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Problem #37 from pg. 61 and problem #1 from pg. 68 are graded..
#3,pg.100: Let $A = \left[ \begin{array}{ll}
2 & -5 \\
3 & -2
\end{array} \right]$. Compute $3I_2-A$ and $(3I_2)A$.
Solution: First we compute
$$\begin{array}{ll}
3I_2-A &= \left[ \begin{array}{ll}
3 & 0 \\
0 & 3
\end{array} \right] - \left[ \begin{array}{ll}
2 & -5 \\
3 & -2
\end{array} \right] \\
&= \left[ \begin{array}{ll}
1 & 5 \\
-3 & 5
\end{array} \right].
\end{array}$$
Now compute
$$\begin{array}{ll}
(3I_2)A &= \left[ \begin{array}{ll}
3 & 0 \\
0 & 3
\end{array} \right] \left[ \begin{array}{ll}
2 & -5 \\
3 & -2
\end{array} \right] \\
&= \left[ \begin{array}{ll}
6 & -15 \\
9 & -6
\end{array} \right]
\end{array}$$
#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!)
