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

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

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