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for-what-value-of-k-the-system-of-equation-has-no-solution-x-2y-3z-1-2x-ky-5z-1-3x-4y-7z-1-

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Question Number 5963 by Ashis last updated on 07/Jun/16
for what value of k the system of equation has no solution x+2y+3z=1 2x+ky+5z=1 3x+4y+7z=1
$$\boldsymbol{\mathrm{for}}\:\boldsymbol{\mathrm{what}}\:\boldsymbol{\mathrm{value}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{k}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{system}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{equation}}\:\boldsymbol{\mathrm{has}}\:\boldsymbol{\mathrm{no}}\:\boldsymbol{\mathrm{solution}} \\ $$$$\boldsymbol{\mathrm{x}}+\mathrm{2}\boldsymbol{\mathrm{y}}+\mathrm{3}\boldsymbol{\mathrm{z}}=\mathrm{1} \\ $$$$\mathrm{2}\boldsymbol{\mathrm{x}}+\boldsymbol{\mathrm{ky}}+\mathrm{5}\boldsymbol{\mathrm{z}}=\mathrm{1} \\ $$$$\mathrm{3}\boldsymbol{\mathrm{x}}+\mathrm{4}\boldsymbol{\mathrm{y}}+\mathrm{7}\boldsymbol{\mathrm{z}}=\mathrm{1} \\ $$
Commented by Yozzii last updated on 07/Jun/16
k=3?
$${k}=\mathrm{3}? \\ $$
Commented by prakash jain last updated on 07/Jun/16
D= determinant ((1,2,3),(2,k,5),(3,4,7))=0 D= determinant ((1,2,3),(1,(k−2),2),(1,(4−k),2))=0 (R2→R2−R1,R3→R3−R2) D= determinant ((1,2,3),(0,(k−4),(−1)),(0,(2−k),(−1)))=0 (R2→R2−R1,R3→R3−R1) =−(k−4)+(2−k)=0 =−k+4+2+−k=0⇒k=3 This itself does not guarantee that equation has no solution we also need to check if a) at least one of D_1 ,D_2 or D_3 is non zero b) if D_1 ,D_2 ,D_3 are all zero then verify that no solution exist by solving the equation. D_1 = determinant ((1,2,3),(1,3,5),(1,4,7)) D_1 = determinant ((1,2,3),(0,1,2),(0,1,2))=0 D_2 = determinant ((1,1,3),(2,1,5),(3,1,7))= determinant ((1,1,3),(1,0,2),(1,0,2))=0 D_3 = determinant ((1,2,1),(2,3,1),(3,4,1))= determinant ((1,2,1),(1,1,0),(1,1,0))=0 x+2y+3z=1⇒2x+4y+6z=2 2x+3y+5z=1 from the above two y+z=1⇒y=1−z x=1−2y−3z⇒x=1−2(1−z)−3z=−1−5z 3x+4y+7z=1 3(−1−5z)+4(1−z)+7z=1 −3−15z+4−4z+7z=1⇒z=0 y=1,x=−1 check solution (−1,1,0)for k=3 −1+2=1 (eqn 1) −2+3=1 (eqn 2) −3+4=1(eqn 3) k=3 gives unique solution.
$$\mathrm{D}=\begin{vmatrix}{\mathrm{1}}&{\mathrm{2}}&{\mathrm{3}}\\{\mathrm{2}}&{{k}}&{\mathrm{5}}\\{\mathrm{3}}&{\mathrm{4}}&{\mathrm{7}}\end{vmatrix}=\mathrm{0} \\ $$$$\mathrm{D}=\begin{vmatrix}{\mathrm{1}}&{\mathrm{2}}&{\mathrm{3}}\\{\mathrm{1}}&{{k}−\mathrm{2}}&{\mathrm{2}}\\{\mathrm{1}}&{\mathrm{4}−{k}}&{\mathrm{2}}\end{vmatrix}=\mathrm{0}\:\:\left(\mathrm{R2}\rightarrow\mathrm{R2}−\mathrm{R1},\mathrm{R3}\rightarrow\mathrm{R3}−\mathrm{R2}\right) \\ $$$$\mathrm{D}=\begin{vmatrix}{\mathrm{1}}&{\mathrm{2}}&{\mathrm{3}}\\{\mathrm{0}}&{{k}−\mathrm{4}}&{−\mathrm{1}}\\{\mathrm{0}}&{\mathrm{2}−{k}}&{−\mathrm{1}}\end{vmatrix}=\mathrm{0}\:\:\left(\mathrm{R2}\rightarrow\mathrm{R2}−\mathrm{R1},\mathrm{R3}\rightarrow\mathrm{R3}−\mathrm{R1}\right) \\ $$$$=−\left({k}−\mathrm{4}\right)+\left(\mathrm{2}−{k}\right)=\mathrm{0} \\ $$$$=−{k}+\mathrm{4}+\mathrm{2}+−{k}=\mathrm{0}\Rightarrow{k}=\mathrm{3} \\ $$$$\mathrm{This}\:\mathrm{itself}\:\mathrm{does}\:\mathrm{not}\:\mathrm{guarantee}\:\mathrm{that}\:\mathrm{equation} \\ $$$$\mathrm{has}\:\mathrm{no}\:\mathrm{solution}\:\mathrm{we}\:\mathrm{also}\:\mathrm{need}\:\mathrm{to}\:\mathrm{check}\:\mathrm{if} \\ $$$$\left.