.00050.00060.00070.00080.0009
10000
2000
3000
4000
50006000700080009000
ex2
e−x2
14
Algebra — Vectors © Wei-Chau Xie
Vectors
Let i, j, k be the unit vectors in the x-, y-, z-directions, respectively.
y
x
z
a
b
c
O
j
k
i
v
A vector v is v = a i + b j + c k
☞ Avector is printed in boldface v, orwritten as ⇀av in handwriting.
The norm (length) of a vector is v = |v| =√
a2 + b2 + c2
The unit vector in direction v is v = v
|v|15
Algebra — Vector Addition © Wei-Chau Xie
Vector Addition
Let v1 = a1 i + b1 j + c1 k, v2 = a2 i + b2 j + c2 k.
v = v1 + v2 = v2 + v1
= (a1 i + b1 j + c1 k) + (a2 i + b2 j + c2 k)
= (a1+a2) i + (b1+b2) j + (c1+c2) k
Graphically, the sum v = v1+v2+v3 is obtained by placing them head to tail
and drawing the vector v from the free tail to the free head.
v = v1+ v2+ v3
v1 v1
v2
v3
v3
v2
The tail of v2 is placed
at the head of v1
The tail of v3 is placed
at the head of v2
16
Algebra — Dot Product of Vectors © Wei-Chau Xie
Dot Product of Vectors
Let v1 = a1 i + b1 j + c1 k, v2 = a2 i + b2 j + c2 k.
v1·v2 = (a1 i + b1 j + c1 k)·(a2 i + b2 j + c2 k)
= a1 a2 + b1 b2 + c1 c2 A scalar
=∣∣v1
∣∣∣∣v2
∣∣ cos θ
v1
v2
θ
The projection of vector v in a given direction, specified by the unit vector
u, is given by
v·u = v cos θ
v
θ uˆv . u = v cos θ
17
Algebra — Cross Product of Vectors © Wei-Chau Xie
Cross Product of Vectors
Let v1 = a1 i + b1 j + c1 k, v2 = a2 i + b2 j + c2 k.
v1 × v2 =
∣∣∣∣∣∣∣∣
i j k
a1 b1 c1
a2 b2 c2
∣∣∣∣∣∣∣∣
i j
a1 b1
a2 b2
= (b1 c2−b2 c1) i + (c1 a2−c2 a1) j + (a1 b2−a2 b1) k Avector
Direction obtained using the right-hand rule: flatten the right hand, four
fingers go along v1, then curl the fingers (palm) towards v2; the direction of
the thumb is the direction of v1×v2.
θ v1
v = v1×v2
v2
Magnitude∣∣v1 × v2
∣∣ =
∣∣v1
∣∣∣∣v2
∣∣ sin θ
18
Linear Algebra — Solutions of Systems of Linear Equations © Wei-Chau Xie
Gaussian Elimination
Solve the following system of linear algebraic equations
3x1 + 4x2 = 10 (1)
2x1 − 5x2 = −1 (2)
To solve for x1, i.e., to eliminate x2,
Eqn. (1) × 5 : 15x1 + 20x2 = 50
Eqn. (2) × 4 : 8x1 − 20x2 = −4 (+23x1 = 46 =⇒ x1 = 2
Similarly, to solve for x2, i.e., to eliminate x1,
Eqn. (1) × 2 : 6x1 + 8x2 = 20
Eqn. (2) × 3 : 6x1 − 15x2 = −3 (−23x2 = 23 =⇒ x2 = 1
Alternatively, having obtained x1, x2 can be found from either Eqn. (1) or (2).
From Eqn. (1): x2 = 10 − 3x1
4= 10 − 3×2
4= 1
19
Linear Algebra — Systems of Linear Equations © Wei-Chau Xie
Consider a system of n linear algebraic equations
a11 x1 + a12 x2 + · · · + a1n xn = b1
a21 x1 + a22 x2 + · · · + a2n xn = b2
· · · · · ·an1 x1 + an2 x2 + · · · + ann xn = bn
where x1, x2, . . . , xn are the n unknowns.
