System of Linear Equations & Gauss Elimination Method

3 Variables SLEs in Matrix Form

a11x + a12y + a13z = b1a21x + a22y + a23z = b2a31x + a32y + a33z = b3

𝑎11 𝑎12 𝑎13𝑎21 𝑎22 𝑎23𝑎31 𝑎32 𝑎33

𝑥𝑦𝑧

=𝑏1𝑏2𝑏3

A X = B

𝐴/𝐵 =𝑎11𝑎21𝑎31

𝑎12𝑎22𝑎32

𝑎13𝑎23𝑎33

𝑏1𝑏2𝑏3

Augmented Matrix,

R1

R2

R3

R1 R1/a11, 𝐴/𝐵 ≅1𝑎21𝑎31

𝑎12′𝑎22𝑎32

𝑎13′𝑎23𝑎33

𝑏1′𝑏2𝑏3

Gauss Elimination Method,

R2 R2 – a21R1, 𝐴/𝐵 ≅10𝑎31

𝑎12′𝑎22′𝑎32

𝑎13′𝑎23′𝑎33

𝑏1′𝑏2′𝑏3

R3 R3 – a31R1, 𝐴/𝐵 ≅100

𝑎12′𝑎22′𝑎32′

𝑎13′𝑎23′𝑎33′

𝑏1′𝑏2′𝑏3′

R2 R2/a22’, 𝐴/𝐵 ≅100

𝑎12′1

𝑎32′

𝑎13′𝑎23′′𝑎33′

𝑏1′𝑏2′′𝑏3′

R3 R3 – a32’R2, 𝐴/𝐵 ≅100

𝑎12′10

𝑎13′𝑎23′′𝑎33′′

𝑏1′𝑏2′′𝑏3′′

Case 1

𝐴/𝐵 ≅100

𝑎12′10

𝑎13′𝑎23′′𝑎33′′

𝑏1′𝑏2′′𝑏3′′

r(A) = r(A/B) = 3

Case 2 r(A) = 2 & r(A/B) = 3 r(A) ≠ r(A/B)

Case 3 r(A) = r(A/B) = 2

a33’’ ≠ 0 Unique Solution

𝐴 ≅100

𝑎12′10

𝑎13′

𝑎23′′

𝑎33′′

a33’’ = 0 b3’’ ≠ 0 No Solution

a33’’ = 0 b3’’ = 0 Infinitely Many Solutions

1 𝑎12′ 𝑎13′0 1 𝑎23′′0 0 𝑎33′′

𝑥𝑦𝑧

=𝑏1′𝑏2′′𝑏3′′

a33’’z = b3’’ z = b3’’/a33’’= L

y + a23’’z = b2’’y = b2’’ - a23’’z = b2’’ - a23’’L = M

x + a12’y + a13’z = b1’ x = b1’ - a12’y - a13’z = b1’ - a12’M - a13’L = N

Consistency of SLEs (Non Homogeneous)

• 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)

Example 1 Test the Consistency and Solve the following SLEs using Gauss Elimination Method if possible:

x + y + z = 6x - y + 2z = 53x + y + z = 8

𝐴/𝐵 =113

1−11

121

658

Augmented Matrix,

1 1 11 −1 23 1 1

𝑥𝑦𝑧

=658

In Matrix form SLEs is, A X = B

R2 R2 – R1, 𝐴/𝐵 ≅103

1−21

111

6−18

R3 R3 – 3R1, 𝐴/𝐵 ≅100

1−2−2

11−2

6−1−10

R3 R3 – R2, 𝐴/𝐵 ≅100

1−20

11−3

6−1−9

R2 R2/(-2), 𝐴/𝐵 ≅100

110

1−1/21

61/23R3 R3/(-3),

Therefore, Solution of the system is unique and it is (x, y , z) = (1, 2, 3).

r(A) = r(A/B) = 3 = n, Therefore, System is Consistent & It has Unique Solution.

1 1 10 1 −1/20 0 1

𝑥𝑦𝑧

=61/23

z = 3

y -1/2 z = 1/2 y = 1/2 + (1/2) (z) = 1/2 + (1/2) (3) = (3+1)/2 = 4/2 = 2

x + y + z = 6 x = 6 - y - z = 6 – 2 – 3 = 1

Example 2 Test the Consistency and Solve the following SLEs using Gauss Elimination Method if possible:

3x + 2y - 5z = 4x + y - 2z = 15x + 3y - 8z = 6

𝐴/𝐵 =315

213

−5−2−8

416

Augmented Matrix,

3 2 −51 1 −25 3 −8

𝑥𝑦𝑧

=416

In Matrix form SLEs is, A X = B

R1 R2, 𝐴/𝐵 ≅135

123

−2−5−8

146

R2 R2 – 3R1, 𝐴/𝐵 ≅105

1−13

−21−8

116

R3 R3 – 5R1, 𝐴/𝐵 ≅100

1−1−2

−212

111

R2 R2/(-1), 𝐴/𝐵 ≅100

11−2

−2−12

1−11

r(A) = 2

Therefore, System is inconsistent & It has no solution.

