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Hi,

Let $a$ be a fixed parameter, and let $(S) : \begin{cases} x - y + a z = a ; \ a+ay-z=-1 ; \ x + y +z = 2 \end{cases}$

I want to determine the solution of this system depending on the value of $a$. "By hand", I find that there is a unique solution if and only if $a \not \in {-1 ; 3 }$, that there is no solution if $a=3$, and that there are infinitely many solutions if $a=-1$. But when try with SageMath, I only get the case where $a \not \in { -1 ; 3 }$...

var('x,y,z,a')
eq1 = x-y+a*z==a
eq2 = x+a*y-z==-1
eq3 = x+y+z==2
solve([eq1,eq2,eq3],x,y,z)

returns

[[x == (a - 1)/(a - 3), y == -1/(a - 3), z == (a - 4)/(a - 3)]]

How could I we do to solve properly this kind of problem with SageMath ?

Thanks in advance ;)

Hi,

Let $a$ be a fixed parameter, and let $(S) : \begin{cases} x - y + a z = a ; \ a+ay-z=-1 ; \ x + y +z = 2 \end{cases}$

I want to determine the solution of this system depending on the value of $a$. "By hand", I find that there is a unique solution if and only if $a \not \in {-1 ; 3 }$, that there is no solution if $a=3$, and that there are infinitely many solutions if $a=-1$. But when try with SageMath, I only get the case where $a \not \in { -1 ; 3 }$...

var('x,y,z,a')
eq1 = x-y+a*z==a
eq2 = x+a*y-z==-1
eq3 = x+y+z==2
solve([eq1,eq2,eq3],x,y,z)

returns

[[x == (a - 1)/(a - 3), y == -1/(a - 3), z == (a - 4)/(a - 3)]]

How could I we do to solve properly this kind of problem with SageMath ?

Thanks in advance ;)

Hi,

Let $a$ be a fixed parameter, and let $(S) : \begin{cases} x - y + a z = a ~ ; \ a+ay-z=-1 ~ ; \ x + y +z = 2 \end{cases}$

I want to determine the solution of this system depending on the value of $a$. "By hand", I find that there is a unique solution if and only if $a \not \in {-1 ; 3 }$, that there is no solution if $a=3$, and that there are infinitely many solutions if $a=-1$. But when try with SageMath, I only get the case where $a \not \in { -1 ; 3 }$...

var('x,y,z,a')
eq1 = x-y+a*z==a
eq2 = x+a*y-z==-1
eq3 = x+y+z==2
solve([eq1,eq2,eq3],x,y,z)

returns

[[x == (a - 1)/(a - 3), y == -1/(a - 3), z == (a - 4)/(a - 3)]]

How could I we do to solve properly this kind of problem with SageMath ?

Thanks in advance ;)

Hi,

Let $a$ be a fixed parameter, and let $(S) $$(S) : \begin{cases} x - y + a z = a ~ ; \ a+ay-z=-1 ~ ; \ x + y +z = 2 \end{cases}$\end{cases}$$

I want to determine the solution of this system depending on the value of $a$. "By hand", I find that there is a unique solution if and only if $a \not \in {-1 ; 3 }$, that there is no solution if $a=3$, and that there are infinitely many solutions if $a=-1$. But when try with SageMath, I only get the case where $a \not \in { -1 ; 3 }$...

var('x,y,z,a')
eq1 = x-y+a*z==a
eq2 = x+a*y-z==-1
eq3 = x+y+z==2
solve([eq1,eq2,eq3],x,y,z)

returns

[[x == (a - 1)/(a - 3), y == -1/(a - 3), z == (a - 4)/(a - 3)]]

How could I we do to solve properly this kind of problem with SageMath ?

Thanks in advance ;)

Hi,

Let $a$ be a fixed parameter, and let $$(S) [(S) : \begin{cases} x - y + a z = a ~ ; \ a+ay-z=-1 ~ ; \ x + y +z = 2 \end{cases}$$\end{cases}]

I want to determine the solution of this system depending on the value of $a$. "By hand", I find that there is a unique solution if and only if $a \not \in {-1 ; 3 }$, that there is no solution if $a=3$, and that there are infinitely many solutions if $a=-1$. But when try with SageMath, I only get the case where $a \not \in { -1 ; 3 }$...

var('x,y,z,a')
eq1 = x-y+a*z==a
eq2 = x+a*y-z==-1
eq3 = x+y+z==2
solve([eq1,eq2,eq3],x,y,z)

returns

[[x == (a - 1)/(a - 3), y == -1/(a - 3), z == (a - 4)/(a - 3)]]

How could I we do to solve properly this kind of problem with SageMath ?

Thanks in advance ;)

Hi,

Let $a$ be a fixed parameter, and let [(S) $$(S) : \begin{cases} x - y + a z = a ~ ; \ a+ay-z=-1 ~ ; \ x + y +z = 2 \end{cases}]\end{cases}$$

I want to determine the solution of this system depending on the value of $a$. "By hand", I find that there is a unique solution if and only if $a \not \in {-1 ; 3 }$, that there is no solution if $a=3$, and that there are infinitely many solutions if $a=-1$. But when try with SageMath, I only get the case where $a \not \in { -1 ; 3 }$...

var('x,y,z,a')
eq1 = x-y+a*z==a
eq2 = x+a*y-z==-1
eq3 = x+y+z==2
solve([eq1,eq2,eq3],x,y,z)

returns

[[x == (a - 1)/(a - 3), y == -1/(a - 3), z == (a - 4)/(a - 3)]]

How could I we do to solve properly this kind of problem with SageMath ?

Thanks in advance ;)