1. ## Linear Equations

Consider the following set of linear equations

-x + y +2z = 2
x + 0y +az = -1
ax + 3y + 10z = 7

where (a = alpha)

for which value of the parameter a will the system have:
i) exactly 1 solution
ii) Many solutions
iii) No solutions
iv) Find solution space for the case when there are many solutions

ive reduced the matrix to
|1 -1 -2 | -2|
|0 0 2a-(a^2) |2a-(a^2)|
|0 -3 10 |7|

and i get x1=x2=x3= 1

can anyone check if this is right as the inclusion of alpha in this question has complicated the answer

2. Originally Posted by asingh88
Consider the following set of linear equations

-x + y +2z = 2
x + 0y +az = -1
ax + 3y + 10z = 7

where (a = alpha)

for which value of the parameter a will the system have:
i) exactly 1 solution
ii) Many solutions
iii) No solutions
iv) Find solution space for the case when there are many solutions

ive reduced the matrix to
|1 -1 -2 | -2|
|0 0 2a-(a^2) |2a-(a^2)|
|0 -3 10 |7|

and i get x1=x2=x3= 1

can anyone check if this is right as the inclusion of alpha in this question has complicated the answer
First, your reduction is completely wrong! Please show how you got that.

If that were correct or you after you reduce it correctly, to get something like "z= a value" you have to divide by the coefficient of z. You cannot do that is a is such that the coefficient is 0. For one value of a, you will get something like 0z= 0, so that z can be anything and there are an infinite number of solutions, while for another you get something like 0z= 7 which is not true for any z- there is no solution. Notice, by the way, that you are NOT required to find the solution in the case that there is a unique solution. That simplifies the arithmetic! Also, although it says "solution space", this is not a homogeneous system and so the solution set is not a subspace.

3. Accidental double post.
(clicked "quote" when I meant to click "edit".)