PROBLEMS IN LINEAR EQUATIONS

PROF. SEBASTIAN VATTAMATTAM

1. Homogeneous System of Linear Equations

Problem 1.1. The vector (c1, c2, c3) is a solution of

a11x1 + a12x2 + a13x3 = 0

a21x1 + a22x2 + a23x3 = 0

a31x1 + a32x2 + a33x3 = 0

Show that, for k 6= 0, (kc1, kc2, kc3) is also a solution.

Since (c1, c2, c3) is a solution of the system of equations,

a11c1 + a12c2 + a13c3 = 0

a21c1 + a22c2 + a23c3 = 0

a31c1 + a32c2 + a33c3 = 0

Multiplying each equation by k,

a11kc1 + a12kc2 + a13kc3 = k(a11c1 + a12c2 + a13c3) = 0

a21kc1 + a22kc2 + a23kc3 = k(a21c1 + a22c2 + a23c3) = 0

a31kc1 + a32kc2 + a33kc3 = k(a31c1 + a32c2 + a33c3) = 0

Hence the conclusion.

Problem 1.2. Given the system of equations,

x + 3y + 2z = 1(1)

2x + 2y − z = −1(2)

x + 4y + 3z = 5(3)

(1) Write down the coefficient matrix.(2) Write down the augmented matrix.(3) Transform the augmented matrix to the triangular form.(4) Solve the system of equations.

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2 PROF. SEBASTIAN VATTAMATTAM

(1) The coefficient matrix is

A =

1 3 22 2 −11 4 3

(2) The augmented matrix is

[A : b] =

1 3 2 12 2 −1 −11 4 3 5

(3) Applying elementary raw transformations,

[A : b] =

1 3 2 12 2 −1 −11 4 3 5

→ 1 3 2 1

0 −4 −5 −30 1 1 4

→

1 3 2 10 1 1 40 −4 −5 −3

→

1 3 2 10 1 1 40 0 −1 13

→

1 0 −1 −110 1 1 40 0 −1 13

→

1 0 −1 −110 1 1 40 0 1 −13

→

1 0 0 −240 1 0 170 0 1 −13

(4) Hence we have,

x1 = −24, x2 = 17, x3 = −13

LINEAR EQUATIONS 3

2. Non-homogeneous System of LinearEquations

Gauss-Jordan Method

Theorem 2.1. If Ax = b is a system of linear equationsand [A : b] is the augmented matrix, then the system isconsistent iff

rankA = rank[A : b]

IfrankA = rank[A : b] =: r

and n denotes the number of unknowns, then

Case 1: n = r. The system has a unique solution.Case 2: n > r. The system has infinite number of so-

lutions.n− r of the unknowns can be given arbitrary values.

Problem 2.2. Use Gauss-Jordan method to examine theconsistency of the following system of linear equations.

x1 + x2 + x4 = 3(4)

x2 + x3 = 4(5)

x1 + 2x2 + x3 + x4 = 8(6)

SolutionDenote the equations by E1, E2, E3

ApplyE3 → E3 − E1

x1 + x2 + x4 = 3(7)

x2 + x3 = 4(8)

x2 + x3 = 5(9)

Here 1.8 and 1.9 contradict each other, and hence the sys-tem is not consistent.

Gauss-Jordan method in the Matrix form

4 PROF. SEBASTIAN VATTAMATTAM

The system is of the form Ax = b where

A =

1 1 0 10 1 1 01 2 1 1

Augmented matrix

[A, b] =

1 1 0 1 . 30 1 1 0 . 41 2 1 1 . 8

Apply the elementary transformation

R3 → R3 −R1

[A : b] =

1 1 0 1 . 30 1 1 0 . 40 1 1 0 . 5

Apply the elementary transformation

R3 → R3 −R2

[A : b] =

1 1 0 1 . 30 1 1 0 . 40 0 0 0 . 1

Now [A : b] is in the triangular form.

rank[A : b] = 3, rankA = 2

rank[A : b] 6= rankA

Therefore the system is not consistent.

