00:01
Hello, so we consider n by n matrices, a, b, c, and d, such that the rank of a is equal to the rank of a, b, c, d, which is equal to n.
00:08
Now, for part a, okay, it's okay, it's just for part a.
00:22
We want to show that d is equal to c times a inverse times b.
00:31
So, since we have the rank of a equal to the rank of a, b, c, d, this means that c and d are going to to be linear combinations of a and b.
00:44
Okay, so therefore, we then have that, again, that d minus c times a inverse times b is going to be equal to zero, which implies that d is equal to c times a inverse times b.
01:07
Okay, then part b, we consider the product here...