Can This Method Solve the Modified System of Linear Equations?

[MathematicalPhysicist]

we have the system a_i1x+a_i2y+a_i3z=b_i i=1,2,3
and we are given that it has a unique solution (1,2,0)
i need to find the set of solutions for the next system:
a_i1x+b_iy+a_i3z=a_i2
what i did is:
we have the solution for the first equation then we have a_i1+2a_i2=b_i
so we put it in the second equation and we have:
a_i1x+(a_i1+2a_i2)y+a_i3z=a_i2
(a_i1x+a_i2y+a_i3z)+(a_i1+a_i2)y=a_i2
b_i+(a_i1+a_i2)y=a_i2
(a_i1+2a_i2)+(a_i1+a_i2)y=a_i2
from here i found the solution, is this method valid?

 

[arildno]

Hint:
Look up Cramer's rule, and see what you can find out with that.

 

[MathematicalPhysicist]

do you know perhaps another method to solve this other than cramer's rule?
cause I am not supposed to use it here.

 

[radou]

Try to look at the corresponding vector equation: \vec{a}_{1}x + \vec{a}_{2}y + \vec{a}_{3}z = \vec{b}.

 

[MathematicalPhysicist]

i don't understand, radou can you elaborate?

 

[radou]

i don't understand, radou can you elaborate?

Well, after plugging the solution (x, y, z) = (1, 2, 0) into the vector equation, you find out that the vector b can be written as a linear combination of vectors a1 and a2, right? So, that means that the vector b lies in the same plane as a1 and a2 do. Hence, you know that the vectors a1, a2 and a3 are linearly independent. Now, after plugging in the linear combination a1 + 2a2 = b into the equation a1x + by + a3z = a2, and rearranging, you can use the fact that a1, a2 and a3 are linearly independent to solve the equation.

 

[MathematicalPhysicist]

when you mean rearranging, you mean like i did i.e:
a1x+a2y+a3z+(a1+a2)y=a2
but how do i find for (x,y,z) with your approach?
i mean if they are independent then any combination of them, its coeffiencts are zero like this a1x+a2y+a3z=0 then x=y=z=0, but how do i use it here?

 

[radou]

I meant a1x + (a1+2a2)y+a3z = a2 =>
a1(x+1) + a2(2y-1)+a3z = 0. Now use the fact that a1, a2 and a3 are linearly independent.

Edit: too bad you're not allowed to use Cramer's rule, as Arildno suggested, since this problem seems to be ideal for its application.

 

[MathematicalPhysicist]

ok, thanks i understand now.