00:01
Okay, so we have a and b are two invertible n by n matrices, and we want to either say some statements are true or false.
00:10
So part a is that 6a is invertible, so this is true.
00:15
If we multiply an invertible matrix by a constant, then the matrix you get is still invertible.
00:21
All we need to do is take 1 over 6 times a inverse, and this will be the inverse.
00:26
For part b, we're looking at a, b, a inverse.
00:31
Equals b is this true? well, if we take this equation and multiply by both sides, by the right -hand side on both sides by a, we get a -b equals b -a, because if we multiply this by a, then we get the identity here.
00:50
But in general, two matrices don't commute, even if they're invertible.
00:54
So this is false.
00:55
This is not always true.
00:59
For part c, we look at a plus b squared.
01:03
Or this can be thought of as a plus b times a plus b and we can expand the bracket set like we would with numbers so the first term would be a times a the second term would be a times b but we have to be careful about the order in which we multiply them a b the next term is going to be b a and the final term will be b so this is a squared plus b squared plus ab ab plus a b -a, but in general, like we said in the previous part, ab is not equal to b -a, right? this is not always true.
01:44
So we can't just write this as ab and then group these terms together, right? this is the general formula, but this does not equal a squared plus b -squared plus 2a -b in general, because like i said, b -a is not always equal to ab, so we can't rewrite this as ab as a -b as a -b and then group these terms.
02:07
So this is in general false, part c.
02:11
For part d is a plus the identity invertible...