ENGINEERING MATHEMATICS 1

[CHAPTER 5: MATRIX]

EXAMPLE 7

Given B  4 6 . Find the determinant of matrix B.
3 8

SOLUTION

|B| = (4 × 8) − (6 × 3) = 32 − 18
|B| = 14

ii. FOR A 3 3MATRIX

 For a 3 3 matrix (3 rows and 3 columns)

 Formula

a b c  The determinant is;
A  d 
e f  A  aei  fh bdi  fg cdh  eg

g h i 

EXAMPLE 8

6 1 1
C  4  2 5
2 8 7
Given;

Find the determinant of matrix C.

SOLUTION:

C  6 14  40  128  10  132  4

C  306

95

[CHAPTER 5: MATRIX]

LET’S PRACTICE 4

1. Find the determinant of matrix A.

A  2 5
1  3

2. Find the determinant of matrix B. Ans: 11
Ans: 10
B  3 2
1 4 Ans:  44
Ans:  4
3. Find the determinant of matrix M.
Ans:  266
 3 0 1
  96
M   2 5 4 

 3 1 3 

4. Find the determinant of matrix N.

1 1 1 
N  2 5 
7 

2 1 1

5. Find the determinant of matrix Z.

 7 5 4
 2 6
Z   4

 2  3 4

[CHAPTER 5: MATRIX]
5.4 INVERSE MATRIX USING MINOR, COFACTOR AND ADJOIN

 Ignore the values on the current row and column
 Calculate the determinant of the remaining values

EXAMPLE 9


3 0 2 
Given A  2 0  2

0 1 1  . Find the inverse matrix of A
SOLUTION:
STEP 1: DETERMINANT

A  30  2 0  22  0  6  4

A  10
STEP 2: MINOR

M indicates the minor of matrices A

m11  0  2  0   21  2 m12  2  2  21   20  2 m13  2 0  21  0  2
1  0  0 1
1  1 

m21  0 2  0  2  2 m22  3 2  31  20  3 m23  3 0  31  0  3
1 1 0 1 0 1

m31  0 2  0  0  0 m32  3 2  3 2  22  10 m33  3 0  30  0  0
0  2 2  2 2 0

 2 2 2
M   2 3 3

 0  10 0

97

[CHAPTER 5: MATRIX]

STEP 3: COFACTOR

 2 2 2   2  2 2 
Cofactor   2 3 3    2 3  3

 0 10 0   0 10 0 

STEP 4: ADJOIN
 “Transpose” all elements in previous matrix.
 The elements of row will be elements of columns.

 2 2 0
AdjA   2 3 10

 2  3 0 

STEP 5: INVERSE MATRIX

 Formula;

A1  1  AdjA
A

1 1 0⎤
12 2 0 ⎡5 5 1⎥⎥
= 10 −2 3 10 = ⎢⎢− 1 5 0⎦⎥
2 −3 3
0 ⎢ 1 5 10
⎣
− 3 10

LET’S PRACTICE 5

1. Find the inverse matrix

1 0  3
C  2  2 
1 

0 1 3 

Using minor, cofactor and adjoin.

 5 3  6
Ans:  6 3  7

 2 1  2
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[CHAPTER 5: MATRIX]

2. Find the inverse matrix

 2 1 0 
 3 1
M   1

 3 0 1 

Using minor, cofactor and adjoin.

3 1 1
Ans: 1942 
4 4 
1 1 
2 2 
3 7 

 4 4 4

3. Find the inverse matrix

 3 1  6
 
B   2 0 4 

1 2  3

Using minor, cofactor and adjoin.

 4 92 2
 32 
Ans:  1 52
0
 2  1


4. Find the inverse matrix

 3 7  5
 
F   4 1 12 

 2 9 1

Using minor, cofactor and adjoin.

 107  38 79 
 
Ans:  327 327 327 
 28 13 16 
 327 327 327 
 38 41  25 

 327 327 327 

99

[CHAPTER 5: MATRIX]

5.5 SIMULTANEOUS LINEAR EQUATION USING MATRIX

REMEMBER:-
i. The system must have the same number of equations as variables, that is, the
coefficient matrices of the system must be square.
ii. The determinant of the coefficient matrices must be non-zero. The reason, of
course, is that the inverse of a matrix exist precisely when its determinant is
non-zero.

