This question was previously asked in

NDA (Held On: 17 April 2016) Maths Previous Year paper

Option 1 : k ≠ 1 and k ≠ -2

Electric charges and coulomb's law (Basic)

41133

10 Questions
10 Marks
10 Mins

__Concept__

Let the system of equations be,

a_{1}x + b_{1}y + c_{1}z = d_{1}

a_{2}x + b_{2}y + c_{2}z = d_{2}

a_{3}x + b_{3}y + c_{3}z = d_{3}

\(\Rightarrow \left[ {\begin{array}{*{20}{c}} {{{\rm{a}}_1}}&{{{\rm{b}}_1}}&{{{\rm{c}}_1}}\\ {{{\rm{a}}_2}}&{{{\rm{b}}_2}}&{{{\rm{c}}_2}}\\ {{{\rm{a}}_3}}&{{{\rm{b}}_3}}&{{{\rm{c}}_3}} \end{array}} \right]{\rm{\;}}\left[ {\begin{array}{*{20}{c}} {\rm{x}}\\ {\rm{y}}\\ {\rm{z}} \end{array}} \right] = {\rm{\;}}\left[ {\begin{array}{*{20}{c}} {{{\rm{d}}_1}}\\ {{{\rm{d}}_2}}\\ {{{\rm{d}}_3}} \end{array}} \right]\)

⇒ AX = B

⇒ X = A^{-1} B = \({\rm{\;}}\frac{{{\rm{adj\;}}\left( {\rm{A}} \right)}}{{\det {\rm{\;}}({\rm{A}})}}{\rm{\;B}}\)

⇒ If det (A) ≠ 0, system is consistent having **unique solution**.

__Calculation:__

Given system of linear equation are kx + y + z = 1, x + ky + z = 1 and x + y + kz = 1

Let A = \(\left[ {\begin{array}{*{20}{c}} {\rm{k}}&1&1\\ 1&{\rm{k}}&1\\ 1&1&{\rm{k}} \end{array}} \right]\)

det (A) = |A| = k (k^{2} – 1) – 1(k -1) + 1 (1 – k)

⇒ |A| = k^{3} – k – k + 1 + 1 – k = k^{3} – 3k + 2

For unique solution,

det (A) ≠ 0

⇒ k^{3} – 3k + 2 ≠ 0

⇒ (k – 1) (k^{2} + k - 2) ≠ 0

⇒ (k – 1) (k – 1) (k + 2) ≠ 0

∴ k ≠ 1 and k ≠ -2