U
Linear Algebra and Matrices
Linear Equations
Suppose x1, x2 ,, xn are n unknowns, and a1 , a 2 ,  , a n and b are scalars, then
a1x1  a2 x2    an xn  b is called a linear equation in the n unknowns.
Example
3x1  4 x2  2 x3  5
System of Linear Equations
A set of simultaneous linear equations of n unknowns forms a system of equations of
the n unknowns.
Example
3x1  2 x2  x3  10
2 x1  x2  x3  1
x1  x2  x3  2
Matrix
Matrix
A system of equations being written in an abbreviated form as a rectangular array of
numbers enclosed by brackets is called a matrix (pl. matrices). Each number in the
array is called an element or entry of the matrix.
Example
 2 3 2
A
 is a 23 matrix.
1 3 4 
Transpose of A Matrix
The transpose of a matrix can be obtained by taking the rows of the matrix and
making them column (or, effectively equal to taking the columns of the matrix and
making them rows).
Example
2 1 
 2 3 2
.
A  3 3 is the transpose of A  
1 3 4

2 4
T
Matrix Equality
Two matrices are equal if they are the same size and if each pair of corresponding
elements is equal.
Addition and Subtraction of Matrices
Addition of Matrices
The sum of two m n matrices X and Y is the m n matrix X  Y in which each
element is the sum of the corresponding elements of X and Y. (It is important to
remember that only matrices those are the same size can be added.)
Example
 1 2   1 2   0 4
 3 4   1  2   4 2

 
 

 1 3  3
1  2 4
Additive Inverse
The additive inverse (or negative) of a matrix X is the matrix  X in which each
element is the additive inverse of the corresponding element of X.
Example
  2 3 2
 2  3  2
If A  
, A

.
 1  3 4
  1 3  4
Zero Matrix
If O is an m n zero matrix, and A is any m n matrix, then
AO  O  A  A .
Example
0 0 0 
0 0 0 is the zero matrix of size 33.


0 0 0
Subtraction of Matrices
For two m n matrices X and Y, the difference X  Y is the m n matrix defined
by
X  Y  X   Y  .
Example
2  1 0 
 1 1 3
Let X  
and Y  

.
4 5  3
 2 0 1
2  1 0   1  1  3 3  2  3
X  Y  X   Y   


.
4 5  3  2 0  1 2 5  4
Multiplication of Matrices
Product of a Matrix and a Scalar
The product of a scalar k and a matrix X is the matrix kX, each of whose elements is k
times the corresponding element of X.
Example
2 0  1 10 0  5
5  1 1 4    5 5 20 
3 2 5  15 10 25 
Product of Two Matrices
Let A be an m n matrix and let B be an n  k matrix. To find the element in the
ith row and jth column of the product matrix AB, multiply each element in the ith row
of A by the corresponding element in the jth column of B, and then add these products.
The product matrix AB is an m  k matrix. (The product AB of two matrices A and
B can be found only if the number of columns of A is the same as the number of rows
of B.)
Example
1 2 1  1
1 2 0
Let A  
and B  0  1 2 1  .

 2 3  1
3 0 0 4 
0 5 1
1
AB  

 5  7 4 1
Identity Matrix
If I is the identity matrix, both of the products AI and IA must equal A.
that an identity matrix exists only for square matrices.
This means
Example
1 0 
0 1  is the 2  2 identity matrix.


Multiplicative Inverse Matrix
For a given square matrix A, A1 is called the multiplicative inverse matrix of A if
AA1  A1A  I . ( A1 does not mean 1 A ; here, A1 is just the notation for the
multiplicative inverse of matrix A. Also, only square matrices can have inverses
because both A1 A and AA1 must exist and be equal to I.)
To Rewrite a System of Equations in Matrix Form
Example
3 x1  2 x2  x3  10
2 x1  x2  x3  1
x1  x2  x3  2

3 2 1   x1  10
2 1  1  x    1 

 2   
1  1 1   x3   2 
Gauss-Jordan Method of Solving Systems of Linear
Equations
Example
3x1  2 x2  x3  10
2 x1  x2  x3  1
x1  x2  x3  2

 3 2 1 10
 2 1 1 1 


1 1 1 2 
To separate the constants in the last column of the matrix from the coefficients of the
variables, we use a vertical line, producing the following augmented matrix.
 3 2 1 10 