5.5.1 SIMULTANEOUS EQUATION USING INVERSE MATRIX
METHOD
EXAMPLE 10

Given;
 x 3y  z 1
2x  5y  3
3x  y  2z  2

Solve the simultaneous equation by using inverse matrix method.

SOLUTION:

STEP 1: Rewrite the system using matrix multiplication

1 3 1 x  1 
  y  
 2 5 0    3 

 3 1  2 z   2

STEP 2: Writing the coefficient matrix as A

x  1  1 3 1 
A   
y    3  Where A   2 5 0 
 
 z   2  3 1  2

100

[CHAPTER 5: MATRIX]

STEP 3: Formula

A1  1 AdjA
A

x  1 
 y  
  A 1  3 

 z   2

STEP 4: Determinant of A

A  15  2  0 1 32  2  0  312 1  5 3

A 9

STEP 5: Minor of A

m11  5 2  01  10 m12  2 2  03  4 m13  21  53  13
m21  3 2  11  7 m22  1 2 13  1 m23  11  33  10
m31  30  15  5 m32  10  12  2 m33  15  32  11

10  4 13
 1 10
Minor    7

  5  2 11

STEP 6: Cofactor of A

10  4 13   10 4 13
  10    
Cofactor    7 1    7 1 10 

  5  2 11     5 2 11

101

[CHAPTER 5: MATRIX]

STEP 7: Adjoin of A

10 7  5 
 
AdjA   4 1 2 

13 10 11

STEP 8: Inverse matrix of A

A1  1 AdjA
A

1  10 7 5
9  1 
A1   4 10 2 

 13  11

STEP 9: Solve the equation

x 1  10 7  5  1 
 y 9  1  
   4 10 2  3 

 z   13  11 2

x 1 10  21  10
 y 9  
   434 

 z  13  30  22

x 1  21
 y 9  3
  39 

 z 

STEP 10: Find the x, y and z

x 7133 
 y 
 

 z   13 
 3 

102

[CHAPTER 5: MATRIX]

LET’S PRACTICE 6

Solve the simultaneous equation below by using inverse matrix method.
x  y  z  3

1. 2x  3y  4z  23
 3x  y  2z  15

Ans: 2,1,4

x  2y  z  7
2. 2x  3y  4z  3

xyz 0

Ans: 1,3,2

4x  2 y  2z  10
3. 2x  8 y  4z  32

30x  12 y  4z  24

Ans:  2,6,3

3x  2 y  z  24
4. 2x  2 y  2z  12

x  5 y  2z  31

Ans: 3,4,7

5x  2y  4x  0
5. 2x  3y  5z  8

3x  4 y  3z  11

Ans:  2,1,3

103

[CHAPTER 5: MATRIX]
5.5.2 SIMULTANEOUS EQUATION USING CRAMER’s RULE
EXAMPLE 11

x  2 y  3z  5
Given 3x  y  3z  4

 3x  4 y  7z  7

Solve the simultaneous equation by using Cramer’s Rule.

SOLUTION:

STEP 1: Rewrite the system using matrix multiplication

 1 2 3 x  5
  3 y  
 3 1   4 

 3 4 7  z   7

STEP 2: Find the determinant

A  117   34 237   3 3 334 1 3

A  19  24  45
A  40

STEP 3: Construct another 3 matrices

 5 2 3   1 5 3   1 2  5
 1  3  4  3  
Ax   4 Ay   3 Az   3 1 4 

 7 4 7   3  7 7   3 4  7

STEP 4: Find the determinant of each matrices

 5 2 3 
 1  3  517   34 247   3 7 344 1 7  95 14  69  40
Ax   4

 7 4 7 

 1 5 3 
 4  3  147  3 7  537  3 3 33 7 4 3  7  60  27  40
Ay   3

 3  7 7 

104

[CHAPTER 5: MATRIX]

 1 2  5
 
Az   3 1 4   11 7  44  23 7 4 3   534 1 3  23 18  75  80

 3 4  7

STEP 5: Solve the equation

x Ax   40  1
A 40

y AY  40 1
A 40

z Az   80  2
A 40

LET’S PRACTICE 7

Solve the simultaneous equation below by using Cramer’s Rule
 2x  2 y  4z  22

1.  3x  5 y  2z  35
 6x  3y  3z  21

4x  y  2z  1 Ans: 2,9,2
2.  2x  y  6z  3
Ans: 1,1,1
4x  6 y  3z  7
Ans:  2,3,1
 5x  9 y  3z  40
3. 9x  4 y  2z  4 105

