Engineering Maths for GATE ME/CE/EC papers. Section 1: Linear Algebra

1. Section 1 : Linear
Algebra
Topic 2 : System of Linear Equations

2. Linear System
 Linear equation: ax1 + bx2 = c
where, a, b, c = constants x1, x2 = variables
 A linear system consists of more than one linear equations.
 For example:- a1x1 + b1x2 + c1x3= k1……………….(1)
a2x1 + b2x2 + c2x2= k2……………….(2)
a3x1 + b3x2 + c3x3= k3……………….(3)
 To find a solution of this system means to find the value/values of x1, x2, x3 which satisfies all
the equations in the linear system.
 A linear system can have a unique solution, more than one solutions or no solution.

3. Linear System
 Rewriting the above equation in form of matrix
AX = B
A =
𝑎1 𝑏1 𝑐1
𝑎2 𝑏2 𝑐2
𝑎3 𝑏3 𝑐3
, X =
𝑥1
𝑥2
𝑥3
, B =
𝑘1
𝑘2
𝑘3
 The matrix A is called the coefficient matrix and the block matrix [A B] , is the augmented
matrix of the linear system.Type equation here.
Augmented matrix :
𝑎1 𝑏1 𝑐1 𝑘1
𝑎2 𝑏2 𝑐2 𝑘2
𝑎3 𝑏3 𝑐3 𝑘3
 If all the elements of B are zero then linear system is called Homogeneous otherwise Non-
Homogeneous.

4. Row Reduced Echelon Form of a
Matrix
 A matrix C is said to be in the row reduced form if
1. The first non-zero entry in each row of C is 1
2. The column containing this 1 has all its other entries zero
e.g. A =
1 1 3
0 3 2
0 0 1
B =
1 0 1 −2
0 1 2 3
0 0 4 1
0 0 −1 −2
C =
1 −1
0 2

5. Gauss Elimination Method
 Gaussian elimination is a method of solving a linear system AX = B (consisting of m
equations in n unknowns) by bringing the augmented matrix
[A B] =
𝑎11 𝑎12 ⋯ 𝑎1𝑛 𝑏1
𝑎21 𝑎22 ⋯ 𝑎2𝑛 𝑏2
⋮ ⋮ ⋱ ⋮ ⋮
𝑎𝑚1 𝑎𝑚2 ⋯ 𝑎𝑚𝑛 𝑏𝑚
 to an upper triangular form (or reduced row echelon form)
𝑐11 𝑐12 ⋯ 𝑐1𝑛 𝑑1
0 𝑐22 ⋯ 𝑐2𝑛 𝑑2
⋮ ⋮ ⋱ ⋮ ⋮
0 0 ⋯ 𝑐𝑚𝑛 𝑑𝑚
 This elimination process is also called the forward elimination method.

6. Gauss Elimination Method
E.X. 1 y + z = 2
2x + 3z = 5
x + y + z = 5
Solution:- Augmented matrix =
0 1 1 2
2 0 3 5
1 1 1 3
 Interchange 1st and 2nd equation
2 0 3 5
0 1 1 2
1 1 1 3

7. Gauss Elimination Method
 Divide the 1st equation by 2
1 0
3
2
5
2
0 1 1 2
1 1 1 3
 Add −1 times the 1st equation to the 3rd equation
1 0
3
2
5
2
0 1 1 2
0 1 −
1
2
1
2
 Add −1 times the 2nd equation to the 3rd equation
1 0
3
2
5
2
0 1 1 2
0 0 −
3
2
−
3
2

8. Gauss Elimination Method
 Multiply the 3rd equation by −2/3
1 0
3
2
5
2
0 1 1 2
0 0 1 1
 Rewriting this matrix in algebraic form:
 x +
3
2
z =
5
2
 y + z = 2
 Z = 1
 The last equation gives z = 1, the second equation now gives y = 1. Finally the first equation
gives x = 1.
Solution : x = 1, y = 1, z = 2

9. Rank of a Matrix
 Unique solution, multiple solutions  Consistent
 No solution  Inconsistent
 Rank of a matrix : The number of non-zero rows in the row reduced form of a matrix is
called the row-rank of the matrix.
e.g. A =
1 2 1
2 3 1
1 1 2
 To find rank of this matrix, first convert it into row echelon form.
 Multiplying 1st row with (-2 and adding in 2nd row, -1 and adding in 3rd row)
1 2 1
0 −1 −1
0 −1 1

10. Rank of a Matrix
 Multiplying 2nd row with -1 and adding 2nd row in 3rd row,
1 2 1
0 1 1
0 0 2
 Multiplying 3rd row with ½ and multiplying 2nd row with -2 and adding in 1st
1 0 −1
0 1 1
0 0 1
 Adding 3rd row in 1st and multiplying 3rd row with -1 and adding in 1st
1 0 0
0 1 0
0 0 1

1 0 2
0 1 −1
0 0 0
(Rank = 2),
1 −3 1
0 0 0
0 0 0
(Rank = 3)
Rank = 3

11. Rank of a Matrix
 Let AX = B be a linear system with m equations and n unknowns.
 row-rank(A) ≥ row-rank([A B])  Consistent
 row-rank(A) < row-rank([A B])  Inconsistent

12. Linear Dependent and Linear
Independent
 Row vector : [x1 x2 x3 x4]
 Column vector :
𝑥
𝑦
𝑧
 Given any set of m vectors a(1), …., a(m) (with the same number of components), a linear
combination of these vectors is an expression of the form
c1a(1) + c2a(2)+….+ cma(m)
where c1, c2,….,cm are any scalars. Now, consider
c1a(1) + c2a(2)+….+ cma(m) = 0
 If all cjs are zero, then a(1), …., a(m) are said to form a linearly independent set. Otherwise
linearly dependent.

13. Linear Dependent and Linear
Independent
 Example:
a(1) = [3 0 2 2]
a(2) = [-6 42 24 54]
a(3) = [21 -21 0 -15]
are linearly dependent because,
6a(1) - ½ a(2) – a(3) = 0