Proving Inverses (Matrix Multiplications)

[jwb]

Let A and B be n x n matrices such that I+AB is invertible.

If B is invertible what can you deduce about the matrix I+BA.

I know that I+BA is invertible but not sure how to go about proving this.

 

[micromass]

What can you tell about the matrix (I+BA)B?

 

[jwb]

it is also invertible due to being a multiplication or addition of only elementary matrices. Its not the theory I am having trouble with its the mathematical steps needed to prove that it is invertible

 

[AlephZero]

it is also invertible due to being a multiplication or addition of only elementary matrices.

I think you are gettng confused between what you "know" (i.e. you know or you have guessed what the answer is) and what you are trying to prove.

(I+BA)B = B + BAB = B(I + AB)

Now think about which of those matrices you KNOW are invertible, and which you want to PROVE are invertible.

 

[blue_raver22]

I would approach this problem by first understanding the properties of inverse matrices and how they relate to matrix multiplication. From there, I would use mathematical reasoning and proofs to support my deductions.

Firstly, we know that for any square matrix A, the inverse matrix A^-1 exists if and only if the determinant of A is non-zero. This means that if I+AB is invertible, then its determinant must be non-zero.

Next, let's consider the matrix I+BA. Since B is invertible, we know that B^-1 exists and its determinant is also non-zero. Using the properties of determinants, we can rewrite the determinant of I+BA as:

det(I+BA) = det(I) * det(BA) = det(BA)

Now, using the property of matrix multiplication, we can expand the product BA as:

det(BA) = det(B) * det(A)

Since both B and A are n x n matrices, their determinants are non-zero and we can conclude that det(BA) is also non-zero. This means that I+BA is invertible, as its determinant is non-zero.

Therefore, we can deduce that if B is invertible, then I+BA is also invertible. This can also be proven by showing that (I+BA)^-1 = I - B(BA+I)^-1A, which can be derived using the properties of inverse matrices.

In conclusion, as a scientist, I can confidently say that if B is invertible, then I+BA is also invertible, based on the properties of inverse matrices and matrix multiplication.