 5x  7 y  4z  15

[CHAPTER 5: MATRIX]

4x  2 y  2z  4
4.  7x  y  2z  22

2x  6 y  z  26

4x  7 y  3z  37 Ans: 2,4,2
5.  2x  3y  6z  61 Ans:  5,5,6

2x  2 y  4z  44

106

[CHAPTER 6: VECTOR AND SCALAR]

VECTOR AND SCALAR

OBJECTIVES:
At the end of this topic, students should be able to:

i. define vector
ii. understand the operation of vector
iii. apply scalar (dot) product of two vectors
iv. apply vector (cross) product of two vectors
v. understand area of parallelogram

6.1 INTRODUCTION OF VECTOR

 Physical quantities can be classified under two main headings – vectors and scalars.

VECTOR
 Vector is a physical quantity which is specified by magnitude or length and a
direction in space.
 For example, displacement and velocity are both specified by a magnitude and a
direction and are therefore examples of a vector quantities.

SCALAR
 Scalar is a physical quantity which is specified by just its magnitude.
 For example, distance and speed are both fully specified by a magnitude and are
therefore examples of scalar quantities.

VECTOR NOTATION

 Vectors are written as Y, y, ỹ or Y .
 The magnitude of a vector Y is written as Y .

107

[CHAPTER 6: VECTOR AND SCALAR]

VECTOR REPRESENTATION
 A vector is represented by a straight line with an arrowhead.

Figure 1

 In the Figure 1, the line OA represents a vector OA .
 You can write:

OA   24
which means that to go from O to A, move 4 units in the positive x
direction and 2 units in the positive y direction. This is called the column
vector or matrix form.
 In Cartesian form, the vector OA can be represented as

OA  4i  2 j

 Then,

OA   4   4i  2 j
2

VECTOR MAGNITUDE

 The magnitude or modulus of the vector OA is represented by the length OA and its
donated by OA .

108

[CHAPTER 6: VECTOR AND SCALAR]
EXAMPLE 1

 By referring Figure 2;

Figure 2 Angle;
The magnitude of a vector;
  tan 1  y 
OA  x 2  y 2  x 
OA  42  22
OA  20units   tan 1  2 
 4 

  26.570

LET’S PRACTICE 1

1. Find the magnitude of each of these vectors.

a. 4i  3 j b. 2i  2 j  k

c. 79 d.  57 

 3 

Ans: 5 , 3 , 11.40 & 9.11

109

[CHAPTER 6: VECTOR AND SCALAR]

EQUALITY OF VECTORS
 Two vectors are said to be equal if they have the
same magnitude and direction.

NEGATIVE VECTOR
 A vector having the same magnitude but opposite
direction.

UNIT VECTOR

 A unit vector is a vector of length 1.

unit vector, uˆ  u
u

EXAMPLE 2
1. Find the unit vector in the direction of v = 5i – 2j +4k.

SOLUTION:

magnitude of v, v  52   22  42  45

if vˆ is a unit vector in the direction of v:

vˆ  5i  2 j  4k 4k
45 45

vˆ  5 i  2 j 
45 45

110

[CHAPTER 6: VECTOR AND SCALAR]
LET’S PRACTICE 2

1. Find a unit vector in the direction of the vector 8i  6 j .

Ans: 4i  3 j
5

2. Find a unit vector in the direction of v  3i  2 j  5k .

3. Find a unit vector in the direction of the vector  7  Ans: 3i  2 j  5k
9 38

Ans:  7i  9 j
130

4. Find a unit vector in the direction of the vector  123

  4

Ans:  3i  12 j  4k
13

111

[CHAPTER 6: VECTOR AND SCALAR]

POSITION VECTORS

 By referring to the Figure 3, the position vector of a point P with respect to a
fixed origin O is the vector OP.

 This is not a free vector, since O is fixed point.

 It can be write as; OP  P
 Then,

PQ  PO  OQ
 PQ  P  Q

PQ  Q  P

Figure 3

EXAMPLE 3

7  5
   
1. Given that a   3  and b   2  , find ab and ab . Hence, find unit vector

 2  3 

in the direction of a  b .

