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Question Number 180682 by Mastermind last updated on 15/Nov/22

Determine the value of k such that the following system has (i) a Unique solution (ii) No solution (iii) More than one solution (a) Kx + y + z = 1 x +Ky + z = 1 x + y + Kz = 1

$$\mathrm{Determine}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{k}\:\mathrm{such}\:\mathrm{that}\:\mathrm{the} \\ $$$$\mathrm{following}\:\mathrm{system}\:\mathrm{has}\:\left(\mathrm{i}\right)\:\mathrm{a}\:\mathrm{Unique}\: \\ $$$$\mathrm{solution}\:\left(\mathrm{ii}\right)\:\mathrm{No}\:\mathrm{solution}\:\left(\mathrm{iii}\right)\:\mathrm{More}\:\mathrm{than} \\ $$$$\mathrm{one}\:\mathrm{solution} \\ $$$$ \\ $$$$\left(\mathrm{a}\right)\:\mathrm{Kx}\:+\:\mathrm{y}\:+\:\mathrm{z}\:=\:\mathrm{1} \\ $$$$\:\:\:\:\mathrm{x}\:+\mathrm{Ky}\:+\:\mathrm{z}\:=\:\mathrm{1} \\ $$$$\:\:\:\:\mathrm{x}\:+\:\mathrm{y}\:+\:\mathrm{Kz}\:=\:\mathrm{1} \\ $$

Answered by manolex last updated on 16/Nov/22

((k,1,1,1),(1,k,1,1),(1,1,k,1) ) ((1,1,k,1),(1,k,1,1),(k,1,1,1) ) −f_1 +f_2 −kf_1 +f_3 ((1,1,k,1),(0,(k−1),(1−k),0),(0,(−k+1),(−k^2 +1),(−k+1)) ) f_2 +f_1 ((1,1,k,1),(0,(k−1),(1−k),0),(0,0,(2−k−k^2 ),(−k+1)) ) k≠−2;1 z=(((−k+1))/((k+2)(−k+1)))=(1/(k+2)) (k−1)y+(1−k)[(1/(k+2))]=0 y−[(1/(k+2))]=0⇒y=(1/(k+2)) x+[(1/(k+2))]+k[(1/(k+2))]=1 x=1−((k+1)/(k+2)) x=(1/(k+2)) C.S={(1/(k+2));(1/(k+2));(1/(k+2))} if k=1 ((1,1,1,1),(0,0,0,(0])),(0,0,0,0) ) 1eq,3 incog; 0+0+0=0 compatible indeterminado z=t y=s x=1−t−s C.S={1−t−s;s;t} if k=−2 ((1,1,(−2),1),(0,(−3),3,0),(0,0,0,3) ) 0+0+0=3⇒incompatible .no solution resumen determinant ((k,x,y,z,(comentario)),((≠−2;1),( (1/(k+2))),(1/(k+2)),(1/(k+2)),(all equal)),(( 1),(1−t−s),( s),( t),(infinitas solution)),((−2),,,,(incompatible solution)))

