Treat and solve the system of equations in dependence of parameter a

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The task gives and asks to treat and solve the system of equations in dependence of parameter $a$. The system given:



$x + y - z = 1$



$2x + 3y + az = -1$



$x + ay + 3z = 13$



I used Gaussian elimination method to get the equation to solve for $a$, but I always end up with wrong value so I can't treat the system any further.



I transformed the original matrix to this:
$$
beginmatrix
1 & 1 & -1 & 1 \
6 & 9 & 3a & -3 \
a & a^2 & 3a & 13a \
endmatrix
$$
which has led me to the equation $6 - a + 9 - a^2 = 13a$, but this is the wrong result.







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  • 2




    Can you show us your working?
    – Siong Thye Goh
    Aug 27 at 15:49










  • Why do you solve for $a?$ As I understand the question, you should solve for $x,y,z$ in terms of $a.$
    – saulspatz
    Aug 27 at 15:54










  • Yes, but in order to solve for $x, y, z$ you must find possible values of $a$ that you can work with. I apologize for misuse of terminology as english is not my first language.
    – ckl
    Aug 27 at 16:00






  • 3




    @cki I upvoted because I think that your approach is logical. However, shouldn't your last equation be $6-a+9-a^2 = -3-13a.$
    – user2661923
    Aug 27 at 16:06










  • Why don't you try to find the value of $x,y,z$ using cross multiplication and put it in one of the equations by treating $a$ as constant?
    – Love Invariants
    Aug 27 at 16:07















up vote
1
down vote

favorite












The task gives and asks to treat and solve the system of equations in dependence of parameter $a$. The system given:



$x + y - z = 1$



$2x + 3y + az = -1$



$x + ay + 3z = 13$



I used Gaussian elimination method to get the equation to solve for $a$, but I always end up with wrong value so I can't treat the system any further.



I transformed the original matrix to this:
$$
beginmatrix
1 & 1 & -1 & 1 \
6 & 9 & 3a & -3 \
a & a^2 & 3a & 13a \
endmatrix
$$
which has led me to the equation $6 - a + 9 - a^2 = 13a$, but this is the wrong result.







share|cite|improve this question


















  • 2




    Can you show us your working?
    – Siong Thye Goh
    Aug 27 at 15:49










  • Why do you solve for $a?$ As I understand the question, you should solve for $x,y,z$ in terms of $a.$
    – saulspatz
    Aug 27 at 15:54










  • Yes, but in order to solve for $x, y, z$ you must find possible values of $a$ that you can work with. I apologize for misuse of terminology as english is not my first language.
    – ckl
    Aug 27 at 16:00






  • 3




    @cki I upvoted because I think that your approach is logical. However, shouldn't your last equation be $6-a+9-a^2 = -3-13a.$
    – user2661923
    Aug 27 at 16:06










  • Why don't you try to find the value of $x,y,z$ using cross multiplication and put it in one of the equations by treating $a$ as constant?
    – Love Invariants
    Aug 27 at 16:07













up vote
1
down vote

favorite









up vote
1
down vote

favorite











The task gives and asks to treat and solve the system of equations in dependence of parameter $a$. The system given:



$x + y - z = 1$



$2x + 3y + az = -1$



$x + ay + 3z = 13$



I used Gaussian elimination method to get the equation to solve for $a$, but I always end up with wrong value so I can't treat the system any further.



I transformed the original matrix to this:
$$
beginmatrix
1 & 1 & -1 & 1 \
6 & 9 & 3a & -3 \
a & a^2 & 3a & 13a \
endmatrix
$$
which has led me to the equation $6 - a + 9 - a^2 = 13a$, but this is the wrong result.







share|cite|improve this question














The task gives and asks to treat and solve the system of equations in dependence of parameter $a$. The system given:



$x + y - z = 1$



$2x + 3y + az = -1$



$x + ay + 3z = 13$



I used Gaussian elimination method to get the equation to solve for $a$, but I always end up with wrong value so I can't treat the system any further.



I transformed the original matrix to this:
$$
beginmatrix
1 & 1 & -1 & 1 \
6 & 9 & 3a & -3 \
a & a^2 & 3a & 13a \
endmatrix
$$
which has led me to the equation $6 - a + 9 - a^2 = 13a$, but this is the wrong result.









share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Aug 27 at 15:59

























asked Aug 27 at 15:49









ckl

274




274







  • 2




    Can you show us your working?
    – Siong Thye Goh
    Aug 27 at 15:49










  • Why do you solve for $a?$ As I understand the question, you should solve for $x,y,z$ in terms of $a.$
    – saulspatz
    Aug 27 at 15:54










  • Yes, but in order to solve for $x, y, z$ you must find possible values of $a$ that you can work with. I apologize for misuse of terminology as english is not my first language.
    – ckl
    Aug 27 at 16:00






  • 3




    @cki I upvoted because I think that your approach is logical. However, shouldn't your last equation be $6-a+9-a^2 = -3-13a.$
    – user2661923
    Aug 27 at 16:06










  • Why don't you try to find the value of $x,y,z$ using cross multiplication and put it in one of the equations by treating $a$ as constant?
    – Love Invariants
    Aug 27 at 16:07













  • 2




    Can you show us your working?
    – Siong Thye Goh
    Aug 27 at 15:49










  • Why do you solve for $a?$ As I understand the question, you should solve for $x,y,z$ in terms of $a.$
    – saulspatz
    Aug 27 at 15:54










  • Yes, but in order to solve for $x, y, z$ you must find possible values of $a$ that you can work with. I apologize for misuse of terminology as english is not my first language.
    – ckl
    Aug 27 at 16:00






  • 3




    @cki I upvoted because I think that your approach is logical. However, shouldn't your last equation be $6-a+9-a^2 = -3-13a.$
    – user2661923
    Aug 27 at 16:06










  • Why don't you try to find the value of $x,y,z$ using cross multiplication and put it in one of the equations by treating $a$ as constant?
    – Love Invariants
    Aug 27 at 16:07








2




2




Can you show us your working?
– Siong Thye Goh
Aug 27 at 15:49




Can you show us your working?
– Siong Thye Goh
Aug 27 at 15:49












Why do you solve for $a?$ As I understand the question, you should solve for $x,y,z$ in terms of $a.$
– saulspatz
Aug 27 at 15:54




Why do you solve for $a?$ As I understand the question, you should solve for $x,y,z$ in terms of $a.$
– saulspatz
Aug 27 at 15:54












Yes, but in order to solve for $x, y, z$ you must find possible values of $a$ that you can work with. I apologize for misuse of terminology as english is not my first language.
– ckl
Aug 27 at 16:00




Yes, but in order to solve for $x, y, z$ you must find possible values of $a$ that you can work with. I apologize for misuse of terminology as english is not my first language.
– ckl
Aug 27 at 16:00




3




3




@cki I upvoted because I think that your approach is logical. However, shouldn't your last equation be $6-a+9-a^2 = -3-13a.$
– user2661923
Aug 27 at 16:06




@cki I upvoted because I think that your approach is logical. However, shouldn't your last equation be $6-a+9-a^2 = -3-13a.$
– user2661923
Aug 27 at 16:06












Why don't you try to find the value of $x,y,z$ using cross multiplication and put it in one of the equations by treating $a$ as constant?
– Love Invariants
Aug 27 at 16:07





Why don't you try to find the value of $x,y,z$ using cross multiplication and put it in one of the equations by treating $a$ as constant?
– Love Invariants
Aug 27 at 16:07
















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