SOLUTION:

→ + → = 7 −5 2
3 + 2 =5

−2 3 1

→ + → = 2 + 5 + 1 = √30

a    a  
b a  b

Unit vector, b

a    2i  5 j  k  2 i 5 j 1k
b 30 30 30
30

112

[CHAPTER 6: VECTOR AND SCALAR]

2. Given that OX  6i  3 j  k and OY  2i  4 j  5k . Find the unit vector in the
direction of XY .

SOLUTION:

XY  OY  OX

 2   6   4
   3  
XY   4     7 

 5  1   6

XY   42  7 2   62  101

Unit vector, XY  XY
XY

XY   4i  7 j  6k
101

XY   4 i  7 j 6k
101 101 101

LET’S PRACTICE 3
1

1. If given B  3 , find the unit vector of B .
2

Ans: 1 i  3 j  2 k
14 14 14
113

[CHAPTER 6: VECTOR AND SCALAR]

2. If given vector A  2i  3 j  6k and B  i  j  2k . Find
a. 2A  B
b. 2A  B
c. Unit vector of 2A  B

Ans: 5i  5 j  14k , 246 & 5 i  5 j  14 k
246 246 246

3. Find the unit vector of BA if given OA  i  5 j  12k and OB  3i  5 j  k .

Ans:  2 i  11 k
125 125

6.2 THE OPERATION OF VECTOR
6.2.1 VECTOR ADDITION

Figure 4
By referring the Figure 4 above;

OR  OP  PR OR  OQ  QR

i) OR  A  B  C ii) OR  A  B  C

From i) and ii)

OR  A  B  C  A  B C

114

[CHAPTER 6: VECTOR AND SCALAR]

6.2.2 ADDITION AND SUBTRACTION OF VECTOR USING PARALLELOGRAM
METHOD
 PARALLELOGRAM is a quadrilateral with both pairs of opposite sites parallel.
 Quadrilateral are four side polygons.
 Congruent refer to a shape (in mathematics) that has the same shape and size as

another.

Properties of Parallelogram

Figure 5

PROPERTIES
i. If a quadrilateral is a parallelogram, then its opposite sides are congruent.

PQ  RS and SP  QR

ii. If a quadrilateral is a parallelogram, then its opposite angles are congruent.

P  R and Q  S

iii. If a quadrilateral is a parallelogram, then its consecutive angles are

supplementary.

P  Q  180 0 R  S  180 0

Q  R  180 0 S  P  180 0

iv. If a quadrilateral is a parallelogram, then its diagonals bisect each other

QM  SM and PM  RM

115

[CHAPTER 6: VECTOR AND SCALAR]

EXAMPLE 4

Figure 6
ABCD is a parallelogram, find the sum of vectors below in unit of vector guide.

i. DA  AC
ii. AD  AB
iii. AB  CB
Solution:
i. DA AC  DC

ii. AD  AB  AC

iii. AB  CB  DB

6.2.3 ADDITION AND SUBTRACTION OF VECTOR USING TRIANGLE RULE

 Another way to define addition of two vectors is by a head-to-tail construction
that creates two sides of a triangle.

 The third side of the triangle determines the sum of the two vectors.

Figure 7

The vector u  v is defined to be the vector OP

116

[CHAPTER 6: VECTOR AND SCALAR]

EXAMPLE 5

Figure 8

ABCD is a triangle where BC  CD  DE . Given AB  6a  4b and BC  a  b . Express
the followings in term of a and b.

i. ED
ii. AC
iii. DA

SOLUTION:

i. ED  CB  BC  a  b  b  a

AC  AB  BC

ii. AC  6a  4b a  b

AC  7a  3b
DA  DB  BA

iii. DA  2a  b  6a  4b

DA  2a  2b  6a  4b
DA  8a  2b

117

[CHAPTER 6: VECTOR AND SCALAR]
LET’S PRACTICE 4

1. Given that FGHJ is a parallelogram, find MH and FH. Given FM  5

Figure 9

Ans: 5,10
2.

