Question about invertible matrix

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In summary: Finally, factor out an X and solve for it. In summary, the conversation discusses the invertibility of matrices and how to solve for X in an equation involving invertible matrices. The final equation is X = BA(I + BA)^(-1), with the explanation that (I + BA) is invertible due to the invertibility of A, B, and X in the original equation. The conversation also mentions the importance of proper ordering in the equation.
  • #1
tysonk
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Hi, I have a question about invertible matrix. If A is invertible does adding the identity matrix to it still make it invertible.

In other words, A +I = B , it B still invertible? Why?

Thanks.
 
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  • #2
-I is invertible. Adding I to it gives you something pretty noninvertible.
 
  • #3
Dick said:
-I is invertible. Adding I to it gives you something pretty noninvertible.

What about
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  • #4
The question actually is:
Solve for X. If you need to invert a matrix explain why that matrix is invertible.

A,B,X, are nbyn matrices. A,X, A-AX are invertible.

(A-AX)^-1 = X^-1 B

So I got X = BA(I + BA) ^-1
I'm just not sure why (I + BA) is invertible.
Thanks for the help
 
  • #5
tysonk said:
The question actually is:
Solve for X. If you need to invert a matrix explain why that matrix is invertible.

A,B,X, are nbyn matrices. A,X, A-AX are invertible.

(A-AX)^-1 = X^-1 B

So I got X = BA(I + BA) ^-1
I'm just not sure why (I + BA) is invertible.
Thanks for the help

You got there by showing BA=(1+BA)X, right? If X-AX is invertible and X is invertible then B is invertible from your original equation, yes? Now if B is invertible, A is invertible and X is invertible, then (1+BA) is invertible. If you have an equation like J*K=L and two of the three matrices are invertible, then the third is invertible. And haven't you got the BA and (1+BA)^(-1) factors in reverse order?
 
Last edited:
  • #6
Makes sense, Thanks!
Edit: yes, the ordering should be the other way.
 
  • #7
Hi, is there anyway you could go over how you got to the form X= BA(I+BA)^-1 from the original equation?
 
  • #8
Rad021 said:
Hi, is there anyway you could go over how you got to the form X= BA(I+BA)^-1 from the original equation?

Starting from (A-AX)^(-1)=X^(-1)B, begin by inverting both sides. Then expand and try to get all the X's on one side.
 

What is an invertible matrix?

An invertible matrix is a square matrix that has a unique inverse matrix. This means that when multiplied together, the inverse matrix and the original matrix will result in the identity matrix, which is a square matrix with 1s along the main diagonal and 0s everywhere else.

How do you determine if a matrix is invertible?

A matrix is invertible if its determinant is not equal to 0. The determinant is a numerical value that can be calculated using a specific formula for square matrices. If the determinant is 0, the matrix does not have an inverse.

Can all matrices be inverted?

No, not all matrices are invertible. Only square matrices can have an inverse, and even then, only matrices with a non-zero determinant are invertible.

What is the purpose of an invertible matrix?

Invertible matrices are useful in solving systems of linear equations, as well as in finding solutions to other types of mathematical problems. They also have applications in computer graphics and cryptography.

How do you find the inverse of a matrix?

The inverse of a matrix can be found by using a specific formula or by using row operations to transform the original matrix into the identity matrix. There are also algebraic methods for finding the inverse, such as the adjugate matrix method.

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