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Tamil Nadu Board of Secondary EducationHSC Science Class 12

Find the value of k for which the equations kx – 2y + z = 1, x – 2ky + z = -2, x – 2y + kz = 1 have no solution - Mathematics

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Question

Find the value of k for which the equations kx – 2y + z = 1, x – 2ky + z = -2, x – 2y + kz = 1 have no solution

Sum

Solution

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

ρ(A) = 2

ρ(A|B) = 3.

ρ(A) ≠ ρ(A|B)

The system is inconsistent and it has no solution.

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Applications of Matrices: Consistency of System of Linear Equations by Rank Method
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Chapter 1: Applications of Matrices and Determinants - Exercise 1.6 [Page 42]

APPEARS IN

Samacheer Kalvi Mathematics - Volume 1 and 2 [English] Class 12 TN Board
Chapter 1 Applications of Matrices and Determinants
Exercise 1.6 | Q 2. (i) | Page 42

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