Related Formula
Δ = vmatrix a₁ & b₁ & c₁ a₂ & b₂ & c₂ a₃ & b₃ & c₃ vmatrix$$\Delta = \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix}$$
If Δ ≠ 0$\Delta \neq 0$, unique solution.
If Δ = 0$\Delta = 0$ and Δₓ = Δy = Δz = 0$\Delta_x = \Delta_y = \Delta_z = 0$, infinitely many solutions.
If Δ = 0$\Delta = 0$ and at least one of Δₓ, Δy, Δz ≠ 0$\Delta_x, \Delta_y, \Delta_z \neq 0$, inconsistent (no solution).
Core Logic
Given system:
x + y + z = 4μ$x + y + z = 4\mu$
x + 2y + 2λ z = 10μ$x + 2y + 2\lambda z = 10\mu$
x + 3y + 4λ² z = μ² + 15$x + 3y + 4\lambda^2 z = \mu^2 + 15$
Compute the main determinant Δ$\Delta$:
Δ = vmatrix 1 & 1 & 1 1 & 2 & 2λ 1 & 3 & 4λ² vmatrix$$\Delta = \begin{vmatrix} 1 & 1 & 1 \\ 1 & 2 & 2\lambda \\ 1 & 3 & 4\lambda^2 \end{vmatrix}$$
Apply operations: R₂ → R₂ - R₁$R_2 \to R_2 - R_1$, R₃ → R₃ - R₁$R_3 \to R_3 - R_1$
Δ = vmatrix 1 & 1 & 1 0 & 1 & 2λ-1 0 & 2 & 4λ²-1 vmatrix = 1 · (4λ² - 1 - 2(2λ - 1))$$\Delta = \begin{vmatrix} 1 & 1 & 1 \\ 0 & 1 & 2\lambda-1 \\ 0 & 2 & 4\lambda^2-1 \end{vmatrix} = 1 \cdot (4\lambda^2 - 1 - 2(2\lambda - 1))$$
= 4λ² - 1 - 4λ + 2 = 4λ² - 4λ + 1 = (2λ - 1)²$$= 4\lambda^2 - 1 - 4\lambda + 2 = 4\lambda^2 - 4\lambda + 1 = (2\lambda - 1)^2$$
Step 1: Analyzing Unique Solution
For a unique solution, Δ ≠ 0 ⇒ 2λ - 1 ≠ 0 ⇒ λ ≠ (1)/(2)$\Delta \neq 0 \Rightarrow 2\lambda - 1 \neq 0 \Rightarrow \lambda \neq \frac{1}{2}$.
Note: For unique solution, μ$\mu$ can be anything. Option (4) states the system is consistent if λ ≠ (1)/(2)$\lambda \neq \frac{1}{2}$, which is purely correct. Option (1) says "unique solution if λ ≠ (1)/(2)$\lambda \neq \frac{1}{2}$ and μ ≠ 1, 15$\mu \neq 1, 15$". While true that it has a unique solution under those conditions, it also has a unique solution for μ = 1, 15$\mu = 1, 15$. Let's check consistency conditions.
Step 2: Checking Delta components
Let Δ = 0$\Delta = 0$, so λ = (1)/(2)$\lambda = \frac{1}{2}$.
Compute Δₓ$\Delta_x$ and Δz$\Delta_z$ (or Δy$\Delta_y$):
Wait, substituting λ = 1/2$\lambda = 1/2$, the equations become:
x + y + z = 4μ$x + y + z = 4\mu$
x + 2y + z = 10μ$x + 2y + z = 10\mu$
x + 3y + z = μ² + 15$x + 3y + z = \mu^2 + 15$
From the first two, (x+2y+z) - (x+y+z) = 10μ - 4μ ⇒ y = 6μ$(x+2y+z) - (x+y+z) = 10\mu - 4\mu \Rightarrow y = 6\mu$.
From the second and third, (x+3y+z) - (x+2y+z) = μ² + 15 - 10μ ⇒ y = μ² - 10μ + 15$(x+3y+z) - (x+2y+z) = \mu^2 + 15 - 10\mu \Rightarrow y = \mu^2 - 10\mu + 15$.
For the system to be consistent (infinite solutions since Δ = 0$\Delta = 0$), the two values of y$y$ must match:
6μ = μ² - 10μ + 15$$6\mu = \mu^2 - 10\mu + 15$$
μ² - 16μ + 15 = 0$$\mu^2 - 16\mu + 15 = 0$$
(μ - 1)(μ - 15) = 0$$(\mu - 1)(\mu - 15) = 0$$
So, if λ = (1)/(2)$\lambda = \frac{1}{2}$, the system is consistent (infinite solutions) ONLY when μ = 1$\mu = 1$ or μ = 15$\mu = 15$.
If λ = (1)/(2)$\lambda = \frac{1}{2}$ and μ ≠ 1, 15$\mu \neq 1, 15$, it is inconsistent.
Step 3: Checking Options
Option (2) states: "The system is inconsistent if λ = (1)/(2)$\lambda = \frac{1}{2}$ and μ ≠ 1$\mu \neq 1$".
If μ = 15$\mu = 15$ (which is ≠ 1$\neq 1$), the system is actually CONSISTENT (infinite solutions). Therefore, Option (2) is NOT strictly correct because μ=15$\mu=15$ makes it consistent.
Thus, statement (2) is the incorrect statement.
Pattern Recognition
Cramer's rule dependencies can be quickly identified using algebraic elimination. When variables align symmetrically (like z$z$ mapping identically), subtracting equations exposes the consistency constraint directly without resolving full 3×3$3\times3$ determinants.
Chapter Mix
Class 12 Maths: Determinants
