Answer
(a) \begin{align*}
A+B&=\left[\begin{array}{rrr}{2} & {9} & {6} \\ {4} & {2} & {-1} \\ {1} & {2} & {5}\end{array}\right].
\end{align*}
(b) \begin{align*}
A-B&=\left[\begin{array}{rrr}{2} & {-3} & {2} \\ {-4} & {0} & {-1} \\ {3} & {-2} & {-3}\end{array}\right].
\end{align*}
(c) \begin{align*}
2A&=\left[\begin{array}{rrr}{4} & {6} & {8} \\ {0} & {2} & {-2} \\ {4} & {0} & {2}\end{array}\right].
\end{align*}
(d)\begin{align*}
2A-B
&=\left[\begin{array}{rrr}{4} & {0} & {6} \\ {-4} & {1} & {-2} \\ {5} & {-2} & {-2}\end{array}\right].
\end{align*}
(e)\begin{align*}
\frac{1}{2}A+B&=\left[\begin{array}{rrr}{1} & {\frac{15}{2}} & {4} \\ {4} & {\frac{3}{2}} & {-\frac{1}{2}} \\ {0} & {2} & {\frac{9}{2}}\end{array}\right].
\end{align*}
Work Step by Step
Given $$ A=\left[\begin{array}{rrr}{2} & {3} & {4} \\ {0} & {1} & {-1} \\ {2} & {0} & {1}\end{array}\right], \quad B=\left[\begin{array}{rrr}{0} & {6} & {2} \\ {4} & {1} & {0} \\ {-1} & {2} & {4}\end{array}\right]
$$
(a) \begin{align*}
A+B&=\left[\begin{array}{rrr}{2} & {3} & {4} \\ {0} & {1} & {-1} \\ {2} & {0} & {1}\end{array}\right]+\left[\begin{array}{rrr}{0} & {6} & {2} \\ {4} & {1} & {0} \\ {-1} & {2} & {4}\end{array}\right]\\
&=\left[\begin{array}{rrr}{2} & {9} & {6} \\ {4} & {2} & {-1} \\ {1} & {2} & {5}\end{array}\right].
\end{align*}
(b) \begin{align*}
A-B&=\left[\begin{array}{rrr}{2} & {3} & {4} \\ {0} & {1} & {-1} \\ {2} & {0} & {1}\end{array}\right]-\left[\begin{array}{rrr}{0} & {6} & {2} \\ {4} & {1} & {0} \\ {-1} & {2} & {4}\end{array}\right]\\
&=\left[\begin{array}{rrr}{2} & {-3} & {2} \\ {-4} & {0} & {-1} \\ {3} & {-2} & {-3}\end{array}\right].
\end{align*}
(c) \begin{align*}
2A&=2\left[\begin{array}{rrr}{2} & {3} & {4} \\ {0} & {1} & {-1} \\ {2} & {0} & {1}\end{array}\right]\\
&=\left[\begin{array}{rrr}{4} & {6} & {8} \\ {0} & {2} & {-2} \\ {4} & {0} & {2}\end{array}\right].
\end{align*}
(d)\begin{align*}
2A-B&=\left[\begin{array}{rrr}{4} & {6} & {8} \\ {0} & {2} & {-2} \\ {4} & {0} & {2}\end{array}\right]-\left[\begin{array}{rrr}{0} & {6} & {2} \\ {4} & {1} & {0} \\ {-1} & {2} & {4}\end{array}\right]\\
&=\left[\begin{array}{rrr}{4} & {0} & {6} \\ {-4} & {1} & {-2} \\ {5} & {-2} & {-2}\end{array}\right].
\end{align*}
(e)\begin{align*}
\frac{1}{2}A+B&=\frac{1}{2}\left[\begin{array}{rrr}{2} & {3} & {4} \\ {0} & {1} & {-1} \\ {2} & {0} & {1}\end{array}\right]+\left[\begin{array}{rrr}{0} & {6} & {2} \\ {4} & {1} & {0} \\ {-1} & {2} & {4}\end{array}\right]\\
&=\left[\begin{array}{rrr}{1} & {\frac{3}{2}} & {2} \\ {0} & {\frac{1}{2}} & {-\frac{1}{2}} \\ {1} & {0} & {\frac{1}{2}}\end{array}\right]+\left[\begin{array}{rrr}{0} & {6} & {2} \\ {4} & {1} & {0} \\ {-1} & {2} & {4}\end{array}\right]\\
&=\left[\begin{array}{rrr}{1} & {\frac{15}{2}} & {4} \\ {4} & {\frac{3}{2}} & {-\frac{1}{2}} \\ {0} & {2} & {\frac{9}{2}}\end{array}\right].
\end{align*}
