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Question Number 88872 by Zainal Arifin last updated on 13/Apr/20

The system of equations kx+y+z=1, x+ky+z=k and x+y+kz=k^2 have no solution if k equals

$$\mathrm{The}\:\mathrm{system}\:\mathrm{of}\:\mathrm{equations}\:\:\:{kx}+{y}+{z}=\mathrm{1}, \\ $$$${x}+{ky}+{z}={k}\:\:\:\mathrm{and}\:\:\:{x}+{y}+{kz}={k}^{\mathrm{2}} \:\mathrm{have} \\ $$$$\mathrm{no}\:\mathrm{solution}\:\mathrm{if}\:\:\:{k}\:\:\mathrm{equals} \\ $$

Commented by john santu last updated on 13/Apr/20

⇒k determinant (((k 1)),((1 k)))− determinant (((1 1)),((1 k)))+ determinant (((1 k)),((1 1)))=0 k(k^2 −1)−(k−1)+1−k = 0 k^3 −k−k+1+1−k = 0 k^3 −3k+2 = 0 (k−1)(k^2 +k−2) =0 (k−1)(k+2)(k−1) = 0 k = { (1),((−2)) :}

$$\Rightarrow{k}\:\begin{vmatrix}{{k}\:\:\:\:\mathrm{1}}\\{\mathrm{1}\:\:\:\:{k}}\end{vmatrix}−\begin{vmatrix}{\mathrm{1}\:\:\:\mathrm{1}}\\{\mathrm{1}\:\:\:{k}}\end{vmatrix}+\begin{vmatrix}{\mathrm{1}\:\:\:\:{k}}\\{\mathrm{1}\:\:\:\:\mathrm{1}}\end{vmatrix}=\mathrm{0} \\ $$$${k}\left({k}^{\mathrm{2}} −\mathrm{1}\right)−\left({k}−\mathrm{1}\right)+\mathrm{1}−{k}\:=\:\mathrm{0} \\ $$$${k}^{\mathrm{3}} −{k}−{k}+\mathrm{1}+\mathrm{1}−{k}\:=\:\mathrm{0} \\ $$$${k}^{\mathrm{3}} \:−\mathrm{3}{k}+\mathrm{2}\:=\:\mathrm{0}\: \\ $$$$\left({k}−\mathrm{1}\right)\left({k}^{\mathrm{2}} +{k}−\mathrm{2}\right)\:=\mathrm{0} \\ $$$$\left({k}−\mathrm{1}\right)\left({k}+\mathrm{2}\right)\left({k}−\mathrm{1}\right)\:=\:\mathrm{0} \\ $$$${k}\:=\:\begin{cases}{\mathrm{1}}\\{−\mathrm{2}}\end{cases} \\ $$

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