Matrices

A(x,y,z)=(2,2,2)
where A=
1 3 -1
0 1 -1
a 0 2
For which values of a does the system have solutions?

If you were trying to find the solution you would multiply both sides of the equation by the inverse of A. So the question is asking you to find for which values of a does A have an inverse

Reply 2

the bear

you can think of this as finding where 3 planes intersect at

i) a single point, or

ii) a line.

i) you would have to show that the scalar triple product of the normal vectors is non-zero.

ii) show that any of the 3 planes is a scalar multiple of any of the other 2.

however in this case because of the zeros ii) can be ruled out

so concentrate on i).

this will help:

https://www.youtube.com/watch?v=duFRYId7kNU

(edited 6 years ago)

Reply 3

RichE

Original post by Faction Paradox

If you were trying to find the solution you would multiply both sides of the equation by the inverse of A. So the question is asking you to find for which values of a does A have an inverse

The system can still have solutions even when the matrix A is not invertible (the solution would just not be unique in that case).

Reply 4

Faction Paradox

Original post by RichE

The system can still have solutions even when the matrix A is not invertible (the solution would just not be unique in that case).

Yes, this is true, but you still need to work out which values of a don't result in an inverse, and then check whether those values of a cause the set of equations to be consistent or inconsistent

Reply 5

RichE

Original post by Faction Paradox

Yes, this is true, but you still need to work out which values of a don't result in an inverse, and then check whether those values of a cause the set of equations to be consistent or inconsistent

But you didn't write that previously - you wrote

"If you were trying to find the solution you would multiply both sides of the equation by the inverse of A."

which isn't generally true and was the point I was addressing.

Reply 6

Faction Paradox

Original post by RichE

Maybe I should have been more specific in my wording, but in this particular case the point is not worth arguing about because in the case where A is singular, the equations are not consistent.

I probably should have been more careful as i was more trying to say that finding the value of a that makes A singular is the first step in the process of finding the answer, but I'm not really in the mood to have arguments about it

(edited 6 years ago)

Reply 7

Yiuyu

Guys I am struggling to do this , can someone help me please?

Reply 8

simon0

We have the system:

\displaystyle Ax = b.

We can have three situations:
1) Unique solutions exist, (determinant of A is not equal to 0),

2) Infinite solutions exist, (

\det(A) = 0, \, \mathrm{rank}(A) < \text{``Number of columns of matrix A"}

and

\mathrm{rank}(A) = \mathrm{rank}(A \rvert \mathbf{b})

),

3) No solutions exist, (

\det(A) = 0, \, \mathrm{rank}(A \lvert \mathbf{b}) > \mathrm{rank}(A).

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First, have you found the determiant of matrix A?

Once you found the determiant, what is the value of a so that: