Skip to content
Exercise 1.7 · Q2

Q.Determine the values of λ\lambda for which the following system of equations x+y+3z=0, 4x+3y+λz=0, 2x+y+2z=0x+y+3z=0,\ 4x+3y+\lambda z=0,\ 2x+y+2z=0 has

(i) a unique solution
(ii) a non-trivial solution.
Tamil Nadu DgeTextbookSubjectiveImportance★★★★★
31% · 37/118 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Step 1. Set up and compute the determinant.

det⁡A=∣11343λ212∣.\det A=\begin{vmatrix} 1 & 1 & 3\\ 4 & 3 & \lambda\\ 2 & 1 & 2\end{vmatrix}.

Expanding along the first row:

det⁡A=1[(3)(2)−λ(1)]−1[(4)(2)−λ(2)]+3[(4)(1)−(3)(2)]\det A=1\big[(3)(2)-\lambda(1)\big]-1\big[(4)(2)-\lambda(2)\big]+3\big[(4)(1)-(3)(2)\big]

=(6−λ)−(8−2λ)+3(4−6)=6−λ−8+2λ−6=λ−8.=(6-\lambda)-(8-2\lambda)+3(4-6)=6-\lambda-8+2\lambda-6=\lambda-8.

Step 2. λ≠8\lambda\ne8: only the trivial solution. Here det⁡A≠0⇒ρ(A)=3=n\det A\ne0\Rightarrow\rho(A)=3=n, so the homogeneous system has ρ(A)=n\rho(A)=n and therefore only the trivial solution x=y=z=0x=y=z=0 — this is what "a unique solution" means for a homogeneous system. …

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.