Figure 10

ABCD is a parallelogram, find the sum of vectors below in unit of vector guide.
a. AB  BD

b. CO  OD Ans: AD
c. CA BC Ans: CD
d. OB  DO Ans: BA

Ans: DB

118

[CHAPTER 6: VECTOR AND SCALAR]

LET’S PRACTICE 5

O, A, B,C and D are five points where OA  a , OB  b , OC  a  2b and OD  2a  b .
Express AB, BC,CD, AC and BD in terms of and .

Ans:
AB  b  a
BC  a  b
CD  a  3b
AC  2b
BD  2a  2b

LET’S PRACTICE 6

is a parallelogram with VW  a and WX  b . Express the vectors below in terms
of and .
i. VX

Ans: a  b

ii. XV

Ans:  b  a

iii. WY

Ans: b  a

iv. YW

Ans: a  b
119

[CHAPTER 6: VECTOR AND SCALAR]
6.3 APPLY SCALAR (DOT) PRODUCT OF TWO VECTORS

Definition of scalar product

 In Figure, ⃗ = and ⃗ = . Figure
 The angle between and is defined as the

angle between ⃗ and ⃗ .
 The scalar product of a vector and is

represented by ∙ , and
 ∙ = | || | cos , with as the angle

between and .

 as the angle between and , then ∙ = | || | cos .

 Therefore, cos = ∙
| || |
+
cos =
( ) +( ) ( ) +( )

Properties of scalar product

i.  
AB  B A

ii. For non zero vector A and B ;

 0; If and only if  is perpendicular to 
B A  A B

    
A BC  AB AC
 iii.

    
A kB  kA  B  k A B
    iv.

v. If vectors  and  are given in term of their component with respect to the
A B

standard vectors i, j and k as

A  a1i  a2 j  a3k and B  b1i  b2 j  b3k

Then;

A B  a1b1  a2b2  a3b3

120

[CHAPTER 6: VECTOR AND SCALAR]

EXAMPLE 6

1. If = 2 + and = − 3 . Find
a. ∙
b. ∙ ( − )
c. (10 + ) ∙
d. (3 ∙ ) ∙

SOLUTION:
a. ∙ = (2 × 1) + [1 × (−3)] = 2 − 3
∙ = −

b. ∙ ( − ) = (2 + ) ∙ [( − 3 ) − (2 + )]
∙ ( − ) = (2 + ) ∙ (− − 4 )
∙ ( − ) = [2 × (−1)] + [1 × (−4)]
∙ ( − ) = −

c. (10 + ) ∙ = [10(2 + ) + ( − 3 )] ∙ ( − 3 )
(10 + ) ∙ = [(20 + 10 ) + ( − 3 )] ∙ ( − 3 )
(10 + ) ∙ = (21 + 7 ) ∙ ( − 3 )
(10 + ) ∙ = (21 × 1) + [7 × (−3)]
( + ) ∙ =

d. (3 ∙ ) ∙ = [3(2 + )] ∙ ( − 3 ) ∙ (2 + )
(3 ∙ ) ∙ = [(6 + 3 ) ∙ ( − 3 )] ∙ (2 + )
(3 ∙ ) ∙ = [(6 × 1) + 3(−3)] ∙ (2 + )
(3 ∙ ) ∙ = (−3) ∙ (2 + )
( ∙ ) ∙ = − −

121

[CHAPTER 6: VECTOR AND SCALAR]
2. A, B and C are three points with ⃗ = 2 + and ⃗ = + 3 . Find angle .

SOLUTION:

⃗ = − ⃗ = −2 −

⃗ ∙ ⃗
cos = ⃗ ⃗

(−2 − ) ∙ ( + 3 )
cos =

(−2) +(−1) (1) +(3)

−2 − 3
cos =

√5√10
−2 − 3

∠ = cos
√5√10

∠ = °
3. The position vectors of points A, B and C, relative to the origin O, are

