Q.Discuss the solutions of the following system of equations for all values of $\lambda$. $x + y + z = 2$, $2x + y - 2z = 2$, $\lambda x + y + 4z = 2$

Compute the coefficient determinant Δ (a multiple of λ) and the three numerator determinants to classify the system by Cramer's rule, then solve explicitly in each case.

1. System: $x+y+z=2$, $2x+y-2z=2$, $\lambda x+y+4z=2$.
2. Coefficient determinant: $\Delta=\begin{vmatrix}1&1&1\\2&1&-2\\\lambda&1&4\end{vmatrix}$.
3. Expand along column 1: $\Delta=1[(1)(4)-(-2)(1)]-2[(1)(4)-(1)(1)]+\lambda[(1)(-2)-(1)(1)]=1(6)-2(3)+\lambda(-3)=6-6-3\lambda=-3\lambda$.
4. So $\Delta=-3\lambda$, which vanishes exactly when $\lambda=0$.
5. Case $\lambda\neq0$: $\Delta\neq0$, so by Cramer's rule the system has a unique solution.
6. $\Delta_x$ (replace the $x$-column, which contains $1,2,\lambda$, by the constants $2,2,2$): $\Delta_x=\begin{vmatrix}2&1&1\\2&1&-2\\2&1&4\end{vmatrix}=2[(1)(4)-(-2)(1)]-1[(2)(4)-(-2)(2)]+1[(2)(1)-(1)(2)]=2(6)-1(12)+1(0)=12-12+0=0$. This determinant never contains $\lambda$ (that column was replaced), so $\Delta_x=0$ for every $\lambda$.
7. $\Delta_y$ (replace the $y$-column by constants): $\Delta_y=\begin{vmatrix}1&2&1\\2&2&-2\\\lambda&2&4\end{vmatrix}=1[(2)(4)-(-2)(2)]-2[(2)(4)-(-2)\lambda]+1[(2)(2)-(2)\lambda]=1(12)-2(8+2\lambda)+1(4-2\lambda)=12-16-4\lambda+4-2\lambda=-6\lambda$.
8. $\Delta_z$ (replace the $z$-column by constants): $\Delta_z=\begin{vmatrix}1&1&2\\2&1&2\\\lambda&1&2\end{vmatrix}=1[(1)(2)-(2)(1)]-1[(2)(2)-(2)\lambda]+2[(2)(1)-(1)\lambda]=1(0)-1(4-2\lambda)+2(2-\lambda)=-4+2\lambda+4-2\lambda=0$; this also never survives with $\lambda$, so $\Delta_z=0$ for every $\lambda$.
9. For $\lambda\neq0$: $x=\dfrac{\Delta_x}{\Delta}=\dfrac{0}{-3\lambda}=0$, $y=\dfrac{\Delta_y}{\Delta}=\dfrac{-6\lambda}{-3\lambda}=2$, $z=\dfrac{\Delta_z}{\Delta}=\dfrac{0}{-3\lambda}=0$. So the unique solution is $(x,y,z)=(0,2,0)$ — independent of the particular nonzero $\lambda$ (it satisfies all three equations for any $\lambda$). …