Menu Close

b-x-y-Kz-2-3x-4y-2z-K-2x-3y-z-1-

Tinku Tara 0 Comments
Pin




Question Number 180683 by Mastermind last updated on 15/Nov/22
(b) x + y + Kz = 2 3x + 4y + 2z = K 2x + 3y − z = 1
$$\left(\mathrm{b}\right)\:\mathrm{x}\:+\:\mathrm{y}\:+\:\mathrm{Kz}\:=\:\mathrm{2} \\ $$$$\:\:\:\:\mathrm{3x}\:+\:\mathrm{4y}\:+\:\mathrm{2z}\:=\:\mathrm{K} \\ $$$$\:\:\:\:\mathrm{2x}\:+\:\mathrm{3y}\:−\:\mathrm{z}\:=\:\mathrm{1} \\ $$
Answered by manxsol last updated on 15/Nov/22
{ (1,1,k,2),(3,4,2,k),(2,3,(−1),1) :} f_1 .(−3)+f_2 f_1 .(−2)+f_3 { (1,1,( k),( 2)),(0,1,(−3k+2),( k−6)),(0,1,(−1−2k),(−3)) :} f_2 .(−1)+f_3 { (1,1,( k),( 2)),(0,1,(−3k+2),(k−6)),(0,0,(−3+k),(3−k)) :} k≠3;(2/3) z=((3−k)/(−3+k))=−1 y+(−3k+2)(−1)=k−6 y=4k−4 x+(4k−4)+k(−1)=2 x=−3k+6 c.s=(−3k+6;4k−4;−1) si k=3 { (1,1,3,2),(0,1,(−7),(−3)),(0,0,0,0) :} 0+0+0=0 compatible indeterminado z=t y−7t=−3 y=7t−3 x=2−3t−7t+3 x=5−10t C.S={5−10t;7t−3;t} si k=(2/3) { (1,1,(2/3),( 2)),(0,1,( 0),((−16)/( 3))),(0,0,((−7)/( 3)),( (7/3))) :} { (3,3,( 2),( 6)),(0,3,( 0),(−16)),(0,0,(−7),( 7)) :} z=(7/(−7))=−1 y=((−16)/3) 3x+3(((−16)/3))+2(−1)=6 3x=24 x=3 CS={3;−((16)/3);−1} resumen determinant ((k,x,y,z,),((≠(2/3);3),(−3k+6),(4k−4),(−1),),(3,(5−10t),(7t−3),t,(2eq,3var 0+0+0=0 compatible indeterminado)),((2/3),3,(−((16)/3)),(−1),))
$$\begin{cases}{\mathrm{1}}&{\mathrm{1}}&{{k}}&{\mathrm{2}}\\{\mathrm{3}}&{\mathrm{4}}&{\mathrm{2}}&{{k}}\\{\mathrm{2}}&{\mathrm{3}}&{−\mathrm{1}}&{\mathrm{1}}\end{cases} \\ $$$${f}_{\mathrm{1}} .\left(−\mathrm{3}\right)+{f}_{\mathrm{2}} \\ $$$${f}_{\mathrm{1}} .\left(−\mathrm{2}\right)+{f}_{\mathrm{3}} \\ $$$$\begin{cases}{\mathrm{1}}&{\mathrm{1}}&{\:\:\:\:\:\:\:{k}}&{\:\:\:\:\mathrm{2}}\\{\mathrm{0}}&{\mathrm{1}}&{−\mathrm{3}{k}+\mathrm{2}}&{\:\:\:\:{k}−\mathrm{6}}\\{\mathrm{0}}&{\mathrm{1}}&{−\mathrm{1}−\mathrm{2}{k}}&{−\mathrm{3}}\end{cases} \\ $$$${f}_{\mathrm{2}} .\left(−\mathrm{1}\right)+{f}_{\mathrm{3}} \\ $$$$\begin{cases}{\mathrm{1}}&{\mathrm{1}}&{\:\:\:\:\:\:{k}}&{\:\mathrm{2}}\\{\mathrm{0}}&{\mathrm{1}}&{−\mathrm{3}{k}+\mathrm{2}}&{{k}−\mathrm{6}}\\{\mathrm{0}}&{\mathrm{0}}&{−\mathrm{3}+{k}}&{\mathrm{3}−{k}}\end{cases} \\ $$$${k}\neq\mathrm{3};\frac{\mathrm{2}}{\mathrm{3}} \\ $$$${z}=\frac{\mathrm{3}−{k}}{−\mathrm{3}+{k}}=−\mathrm{1} \\ $$$${y}+\left(−\mathrm{3}{k}+\mathrm{2}\right)\left(−\mathrm{1}\right)={k}−\mathrm{6} \\ $$$${y}=\mathrm{4}{k}−\mathrm{4} \\ $$$${x}+\left(\mathrm{4}{k}−\mathrm{4}\right)+{k}\left(−\mathrm{1}\right)=\mathrm{2} \\ $$$${x}=−\mathrm{3}{k}+\mathrm{6} \\ $$$${c}.{s}=\left(−\mathrm{3}{k}+\mathrm{6};\mathrm{4}{k}−\mathrm{4};−\mathrm{1}\right) \\ $$$$ \\ $$$${si}\:{k}=\mathrm{3} \\ $$$$\begin{cases}{\mathrm{1}}&{\mathrm{1}}&{\mathrm{3}}&{\mathrm{2}}\\{\mathrm{0}}&{\mathrm{1}}&{−\mathrm{7}}&{−\mathrm{3}}\\{\mathrm{0}}&{\mathrm{0}}&{\mathrm{0}}&{\mathrm{0}}\end{cases} \\ $$$$\mathrm{0}+\mathrm{0}+\mathrm{0}=\mathrm{0}\:\:\:{compatible}\:{indeterminado} \\ $$$${z}={t} \\ $$$${y}−\mathrm{7}{t}=−\mathrm{3} \\ $$$${y}=\mathrm{7}{t}−\mathrm{3} \\ $$$${x}=\mathrm{2}−\mathrm{3}{t}−\mathrm{7}{t}+\mathrm{3} \\ $$$${x}=\mathrm{5}−\mathrm{10}{t} \\ $$$${C}.{S}=\left\{\mathrm{5}−\mathrm{10}{t};\mathrm{7}{t}−\mathrm{3};{t}\right\} \\ $$$${si}\:{k}=\frac{\mathrm{2}}{\mathrm{3}} \\ $$$$\begin{cases}{\mathrm{1}}&{\mathrm{1}}&{\frac{\mathrm{2}}{\mathrm{3}}}&{\:\:\:\:\:\:\mathrm{2}}\\{\mathrm{0}}&{\mathrm{1}}&{\:\:\:\:\mathrm{0}}&{\frac{−\mathrm{16}}{\:\:\:\:\:\:\mathrm{3}}}\\{\mathrm{0}}&{\mathrm{0}}&{\frac{−\mathrm{7}}{\:\:\:\mathrm{3}}}&{\:\:\:\:\frac{\mathrm{7}}{\mathrm{3}}}\end{cases} \\ $$$$\begin{cases}{\mathrm{3}}&{\mathrm{3}}&{\:\:\:\:\mathrm{2}}&{\:\:\:\:\:\:\:\mathrm{6}}\\{\mathrm{0}}&{\mathrm{3}}&{\:\:\:\:\mathrm{0}}&{−\mathrm{16}}\\{\mathrm{0}}&{\mathrm{0}}&{−\mathrm{7}}&{\:\:\:\:\:\:\mathrm{7}}\end{cases} \\ $$$${z}=\frac{\mathrm{7}}{−\mathrm{7}}=−\mathrm{1} \\ $$$${y}=\frac{−\mathrm{16}}{\mathrm{3}} \\ $$$$\mathrm{3}{x}+\mathrm{3}\left(\frac{−\mathrm{16}}{\mathrm{3}}\right)+\mathrm{2}\left(−\mathrm{1}\right)=\mathrm{6} \\ $$$$\mathrm{3}{x}=\mathrm{24} \\ $$$${x}=\mathrm{3} \\ $$$$\:\: \\ $$$${CS}=\left\{\mathrm{3};−\frac{\mathrm{16}}{\mathrm{3}};−\mathrm{1}\right\} \\ $$$$ \\ $$$${resumen} \\ $$$$\begin{array}{|c|c|c|c|}{{k}}&\hline{{x}}&\hline{{y}}&\hline{{z}}&\hline{}\\{\neq\frac{\mathrm{2}}{\mathrm{3}};\mathrm{3}}&\hline{−\mathrm{3}{k}+\mathrm{6}}&\hline{\mathrm{4}{k}−\mathrm{4}}&\hline{−\mathrm{1}}&\hline{}\\{\mathrm{3}}&\hline{\mathrm{5}−\mathrm{10}{t}}&\hline{\mathrm{7}{t}−\mathrm{3}}&\hline{{t}}&\hline{\mathrm{2}{eq},\mathrm{3}{var}\:\:\mathrm{0}+\mathrm{0}+\mathrm{0}=\mathrm{0}\:{compatible}\:{indeterminado}}\\{\frac{\mathrm{2}}{\mathrm{3}}}&\hline{\mathrm{3}}&\hline{−\frac{\mathrm{16}}{\mathrm{3}}}&\hline{−\mathrm{1}}&\hline{}\\\hline\end{array} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *