Question

Test for consistency and if possible, solve the following systems of equations by rank method:

3x + y + z = 2, x – 3y + 2z = 1, 7x – y + 4z = 5

Answer 1

Matrix form `[(3, 1, 1),(1, -3, 2),(7, -1, 4)][(x),(y),(z)] = [(2),(1),(5)]`

AX = B

Augmented martix

[A|B] = `[(3, 1, 1, |, 2),(1, -3, 2, |, 1),(7, -1, 4, |, 5)]`

`{:("R"_1 ↔ "R"_2),(->):} [(1, -3, 2, |, 1),(3, 1, 1, |, 2),(7, -1, 4, |, 5)]`

`{:("R"_2 -> "R"_2 - 3"E"_1),("R"_3 -> "R"_3 - 7"R"_1),(->):} [(1, -3, 2, |, 1),(0, 10, -5, |, -1),(0, 20, -10, |, -2)]`

`{:("R"_3 -> "R"_2 - 2"R"_2),(->):} [(1, -3, 2, |, 1),(0, 10, -5, |, 1),(0, 0, 0, |, 0)]`

ρ(A) = 2

ρ[A|B] = 2

ρ(A) = ρ[A|B] = 2 < n

The system is consistent. It has infinitely many solution.

Writing the equivalent equations from echelon form.

x – 3y + 2z = 1  ........(1)

10y – 5z = – 1  ........(2)

Put z = t.

(2) ⇒ 10y – 5z = – 1

10y = – 1 + 5z

= 5t – 1

y = `(5"t" - 1)/10`

(1) ⇒ x – 3y + 2z = 1

`x - 3((5"t" - 1)/10) + 2"t"` = 1

x = `1 - 2"t" + 3((5"t" - 1)/10)`

= `(10 - 20"t" + 15"t" - 3)/10`

x = `(7 - 5"t")/10`

(x, y, z) = `((7 - 5"t")/10, (5"t" - 1)/10, "t") ∀ t ∈ R.