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### Solving a linear system of equations depending of a parameter

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 ;)

### Solving a linear system of equations depending of a parameter

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 ;)

### Solving a linear system of equations depending of a parameter

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 ;) ### Solving a linear system of equations depending of a parameter 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 ;) ### Solving a linear system of equations depending of a parameter 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 ;)