\mathrm{a}\right)\:\mathrm{at}\:\mathrm{least}\:\mathrm{one}\:\mathrm{of}\:\mathrm{D}_{\mathrm{1}} ,\mathrm{D}_{\mathrm{2}} \:\mathrm{or}\:\mathrm{D}_{\mathrm{3}} \:\mathrm{is}\:\mathrm{non}\:\mathrm{zero} \\ $$$$\left.\mathrm{b}\right)\:\mathrm{if}\:\mathrm{D}_{\mathrm{1}} ,\mathrm{D}_{\mathrm{2}} ,\mathrm{D}_{\mathrm{3}} \:\mathrm{are}\:\mathrm{all}\:\mathrm{zero}\:\mathrm{then}\:\mathrm{verify}\:\mathrm{that}\:\mathrm{no} \\ $$$$\mathrm{solution}\:\mathrm{exist}\:\mathrm{by}\:\mathrm{solving}\:\mathrm{the}\:\mathrm{equation}. \\ $$$$\mathrm{D}_{\mathrm{1}} =\begin{vmatrix}{\mathrm{1}}&{\mathrm{2}}&{\mathrm{3}}\\{\mathrm{1}}&{\mathrm{3}}&{\mathrm{5}}\\{\mathrm{1}}&{\mathrm{4}}&{\mathrm{7}}\end{vmatrix} \\ $$$$\mathrm{D}_{\mathrm{1}} =\begin{vmatrix}{\mathrm{1}}&{\mathrm{2}}&{\mathrm{3}}\\{\mathrm{0}}&{\mathrm{1}}&{\mathrm{2}}\\{\mathrm{0}}&{\mathrm{1}}&{\mathrm{2}}\end{vmatrix}=\mathrm{0} \\ $$$$\mathrm{D}_{\mathrm{2}} =\begin{vmatrix}{\mathrm{1}}&{\mathrm{1}}&{\mathrm{3}}\\{\mathrm{2}}&{\mathrm{1}}&{\mathrm{5}}\\{\mathrm{3}}&{\mathrm{1}}&{\mathrm{7}}\end{vmatrix}=\begin{vmatrix}{\mathrm{1}}&{\mathrm{1}}&{\mathrm{3}}\\{\mathrm{1}}&{\mathrm{0}}&{\mathrm{2}}\\{\mathrm{1}}&{\mathrm{0}}&{\mathrm{2}}\end{vmatrix}=\mathrm{0} \\ $$$$\mathrm{D}_{\mathrm{3}} =\begin{vmatrix}{\mathrm{1}}&{\mathrm{2}}&{\mathrm{1}}\\{\mathrm{2}}&{\mathrm{3}}&{\mathrm{1}}\\{\mathrm{3}}&{\mathrm{4}}&{\mathrm{1}}\end{vmatrix}=\begin{vmatrix}{\mathrm{1}}&{\mathrm{2}}&{\mathrm{1}}\\{\mathrm{1}}&{\mathrm{1}}&{\mathrm{0}}\\{\mathrm{1}}&{\mathrm{1}}&{\mathrm{0}}\end{vmatrix}=\mathrm{0} \\ $$$${x}+\mathrm{2}{y}+\mathrm{3}{z}=\mathrm{1}\Rightarrow\mathrm{2}{x}+\mathrm{4}{y}+\mathrm{6}{z}=\mathrm{2} \\ $$$$\mathrm{2}{x}+\mathrm{3}{y}+\mathrm{5}{z}=\mathrm{1} \\ $$$${from}\:{the}\:{above}\:{two}\:{y}+{z}=\mathrm{1}\Rightarrow{y}=\mathrm{1}−{z} \\ $$$${x}=\mathrm{1}−\mathrm{2}{y}−\mathrm{3}{z}\Rightarrow{x}=\mathrm{1}−\mathrm{2}\left(\mathrm{1}−{z}\right)−\mathrm{3}{z}=−\mathrm{1}−\mathrm{5}{z} \\ $$$$\mathrm{3}{x}+\mathrm{4}{y}+\mathrm{7}{z}=\mathrm{1} \\ $$$$\mathrm{3}\left(−\mathrm{1}−\mathrm{5}{z}\right)+\mathrm{4}\left(\mathrm{1}−{z}\right)+\mathrm{7}{z}=\mathrm{1} \\ $$$$−\mathrm{3}−\mathrm{15}{z}+\mathrm{4}−\mathrm{4}{z}+\mathrm{7}{z}=\mathrm{1}\Rightarrow{z}=\mathrm{0} \\ $$$${y}=\mathrm{1},{x}=−\mathrm{1} \\ $$$${check}\:{solution}\:\left(−\mathrm{1},\mathrm{1},\mathrm{0}\right){for}\:{k}=\mathrm{3} \\ $$$$−\mathrm{1}+\mathrm{2}=\mathrm{1}\:\left(\mathrm{eqn}\:\mathrm{1}\right) \\ $$$$−\mathrm{2}+\mathrm{3}=\mathrm{1}\:\left(\mathrm{eqn}\:\mathrm{2}\right) \\ $$$$−\mathrm{3}+\mathrm{4}=\mathrm{1}\left(\mathrm{eqn}\:\mathrm{3}\right) \\ $$$${k}=\mathrm{3}\:\mathrm{gives}\:\mathrm{unique}\:\mathrm{solution}. \\ $$
Answered by Ashis last updated on 07/Jun/16
yes 3 but how ?
$${yes}\:\mathrm{3}\:{but}\:{how}\:? \\ $$

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