Question

Prove that the equation kx + y + z = 1 , x+ ky + z = k and x + y + kz = k²​ will have unique solution when k​ ​​≠ -2. Solve the equations in this case

Answer 1

Dear student
$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ \mathrm{D}=\left|\begin{array}{ccc}\mathrm{k}& 1& 1\\ 1& \mathrm{k}& 1\\ 1& 1& \mathrm{k}\end{array}\right| \\ =\mathrm{k}\left({\mathrm{k}}^2-1\right)-1\left(\mathrm{k}-1\right)+1\left(1-\mathrm{k}\right) \\ ={\mathrm{k}}^3-\mathrm{k}-\mathrm{k}+1+1-\mathrm{k} \\ ={\mathrm{k}}^3-3\mathrm{k}+2 \\ =\left(\mathrm{k}-1\right)\left({\mathrm{k}}^2+\mathrm{k}-2\right) \\ =\left(\mathrm{k}-1\right)\left(\mathrm{k}-1\right)\left(\mathrm{k}+2\right) \\ \mathrm{Since}\mathrm{k}\ne -2.\mathrm{So}\mathrm{k}=1 \\ \mathrm{So},\mathrm{our}\mathrm{equation}\mathrm{will}\mathrm{become} \\ \mathrm{x}+\mathrm{y}+\mathrm{z}=1,\mathrm{x}+\mathrm{y}+\mathrm{z}=1\mathrm{and}\mathrm{x}+\mathrm{y}+\mathrm{z}=1 \end{gathered}$$
Regards