Question

Show that the equations 5x + 3y + 7z = 4, 3x + 26y + 2z = 9, 7x + 2y + 10z = 5 are consistent and solve them by rank method

Answer 1

5x + 3y + 7z = 4

3x + 26y + 2z = 9

7x + 2y + 10z = 5

The matrix equation corresponding to the given systematic

`[(5, 3, 7),(3, 26, 2),(7, 2, 10)][(x),(y),(z)] = [(4),(9),(5)]`
         A              X     =    B

Augmented Matrix [A, B]	Elementary Tranformation
`[(5, 3, 7, 4),(3, 26, 2, 9),(7, 2, 10, 5)]`
`∼[(7, 2, 10, 5),(3, 26, 2, 9),(5, 3, 7, 4)]`	`{:"R"_1 ↔ "R"_3:}`
`∼[(2, -1, 3, 1),(3, 26, 2, 9),(5, 3, 7, 4)]`	`{:"R"_1 -> "R"_1 - "R"_3:}`
`∼[(2, -1, 3, 1),(1, 2, -1, 8),(1, 5, 1, 2)]`	`{:("R"_2 -> "R"_2 - "R"_1),("R"_3 -> "R"_3 - 2"R"_1):}`

`∼[(1, 5, 1, 2),(1, 2, -1, 8),(2, -1, 3, 1)]`	`{:"R"_1 ↔ "R"_3:}`
`∼[(1, 5, 1, 2),(0, 22, -2, 6),(0, -11, 1, -3)]`	`{:("R"_2 -> "R"_2 - "R"_1),("R"_3 -> "R"_3 - 2"R"_1):}`
`∼[(1, 5, 1, 2),(0, -11, 1, -3),(0, 22, -2, 6)]`	`{:"R"_2 ↔ "R"_3:}`
`∼[(1, 5, 1, 2),(0, -11, 1, -3),(0, 0, 0, 0)]`	`{:"R"_3 -> "R"_ 3 + "R"_2:}`
p(A) = 2; p(A, B) = 2

Obviously, the last equivalent matrix is in the echelon form.

It has two non-zero rows.

p([A, B]) = 2, p(A) = 2

p(A) = p([A, B]) = 2 < Number of unknowns

The given system is consistent and has infinitely many solutions.

The given system is equivalent to the matrix equation

`[(1, 5, 1),(0, -11, 1),(0, 0, 0)] [(x),(y),(z)] = [(2),(-3),(0)]`

x + 5y + z = 2  .......(1)

– 11y + z = – 3  .......(2)

Let z = k

Equation (1) ⇒

`x + 5[1/11 (3 + "k")] + "k"` = 2

x = `2 - "k" - 5/11 (3 + "k")`

x = `(22 - 11"k" - 15 - 5"k")/11`

x = `1/11 (7 - 16"k")`

Equation (2) ⇒

– 11y + k = – 3

3 + k = 11y

y = `1/11 (3 + "k")`

(x, y, z) = `(1/11 (7 - 16"k"), 1/11 (3 + "k"), "k")`

Where K ε R