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Problems #2,18 on pg.109-110 are graded.
#2, pg.109: Find the inverse of the matrix [3285].
Solution: Using the formula in Theorem 4, pg.103 we see that
[3285]−1=115−16[5−2−83]=[−528−3].
Click here to see that this is indeed the correct inverse.
#18,pg.110: Solve the eqution AB=BC for A, assuming that A,B,C are square and B is invertible.
Solution: From the equation AB=BC and the fact that B is invertible, we will multiply the equaton on the right by B−1 to get
ABB−1=BCB−1.
By the definition of inverse, we know that ABB−1=AI=A and so we have shown that
A=BCB−1.
(note: we cannot go further and say that BCB−1=CBB−1=C because that would assume that CB−1=B−1C which is not true in general!)
