• Cyborg31
In summary, the problem involves solving for Y when given A, B, I-Y^-1, X, and Y as n x n invertible matrices. The solution involves manipulating the given equation (A(I-Y^-1))^-1 = YB and ultimately results in the solution Y = A^-1B^-1 + I. There is some confusion in the solution attempt, but the correct solution is Y = A^-1B^-1 + I.
Cyborg31

## Homework Statement

A, B, I-Y^-1, X, Y are n x n invertible matrices

(A(I-Y^-1))^-1 = YB

Solve for Y

## The Attempt at a Solution

(I - Y^-1)^-1A^-1 = YB
(I - Y^-1)^-1A^-1B^-1 = Y
A^-1B^-1 = (I - Y^-1)Y
A^-1B^-1 = Y - I
A^-1B^-1*(A^-1B^-1)^-1 = Y - I(A^-1B^-1)^-1
I = Y - (A^-1B^-1)^-1
Y^-1*I = Y^-1*Y - (A^-1B^-1)^-1
Y^-1 = I - (A^-1B^-1)^-1
Inverse both sides
Y = (I - (A^-1B^-1)^-1)^-1
or is it
Y = I^-1 - A^-1B^-1
Y = I - A^-1B^-1

Is this correct?

Cyborg31 said:

## Homework Statement

A, B, I-Y^-1, X, Y are n x n invertible matrices

(A(I-Y^-1))^-1 = YB

Solve for Y

## The Attempt at a Solution

(I - Y^-1)^-1A^-1 = YB
(I - Y^-1)^-1A^-1B^-1 = Y
A^-1B^-1 = (I - Y^-1)Y
A^-1B^-1 = Y - I
You can solve for Y from here immediately:
A^-1B^-2+ I= Y

A^-1B^-1*(A^-1B^-1)^-1 = Y - I(A^-1B^-1)^-1
Don't you mean = (Y- I)(A^-1B^-1)^-1

I = Y - (A^-1B^-1)^-1
This is wrong now.

Y^-1*I = Y^-1*Y - (A^-1B^-1)^-1
Y^-1 = I - (A^-1B^-1)^-1
Inverse both sides
Y = (I - (A^-1B^-1)^-1)^-1
or is it
Y = I^-1 - A^-1B^-1
Y = I - A^-1B^-1

Is this correct?

Didn't know you could move -I to the other side like normal algebra.

Is the B^-2 a typo? Cause I don't know what that's supposed to represent.

## 1. What is an invertible matrix?

An invertible matrix, also known as a non-singular matrix, is a square matrix that has an inverse matrix. This means that it can be multiplied by another matrix to produce the identity matrix. In simpler terms, an invertible matrix is a matrix that can be "undone", or reversed, by another matrix.

## 2. How do you determine if a matrix is invertible?

To determine if a matrix is invertible, you can use the determinant. If the determinant is non-zero, then the matrix is invertible. Another way is to check if the matrix has a pivot in every row, which indicates that the columns are linearly independent and the matrix is invertible.

## 3. What is the inverse of a matrix?

The inverse of a matrix is a matrix that when multiplied by the original matrix, results in the identity matrix. The inverse of a matrix is denoted by A-1. In other words, the inverse of a matrix "undoes" the effects of the original matrix.

## 4. Can a non-square matrix be invertible?

No, a non-square matrix cannot be invertible. In order for a matrix to have an inverse, it must be a square matrix, meaning it has the same number of rows and columns.

## 5. How is an invertible matrix used in real life?

Invertible matrices have many applications in various fields such as engineering, physics, and economics. They are used to solve systems of linear equations, find the inverse of a function, and perform transformations in computer graphics. Invertible matrices also play a crucial role in error correction in data transmission and cryptography.

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