Problem 2.3. Use Gauss-Jordan method to examine theconsistency of the following system of linear equations.

x1 + x2 + x3 = 4(10)

x1 + 2x2 = 6(11)

SolutionThe system is of the form Ax = b Augmented matrix

[A : b] =

(1 1 1 . 41 2 0 . 6

)

LINEAR EQUATIONS 5

Apply the elementary transformation

R2 → R2 −R1

[A : b] =

(1 1 1 . 40 1 −1 . 2

)Now [A : b] is in the triangular form.

rank[A : b] = 2 = rankA

Therefore the system is consistent.The equations in the reduced form are

x1 + x2 + x3 = 4(12)

x2 − x3 = 2(13)

n = 3, n− r = 1Let x3 = aThen

x2 = a + 2, x1 = −2a + a

The general solution is −2a + 2a + 2a

=

220

+ a

−211

Problem 2.4. Use Gauss-Jordan method to examine theconsistency of the following system of linear equations.

x1 + x2 = 1(14)

2x1 + x2 = 3(15)

3x1 + 2x2 = 4(16)

SolutionThe system is of the form Ax = b Augmented matrix

[A : b] =

1 1 . 12 1 . 33 2 . 4

Apply the elementary transformations

R2 → R2 − 2R1, R3 → R3 − 3R1

6 PROF. SEBASTIAN VATTAMATTAM

[A : b] =

1 1 . 10 −1 . 10 −1 . 1

Apply the elementary transformation

R3 → R3 −R2

[A : b] =

1 1 . 10 −1 . 10 0 . 0

Now [A : b] is reduced to the triangular form.

rank[A : b] = 2 = rankA

Therefore the system is consistent.n = 2 = rThere is a unique solution.The equations in the reduced form are

x1 + x2 = 1(17)

−x2 = 1(18)

The solution is (2−1

)Problem 2.5. Use Gauss-Jordan method to examine theconsistency of the following system of linear equations.

2x1 − x2 + x3 + x4 = 6(19)

x1 + x2 + x3 = 4(20)

SolutionThe system is of the form Ax = b Augmented matrix

[A : b] =

(2 −1 1 1 . 61 1 1 0 . 4

)Apply the elementary transformations

R1 ↔ R2

LINEAR EQUATIONS 7

[A : b] =

(1 1 1 0 . 42 −1 1 1 . 6

)Apply the elementary transformation

R2 → R2 − 2R1

[A : b] =

(1 1 1 0 . 40 −3 −1 1 . −2

)Now [A : b] is reduced to the triangular form.

rank[A : b] = 2 = rankA

Therefore the system is consistent.n = 4, r = 2There are infinite number of solutions.n− r = 2 of the unknowns can be given arbitrary values.The equations in the reduced form are

−3x2 − x3 + x4 = −2(21)

x1 + x2 + x3 = 4(22)

Let

x3 = a, x4 = b

The solution is −2a−b+10

3−a+b+2

3ab

= a

−2/3−1/3

10

+ b

−1/31/301

+

10/32/300

8 PROF. SEBASTIAN VATTAMATTAM

Problem 2.6. Use Gauss-Jordan method to examine theconsistency of the following system of linear equations.

x1 + x4 = 5(23)

x2 + 2x4 = 5(24)

x3 + 0.5x4 = 1(25)

2x3 + x4 = 3(26)

SolutionThe system is of the form Ax = b Augmented matrix

[A : b] =

1 0 0 1 . 50 1 0 2 . 50 0 1 0.5 . 10 0 2 1 . 3

Apply the elementary transformation

R4 → R4 − 2R3

[A : b] =

1 0 0 1 . 50 1 0 2 . 50 0 1 0.5 . 10 0 0 0 . 1

Now [A : b] is in the triangular form.

rank[A : b] = 4, rankA = 3

rank[A : b] 6= rankA

Therefore the system is not consistent.

Problem 2.7. Use Gauss-Jordan method to examine theconsistency of the following system of linear equations.

2x2 + 2x3 = 4(27)

x1 + 2x2 + x3 = 4(28)

x2 − x3 = 0(29)

LINEAR EQUATIONS 9

SolutionThe system is of the form Ax = b Augmented matrix

[A : b] =

0 2 2 . 41 2 1 . 40 1 −1 . 0

Apply the elementary transformation

R1 ↔ R2

[A : b] =

1 2 1 . 40 2 2 . 40 1 −1 . 0

Apply the elementary transformation

R2 → (1/2)R2

[A : b] =

1 2 1 . 40 1 1 . 20 1 −1 . 0

Apply the elementary transformation

R3 → R3 −R2

[A : b] =

1 2 1 . 40 1 1 . 20 0 −2 . −2

Now [A : b] is in the triangular form.

rank[A : b] = 3, rankA = 3

The solution isx1 = x2 = x3 = 1

For any clarification please contact the author.E-mail address: vattamattam@gmail.com