Suppose A is invertible and you exchange its first two rows

Suppose A is invertible and you exchange its first two rows to reach B. Is the new B invertible? If it is, how would you find B^-1 from A^-1 Suppose A is invertible. Under what conditions is M=A-munu^T invertible? Show that if M is invertible, its inverse is given by A^-1+A^-1munu^T A^-1/(1-nu^T A^-1mu). (Note that mu and nu are column vectors.)

Solution

a) Yes, the new matrix B is invertible. Explanation >> Given that A is invertible, so it has a non-zero determinant. Now we are exchanging its 2 rows only and that would change the sign of the determinant only,
so det(A) = -det(B). Now since det(A) is not = 0, so det(B) is also not = 0.
Also, the inverse of B would be the inverse of A with the first two columns interchanged.

b) We are given that A is invertible. So A must be a square matrix of order n. Also, u and v are column vectors so they must have dimensions n by 1:

=> uv^T is also a square matrix of order n.
=> Now to find that under what condition M is invertible, means there exists a matrix M^-1 such that,
    
     M^-1(Auv^T) = (Auv^T)M^-1 = I         …….(where I is an Identity matrix)

=> So, condition is uv^Tis also invertible as we know that the difference of two invertible matrices is also invertible.