Let A and B be square matrices. Show that if ABequals=I then A and B are both invertible, with Bequals=Upper A Superscript negative 1A−1 and Aequals=Upper B Superscript negative 1B−1.

Step-by-step explanation:

Since AB=I, we have

det(A)det(B)=det(AB)=det(I)=1.

This implies that the determinants det(A) and det(B) are not zero.

Hence A,B are invertible matrices: A−1,B−1 exist.

Now we compute

I=BB−1=BIB−1=B(AB)B−1=BAI=BA.since AB=I

Hence we obtain BA=I.

Since AB=I and BA=I, we conclude that B=A−1.

the equation cos(35) =a/25 can be used to find the length of bc what is the length of bc? round to the nearest tenth

step-by-step explanation: