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Question Number 182900 by greougoury555 last updated on 16/Dec/22

 Determine the values of b so that   the system of linear equations    { ((x+2y+z=1)),((2x+by+2z=2)),((4x+8y+b^2  z=2b)) :}   has (a) no solution    (b) a unique solution   (c) infinitely many solutions

$$\:{Determine}\:{the}\:{values}\:{of}\:{b}\:{so}\:{that} \\ $$$$\:{the}\:{system}\:{of}\:{linear}\:{equations} \\ $$$$\:\begin{cases}{{x}+\mathrm{2}{y}+{z}=\mathrm{1}}\\{\mathrm{2}{x}+{by}+\mathrm{2}{z}=\mathrm{2}}\\{\mathrm{4}{x}+\mathrm{8}{y}+{b}^{\mathrm{2}} \:{z}=\mathrm{2}{b}}\end{cases} \\ $$$$\:{has}\:\left({a}\right)\:{no}\:{solution}\: \\ $$$$\:\left({b}\right)\:{a}\:{unique}\:{solution} \\ $$$$\:\left({c}\right)\:{infinitely}\:{many}\:{solutions} \\ $$

Answered by cortano1 last updated on 16/Dec/22

(a)  (((1       2            1             1)),((0    b−4        0             0)),((0       0       b^2 −4    2b−4)) )    { ((x=1−2y−z)),(((b−4)y = 0)),(((b^2 −4)z = 2b−4)) :}    has no solution if  { ((b^2 −4 =0)),((2b−4 ≠ 0)) :}   i.e b =− 2

$$\left({a}\right)\:\begin{pmatrix}{\mathrm{1}\:\:\:\:\:\:\:\mathrm{2}\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}}\\{\mathrm{0}\:\:\:\:{b}−\mathrm{4}\:\:\:\:\:\:\:\:\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{0}}\\{\mathrm{0}\:\:\:\:\:\:\:\mathrm{0}\:\:\:\:\:\:\:{b}^{\mathrm{2}} −\mathrm{4}\:\:\:\:\mathrm{2}{b}−\mathrm{4}}\end{pmatrix} \\ $$$$\:\begin{cases}{{x}=\mathrm{1}−\mathrm{2}{y}−{z}}\\{\left({b}−\mathrm{4}\right){y}\:=\:\mathrm{0}}\\{\left({b}^{\mathrm{2}} −\mathrm{4}\right){z}\:=\:\mathrm{2}{b}−\mathrm{4}}\end{cases} \\ $$$$\:\:{has}\:{no}\:{solution}\:{if}\:\begin{cases}{{b}^{\mathrm{2}} −\mathrm{4}\:=\mathrm{0}}\\{\mathrm{2}{b}−\mathrm{4}\:\neq\:\mathrm{0}}\end{cases} \\ $$$$\:{i}.{e}\:{b}\:=−\:\mathrm{2} \\ $$

Commented by manolex last updated on 16/Dec/22

analizando[resultados}de sir Cortano  b=4      ((1,2,1,1),(0,0,0,0),(0,0,(12),4) )  12z=4⇒z=3  x+2y+3=4  y=t  x=1−2t  solution   b=4⇒(1−2t;t;3) infinite solution  b=2   ((1,2,1,1),(0,(−2),0,0),(0,0,0,0) )  −2y=0 ⇒y=0  x+z=1  z=t  x=1−t  solution  b=2⇒(1−t;0;t)infinite solution  b=−2     ((1,2,1,1),(0,(−6),0,0),(0,0,0,8) )  0=8⇒no solution  result     b=−2   no solution  b≠2,−2,4   ((1,2,1,1),(0,1,0,0),(0,0,(b+2),2) )  (b+2)z=2⇒z=(2/(b+2))  y=0  x+2y+z=1  x=1−(2/(b+2))  result     (1−(2/(b+2));0;(2/(b+2)))  unique solution  conclusion   determinant ((b,x,y,z,(conclusion)),(4,(1−2t),t,3,(infinite solution)),(2,(1−t),0,t,(infinite solution)),((−2),,,(0=8),(no solution)),((≠2,−2,4),(1−(2/(b+2))),0,(2/(b+2)),(unique solution)))

