# Homework Help: Why this equation can't solve

1. Aug 19, 2010

### jacky_pcs

I would like to solve equation with x=a-1*b but when I inverse matrix A . It is bad solution to solve (I use mathematica to solve it)(Attach matrix A in excel file).
I want to know why matrix A is bad solution when I inverse???????
If I want to solve this equation. How i can do?

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• ###### Matrix B.xls
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Last edited: Aug 19, 2010
2. Aug 19, 2010

Is your matrix a square? If not - no inverse. (Although being square is not itself sufficient for an inverse to exist.)

3. Aug 19, 2010

### jacky_pcs

x=(aT*a)-1*B

4. Aug 19, 2010

Okay, for which item are you trying to solve.

$$x = a - 1b$$

That has nothing to do at all with your "solution" - your result doesn't follow from the equation.

5. Aug 19, 2010

### hunt_mat

Roger Penrose worked on the inverses on non-square matries before relativity.

6. Aug 19, 2010

### Staff: Mentor

I believe the OP meant x = A-1b.

Then he/she revised the above to this: x=(aT*a)-1*B,
which I believe means this: x = (ATA)-1b.

7. Aug 20, 2010

Aha - good catch. Of course, this means that the original equation is nowhere to be seen, so it remains impossible to see why this approach fails.

8. Aug 20, 2010

### hotvette

Need to pay attention to dimension. The matrix A has 35 rows and 20 columns (i.e. 35 x 20) and B has 35 rows (i.e. 35 x 1). ATA is a 20 x 20 matrix and so is (ATA)-1. Multiplying a 35 x 1 matrix by a 20 x 20 matrix isn't valid.

This smells like a least squares problem, which means the solution would be x = (ATA)-1 ATb.

9. Aug 21, 2010

### HallsofIvy

What in the world do you mean by this? Roger Penrose wasn't born until 26 years after relativity was developed! Do you mean he worked on them before they were used in relativity?

10. Aug 21, 2010

### willem2

Penrose worked those inverses before he worked on relativity