ngular mat AX=

Substitution meuid

ngular mat

Note. This method

of making

^{the matrix A as upper trian}

been taught

in lower classes while

finding the rank of the ma

This method is a modification of the above Gause

method. In this method, the coefficient matrix A of the svsta AX=

atrix A

brought to a diagonal matrix or unit matrix by making the m.

em

syst

4-21 Gauss-Jordan elimination

^method

(Direct method

Gauss elimi

m

This

only upper triangular but also lower triangular by making all el above the leading diagonal of A also

^as zeros.

By this way, the

_s

AX= B will reduce to the form.

o 0 0 bi

a11 0

0 b22 b 0 0 0 C2

0 0 0 0 Cn d

From (7)

2b22 b

K

₌

Note. By using this the method, process the of values back of x1, substitution. X2,

^{. . . ,}

X,

^are

got immediately wiul

Example 1. Solve the system of equations by (i) Gauss elmin

method (ii) Gauss-Jordan method.

[MKU 198

x+2y+z=3, 2x+3y + 3z= 10;

3x- y+ 2z= 13.

Solution. (By Gauss method)

The given system is

equivalent to

2 3

2 3 3

10

- 1

2 13

A X=B

2 3

(A, B)= 3 3 10

2 13

Numerical M e t h o d s - / V

115

Now, we will make the matrix A uPper triangular.

3

(A, B)=

-12

^{2 13 10}

R+(-2)R,

R+(-3) R, ie., R (-3)

ie., R21 -2)

o-7- |4

Now take b22=-1 as the pivot and make by2 as zero.

(1 2

(A, B)-|

3

4 Rg2-7)

...(2)

-24 From this, we get

x+2y + z=3

- y + = 4 - 8 z = - 2 4

z= 3, y=-1,x=2 by back substitution.

= 2, y = - 1, z=3

i.e.

Solution. (Gauss-Jordan method)

In stage 2, make the element, in the position (1, 2), also zero.

(A, B)-| (1

^{0 - 1}

2

1 4

-8-24

0 3

0 -1 ¹ 4

Ri2 (2)

o -8-24

0-1

(1 R, Ry (3), R2 ()

.

X=2,-y =

^{I, - z = - 3}

= 2, y =- 1,z=3

Eample 2. Solve the system by Gauss-Elimination method

2x+3y-z= 5; 4x+4y - 32 = 3 and 2x - 3y+ 22 =2. {MKU 1980]

Solution. The system is equivalent to

all ot them s T ultancously

Example 1. Find, by GausSian el1m1nation method. ne

A - 15

6

5 2 2

Solution. Step 1. We write

down the augmented sys

Numerical Methods-/V

123 3 -1

0 0

(A, )=

^{I 5 6} _{5 0 0}

2

0

⁰

I ..(1)

5 2

Cton 2. Our aim is to

reduce

the matrix _{A to an}

upper

trianguiar

matrix. Now

we will

reduce

all

elements

_below

a

to zero.

System () becomes

3 - 1 1 0 0

R+5R

0 1 0

5 0

.(2)

- - o -3 )8,

Note. When we reduce the elements below

d

^{in A to zero,}

only

^the first column of I is

changed

while the second and third column remain

unchanged.

Step

^3.

Now,

^we will reduce the

elemen1s

oeiow tne

position (2,

²⁾

to zero.

Now the system (1), reduces to

3 - 1 I 0

0 5 0. R,+ ..(3

0 0

Note. When the elements below the position (2, 2) are reduced to zero, only the second column of I is changed whercas the third column of I is

unchanged.

Step 4. Now the system is equivalent to the three systems,

3 - I

0 0 5 ..(4)

3 - I I 0

0 0 1

..(5)

3 - I I 0

and ^...(6)

Solution

of Simultaneous Linear

Algebraie a

124

That is

3 + =

A=5 2

+ =0

i2=

and

3

20 1

Hence A'=|5

Example

^2.

By

Gaussian

eliminution, find

^the inverse of

A =

^{2 0}

4 Solution. The

augmented system (A.

) ^is

A, )= 2 0 | 0 1 0

3 -1-4 0 0

Since the element _aj ⁼ 0, ^we will

interchange

^{the first}

<anu

row. The reduced

system

^is

2 0 0 I 0

3 - I - 4

0 0

By perlorning R, +(-3) R,,

^we

get

2 ₀

0 0

0 0

0 -7 -4

0 -3

Numerical M e t h o d s V

P e r f o r m i n g

R,

⁺

7R,

2 0

0 0

0 3

^{7 -3 11}

+ 2 1 =0

Thus.

21-

21t = |

3 7

i + 2 I 2 t2=0

3xg-3

3 2 - 1

22= 1

1 2 - 1

3 +221=0

2 t 3 =0

3x1=

23

- I

Hence A=|

Example 3. By Gaussian elimination, find A' if

A =2 3 - 2

Solution. (A.

|2 3

i

-2

4 1 2

-2 2 o 0 .(1)

Stage 1.

Perform R2+

to

R and R +| R. Then (1) reduces

2 o )