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The system of equations

x + y + z = 6;

x + 4y + 6z = 20;

x + 4y + λz = μ

has NO solution for values of λ and μ given by


1. λ = 6, μ = 20
2. λ = 6, μ ≠ 20
3. λ ≠ 6, μ = 20
4. λ ≠ 6, μ ≠ 20

1 Answer

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Best answer
Correct Answer - Option 2 : λ = 6, μ ≠ 20

Concept:

The number of solutions can be determined by finding out the rank of the Augmented matrix and the rank of the Coefficient matrix.

  • If rank(Augmented matrix) = rank(Coefficient matrix) = no. of variables then no of solutions = 1.
  • If rank(Augmented matrix)  ≠ rank(Coefficient matrix) then no of solutions = 0.
  • If rank(Augmented matrix) = rank(Coefficient matrix) < no. of variables, no of solutions = infinite.

 

Calculation:

The augmented matrix for the system of equations is

\(\left[ {A{\rm{|}}B} \right] = \left[ {\left. {\begin{array}{*{20}{c}} 1&1&1\\ 1&4&6\\ 1&4&\lambda \end{array}} \right|\begin{array}{*{20}{c}} 6\\ {20}\\ \mu \end{array}} \right]\)

Performing: R3 → R3 – R2

\(\left[ {A{\rm{|}}B} \right] = \left[ {\left. {\begin{array}{*{20}{c}} 1&1&1\\ 1&4&6\\ 0&0&{\lambda - 6} \end{array}} \right|\begin{array}{*{20}{c}} 6\\ {20}\\ {\mu - 20} \end{array}} \right]\)      …

If λ = 6 and μ ≠ 20 then

Rank (A | B) = 3 and Rank (A) = 2

∵ Rank (A | B) ≠ Rank (A)

∴ Given the system of equations has no solution for λ = 6 and μ ≠ 20

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