 2 1 1 1 
 1 1 1 2 
Row Operations
For any augmented matrix of a system of equations, the following operations produce
the augmented matrix of an equivalent system:
1.
2.
3.
interchanging any two rows;
multiplying the elements of a row by any nonzero real number;
adding a nonzero multiple of the elements of one row to the corresponding
elements of a nonzero multiple of some other rows.
Gauss-Jordan Method of Solving a Linear System
1.
2.
3.
4.
5.
Write each equation so that variable terms are in the same order on the left side
of the equals sign and constants are on the right.
Write the augmented matrix that corresponds to the system.
Use row operations to transform the first column so that all elements except the
element in the first row are zero.
Use row operations to transform the second column so that all elements except
the element in the second row are zero.
Use row operations to transform the third column so that all elements except the
element in the third row are zero.
6.
Continue in this way until the last row is written in the form
0 0 0
7.
0
j k  ,
where j and k are constants.
Multiply each row by the reciprocal of the nonzero element in that row.
Example
Solve
1.
 x1  3x2  x3  6

 x3  3
 x1
 x  x  2 x  4
3
 1 2
Write each equation so that variable terms are in the same order on the left side
of the equals sign and constants are on the right.
 x1  3x2  x3  6

 x3  3
 x1
 x  x  2 x  4
3
 1 2
2.
Write the augmented matrix that corresponds to the system.
1 3 1 6 
 x1  3x2  x3  6



 x3  3 
 1 0 1 3 
 x1
 x  x  2 x  4
1 1 2 4 
3
 1 2
3.
Use row operations to transform the first column so that all elements except the
element in the first row are zero.
1 3 1 6 
1 3 1 6 




R3:R3  R1
 0 3 0 3  0 3 0 3 
1 1 2 4 
0 4 3 10 
R2 :R2  R1
4.
Use row operations to transform the second column so that all elements except
the element in the second row are zero.
1 0 1 3 
1 0 1 3 




R3:3 R3  4 R2
 0 3 0 3   0 3 0 3 
0 4 3 10 
0 0 9 18
R1:R1  R2
5.
Use row operations to transform the third column so that all elements except the
element in the third row are zero.
9 0 0 9 



  0  3 0 3 
0 0 9 18
R1:9 R1  R3
6.
Continue in this way until the last row is written in the form
0 0 0
0
j k  ,
where j and k are constants.
9 0 0 9 



  0  3 0 3 
0 0 9 18
R1:9 R1  R3
7.
Multiply each row by the reciprocal of the nonzero element in that row.
 19 R1
R2 : 13  R2
1
R3: 19  R3


 0
R1:
0 0 1

 1 0 1
0 0 1 2 
The solution of the system is x1  1, x2  1 and x3  2.
Solving a System AX  B Using Matrix Inverses
To solve a system of equations AX  B , where A is the matrix of coefficients, X is
the matrix of variables, and B is the matrix of constants, first find A1 . Then
X  A1B .
Finding a Multiplicative Inverse Matrix
To obtain A1 for any n  n matrix A for which A1 exists, follow these steps.
1.
Form the augmented matrix  A I  , where I is the n  n identity matrix.
2.
Perform row operations on  A I  to get a matrix of the form  I B  .
3.
Matrix B is A1 .
Example
1 2 0 
A  2 1  2
2 0 4 
0
1 0 0
1 2 0 1 0 0 R2:R2 2 R1 1 2
:
R

2
R
2 1  2 0 1 0 R

3 3
1


0  3  2  2 1 0


2 0 4 0 0 1
0  4 4  2 0 1
0  3 6 3
3 0  4  1 2 0 R1:5 R1  R3
15 0
R2:10 R2  R3 


  0  3  2  2 1 0    0  30 0  18 6 3
0 0 20 2  4 3
 0
0
20 2  4 3
R1:3R1  2 R2
R3:3R3 4 R2
1 R
R1:15
1
1 R
R2 : 30
2
1 R
R3: 20
3
1 0 0

 0 1 0
0 0 1

1
5
3
5
1
10
2
5
1
5
1
5
1 
5
1 
10 
3
20 
 51

 A1   53
1
 10
2
5
1
5
1
5
1 
5
1 
10 
3
20 
.
Formulation of Systems of Equations
I.
Ans:
 x1  x5  800
 x  x  x  400
4
 1 2
 x2  x3  600

 x3  x7  1200
 x4  x6  x7  0

 x5  x6  1000
II.
A coffee merchant has a supply of 300 kg of Type A coffee, 2400 kg of Type B coffee,
and 4800 kg of Type C coffee. He makes a Mocha blend consisting of 25% Type A
and 75% Type B coffee. He makes a Columbian blend consisting of 10% Type A,
60% Type B, and 30% Type C coffee. He makes a Brazilian blend consisting of
20% Type B and 80% Type C coffee. How many kg of each of the three blends
should he prepare?