$$\begin{pmatrix}{{k}}&{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}\\{\mathrm{1}}&{{k}}&{\mathrm{1}}&{\mathrm{1}}\\{\mathrm{1}}&{\mathrm{1}}&{{k}}&{\mathrm{1}}\end{pmatrix} \\ $$$$\begin{pmatrix}{\mathrm{1}}&{\mathrm{1}}&{{k}}&{\mathrm{1}}\\{\mathrm{1}}&{{k}}&{\mathrm{1}}&{\mathrm{1}}\\{{k}}&{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}\end{pmatrix} \\ $$$$−{f}_{\mathrm{1}} +{f}_{\mathrm{2}} \\ $$$$−{kf}_{\mathrm{1}} +{f}_{\mathrm{3}} \\ $$$$\begin{pmatrix}{\mathrm{1}}&{\mathrm{1}}&{{k}}&{\mathrm{1}}\\{\mathrm{0}}&{{k}−\mathrm{1}}&{\mathrm{1}−{k}}&{\mathrm{0}}\\{\mathrm{0}}&{−{k}+\mathrm{1}}&{−{k}^{\mathrm{2}} +\mathrm{1}}&{−{k}+\mathrm{1}}\end{pmatrix} \\ $$$${f}_{\mathrm{2}} +{f}_{\mathrm{1}} \\ $$$$\begin{pmatrix}{\mathrm{1}}&{\mathrm{1}}&{{k}}&{\mathrm{1}}\\{\mathrm{0}}&{{k}−\mathrm{1}}&{\mathrm{1}−{k}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{0}}&{\mathrm{2}−{k}−{k}^{\mathrm{2}} }&{−{k}+\mathrm{1}}\end{pmatrix} \\ $$$${k}\neq−\mathrm{2};\mathrm{1} \\ $$$${z}=\frac{\left(−{k}+\mathrm{1}\right)}{\left({k}+\mathrm{2}\right)\left(−{k}+\mathrm{1}\right)}=\frac{\mathrm{1}}{{k}+\mathrm{2}} \\ $$$$\left({k}−\mathrm{1}\right){y}+\left(\mathrm{1}−{k}\right)\left[\frac{\mathrm{1}}{{k}+\mathrm{2}}\right]=\mathrm{0} \\ $$$${y}−\left[\frac{\mathrm{1}}{{k}+\mathrm{2}}\right]=\mathrm{0}\Rightarrow{y}=\frac{\mathrm{1}}{{k}+\mathrm{2}} \\ $$$${x}+\left[\frac{\mathrm{1}}{{k}+\mathrm{2}}\right]+{k}\left[\frac{\mathrm{1}}{{k}+\mathrm{2}}\right]=\mathrm{1} \\ $$$${x}=\mathrm{1}−\frac{{k}+\mathrm{1}}{{k}+\mathrm{2}} \\ $$$${x}=\frac{\mathrm{1}}{{k}+\mathrm{2}} \\ $$$${C}.{S}=\left\{\frac{\mathrm{1}}{{k}+\mathrm{2}};\frac{\mathrm{1}}{{k}+\mathrm{2}};\frac{\mathrm{1}}{{k}+\mathrm{2}}\right\} \\ $$$${if}\:{k}=\mathrm{1} \\ $$$$\begin{pmatrix}{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}\\{\mathrm{0}}&{\mathrm{0}}&{\mathrm{0}}&{\left.\mathrm{0}\right]}\\{\mathrm{0}}&{\mathrm{0}}&{\mathrm{0}}&{\mathrm{0}}\end{pmatrix} \\ $$$$\mathrm{1}{eq},\mathrm{3}\:{incog}; \\ $$$$\mathrm{0}+\mathrm{0}+\mathrm{0}=\mathrm{0}\:{compatible}\:{indeterminado} \\ $$$${z}={t} \\ $$$${y}={s} \\ $$$${x}=\mathrm{1}−{t}−{s} \\ $$$${C}.{S}=\left\{\mathrm{1}−{t}−{s};{s};{t}\right\} \\ $$$${if}\:{k}=−\mathrm{2} \\ $$$$\begin{pmatrix}{\mathrm{1}}&{\mathrm{1}}&{−\mathrm{2}}&{\mathrm{1}}\\{\mathrm{0}}&{−\mathrm{3}}&{\mathrm{3}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{0}}&{\mathrm{0}}&{\mathrm{3}}\end{pmatrix} \\ $$$$\mathrm{0}+\mathrm{0}+\mathrm{0}=\mathrm{3}\Rightarrow{incompatible}\:.{no}\:{solution} \\ $$$${resumen} \\ $$$$\begin{array}{|c|c|c|c|}{{k}}&\hline{{x}}&\hline{{y}}&\hline{{z}}&\hline{{comentario}}\\{\neq−\mathrm{2};\mathrm{1}}&\hline{\:\:\frac{\mathrm{1}}{{k}+\mathrm{2}}}&\hline{\frac{\mathrm{1}}{{k}+\mathrm{2}}}&\hline{\frac{\mathrm{1}}{{k}+\mathrm{2}}}&\hline{{all}\:{equal}}\\{\:\:\:\:\mathrm{1}}&\hline{\mathrm{1}−{t}−{s}}&\hline{\:\:\:{s}}&\hline{\:\:\:{t}}&\hline{{infinitas}\:{solution}}\\{−\mathrm{2}}&\hline{}&\hline{}&\hline{}&\hline{{incompatible}\:{solution}}\\\hline\end{array} \\ $$

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