Question

Find the value of k for which the equations kx – 2y + z = 1, x – 2ky + z = -2, x – 2y + kz = 1 have unique solution

Answer 1

Martix form `[("k", -2, 1),(1, -2"k", 1),(1, -2, "k")][(x),(y),(z)] = [(1),(-2),(1)]`

AX = B

Augmented martix [A|B] = `[("k", -2, 1, |, 1),(1, -2"k", 1, |, -2),(1, -2, "k", |, 1)]`

`{:("R"_1 ↔ "R"_3),(->):} [(1, -2, "k", |, 1),(1, -2"k", 1, |, -2),("k", -2, 1, |, 1)]`

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

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

`-> [(1, -2, "k", |, 1),(0, 2(1 - "k"), (1 - "k"), |, -3),(0, 0, ("k" + 2)(1 - "k"), |, ("k" + 2))]`

∵ 2 – k – k² = – (k² + k – 2)

= – (k + 2)(k – 1)

= (k + 2)(1 – k)

Case:

If k ≠ 1

k ≠ – 2

ρ(A) = 3

ρ(A|B) = 3 = n

The system is consistent and it has unique solution.