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Given matrices A, B, and C below, perform the indicated operations if possible. If the operation is not possible, explain

Given matrices A, B, and C below, perform the indicated operations if possible. If-example-1

2 Answers

4 votes

a)


3\left[\begin{array}{ccc}2&amp;-1&amp;0\\0&amp;5&amp;0.3\\1&amp;4&amp;10\end{array}\right][tex] + [tex]\left[\begin{array}{ccc}5&amp;0&amp;2\\1&amp;-3&amp;9\\2&amp;0&amp;4\end{array}\right][tex]</p><p>= [tex]\left[\begin{array}{ccc}6&amp;-3&amp;0\\0&amp;15&amp;0.9\\3&amp;12&amp;30\end{array}\right] [tex]+ [tex]\left[\begin{array}{ccc}5&amp;0&amp;2\\1&amp;-3&amp;9\\2&amp;0&amp;4\end{array}\right][tex]</p><p>= [tex]\left[\begin{array}{ccc}11&amp;-3&amp;2\\1&amp;12&amp;9.9\\5&amp;12&amp;34\end{array}\right][tex]</p><p>b)</p><p>not possible because matrices B and C do not have the same dimensions</p><p>c)</p><p>[tex][1 &nbsp;3 &nbsp;5] x \left[\begin{array}{ccc}2&amp;-1&amp;0\\0&amp;5&amp;0.3\\1&amp;4&amp;10\end{array}\right][tex]</p><p>= [tex]\left[\begin{array}{ccc}1(2) + 3(0) + 5(1)\\1(-1) + 3(5) + 5(4)\\1(0) + 3(0.3) + 5(10)\end{array}\right]

=
\left[\begin{array}{ccc}7\\34\\50.9\end{array}\right]

User WestDiscGolf
by
5.9k points
5 votes

Answer:

Explanation:

(a) We first compute 3A and then 3A+B. To compute 3A, we multiply each entry of A by 3


3A=\left[\begin{array}{ccc}3 \cdot 2&amp; 3\cdot -1&amp;3\cdot 0\\3 \cdot 0&amp;3 \cdot 5&amp; 3 \cdot 0.3\\3 \cdot 1 &amp;3 \cdot 4 &amp;3 \cdot 10\end{array}\right] =\left[\begin{array}{ccc}6&amp;-3&amp;0\\0&amp;15&amp;0.9\\3&amp;12&amp;30\end{array}\right]

the we sum the entries of 3A and B that are in the same position


3A+B=\left[\begin{array}{ccc}6&amp;-3&amp;0\\0&amp;15&amp;0.9\\3&amp;12&amp;30\end{array}\right] +\left[\begin{array}{ccc}5&amp;0&amp;2\\1&amp;-3&amp;9\\2&amp;0&amp;4\end{array}\right]=\left[\begin{array}{ccc}6+5&amp;0-3&amp;0+2\\0+1&amp;15-3&amp;0.9+9\\3+2&amp;12+0&amp;30+4\end{array}\right]=\left[\begin{array}{ccc}11&amp;-3&amp;2\\1&amp;12&amp;9.9\\5&amp;12&amp;34\end{array}\right]

(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.


CA=\left[\begin{array}{ccc}1&amp;3&amp;5\end{array}\right]\left[\begin{array}{ccc}2&amp;-1&amp;0\\0&amp;5&amp;0.3\\1&amp;4&amp;10\end{array}\right]\\= \left[\begin{array}{ccc}2 \cdot 1 + 3 \cdot 0 + 5 \cdot 1&amp; -1\cdot1+3 \cdot 5 + 5 \cdot 4 &amp; 0 \cdot1 + 3 \cdot 0.3 +5 \cdot 10 \end{array}\right]\\=\left[\begin{array}{ccc}2+5 &amp;-1+15+20 &amp; 0.9 +50\end{array}\right]=\left[\begin{array}{ccc}7&amp;34&amp;50.9\end{array}\right]

User Chris Horner
by
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