Treat and solve the system of equations in dependence of parameter a
Clash Royale CLAN TAG#URR8PPP
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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.
linear-algebra
 |Â
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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.
linear-algebra
2
Can you show us your working?
â 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 $a.$
â 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.
â 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 $6-a+9-a^2 = -3-13a.$
â 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?
â Love Invariants
Aug 27 at 16:07
 |Â
show 2 more comments
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.
linear-algebra
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.
linear-algebra
edited Aug 27 at 15:59
asked Aug 27 at 15:49
ckl
274
274
2
Can you show us your working?
â 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 $a.$
â 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.
â 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 $6-a+9-a^2 = -3-13a.$
â 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?
â Love Invariants
Aug 27 at 16:07
 |Â
show 2 more comments
2
Can you show us your working?
â 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 $a.$
â 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.
â 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 $6-a+9-a^2 = -3-13a.$
â 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?
â Love Invariants
Aug 27 at 16:07
2
2
Can you show us your working?
â Siong Thye Goh
Aug 27 at 15:49
Can you show us your working?
â 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 $a.$
â 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 $a.$
â 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.
â 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.
â 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 $6-a+9-a^2 = -3-13a.$
â user2661923
Aug 27 at 16:06
@cki I upvoted because I think that your approach is logical. However, shouldn't your last equation be $6-a+9-a^2 = -3-13a.$
â 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?
â 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?
â Love Invariants
Aug 27 at 16:07
 |Â
show 2 more comments
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2
Can you show us your working?
â 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 $a.$
â 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.
â 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 $6-a+9-a^2 = -3-13a.$
â 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?
â Love Invariants
Aug 27 at 16:07