SOLVING SYSTEM OF LINEAR EQUATIONS BY RANK METHOD

Example 1 :

Solve the following linear equation by rank method

2x + 5y + 7z  =  52, x + y + z  =  9, 2x + y - z  =  0

Solution :

Number of non zero rows are 3.

ρ(A)  =  ρ([A|B])  =  3. The system is consistent and it has unique solution.

From 1^st row,

x+y+z  =  9 -----(1)

From 2^nd row,

3y+5z  =  34 -----(2)

From 3^rd row,

-4z  =  -20 -----(3)

From (3)

z  =  5

By applying the value of z in (2), we get 

3y+5(5)  =  34

3y + 25  =  34

3y  =  34-25

3y  =  9

y  =  3

By applying the value of y and z in (1), we get

x+3+5  =  9

x+8  =  9

x  =  1

So, the solution is (1, 3, 5).

Example 2 :

Solve the following linear equation by rank method

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

Solution :

ρ(A)  =   2 and ρ([A|B])  =  3. The system is inconsistent and it has no solution.

Example 3 :

Solve the following linear equation by rank method

x+9y-z  =  27, x-8y+16z  =  10, 2x+y+15z  =  37 

Solution :

Here ρ(A)  =  ρ([A|B])  =  2 < 3, then the system is consistent and it has infinitely many solution.

From the 1^st row,

x + 9y-z  =  27    ---(1)

From the 2^nd row,

17y + 17z  =  -17   ---(2)

Dividing by 17, we get

y + z  =  -1

Put z  =  t

y  =  -1 - t

By applying the value of y and z in (1), we get

x + 9(-1-t)-t  =  27

x - 9+9t-t  =  27

x  =  27+9-8t

x  =  36-8t

Solution :

x  =  36-8t, y  =  -1-t and z  =  t where t ∈ Real numbers.

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