Row Echelon Form &

Rank of Matrix

a11x + a12y = b1a21x + a22y = b2

𝑎11 𝑎12𝑎21 𝑎22

𝑥𝑦 =

𝑏1𝑏2

A X = B𝐴/𝐵 =

𝑎11 𝑎12 𝑏1𝑎21 𝑎22 𝑏2

R1 R1/a11,

𝐴/𝐵 ≅1 𝑎12/𝑎11 𝑏1/𝑎11𝑎21 𝑎22 𝑏2

1 𝑎12′0 𝑎22′

𝑥𝑦 = 𝑏1′

𝑏2′

R2 R2 – a21R1,

a11x + a12y = b1a21x + a22y = b2

𝐴/𝐵 ≅1 𝑎12/𝑎11 𝑏1/𝑎110 𝑎22 − 𝑎21𝑎12/𝑎11 𝑏2 − 𝑎21𝑏1/𝑎11

General Example for Solving 2x2 SLES using Matrix

Case 1

𝐴 ≅ 1 𝑎12′0 𝑎22′

𝐴/𝐵 ≅ 1 𝑎12′ 𝑏1′0 𝑎22′ 𝑏2′

Case 2

Case 3

a22’ ≠ 0 Unique Solution

𝐴 ≅ 1 𝑎12′0 𝑎22′

𝐴/𝐵 ≅ 1 𝑎12′ 𝑏1′0 𝑎22′ 𝑏2′

a22’ = 0 b2’ ≠ 0 No Solution

𝐴 ≅ 1 𝑎12′0 𝑎22′

𝐴/𝐵 ≅ 1 𝑎12′ 𝑏1′0 𝑎22′ 𝑏2′

a22’ = 0 b2’ = 0 Infinitely Many Solutions

Row Echelon Form (REF) of Matrix• Row echelon form of a matrix is obtained by applying row operations on

matrix which satisfy following conditions:

1. The first non zero number in the row (called a leading coefficient) is 1.

2. Every leading 1 is to the right of the one above it.

3. Any non-zero rows are always above rows with all zeros.

1 10 2

Upper Triangular Matrix

1 10 0

Row Echelon Form of Matrix

REFNot REF

Rank of the Matrix = r(A)

• Rank of a Matrix is number of non zero rows in row echelon form of it.

𝐴 =1 10 2

𝐴 =1 10 0

r(A) = 2 r(A) = 1

𝐴/𝐵 ≅1 1 10 2 2

r(A/B) = 2

𝐴/𝐵 ≅1 1 10 0 1

r(A/B) = 2

𝐴/𝐵 ≅1 1 10 0 0

r(A/B) = 1

Example 1 Find the Rank of Matrix after reducing it to Row Echelon Form.

𝐴 =113

1−11

121

𝐴 =1 1 11 −1 23 1 1

R2 R2 – R1, 𝐴 ≅103

1−21

111

R3 R3 – 3R1, 𝐴 ≅100

1−2−2

11−2

R3 R3 – R2, 𝐴 ≅100

1−20

11−3

R2 R2/(-2), 𝐴 ≅100

110

1

−1

21

As, there are three non zero rows in the row echelon form of given matrix Arank (A)= r(A) = 3

R3 R3/(-3),

Example 2 Find the Rank of Matrix after reducing it to Row Echelon Form.

𝐴 =315

213

−5−2−8

𝐴 =3 2 −51 1 −25 3 −8

R1 R2, 𝐴 ≅135

123

−2−5−8

R2 R2 – 3R1, 𝐴 ≅105

1−13

−21−8

R3 R3 – 5R1, 𝐴 ≅100

1−1−2

−212

R2 R2/(-1), 𝐴 ≅100

11−2

−2−12

R3 R3 + 2R2, 𝐴 ≅100

110

−2−10

As, there are two non zero rows in the row echelon form of given matrix Arank (A)= r(A) = 2

𝐴 =322

210

0312

10−5

Example 3 Find the Rank of Matrix after reducing it to Row Echelon Form.

