Consider the system of linear equation x + y + z = 4mu$x + y + z = 4\mu$, x + 2y + 2lambda z = 10mu$x + 2y + 2\lambda z = 10\mu$, x + 3y + 4lambda^2 z = mu^2 + 15$x + 3y + 4\lambda^2 z = \mu^2 + 15$, where lambda, mu in mathbbR$\lambda, \mu \in \mathbb{R}$. Which one of the following statements is NOT correct?

### Related Formula Delta = beginvmatrix a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 endvmatrix$$\Delta = \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix}$$ If Delta neq 0$\Delta \neq 0$, unique solution. If Delta = 0$\Delta = 0$ and Delta_x = Delta_y = Delta_z = 0$\Delta_x = \Delta_y = \Delta_z = 0$, infinitely many solutions. If Delta = 0$\Delta = 0$ and at least one of Delta_x, Delta_y, Delta_z neq 0$\Delta_x, \Delta_y, \Delta_z \neq 0$, inconsistent (no solution). ### Core Logic Given system: x + y + z = 4mu$x + y + z = 4\mu$ x + 2y + 2lambda z = 10mu$x + 2y + 2\lambda z = 10\mu$ x + 3y + 4lambda^2 z = mu^2 + 15$x + 3y + 4\lambda^2 z = \mu^2 + 15$ Compute the main determinant Delta$\Delta$: Delta = beginvmatrix 1 & 1 & 1 \\ 1 & 2 & 2lambda \\ 1 & 3 & 4lambda^2 endvmatrix$$\Delta = \begin{vmatrix} 1 & 1 & 1 \\ 1 & 2 & 2\lambda \\ 1 & 3 & 4\lambda^2 \end{vmatrix}$$ Apply operations: R_2 to R_2 - R_1$R_2 \to R_2 - R_1$, R_3 to R_3 - R_1$R_3 \to R_3 - R_1$ Delta = beginvmatrix 1 & 1 & 1 \\ 0 & 1 & 2lambda-1 \\ 0 & 2 & 4lambda^2-1 endvmatrix = 1 cdot (4lambda^2 - 1 - 2(2lambda - 1))$$\Delta = \begin{vmatrix} 1 & 1 & 1 \\ 0 & 1 & 2\lambda-1 \\ 0 & 2 & 4\lambda^2-1 \end{vmatrix} = 1 \cdot (4\lambda^2 - 1 - 2(2\lambda - 1))$$ = 4lambda^2 - 1 - 4lambda + 2 = 4lambda^2 - 4lambda + 1 = (2lambda - 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, Delta neq 0 Rightarrow 2lambda - 1 neq 0 Rightarrow lambda neq frac12$\Delta \neq 0 \Rightarrow 2\lambda - 1 \neq 0 \Rightarrow \lambda \neq \frac{1}{2}$. Note: For unique solution, mu$\mu$ can be anything. Option (4) states the system is consistent if lambda neq frac12$\lambda \neq \frac{1}{2}$, which is purely correct. Option (1) says "unique solution if lambda neq frac12$\lambda \neq \frac{1}{2}$ and mu neq 1, 15$\mu \neq 1, 15$". While true that it has a unique solution under those conditions, it also has a unique solution for mu = 1, 15$\mu = 1, 15$. Let's check consistency conditions. ### Step 2: Checking Delta components Let Delta = 0$\Delta = 0$, so lambda = frac12$\lambda = \frac{1}{2}$. Compute Delta_x$\Delta_x$ and Delta_z$\Delta_z$ (or Delta_y$\Delta_y$): Wait, substituting lambda = 1/2$\lambda = 1/2$, the equations become: x + y + z = 4mu$x + y + z = 4\mu$ x + 2y + z = 10mu$x + 2y + z = 10\mu$ x + 3y + z = mu^2 + 15$x + 3y + z = \mu^2 + 15$ From the first two, (x+2y+z) - (x+y+z) = 10mu - 4mu Rightarrow y = 6mu$(x+2y+z) - (x+y+z) = 10\mu - 4\mu \Rightarrow y = 6\mu$. From the second and third, (x+3y+z) - (x+2y+z) = mu^2 + 15 - 10mu Rightarrow y = mu^2 - 10mu + 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 Delta = 0$\Delta = 0$), the two values of y$y$ must match: 6mu = mu^2 - 10mu + 15$$6\mu = \mu^2 - 10\mu + 15$$ mu^2 - 16mu + 15 = 0$$\mu^2 - 16\mu + 15 = 0$$ (mu - 1)(mu - 15) = 0$$(\mu - 1)(\mu - 15) = 0$$ So, if lambda = frac12$\lambda = \frac{1}{2}$, the system is consistent (infinite solutions) ONLY when mu = 1$\mu = 1$ or mu = 15$\mu = 15$. If lambda = frac12$\lambda = \frac{1}{2}$ and mu neq 1, 15$\mu \neq 1, 15$, it is inconsistent. ### Step 3: Checking Options Option (2) states: "The system is inconsistent if lambda = frac12$\lambda = \frac{1}{2}$ and mu neq 1$\mu \neq 1$". If mu = 15$\mu = 15$ (which is neq 1$\neq 1$), the system is actually CONSISTENT (infinite solutions). Therefore, Option (2) is NOT strictly correct because mu=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 3times3$3\times3$ determinants. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Determinants