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NDA 2021
Exam
NDA
Year
2021
Subject
Mathematics
Chapter
Determinants

NDA 2021 · Mathematics · Determinants

For what values of k is the system of equations 2 k^2 x+3 y-1=0,7...

For what values of kk is the system of equations 2k2x+3y−1=0,7x−2y+3=0,6kx+y+1=02 k^2 x+3 y-1=0,7 x-2 y+3=0,6 k x+y+1=0 consistent?

Options

  1. A3±1110\frac{3 \pm \sqrt{11}}{10}
  2. B21±16110\frac{21 \pm \sqrt{161}}{10}
  3. C3±710\frac{3 \pm \sqrt{7}}{10}
  4. D4±1110\frac{4 \pm \sqrt{11}}{10}

Correct answer

B. 21±16110\frac{21 \pm \sqrt{161}}{10}

Step-by-step solution

We know that three lines a1x+b1y+c=0,a2x+b2y+c2=0a_1 x+b_1 y+c=0, a_2 x+b_2 y+c_2=0 and a3x+b3y+c3=0a_3 x+b_3 y+c_3=0 are concurrerce if ∣a1b1c1a2b2c2a3b3c3∣=0∴∣2k23−17−236k11∣=0⇒2k2(−2−3)−7(3+1)+6k(9−2)=0⇒5k2−21k+14=0∴k=21∓16110\begin{aligned} & \left|\begin{array}{lll} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{array}\right|=0 \\ & \therefore\left|\begin{array}{ccc} 2 k^2 & 3 & -1 \\ 7 & -2 & 3 \\ 6 k & 1 & 1 \end{array}\right|=0 \\ & \Rightarrow 2 k^2(-2-3)-7(3+1)+6 k(9-2)=0 \\ & \Rightarrow 5 k^2-21 k+14=0 \\ & \therefore k=\frac{21 \mp \sqrt{161}}{10} \end{aligned}
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