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Question Number 81876 by zainal tanjung last updated on 16/Feb/20

The system of linear equations x+y+z=2,  2x+y−z=3, 3x+2y+kz=4 has a unique  solution if

$$\mathrm{The}\:\mathrm{system}\:\mathrm{of}\:\mathrm{linear}\:\mathrm{equations}\:{x}+{y}+{z}=\mathrm{2}, \\ $$$$\mathrm{2}{x}+{y}−{z}=\mathrm{3},\:\mathrm{3}{x}+\mathrm{2}{y}+{kz}=\mathrm{4}\:\mathrm{has}\:\mathrm{a}\:\mathrm{unique} \\ $$$$\mathrm{solution}\:\mathrm{if} \\ $$

Commented by Tony Lin last updated on 16/Feb/20

Δ= determinant ((1,1,1),(2,1,(−1)),(3,2,k))  =k−3+4−3+2−2k  =−k≠0  ⇒k≠0  Δ_x = determinant ((2,1,1),(3,1,(−1)),(4,2,0))  =6−4−4+4=2≠0  ⇒when k=0, the equation has no solution  k≠0, the equation has a unique solution

$$\Delta=\begin{vmatrix}{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}\\{\mathrm{2}}&{\mathrm{1}}&{−\mathrm{1}}\\{\mathrm{3}}&{\mathrm{2}}&{{k}}\end{vmatrix} \\ $$$$={k}−\mathrm{3}+\mathrm{4}−\mathrm{3}+\mathrm{2}−\mathrm{2}{k} \\ $$$$=−{k}\neq\mathrm{0} \\ $$$$\Rightarrow{k}\neq\mathrm{0} \\ $$$$\Delta_{{x}} =\begin{vmatrix}{\mathrm{2}}&{\mathrm{1}}&{\mathrm{1}}\\{\mathrm{3}}&{\mathrm{1}}&{−\mathrm{1}}\\{\mathrm{4}}&{\mathrm{2}}&{\mathrm{0}}\end{vmatrix} \\ $$$$=\mathrm{6}−\mathrm{4}−\mathrm{4}+\mathrm{4}=\mathrm{2}\neq\mathrm{0} \\ $$$$\Rightarrow{when}\:{k}=\mathrm{0},\:{the}\:{equation}\:{has}\:{no}\:{solution} \\ $$$${k}\neq\mathrm{0},\:{the}\:{equation}\:{has}\:{a}\:{unique}\:{solution} \\ $$

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