Answer:
Step-by-step explanation:
(a) We first compute 3A and then 3A+B. To compute 3A, we multiply each entry of A by 3
[tex]3A=\left[\begin{array}{ccc}3 \cdot 2& 3\cdot -1&3\cdot 0\\3 \cdot 0&3 \cdot 5& 3 \cdot 0.3\\3 \cdot 1 &3 \cdot 4 &3 \cdot 10\end{array}\right] =\left[\begin{array}{ccc}6&-3&0\\0&15&0.9\\3&12&30\end{array}\right][/tex]
the we sum the entries of 3A and B that are in the same position
[tex]3A+B=\left[\begin{array}{ccc}6&-3&0\\0&15&0.9\\3&12&30\end{array}\right] +\left[\begin{array}{ccc}5&0&2\\1&-3&9\\2&0&4\end{array}\right]=\left[\begin{array}{ccc}6+5&0-3&0+2\\0+1&15-3&0.9+9\\3+2&12+0&30+4\end{array}\right]=\left[\begin{array}{ccc}11&-3&2\\1&12&9.9\\5&12&34\end{array}\right][/tex]
(b) Although it's possible to compute 3A, it's not possible to compute 2B+C. We can only sum matrices of the same size, 2B is a 3x3 matrix. However, C is a 1x3 matrix.
(c) C is a 1x3 matrix and A is a 3x3 matrix, hence CA is a 1x3 matrix. To compute CA, we multiply by left the culumns of A with the row of the matrix C.
[tex]CA=\left[\begin{array}{ccc}1&3&5\end{array}\right]\left[\begin{array}{ccc}2&-1&0\\0&5&0.3\\1&4&10\end{array}\right]\\= \left[\begin{array}{ccc}2 \cdot 1 + 3 \cdot 0 + 5 \cdot 1& -1\cdot1+3 \cdot 5 + 5 \cdot 4 & 0 \cdot1 + 3 \cdot 0.3 +5 \cdot 10 \end{array}\right]\\=\left[\begin{array}{ccc}2+5 &-1+15+20 & 0.9 +50\end{array}\right]=\left[\begin{array}{ccc}7&34&50.9\end{array}\right][/tex]
