Why Are There More Solutions for A^2-A-6I=0?

[Jest3r]

Thanks in advance for your help (and sorry about the lack of Latex...I didn't know how to type matrices in that format!)

The textbook question itself is quite straightforward:
Verify that A^2 - A - 6I = 0, given that A is a 2x2 matrix. (I is the identity matrix)

It then goes on to provide two examples of A which satisfy that equation.

However, I tried to 'play around' with the question. Since we know that AI = A, and I^2 = I, i changed the equation to:
A^2 -AI - 6I = 0.
This can then be factored into:
(A-3I)(A+2I) = 0.

However, there are more solutions of 'A' that solve the equation (the ones the textbook provided). My question is...why are there more than 2 solutions? Where have I gone wrong in my reasoning? (I'm self teaching myself Linear Algebra, so I dont' have a teacher to ask about this).

Thanks again!

 

[Dick]

Very creative! The problem is that the product of two matrices being zero does not imply that either of the matrices is zero. Put your examples into your factorization and you'll see this is so.

 

[Jest3r]

Very creative! The problem is that the product of two matrices being zero does not imply that either of the matrices is zero. Put your examples into your factorization and you'll see this is so.

THanks for the prompt reply, but I'm not sure I still understand.
From the factorization, I get the solutions A = 3I, and A = -2I (and they happen to satisfy the equation). However, there are solutions the aforementioned ones that also work. How can there be more than two solutions to a quadratic equation?

Edit: An example of the solutions the textbook provides is Row 1 (A) = [2 2], Row 2(A)= [2 -1]. (2x2 matrix, sorry for formatting again!)

 

[Dick]

I told you. When you are saying (x-A)(x-B)=0 implies x=A or x=B, the reason you can do this is that (x-A)(x-B)=0 implies (x-A)=0 or (x-B)=0 IN THE REAL NUMBERS. Matrices don't have this property. (x-A)(x-B)=0 does not imply that x-A=0 or x-B=0 for matrices.