Related Formula
Δ = vmatrix a₁ & b₁ & c₁ a₂ & b₂ & c₂ a₃ & b₃ & c₃ vmatrix$$\Delta = \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix}$$
Core Logic
For the given system, we find the determinant of the coefficient matrix:
Δ = vmatrix 1 & 1 & 1 1 & 2 & λ² 1 & 3 & λ vmatrix$$\Delta = \begin{vmatrix} 1 & 1 & 1 \\ 1 & 2 & \lambda^2 \\ 1 & 3 & \lambda \end{vmatrix}$$
Expanding along the first row:
Δ = 1(2λ - 3λ²) - 1(λ - λ²) + 1(3 - 2)$$\Delta = 1(2\lambda - 3\lambda^2) - 1(\lambda - \lambda^2) + 1(3 - 2)$$
Δ = 2λ - 3λ² - λ + λ² + 1$$\Delta = 2\lambda - 3\lambda^2 - \lambda + \lambda^2 + 1$$
Δ = -2λ² + λ + 1 = 0$$\Delta = -2\lambda^2 + \lambda + 1 = 0$$
⇒ 2λ² - λ - 1 = 0$$\Rightarrow 2\lambda^2 - \lambda - 1 = 0$$
⇒ (2λ + 1)(λ - 1) = 0 ⇒ λ = 1, -(1)/(2)$$\Rightarrow (2\lambda + 1)(\lambda - 1) = 0 \Rightarrow \lambda = 1, -\frac{1}{2}$$
Step 1: Analyzing Cases
For infinite solutions or inconsistency, we check Δz$\Delta_z$ (or others):
Δz = vmatrix 1 & 1 & 5 1 & 2 & 9 1 & 3 & μ vmatrix = 0$$\Delta_z = \begin{vmatrix} 1 & 1 & 5 \\ 1 & 2 & 9 \\ 1 & 3 & \mu \end{vmatrix} = 0$$
Evaluating the determinant:
1(2μ - 27) - 1(μ - 9) + 5(3 - 2) = 0$$1(2\mu - 27) - 1(\mu - 9) + 5(3 - 2) = 0$$
2μ - 27 - μ + 9 + 5 = 0 ⇒ μ - 13 = 0 ⇒ μ = 13$$2\mu - 27 - \mu + 9 + 5 = 0 \Rightarrow \mu - 13 = 0 \Rightarrow \mu = 13$$
Thus, if λ = 1$\lambda = 1$ and μ = 13$\mu = 13$, the system has an infinite number of solutions.
If λ = 1$\lambda = 1$ and μ ≠ 13$\mu \neq 13$, the system is inconsistent.
Step 2: Checking Options
Option 1 is correct (True statement).
Option 2 is correct (True statement).
Option 3: If λ ≠ 1$\lambda \neq 1$ and λ ≠ -(1)/(2)$\lambda \neq -\frac{1}{2}$, the system has a unique solution (which is consistent) regardless of μ$\mu$. If λ = -(1)/(2)$\lambda = -\frac{1}{2}$ and μ = 13$\mu = 13$, it may have infinite solutions or be inconsistent, but typically it is consistent for unique solutions if λ ≠ 1$\lambda \neq 1$ and λ ≠ -1/2$\lambda \neq -1/2$. Wait, the question option 4 says "System has unique solution if λ ≠ 1$\lambda \neq 1$ and μ ≠ 13$\mu \neq 13$". This is NOT correct because if λ = -(1)/(2)$\lambda = -\frac{1}{2}$ and μ ≠ 13$\mu \neq 13$, the determinant is zero and it does not have a unique solution.
Pattern Recognition
Look for the roots of Δ = 0$\Delta = 0$. Since there are two roots (λ=1, -1/2$\lambda=1, -1/2$), a blanket statement about "unique solution if λ ≠ 1$\lambda \neq 1$" ignores the other root and is inherently false.
Chapter Mix
Class 12 Maths: Matrices and Determinants