Why Does Inverting Matrix A Fail to Solve the Equation?

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Homework Help Overview

The discussion revolves around solving a matrix equation involving matrix A and vector B. The original poster expresses difficulty in obtaining a valid solution after attempting to invert matrix A, leading to questions about the properties of the matrices involved.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants question whether matrix A is square and discuss the implications of matrix dimensions on the existence of an inverse. There are mentions of using pseudo-inverses and least squares methods, as well as clarifications on the original equation's formulation.

Discussion Status

The discussion is active, with participants providing insights into matrix properties and potential approaches to the problem. There is a recognition of the need to clarify the original equation and the dimensions of the matrices involved, which may lead to a better understanding of the issue at hand.

Contextual Notes

Matrix A has 35 rows and 20 columns, while vector B has 35 rows. This raises questions about the validity of certain matrix operations being attempted, particularly regarding the multiplication of matrices of incompatible dimensions.

jacky_pcs
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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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Is your matrix a square? If not - no inverse. (Although being square is not itself sufficient for an inverse to exist.)
 
-I use Pseudo method but answer is bad solution
x=(aT*a)-1*B
 
Okay, for which item are you trying to solve.

your equation was
[tex] x = a - 1b[/tex]

That has nothing to do at all with your "solution" - your result doesn't follow from the equation.
 
Roger Penrose worked on the inverses on non-square matries before relativity.
 
statdad said:
Okay, for which item are you trying to solve.

your equation was
[tex] x = a - 1b[/tex]

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

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.
 
Mark44 said:
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.

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.
 
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.
 
hunt_mat said:
Roger Penrose worked on the inverses on non-square matries before relativity.
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
HallsofIvy said:
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?

Penrose worked those inverses before he worked on relativity
 

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