Question

Solve the following system of equations by rank method

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

Answer 1

x + y + z = 9

2x + 5y + 7z = 52

2x – y – z = 0

The matrix equation corresponding to the given system is

`[(1, 1, 1),(2, 5, 7),(2, -1, -1)] [(x),(y),(z)] = [(9),(52),(0)]`

        A                X    =      B

Augmented Martix [A, B]	Elementary Tranformation
`[(1, 1, 1, 9),(2, 5, 7, 52),(2, -1, -1, 0)]`
`˜[(1, 1, 1, 9),(0, 3, 5, 34),(0, -3, -3, -18)]`	`{:("R"_2 -> "R"_2 - 2"R"_1),("R"_3 -> "R"_3 - 2"R"_1):}`
`˜[(1, 1, 1, 9),(0, 3, 5, 34),(0, 0, 2, 16)]`	`{:"R"_3 ->"R"_3 + "R"_2:}`
p(A) = 3; p(A, B) = 3

Obviously the last equivalent matrix is in the echelon form. It has three non-zero rows.

p(A) = p(A, B)

= 3

= Number of unknowns

The given system is consistent and has unique solution.

To find the solution, let us rewrite the above echelon form into the matrix form.

`[(1, 1, 1),(0, 3,5),(0, 0, 2)] [(x),(y),(z)] = [(9),(34),(16)]`

x + y + z = 9    .....(1)

3y + 5z = 34    .....(2)

2z = 16     .....(3)

2z = 16

`z = 16/2`

z = 8

Substitute z = 8 in eqn (2)

3y + 5(8) = 34

3y + 40 = 34

3y = 34 – 40

3y = –6

y = –2

Substitute y = – 2 and z = 8 in equation (1)

x = (– 2) + 8 = 9

x + 6 = 9

x = 9 – 6

x = 3

∴ x = 3, y = – 2, z = 8