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Problems #2,18 on pg.109-110 are graded.

#2, pg.109: Find the inverse of the matrix $\begin{bmatrix} 3&2 \\ 8&5 \end{bmatrix}$.
Solution: Using the formula in Theorem 4, pg.103 we see that $$\begin{bmatrix} 3&2 \\ 8&5 \end{bmatrix}^{-1} = \dfrac{1}{15-16} \begin{bmatrix} 5 & -2 \\ -8 & 3 \end{bmatrix} = \begin{bmatrix} -5 & 2 \\ 8 & -3 \end{bmatrix}.$$ 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 $CB^{-1}=B^{-1}C$ which is not true in general!)