The system can be written in the matrix form
a11 a12 · · · a1n
a21 a22 · · · a2n... ... ... ...
an1 an2 · · · ann
x1
x2...
xn
=
b1
b2...
bn
=⇒ A x = b
︸ ︷︷ ︸ ︸ ︷︷ ︸ ︸ ︷︷ ︸
A x b
where A is the coefficient matrix, x is the column vector of unknowns, and b is
the column vector of right-hand side constants.
20
Linear Algebra — Matrix © Wei-Chau Xie
Operations of Matrices
Addition of Matrices
Let A = [ aij ]m×n
, B = [ bij ]m×n
. A and B must be the same size.
C = A + B =⇒ cij = aij+bij
A + B = B + A, (A + B) + C = A + (B + C)
[
a11 a12 a13
a21 a22 a23
]
+[
b11 b12 b13
b21 b22 b23
]
=[
a11+b11 a12+b12 a13+b13
a21+b21 a22+b22 a23+b23
]
Multiplication by a Scalar
Let A = [ aij ]m×n
. C = αA =⇒ cij = αaij, α is a scalar.
α
[
a11 a12 a13
a21 a22 a23
]
=[
αa11 αa12 αa13
αa21 αa22 αa23
]
21
Linear Algebra — Matrix © Wei-Chau Xie
Multiplication of Matrices
Let A = [ aik ]m×n
, B = [ bkj ]n×l
C = AB =⇒ cij = ai1b1j + ai2b2j + · · · + ainbnj =n
∑
k=1
aikbkj
... ... · · · ...
ai1 ai2 · · · ain... ... · · · ...
· · · b1j · · ·· · · b2j · · ·· · · ... · · ·· · · bnj · · ·
=
...
· · · cij · · ·...
ith row jth column ijth element
AB is usually not the same as BA.
a11 a12
a21 a22
a31 a32
[
b11 b12
b21 b22
]
=
a11b11+a12b21 a11b12+a12b22
a21b11+a22b21 a21b12+a22b22
a31b11+a32b21 a31b12+a32b22
22
Linear Algebra — Matrix © Wei-Chau Xie
Transpose of Matrices
Let A = [ aij ]m×n
=⇒ Transpose of A : AT = [ aji ]n×m
a11 a12
a21 a22
a31 a32
T
3×2
=[
a11 a21 a31
a12 a22 a32
]
2×3
Properties
(AT)T = A
(A + B)T = AT + BT
(AB)T = BTAT
(αA)T = αAT, α is a scalar
23
Linear Algebra — Determinant © Wei-Chau Xie
Determinant
The determinant of a square matrix A is denoted as
∣∣A
∣∣ = det(A) =
∣∣∣∣∣∣∣∣∣
a11 a12 · · · a1n
a21 a22 · · · a2n... ... ... ...
an1 an2 · · · ann
∣∣∣∣∣∣∣∣∣
= det
a11 a12 · · · a1n
a21 a22 · · · a2n... ... ... ...
an1 an2 · · · ann
Evaluation of Determinants
a11 a12
a21
|A| =a22
= +a11a22 − a21a12 2×2 determinant
a11 a12 a13
a21|A| = a22 a23
a31 a32
a11 a12
a21 a22
a31 a32a33
To evaluate the 3×3 determinant,
copy first two columns at right.
= +a11a22a33 + a12a23a31 + a13a21a32 − a31a22a13 − a32a23a11 − a33a21a12
24
Linear Algebra — Solutions of Systems of Linear Equations © Wei-Chau Xie
Cramer’s Rule
For the following system of n linear algebraic equations
a11 x1 + a12 x2 + · · · + a1n xn = b1
a21 x1 + a22 x2 + · · · + a2n xn = b2
· · · · · ·an1 x1 + an2 x2 + · · · + ann xn = bn
the solutions are given by
xi = 1i
1i = 1, 2, . . . , n, 1 6= 0
where 1 is the determinant of coefficient matrix, 1i is the determinant of the
coefficient matrix with the ith column replaced by the right-hand side vector, i.e.,
1 =
∣∣∣∣∣∣∣∣∣
a11 a12 · · · a1n
a21 a22 · · · a2n... ... · · · ...
