Question

Let the system of linear equations 4x + λy + 2z = 0 ; 2x - y + z = 0 ; μx + 2y + 3z = 0, λ, μ ∈ R has a non-trivial solution. Then which of the following is true ?

• A. λ = 3, μ ∈ R
• B. μ = -6, λ ∈ R
• C. λ = 2, μ ∈ R
• D. μ = 6, λ ∈ R

Hint:

A homogeneous system $AX=0$ has a non-trivial solution if and only if $|A| = 0$.

Answer 1

The Correct Option is B

Step 1: For a non-trivial solution, the determinant of coefficients must be zero. 

Step 2: Expand along $R_1$: $4(-3-2) - \lambda(6-\mu) + 2(4+\mu) = 0$. $-20 - 6\lambda + \lambda\mu + 8 + 2\mu = 0$. $\mu(2+\lambda) - 6\lambda - 12 = 0 \implies \mu(2+\lambda) - 6(2+\lambda) = 0$. 
Step 3: Factoring gives $(\mu - 6)(\lambda + 2) = 0$. This means either $\mu = 6$ or $\lambda = -2$. Checking options: (D) states $\mu = 6, \lambda \in R$.