Let a B be a matrix of order n and a B ab-i be invertible. It is proved that a - (the inverse of B) is invertible

Let a B be a matrix of order n and a B ab-i be invertible. It is proved that a - (the inverse of B) is invertible

Ab-i = ab - (b ^ - 1) * b = (a-b ^ - 1) * B, so both sides of the above formula are right multiplied by (ab-i) ^ - 1 to get I = (a-b ^ - 1) * b * (ab-i) ^ - 1 = (a-b ^ - 1) * (b * (ab-i) ^ - 1)

Inverse of linear algebraic matrix
 

To teach you a method to find the inverse of a matrix, you first increase the original matrix and add an identity matrix of the same order, as shown in the figure
Then do the elementary row transformation, change the left side into the identity matrix, and multiply the third row by - 1 and add it to the first row
So I got it
The square matrix on the right is the inverse of the original matrix

Linear algebraic matrix problem
Let a = diag (1, - 2,1), a * Ba = 2ba-8e, find B
A * is an adjoint matrix

A * = the determinant of a multiplied by the inverse of a
So a * Ba = 2ba-8e can be transformed into
The determinant of a multiplied by the inverse of a BA = 2ba-8e, at the same time, the left multiplied by a, the right multiplied by the inverse of a, we can get: 8e = (2a-a determinant) B, a = diag (1, - 2,1), its determinant is equal to - 2, it is easy to get b = diag (2, - 4,2)

If a matrix multiplied by B matrix multiplied by X matrix is equal to C matrix, then how much is x matrix equal to?

ABX = C so?
Then x = (AB) ^ - 1C
AXB=C
Then x = a ^ - 1cb ^ - 1