1 1 −20 1 −10 0 0

𝑥𝑦𝑧

=1−1−1

R3 R3 + 2R2, 𝐴/𝐵 ≅100

110

−2−10

1−1−1

by row (3), 0 z = -10 = -1 which Is not possible

Therefor, Solution of given SLE is not possible.

r(A/B) = 3 ∴r(A) ≠ r(A/B)&

Example 3 Test the Consistency and Solve the following SLEs using Gauss Elimination Method if possible:

2x + 2y + 2z = 0-2x + 5y + 2z = 18x + y + 4z = -1

𝐴/𝐵 =2−28

251

224

01−1

Augmented Matrix,

2 2 2−2 5 28 1 4

𝑥𝑦𝑧

=01−1

In Matrix form SLEs is, A X = B

R1 R1/2,

R2 R2 + 2R1,

R3 R3 – 8R1,

R3 R3 + R2,

𝐴/𝐵 ≅1−28

151

124

01−1

𝐴/𝐵 ≅108

171

144

01−1

𝐴/𝐵 ≅100

17−7

14−4

01−1

𝐴/𝐵 ≅100

170

140

010

∴(x, y , z) = {(−𝟏

𝟕−

𝟑

𝟕𝒌,

𝟏

𝟕−

𝟒

𝟕𝒌, k) / k є R}.

r(A) = r(A/B) = 2 < n,

Therefore, System is Consistent & It has Infinitely Many Solutions.1 1 10 1 4/70 0 0

𝑥𝑦𝑧

=01/70

z = k, k є R y + 4/7 z = 1/7

y = (1/7) - (4/7) (z) = (1/7) - (4/7) (k) = 𝟏

𝟕−

𝟒

𝟕𝒌, k є R

x + y + z = 0 x = 0 - y - z = 0 – (𝟏

𝟕−

𝟒

𝟕𝒌 ) – k = −

𝟏

𝟕+

𝟒

𝟕𝒌 − 𝒌 = −

𝟏

𝟕−

𝟑

𝟕𝒌

R3 R2/7, 𝐴/𝐵 ≅100

110

14/70

01/70

Example 4 Test the Consistency and Solve the following SLEs using Gauss Elimination Method if possible:

3x + y = -5-6x - 2y = 104x + 5y = 8

𝐴/𝐵 =3 1 −5−6 −2 104 5 8

Augmented Matrix,

3 1−6 −24 5

𝑥𝑦 =

−5108

In Matrix form SLEs is, A X = B

R1 R1/3,

R2 R2 + 6R1,

R3 R3 – 4R1,

R2 R3,

𝐴/𝐵 ≅1 1/3 −5/3−6 −2 104 5 8

𝐴/𝐵 ≅1 1/3 −5/30 0 04 5 8

𝐴/𝐵 ≅1 1/3 −5/30 0 00 11/3 44/3

𝐴/𝐵 ≅1 1/3 −5/30 11/3 44/30 0 0

Therefore, Solution of the system is unique and it is (x, y) = (-3, 4).

r(A) = r(A/B) = 2 = n, Therefore, System is Consistent & It has Unique Solution.

y = 4

x + y/3 = -5/3

R2 R2/(11/3), 𝐴/𝐵 ≅1 1/3 −5/30 1 40 0 0

1 1/30 10 0

𝑥𝑦 =

−5/340

x = -5/3 - y/3 x = -5/3 - 4/3 x = -9/3 = -3

Example 5 Test the Consistency and Solve the following SLEs using Gauss Elimination Method if possible:

x - 2y + w = 3-x + 2y + z – ½ w = -74x - 8y + 6z + 7w = -3

𝐴/𝐵 =1−14

−22−8

016

1 3−1/2 −7

7 −3

Augmented Matrix,

1−14

−22−8

016

1−1/27

𝑥𝑦𝑧𝑤

=3−7−3

In Matrix form SLEs is, A X = B

R2 R2 + R1,

R3 R3 – 4R1,

𝐴/𝐵 ≅104

−20−8

016

1 31/2 −4

7 −3

𝐴/𝐵 ≅100

−200

016

1 31/2 −4

3 −15

R3 R3 – 6R2, 𝐴/𝐵 ≅100

−200

010

1 31/2 −4

0 9

r(A) = 2 r(A/B) = 3 ∴r(A) ≠ r(A/B)&

Therefore, System is inconsistent & It has no solution.

Example 6 Test the Consistency and Solve the following SLEs using Gauss Elimination Method if possible:

3x + 2y + w = 162x + y + 3z = 162x + 12z -5w = 5

Augmented Matrix,

In Matrix form SLEs is, A X = B322

210

0312

10−5

𝑥𝑦𝑧𝑤

=16165

𝐴/𝐵 =322

210

0312

1 160 16−5 5

R1 R1 - R2,

R2 R2 - 2R1,

R3 R3 – 2R1,

R2 R2/(-1),

𝐴/𝐵 ≅122

110

−3312

1 00 16−5 5

𝐴/𝐵 ≅102

1−10

−3912

1 0−2 16−5 5

𝐴/𝐵 ≅100

1−1−2

−3918

1 0−2 16−7 5

𝐴/𝐵 ≅100

11−2

−3−918

1 02 −16−7 5

∴(x, y , z, w) = {(−𝟔𝒌 + 𝟐𝟓, 𝟗𝒌 − 𝟑𝟒, k, 9) / k є R}.

r(A) = r(A/B) = 3 < 4 = n,

Therefore, System is Consistent & It has Infinitely Many Solutions.

z = k, k є R

y - 9z + 2w = -16 y = (-16) + (9)(z) – 2w = (-16) + (9)(k) – 18 = 𝟗𝒌 − 𝟑𝟒, k є R

x + y - 3z + w = 0 x = 0 - y + 3z – w = 0 – (9k-34) + 3k – 9 = −𝟔𝒌 + 𝟐𝟓, 𝐤є R

R3 R3 + 2R2, 𝐴/𝐵 ≅100

110

−3−90

1 02 −16−3 −27

100

110

−3−90

121

𝑥𝑦𝑧𝑤

=0

−169

& w = 9

Next Lecture : Homogeneous SLES