Elementary Linear Algebra 7th Edition

Published by Cengage Learning
ISBN 10: 1-13311-087-8
ISBN 13: 978-1-13311-087-3

Chapter 2 - Matrices - 2.1 Operations with Matrices - 2.1 Exercises: 5


a) $$A+B = \left[ \begin{matrix} 3&-2\\ 1&7 \end{matrix}\right].$$ b) $$A-B = \left[ \begin{matrix} -1&0\\ 3&-9 \end{matrix}\right]. $$ c) $$2A =\left[ \begin{matrix} 2&-2\\ 4&-2 \end{matrix}\right].$$ d) $$2A-B = \left[ \begin{matrix} 0&-1\\ 5&-10 \end{matrix}\right].$$ e) $$B+\frac{1}{2}A = \left[ \begin{matrix} \frac{5}{2}&-\frac{3}{2}\\ 0&\frac{15}{2} \end{matrix}\right]$$

Work Step by Step

a) To add two matrices the number of their columns must match as well as the number of their rows. These matrices have equal number of rows and columns (Both have $2$ rows and both have $2$ columns) so we can add them. The result is the matrix $C$ of the same type (2 rows and two columns) and its element in the $i$-th row and $j$-th column is simply $A_{ij}+B_{ij}$: $$C=A+B =\left[ \begin{matrix} 1+2&-1+(-1)\\ 2+(-1)&-1+8 \end{matrix}\right] = \left[ \begin{matrix} 3&-2\\ 1&7 \end{matrix}\right]. $$ b) If we can add two matrices we can always subtract one from another since exactly the same conditions are required to be satisfied as for addition. Only the result is the matrix $D$ and its element at the position $ij$ is $A_{ij}-B_{ij}$: $$D=A-B =\left[ \begin{matrix} 1-2&-1-(-1)\\ 2-(-1)&-1-8 \end{matrix}\right] = \left[ \begin{matrix} -1&0\\ 3&-9 \end{matrix}\right]. $$ c) Any matrix $A$ can be multiplied by a number and the result will be the matrix $E$ and its element at the place $ij$ will just be that number times $A_{ij}$: $$E = 2A =2\left[ \begin{matrix} 1&-1\\ 2&-1 \end{matrix}\right] = \left[ \begin{matrix} 2\times 1&2\times (-1)\\ 2\times 2&2\times (-1) \end{matrix}\right] = \left[ \begin{matrix} 2&-2\\ 4&-2 \end{matrix}\right].$$ d) $2A$ and $B$ are of the same type so we can do the subtracting and we will use the already calculated $E=2A$: $$F=2A-B = E-B=\left[ \begin{matrix} 2-2&-2-(-1)\\ 4-(-1)&-2-8 \end{matrix}\right] = \left[ \begin{matrix} 0&-1\\ 5&-10 \end{matrix}\right].$$ e) This is possible since both $B$ and $\frac{1}{2}A$ are of the same type: $$G=B+\frac{1}{2}A = \left[ \begin{matrix} 2&-1\\ -1&8 \end{matrix}\right] + \frac{1}{2}\left[ \begin{matrix} 1&-1\\ 2&-1 \end{matrix}\right] = \left[ \begin{matrix} 2&-1\\ -1&8 \end{matrix}\right]+ \left[ \begin{matrix} \frac{1}{2}&-\frac{1}{2}\\ 1&-\frac{1}{2} \end{matrix}\right] = \left[ \begin{matrix} 2+\frac{1}{2}&-1-\frac{1}{2}\\ -1+1&8-\frac{1}{2} \end{matrix}\right] = \left[ \begin{matrix} \frac{5}{2}&-\frac{3}{2}\\ 0&\frac{15}{2} \end{matrix}\right].$$
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