𝐴 =322

210

0312

10−5

R1 R1 - R2,

R2 R2 - 2R1,

𝐴 ≅122

110

−3312

10−5

𝐴 ≅102

1−10

−3912

1−2−5

R3 R3 – 2R1,

R2 R2/(-1),

𝐴 ≅100

1−1−2

−3918

1−2−7

𝐴 ≅100

11−2

−3−918

12−7

R3 R3 + 2R2, 𝐴 ≅100

110

−3−90

12−3

As, there are three non zero rows in the row echelon form of given matrix Arank (A)= r(A) = 3

R3 R3/(-3), 𝐴 ≅100

110

−3−90

121

Example 4 Find the Rank of Matrix after reducing it to Row Echelon Form.

𝐴 =

1 −3 7 62 −2 5 334

−10

31

0−3

R2 R2 - 2R1, 𝐴 ≅

1 −3 7 60 4 −9 −934

−10

31

0−3

R3 R3 - 3R1, 𝐴 ≅

1 −3 7 60 4 −9 −904

80

−181

−18−3

R4 R4 - 4R1, 𝐴 ≅

1 −3 7 60 4 −9 −900

812

−18−27

−18−27

R3 R3 - 2R2,𝐴 ≅

1 −3 7 60 4 −9 −900

00

00

00

R4 R4 - 3R2,

As, there are two non zero rows in the row echelon form of given matrix Arank (A)= r(A) = 2

R2 R2/4,𝐴 ≅

1 −3 7 60 1 −9/4 −9/400

00

00

00

a11x + a12y = b1a21x + a22y = b2

𝑎11 𝑎12𝑎21 𝑎22

𝑥𝑦 =

𝑏1𝑏2

A X = B𝐴/𝐵 =

𝑎11 𝑎12 𝑏1𝑎21 𝑎22 𝑏2

R1 R1/a11,

𝐴/𝐵 ≅1 𝑎12/𝑎11 𝑏1/𝑎11𝑎21 𝑎22 𝑏2

1 𝑎12′0 𝑎22′

𝑥𝑦 = 𝑏1′

𝑏2′

R2 R2 – a21R1,

a11x + a12y = b1a21x + a22y = b2

𝐴/𝐵 ≅1 𝑎12/𝑎11 𝑏1/𝑎110 𝑎22 − 𝑎21𝑎12/𝑎11 𝑏2 − 𝑎21𝑏1/𝑎11

General Example for Solving 2x2 SLES using Matrix

Case 1

𝐴 ≅ 1 𝑎12′0 𝑎22′

𝐴/𝐵 ≅ 1 𝑎12′ 𝑏1′0 𝑎22′ 𝑏2′

r(A) = r(A/B) = 2

Case 2

r(A) = 1 & r(A/B) = 2 r(A) ≠ r(A/B)

Case 3

r(A) = r(A/B) = 1

a22’ ≠ 0 Unique Solution

𝐴 ≅ 1 𝑎12′0 𝑎22′

𝐴/𝐵 ≅ 1 𝑎12′ 𝑏1′0 𝑎22′ 𝑏2′

a22’ = 0 b2’ ≠ 0 No Solution

𝐴 ≅ 1 𝑎12′0 𝑎22′

𝐴/𝐵 ≅ 1 𝑎12′ 𝑏1′0 𝑎22′ 𝑏2′

a22’ = 0 b2’ = 0 Infinitely Many Solutions

Consistency of SLEs

• SLE is Consistent if r(A) = r(A/B)• If r(A) = r(A/B) = n (number of variables) then SLE has unique solution

• If r(A) = r(A/B) < n then SLE has infinitely many solutions

• SLE is inconsistent (no solution) if r(A) ≠ r(A/B)

Next Lecture : Reduced Row Echelon Form & Rank of Matrix