−4 , 0 0 respectively, with a > 0. Calculate:
−2 2

a. The scalar product of ⃗ . ⃗

b. Angle

SOLUTION:

a. The scalar product of ⃗ . ⃗

⃗ = ⃗ − ⃗

⃗ = 0 − −4 = 4
−2 −3

⃗ = ⃗ − ⃗

⃗ = 0 − −4 = 4
2

⃗ . ⃗ = (4 )(4 ) + (−3 )( )
⃗ . ⃗ = 16a

122

[CHAPTER 6: VECTOR AND SCALAR]

b. Angle
⃗ . ⃗

cos = ⃗ ⃗
16a

cos =
(4 ) +(−3 ) (4 ) +( )
16a

cos =
√25 √17
16a

cos = (5 )(4.123 )
∠ = cos 0.6306 = . °

LET’S PRACTICE 7


1. If P  2i  4 j  6k and U  2i  4 j  3k . Find

a.     
P U
b. P U P

2. Given A4,4,8, B5,1,8 and C 2,4,3. Find: Ans: 38 , 76i 152 j  228k

i.  Ans: 80
A B Ans: 32

ii. 
AC

iii. 
BC

Ans: 10

123

[CHAPTER 6: VECTOR AND SCALAR]
6.4 APPLY VECTOR (CROSS) PRODUCT OF TWO VECTORS

Properties of vector product
If A, B and C are vectors and d is a scalar, then

i. A  B  B  A

ii. dA B  d A  B  A  dB
iii. A B  C  A B  A C
iv. A  B C  A C  B  C
v A BC  A B C
vi. A B  C  A  CB  A  BC

EXAMPLE 7


1. Find the vector product for vector A  2i  2 j  2k and B  2i  2 j  3k . Find

SOLUTION:
i j k 

A  B  2 2  2
2  2 3 

 23  2 2i  23  22j  2 2 22k

 2i 10 j  8k
2. Find the unit vector of u  v . Given u  2i  2 j  3k and v  i  3 j  k .

SOLUTION:
i j k 

u  v  2 2  3  11i  5 j  4k
1 3 1 

124

[CHAPTER 6: VECTOR AND SCALAR]

u  v  112   52  42  162

Vector unit of u  v is;  11 i  5 j 4k
162 162 162

LET PRACTICE 8

1. Given vector OA  2i  j  3k , OB  3i  2 j  4k and OC  i  3 j  2k . Determine
a. AB

 b. OA OBOC Ans: i  3 j  7k

 c. OAOB OC Ans: 39

 Ans:  55i 11j 11k
2. If P 8i  5 j  4k and Q  2i  7 j  4k , find:
Ans:  48i  24 j  66k
a. Q  P Ans: 48i  24 j  66k
Ans:  8i 13 j 10k
 Ans: 8i  13 j  10k
b. P  Q
125
3. If a  3i  2 j  5k and b  i  4 j  6k , find:
a. a  b

b. b  a

[CHAPTER 6: VECTOR AND SCALAR]
6.4.1 APPLICATION OF THE VECTOR (CROSS) PRODUCT

Area of parallelogram

 AB AC sin
 AB BC

Area of triangle ABC

 1 AB BC
2

EXAMPLE 8

1. Find the area of the parallelogram with vertices A0,5, B2,0 , C8,1 and

D 6,4.

SOLUTION:

AB  2,5  2i  5 j  0k

BC  6,1  6i  j  0k

Area of parallelogram Area  AB  BC

i jk
AB  BC  2  5 0

6 1 0

AB  BC  0i  0 j  28k
AB  BC  02  02  282  28

2. Calculate the area of the parallelogram spanned by the vectors

  3 and  4
P  3 Q 9

 1  2

126

[CHAPTER 6: VECTOR AND SCALAR]

SOLUTION:

P  3,3,1  3i  3 j  k

Q  4,9,2  4i  9 j  2k

Area of parallelogram Area  P Q
i jk

PQ  3 3 1
492

P  Q   6  9i  6  4 j  27 12k

P  Q  15i  2 j  39k

P  Q  152   22  392  41.83

LET PRACTICE 9

1. Find the area of parallelogram with vertices A0,0, B2,1 ,  3,6 and D 1,5.
C

Ans: 9
2. Find the area of parallelogram with U  i  j  3k and V  6 j  5k .

Ans: 230

3. Find the area of triangle with vertices P1,1,0, Q 2,0,1 and R0,2,3.

Ans: 4.899

127

[CHAPTER 6: VECTOR AND SCALAR]

4. Find the area of parallelogram with vertices P1,5,3, Q0,0,0 and R3,5,1.

Ans: 24.5

5. Find the area of parallelogram with vertices x2,0,3, y1,4,5, and z7,2,9.

Ans: 64.9

128

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Learning.
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