Solution of linear system of equations by gaussian elimination and gauss-jordon methods

SOLUTION OF LINEAR SYSTEM OF EQUATIONS BY GAUSSIAN ELIMINATION AND GAUSS-JORDON METHODS

There are two types of methods to solve simultaneous linear algebraic equations with many unknowns.

 

(a) Gaussian Elimination method - Pivoting

Gauss elimination method is a direct method which consists of transforming the given system of simultaneous equations to an equivalent upper triangular system. From this system the required solution can be obtained by the method of back substitution.

Consider the n linear equations in n unknowns, viz.

Now, multiply the first row of (3), (if a₁₁ ≠ 0) by a_i1 / a ₁₁ and add to the i^th row of [A, B], where i = 2, 3, ..., n. By this, all elements in the first column of [A, B] except a11 are made to zero. Now (3) is of the form

Now, considering b₂₂ as the pivot, we will make all elements below b₂₂ in the second column of (4) as zeros. That is, multiply second row of (4) by - b _i2 / b₂₂ and add to the corresponding elements of the ith row (i = 3, 4, ... n).

Now all elements below b22 are reduced to zero. Now (4) reduces to

Now taking c₃₃ as the pivot, using elementary operations, we make all elements below C₃₃ as zeros. Continuing the process, all elements below the leading diagonal elements of A are made to zero.

Hence, we get [A, B] after all these operations as

From (6), the given system of linear equations is equivalent to

Taking from the bottom to top of these equations, we solve for x_n = K_n / a_nn Using this in the above system of equations, we get x_n-1 and  so. By this back substitution method, we solve for x_n, x_n-1, x_n-2, ..., x₂, x₁

Pivoting

In the elimination process, if any one of the pivot elements a11, a22, .. ann vanishes or becomes very small compared to other elements in that column, then we attempt to rearrange the remaining rows so as to obtain a non-vanishing pivot or to avoid the multiplication by a large number. This strategy is called pivoting.

Pivoting is of two types.

1. Partial pivoting,

2. Complete pivoting.

 

(b) Gauss-Jordan elimination method

This method is a modification of the above Gauss elimination method. In this method, the coefficient matrix A of the system AX = B is brought to a diagonal matrix or unit matrix by making the matrix A not only upper triangular matrix, but also lower triangular matrix by making all elements above the leading diagonal of A also are zeros.

Note: In this method, the values are got immediately without using the process of back substitution.

1. Solve the system of equations by (i) Gauss elimination method (ii) Gauss-Jordan method. [M.U. Oct., 1999]

10x - 2y + 3z = 23

2x + 10y - 5z = -33

3x - 4y + 10z = 41

Solution :

Note : Solve by using Calculator and cross check your answer.

Here, we get x = 1, y = -2, z = 3

(i) Gauss elimination method

The given system is equivalent to

Now, we will make the matrix A as a upper triangular.

Fix the first row, change 2 and 3 row with row 1

Fix 1 and 2 row, change 3 row with 2nd row.

This is an upper triangular matrix,

Now, using back substitution method.

Now, we will make the matrix A a diagonal matrix.

Fix the third row and change 2nd row and first row

Hence, the solution is x = 1, y = -2, z = 3

 

2. Solve the system of equations by Gauss elimination method

x + 2y + z = 3, 2x + 3y + 3z = 10, 3x - y + 2z = 13[A.U N/D 2019 (R17)]

Solution :

Note :

Solve by using Calculator and cross check your answer. Here, we get x = 2, y=1, z = 3

The given system is equivalent to

3. Solve the system of equations by Gauss elimination method.

 5x₁ + x₂ + x₃ + x₄ = 4 ;

x₁ + 7x₂ + x₃ + x₄ = 12

x₁ + x₂ + 6x₃ + x₄ = -5;

x₁ + x₂ + x₃ + 4x₄ = -6

Solution: The given system is equivalent to

Hence, the solution is x₁ = 1, x₂ = 2, x₃ = -1, x₄ = -2.

 

4. Using Gauss elimination method, solve the system.

3.15x -1.96y + 3.85z = 12.95

2.13x + 5.12y - 2.89z = -8.61

5.92x + 3.05y + 2.15z = 6.88[M.K.U. 1981] [A.U N/D 2011]

Solution :

Note :

Solve by using Calculator and cross check your answer.

Here, we get x = 1.709, y = -1.8, z = 1.049

The given system is equivalent to

Hence, the solution is x= 1.709, y = -1.800, z = 1.049

 

5. Solve the following system by Gauss-Jordan method.

5x₁ + x₂ + x₃ + x₄ = 4, x₁ + 7x₂ + x₃ + x₄ = 12,

x₁ + x₂ + 6x₃ + x₄ = -5, x₁ + x₂ + x₃ + 4x₄ = -6.

Solution: Interchange the first and the last equation, so that the coefficient of x₁ in the first equation is 1.

Then, we have,

Fix the pivot element row. In the pivot element column make them the other elements are zero.

Hence, the solution is x₁ = 1, x₂ = 2, x₃ = -1, x₄ = -2

 

6. Using the Gauss-Jordan method solve the following equations[A.U M/J 2016 R13]

[A.U CBT M/J 2010]

 [A.U. Nov./Dec. 2004] [A.U A/M 2008]

10x + y + z = 12

2x+10y+z = 13

x+y+5z = 7

Solution :

Note :

Solve by using Calculator and cross check your answer. Here, we get x = 1, y = 1, z = 1

Interchanging the first and the last equation then

Fix the pivot element row and make the other elements zero in the pivot element column.

 

7. Solve the system of equations by Gauss-Jordan method  [A.U M/J 2013] [A.U N/D 2021 R-17]

x + y + z + w = 2

2xy + 2z-w = -5,

3x + 2y + 3z + 4w = 7;

x-2y-3z + 2w = 5. SI

Solution :

Note : Solve by using Calculator and cross check your answer. Here, we get x = 1, y = 1, z = 1

8. Solve the Gauss – Jordan method, the equations  [A.U A/M 2019 R-17]

2x + y + 4z = 12, 8x – 3y + 2z = 20, 4x + 11y – z = 33

Solution :

Note: Solve by using Calculator and cross check your answer.

Here, we get x = 3, y = 2, z = 1

Interchanging the second and third equation then