Solve the following set of equations using gausselimination

Solve the following set, of equations using gauss-elimination method. (a) 4x_1 + 2x_2 + 4x_3 = 24 x_1 + x_2 + 2x_3 = 10 2x_1 + lx_2 + x_3 = 10 (b) 4x_1 + 2x_2 - x_3 = 2 2x_1 - 2x_2 - 3x_3 = 0 x_1 + 2x_2 + x_3 = 1 (c) 2x_1 + x_2 + x_3 = 2 x_1 + x_2 + 2x_3 = 1 - 5x_1 - 2x_2 - 5x_3 = - 2

Solution

(a) The given system of linear equations can be represented in Matrix form as AX = b, where, b = (24,10,10)^T , X = (x₁,x₂,x)^T and A =

4

4

2

1

1

2

2

1

1

Let B = [A,b] =

4

4

2

24

1

1

2

10

2

1

1

10

To solve the given linear equations, we will reduce the augmented matrix B to its RREF as under:

Multiply the 1st row by ¼; Add -1 times the 1st row to the 2nd row

Add -2 times the 1st row to the 3rd row; Interchange the 2nd row and the 3rd row         

Multiply the 2nd row by -1; Multiply the 3rd row by 2/3

Add -1/2 times the 3rd row to the 1st row; Add -1 times the 2nd row to the 1st row

Then the RREF of B is

1

0

0

8/3

0

1

0

2

0

0

1

8/3

Hence x₁ = 8/3,x₂ = 2 and x₃ = 8/3.

(b) The given system of linear equations can be represented in Matrix form as AX = b, where, b = (2,0,1)^T , X = (x₁,x₂,x)^T and A =

4

2

-1

2

-2

-3

1

2

1

Let B = [A,b] =

4

2

-1

2

2

-2

-3

0

1

2

1

1

To solve the given linear equations, we will reduce the augmented matrix B to its RREF as under:

Multiply the 1st row by ¼ ; Add -2 times the 1st row to the 2nd row

Add -1 times the 1st row to the 3rd row; Multiply the 2nd row by -1/3

Add -3/2 times the 2nd row to the 3rd row; Add -1/2 times the 2nd row to the 1st row

Then the RREF of B is

1

0

-2/3

1/3

0

1

5/6

1/3

0

0

0

0

Hence x₁ -2x₃/3 =1/3 , and x₂ +5x₃/6=1/3. Now, let x₃ = 6t. Then x₁= 1/3 +4t and x₂= 1/3 -5t, where t is an arbitrary real number. Thus, the given linear system has infinite solutions.

(c ) The given system of linear equations can be represented in Matrix form as AX = b,where,b =(2,1,-2)^T , X = (x₁,x₂,x)^T and A =

2

1

1

1

1

2

-5

-2

-5

Let B = [A,b] =

2

1

1

2

1

1

2

1

-5

-2

-5

-2

To solve the given linear equations, we will reduce the augmented matrix B to its RREF as under:

Multiply the 1st row by ½

Add -1 times the 1st row to the 2nd row

Add 5 times the 1st row to the 3rd row

Multiply the 2nd row by 2

Add -1/2 times the 2nd row to the 3rd row

Multiply the 3rd row by -1/4

Add -3 times the 3rd row to the 2nd row

Add -1/2 times the 3rd row to the 1st row

Add -1/2 times the 2nd row to the 1st row

Then the RREF of B is

1

0

0

1/4

0

1

0

9/4

0

0

1

-3/4

Hence x₁ = 1/4,x₂ = 9/4 and x₃ = -3/4.

4	4	2
1	1	2
2	1	1