an1 an2 · · · ann
∣∣∣∣∣∣∣∣∣
, 1i =
∣∣∣∣∣∣∣∣∣∣
a11 · · · a1, i−1 b1 a1, i+1 · · · a1n
a21 · · · a2, i−1 b2 a2, i+1 · · · a2n... · · · ... · · · ... · · · ...
an1 · · · an, i−1 bn an, i+1 · · · ann
∣∣∣∣∣∣∣∣∣∣
︸︷︷︸
ith column25
Linear Algebra — Cramer’s Rule © Wei-Chau Xie
Example
Solve the following system of linear algebraic equations
3x1 + 4x2 = 10
2x1 − 5x2 = −1
1 =∣∣∣∣∣
3 4
2 −5
∣∣∣∣∣= 3 · (−5) − 2 · 4 = −23
11 =∣∣∣∣∣
10 4
−1 −5
∣∣∣∣∣= 10 · (−5) − (−1) · 4 = −46
12 =∣∣∣∣∣
3 10
2 −1
∣∣∣∣∣= 3 · (−1) − 2 · 10 = −23
Apply Cramer’s Rule
x1 = 11
1= −46
−23= 2, x2 = 12
1= −23
−23= 1
26
Linear Algebra — Cramer’s Rule © Wei-Chau Xie
Example
Solve the following system of linear algebraic equations
4 y − 3z = 3
−x + 7 y − 5z = 4
−x + 8 y − 6z = 5
Determinant of the coefficient matrix
1 =
∣∣∣∣∣∣∣
0 4 −3
−1 7 −5
−1 8 −6
∣∣∣∣∣∣∣
0 4
−1 7
−1 8
= 0·7·(−6) + 4·(−5)·(−1) + (−3)·(−1)·8− (−1)·7·(−3) − 8·(−5)·0 − (−6)·(−1)·4
= 0 + 20 + 24 − 21 − 0 − 24 = −1
27
Linear Algebra — Cramer’s Rule © Wei-Chau Xie
11 =
∣∣∣∣∣∣∣
3 4 −3
4 7 −5
5 8 −6
∣∣∣∣∣∣∣
3 4
4 7
5 8
Replace the first column by RHSvector.
= 3·7·(−6) + 4·(−5)·5 + (−3)·4·8 − 5·7·(−3) − 8·(−5)·3 − (−6)·4·4
= −126 − 100 − 96 + 105 + 120 + 96 = −1
12 =
∣∣∣∣∣∣∣
0 3 −3
−1 4 −5
−1 5 −6
∣∣∣∣∣∣∣
0 3
−1 4
−1 5
Replace the second column by RHSvector.
= 0·4·(−6) + 3·(−5)·(−1) + (−3)·(−1)·5− (−1)·4·(−3) − 5·(−5)·0 − (−6)·(−1)·3
= 0 + 15 + 15 − 12 − 0 − 18 = 0
28
Linear Algebra — Cramer’s Rule © Wei-Chau Xie
13 =
∣∣∣∣∣∣∣
0 4 3
−1 7 4
−1 8 5
∣∣∣∣∣∣∣
0 4
−1 7
−1 8
Replace the third column by RHSvector.
= 0·7·5 + 4·4·(−1) + 3·(−1)·8 − (−1)·7·3 − 8·4·0 − 5·(−1)·4
= 0 − 16 − 24 + 21 − 0 + 20 = 1
Apply Cramer’s Rule
x = 11
1= −1
−1= 1, y = 12
1= 0
−1= 0, z = 13
1= 1
−1= −1
29
Linear Algebra — Systems of Homogeneous Linear Equations © Wei-Chau Xie
Systems of Homogeneous Linear Equations
When the right-hand side constants b1 = b2 = · · · = bn = 0, the system of linear
algebraic equations is homogeneous
a11 x1 + a12 x2 + · · · + a1n xn = 0
a21 x1 + a22 x2 + · · · + a2n xn = 0
· · · · · ·an1 x1 + an2 x2 + · · · + ann xn = 0
The system of homogeneous linear equations has zero solution, i.e.
x1 = x2 = · · · = xn = 0
If the determinant of the coefficient matrix 1 6= 0, then the system of
homogeneous linear equations does not have non-zero solutions.
For the system of homogeneous linear equations to have non-zero solutions,
the determinant of the coefficient matrix 1 = 0.
