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Problem #6 from pg.115 and problem #2 from pg.121.
#6, pg.115: Determine if the following matrix is invertible or not:
$$\left[ \begin{array}{lll}
1 & -3 & -6 \\
0 & 4 & 3 \\
-3 & 6 & 0
\end{array} \right].$$
Solution: We compute the reduced echelon form
$$\begin{array}{ll}
\left[ \begin{array}{lll}
1 & -3 & -6 \\
0 & 4 & 3 \\
-3 & 6 & 0
\end{array} \right] &\stackrel{r_2 \leftrightarrow r_3}{\sim} \left[ \begin{array}{lll}
1 & -3 & -6 \\
-3 & 6 & 0 \\
0 & 4 & 3
\end{array} \right] \\
&\stackrel{r_2^*=r_2+3r_1}{\sim} \left[ \begin{array}{lll}
1 & -3 & -6 \\
0 & -3 & -18 \\
0 & 4 & 3
\end{array} \right] \\
&\stackrel{r_1^*=r_1-r_2}{\stackrel{r_3^*=r_3+\frac{4}{3}r_2}{\sim}} \left[ \begin{array}{lll}
1 & 0 & 12 \\
0 & -3 & -18 \\
0 & 0 & -21
\end{array} \right] \\
&\stackrel{r_1^* = r_1+\frac{12}{21}r_3}{\stackrel{r_2^*=r_2-\frac{18}{21}r_3}{\sim}} \left[ \begin{array}{lll}
1 & 0 & 0 \\
0 & -3 & 0 \\
0 & 0 & -21
\end{array} \right] \\
&\stackrel{r_2^*=\frac{-1}{3}r_2}{\stackrel{r_3^*=\frac{-1}{21}r_3}{\sim}} \left[ \begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array} \right].
\end{array}$$
Hence we see by Theorem 8, pg.112 (a) and (b) that the given matrix is invertible.
Note: there are many ways to do this problem using the Invertible Matrix Theorem!
#2, pg.121: Compute the product of the following block matrices:
$$\left[ \begin{array}{ll}
E & 0 & \\
0 & F
\end{array} \right] \left[ \begin{array}{ll}
P & Q \\
R & S
\end{array} \right].$$
Solution: Compute as if it were normal matrix multiplication:
$$\left[ \begin{array}{ll}
E & 0 \\
0 & F
\end{array} \right] \left[ \begin{array}{ll}
P & Q \\
R & S
\end{array} \right] = \left[ \begin{array}{ll}
EP & EQ \\
FR & FS
\end{array} \right].$$
