For the system of linear equations: x - 2 y = 1 ,   x - y + k z = - 2 ,   k y + 4 z = 6 ,   k…

For the system of linear equations:

x-2y=1, x-y+kz=-2, ky+4z=6, k∈R

Consider the following statements:
(A) The system has unique solution if k≠2,k≠-2.
(B) The system has unique solution if k=-2.
(C) The system has unique solution if k=2.
(D) The system has no-solution if k=2.
(E) The system has infinite number of solutions if k≠-2. 

Which of the following statements are correct?

  1. A and E only
  2. B and E only
  3. A and D only
  4. C and D only

Solution

Given system \(\left\{\begin{array}{l} x-2 y=1 \\ x-y+k z=-2 \\ k y+4 z=6 \end{array}\right.\) Coefficient matrix \(A=\left(\begin{array}{ccc} 1 & -2 & 0 \\ 1 & -1 & k \\ 0 & k & 4 \end{array}\right)\) Determinant \(\operatorname{det} A=1\left|\begin{array}{cc} -1 & k \\ k & 4 \end{array}\right|-(-2)\left|\begin{array}{ll} 1 & k \\ 0 & 4 \end{array}\right|=\left(-4-k^2\right)+8=4-k^2\) - Unique solution if \(\operatorname{det} A \neq 0 \Rightarrow k \neq \pm 2\). - Check \(k=2\) : equations become inconsistent ⇒ no solution. - Check \(k=-2\) : equations become dependent ⇒ infinitely many solutions. Correct statements: \((A),(D)\)

Asked in: MHT CET Full Test 7

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