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.

idk but i found this "the graph of y = f(x-2) is the graph of y = f(x) translated two units rightward.  

so, since the point (9,2) lies on the graph of y = f(x), the point (11,2) lies on the graph of y = f(x-2)."

step-by-step explanation:

link is

correct answer is infinitely many solutions

step-by-step explanation:

2(2x + 3) = -4 + 2(2x + 5)

solving the above equation:

4x + 6 = -4 +4x +10

combining like terms

4x -4x = -4+10-6

0=0

since both sides are 0=0, we end up into true statement it means it has infinite many solutions.

so correct answer is infinitely many solutions