$${analizando}\left[{resultados}\right\}{de}\:{sir}\:{Cortano} \\ $$$${b}=\mathrm{4}\:\:\:\:\:\begin{pmatrix}{\mathrm{1}}&{\mathrm{2}}&{\mathrm{1}}&{\mathrm{1}}\\{\mathrm{0}}&{\mathrm{0}}&{\mathrm{0}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{0}}&{\mathrm{12}}&{\mathrm{4}}\end{pmatrix} \\ $$$$\mathrm{12}{z}=\mathrm{4}\Rightarrow{z}=\mathrm{3} \\ $$$${x}+\mathrm{2}{y}+\mathrm{3}=\mathrm{4} \\ $$$${y}={t} \\ $$$${x}=\mathrm{1}−\mathrm{2}{t} \\ $$$${solution}\:\:\:{b}=\mathrm{4}\Rightarrow\left(\mathrm{1}−\mathrm{2}{t};{t};\mathrm{3}\right)\:{infinite}\:{solution} \\ $$$${b}=\mathrm{2} \\ $$$$\begin{pmatrix}{\mathrm{1}}&{\mathrm{2}}&{\mathrm{1}}&{\mathrm{1}}\\{\mathrm{0}}&{−\mathrm{2}}&{\mathrm{0}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{0}}&{\mathrm{0}}&{\mathrm{0}}\end{pmatrix} \\ $$$$−\mathrm{2}{y}=\mathrm{0}\:\Rightarrow{y}=\mathrm{0} \\ $$$${x}+{z}=\mathrm{1} \\ $$$${z}={t} \\ $$$${x}=\mathrm{1}−{t} \\ $$$${solution}\:\:{b}=\mathrm{2}\Rightarrow\left(\mathrm{1}−{t};\mathrm{0};{t}\right){infinite}\:{solution} \\ $$$${b}=−\mathrm{2} \\ $$$$ \\ $$$$\begin{pmatrix}{\mathrm{1}}&{\mathrm{2}}&{\mathrm{1}}&{\mathrm{1}}\\{\mathrm{0}}&{−\mathrm{6}}&{\mathrm{0}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{0}}&{\mathrm{0}}&{\mathrm{8}}\end{pmatrix} \\ $$$$\mathrm{0}=\mathrm{8}\Rightarrow{no}\:{solution} \\ $$$${result}\:\:\:\:\:{b}=−\mathrm{2}\:\:\:{no}\:{solution} \\ $$$${b}\neq\mathrm{2},−\mathrm{2},\mathrm{4} \\ $$$$\begin{pmatrix}{\mathrm{1}}&{\mathrm{2}}&{\mathrm{1}}&{\mathrm{1}}\\{\mathrm{0}}&{\mathrm{1}}&{\mathrm{0}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{0}}&{{b}+\mathrm{2}}&{\mathrm{2}}\end{pmatrix} \\ $$$$\left({b}+\mathrm{2}\right){z}=\mathrm{2}\Rightarrow{z}=\frac{\mathrm{2}}{{b}+\mathrm{2}} \\ $$$${y}=\mathrm{0} \\ $$$${x}+\mathrm{2}{y}+{z}=\mathrm{1} \\ $$$${x}=\mathrm{1}−\frac{\mathrm{2}}{{b}+\mathrm{2}} \\ $$$${result}\:\:\:\:\:\left(\mathrm{1}−\frac{\mathrm{2}}{{b}+\mathrm{2}};\mathrm{0};\frac{\mathrm{2}}{{b}+\mathrm{2}}\right)\:\:{unique}\:{solution} \\ $$$${conclusion} \\ $$$$\begin{array}{|c|c|c|c|c|}{{b}}&\hline{{x}}&\hline{{y}}&\hline{{z}}&\hline{{conclusion}}\\{\mathrm{4}}&\hline{\mathrm{1}−\mathrm{2}{t}}&\hline{{t}}&\hline{\mathrm{3}}&\hline{{infinite}\:{solution}}\\{\mathrm{2}}&\hline{\mathrm{1}−{t}}&\hline{\mathrm{0}}&\hline{{t}}&\hline{{infinite}\:{solution}}\\{−\mathrm{2}}&\hline{}&\hline{}&\hline{\mathrm{0}=\mathrm{8}}&\hline{{no}\:{solution}}\\{\neq\mathrm{2},−\mathrm{2},\mathrm{4}}&\hline{\mathrm{1}−\frac{\mathrm{2}}{{b}+\mathrm{2}}}&\hline{\mathrm{0}}&\hline{\frac{\mathrm{2}}{{b}+\mathrm{2}}}&\hline{{unique}\:{solution}}\\\hline\end{array} \\ $$$$ \\ $$

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