For what values of p and q, the system of equations 2x + py + 6z = 8, x + 2y + qz = 5, x + y + 3z = 4 has (i) no solution (ii) a unique solution (iii) infinitely many solutions.

Engineering

Mathematics

Introduction to Determinants

Question

LiveFree JEE Main Previous Year Online Papers Live Free JEE Advance Previous Year Online Papers

College PredictorLive

Know your College Admission Chances Based on your Rank/Percentile, Category and Home State.

Get your JEE Main Personalised Report with Top Predicted Colleges in JoSA

$\Rightarrow D = |\begin{array}{ccc} 2 & p & 6 \\ 1 & 2 & q \\ 1 & 1 & 3 \end{array}| = |\begin{array}{ccc} 2 & p & 6 \\ 0 & 1 & b - 3 \\ 1 & 1 & 3 \end{array}| = |\begin{array}{ccc} 0 & p - 2 & 0 \\ 0 & 1 & q - 3 \\ 1 & 1 & b \end{array}|$

R₁ – R₂ R₁ – 2R₃

= (q – 3) (P – 2)

$D_{1} = |\begin{array}{lll} 2 D 6 \\ 5 2 0 \\ 4 1 3 \end{array}|$

= 8(6 – 9) – P(15 – 4q) + 6(5 – 8)

= 48 – 8q – 15P + 4Pq + 6 × –3

= (P – 2) (4q – 15)

$D_{2} = |\begin{array}{lll} 2 8 6 \\ 1 5 4 \\ 1 4 3 \end{array}| = 0$

$D_{3} = |\begin{array}{lll} 2 a 8 \\ 1 2 5 \\ 1 1 4 \end{array}| = (a - 2)$

(ii) D ≠ 0

(iii) D = D₁ = D₂ = D₃ = 0 .....(3)

(p – 2) (q – 3) ≠ 0 p = 2, q = 3 → infinitely many solution.

⇒ p ≠ 2, q ≠ 3 → unique solution

(i) D = 0, a = 2 or b = 3

q = 3, no solution D₃ ≠ 0 ....(1)

Please subscribe our Youtube channel to unlock this solution.