30
Linear Algebra — Eigenvalues and Eigenvectors © Wei-Chau Xie
Eigenvalues and Eigenvectors
Consider the following system of homogeneous linear equations
A x = λ x or (A − λI) x = 0 (*)
where I is the n×n unit matrix, i.e.,
I =
1 0 0 · · · 0
0 1 0 · · · 0... ... ... · · · ...
0 0 0 · · · 1
For the system of homogeneous linear equations (*) to have non-zero
solutions, the determinant of the coefficient matrix must be zero, i.e.,
∣∣A − λ I
∣∣ = 0
which is called the characteristic equation.
The solutions (roots) λ of the characteristic equation are called eigenvalues.
The corresponding solutions of system (*) are called eigenvectors.
31
Linear Algebra — Eigenvalues © Wei-Chau Xie
Example
Find the value of λ such that the equations
(3 − λ) x1 − 3 x2 + x3 = 0
2 x1 − (2 + λ) x2 + 2 x3 = 0
−x1 + 2 x2 − λ x3 = 0
have non-zero solutions.
Characteristic equation (setting the determinant of the coefficient matrix to zero)
1 =
∣∣∣∣∣∣∣
3−λ −3 1
2 −2−λ 2
−1 2 −λ
∣∣∣∣∣∣∣
3−λ −3
2 −2−λ
−1 2
= (3−λ)·(−2−λ)·(−λ) + (−3)·2·(−1) + 1·2·2− (−1)·(−2−λ)·1 − 2·2·(3−λ) − (−λ)·2·(−3)
= −λ3 + λ2 − 2 = −λ3 − λ2 + 2λ2 − 2
= −λ2(λ+1) + 2(λ−1)(λ+1) = −(λ+1)(λ2−2λ+2) = 0
32
Linear Algebra — Eigenvalues © Wei-Chau Xie
∴ (λ + 1)(λ2 − 2λ + 2) = 0
λ + 1 = 0 =⇒ λ = −1
λ2 − 2λ + 2 = 0:
λ = −(−2) ±√
(−2)2−4·1·22·1 = 2 ±
√−4
2= 1 ± i, i =
√−1
33
Linear Algebra — Eigenvalues © Wei-Chau Xie
Alternative Method for Solving the Characteristic Equation −λ3+λ2−2 = 0
By trial-and-error, find a root of the characteristic equation
−(−1)3 + (−1)2 − 2 = 0 =⇒ λ=−1 is a root =⇒ (λ+1) is a factor.
Use long division to find the other factor as follows
−λ2 + 2λ − 2
λ + 1∣∣∣ −λ3 + λ2 − 2
−λ3 − λ2 (−2λ2 − 2
2λ2 + 2λ (−− 2λ − 2
− 2λ − 2 (−0
The characteristic equation becomes (λ + 1)(−λ2 + 2λ − 2) = 0.
34
Linear Algebra — Eigenvalues and Eigenvectors © Wei-Chau Xie
Example
Find the eigenvalues λ and the corresponding eigenvectors of
(A − λI)x =[
1−λ 4
1 −2−λ
] {
x1
x2
}
={
0
0
}
The characteristic equation is∣∣∣∣∣
1−λ 4
1 −2−λ
∣∣∣∣∣= (1−λ)(−2−λ) − 1·4 = λ2 + λ − 6 = (λ+3)(λ−2) = 0
The two eigenvalues are λ1 = −3, λ2 = 2.
λ = λ1 = −3:
(A−λ1I)v1 =[
4 4
1 1
] {
v11
v21
}
= 0 =⇒ v11 + v21 = 0
Taking v21 = −1, then v11 = −v21 = 1 =⇒ v1 ={
v11
v21
}
={
1
−1
}
There is one equation for two unknowns.
One unknown can be solved in terms of the other unknown.35
Linear Algebra — Eigenvalues and Eigenvectors © Wei-Chau Xie
λ = λ2 = 2:
(A−λ2I)v2 =[
−1 4
1 −4
] {
v12
v22
}
= 0 =⇒ v12 − 4v22 = 0
Taking v22 = 1, then v12 = 4v22 = 4 =⇒ v2 ={
v12
v22
}
={
4
1
}
36
