Consistent ?? Think again -2

Algebra Level 4

{ x 2 − 2 x 3 − h = 0 x 1 − x 2 + x 3 = 2 x 1 = 3 − k x 2 \begin{cases} x_2 - 2x_3 - h = 0 \\ x_1 - x_2 + x_3 = 2 \\ x_1 = 3 - k x_2 \end{cases}

Given the system of equations above. What are the values of h h and k k so that the system is inconsistent?

A) k = − 1 2 k = -\dfrac12 , h ⊂ R − { 2 } h\subset \mathbb R -\{2 \} .
B) k ⊂ R − { − 1 2 } , h = 2 k \subset \mathbb R - \left \{ - \dfrac12 \right\} , h = 2 .
C) The system is always inconsistent.
D) k ⊂ R , h ⊂ R k \subset \mathbb R , h \subset \mathbb R .
E) k = 0 k = 0 or k = 1 k=1 , h = ± 2 h = \pm \; 2 .

C) A) D) E) B)

Ahmad Hesham by Ahmad Hesham , 25, Egypt

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1 solution

Chew-Seong Cheong
Jul 16, 2015

The system of equation is matrix form is as follows:

( 0 1 − 2 1 − 1 1 1 k 0 ) ( x 1 x 2 x 3 ) = ( h 2 3 ) \begin{pmatrix} 0 & 1 & -2 \\ 1 & -1 & 1 \\ 1 & k & 0 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = \begin{pmatrix} h \\ 2 \\ 3 \end{pmatrix}

The system of equation is inconsistent when the determinant of the coefficient matrix is equal to 0 0 . That is:

∣ 0 1 − 2 1 − 1 1 1 k 0 ∣ = 1 − 2 k − 2 = − 2 k − 1 = 0 \begin{aligned} \begin{vmatrix} 0 & 1 & -2 \\ 1 & -1 & 1 \\ 1 & k & 0 \end{vmatrix} & = 1 - 2k -2 = -2k -1 = 0 \end{aligned}

⇒ k = − 1 2 \Rightarrow k = -\frac{1}{2} and the answer is A \boxed{A} .

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Moderator note:

Simple standard approach

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