1. Sep 18, 2011

### athrun200

1. The problem statement, all variables and given/known data
See question 5

2. Relevant equations

3. The attempt at a solution

For part a, it is very easy.
Multiply the inverse of A 2 times on both side, we can see the B=inverse of A.
i.e. The required B is inverse of A, then the proof is finished.

It seems it is the same part a.

Is part b also correct?

#### Attached Files:

• ###### 11aAss1.pdf
File size:
131.2 KB
Views:
98
2. Sep 18, 2011

### HallsofIvy

Part b is quite different from part a- and the difference is important to learn. Mathematics must be very very precise in its wording- unlike science we don't have observations and experiments to fall back on. In other words, we can't just look at the real world- words are everything!

In part a it ask if, given a non-singular matrix A, there exist a matrix B such that $AB^2= A$. You are right- just multiply, on the left, on both sides by $A^{-1}$, which exists because A is non-singular, and the equation becomes AB= I. Yes, B exists and is the inverse of A.

In part B, it asks if there exists a matrix B such that, for any non-singular matrix, A, $A^2B= A$. "Any" is the crucial word there. Is there a single matrix B that is the inverse of all invertible matrices?

3. Sep 18, 2011

### athrun200

Well, after listening to your explaination, I know part b is obvious wrong.
However, I wonder how to write it out.

4. Sep 18, 2011

### HallsofIvy

"No, there does not exist a single matrix, B, such that $A^2B= A$ for all non-singular matrices, A."

5. Sep 18, 2011

### athrun200

Oh, this is the prove?

6. Sep 18, 2011

### athrun200

Let me try for part d.

Since $A$ is non-singular, $A^{-1}$ exists.
So $\vec{x}$=$A^{-1}$$\vec{y}$ exists.

So, there exists $\vec{x}$ s.t. $A$$\vec{x}$=$\vec{y}$

Again, how to disprove part c?
By simply saying NO, there doesn't exist?