## Elementary Linear Algebra 7th Edition

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) $$B+\frac{1}{2}A = \left[ \begin{matrix} \frac{5}{2}&-\frac{3}{2}\\ 0&\frac{15}{2} \end{matrix}\right]$$
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: $$G=B+\frac{1}{2}A = \left[ \begin{matrix} 2&-1\\ -1&8 \end{matrix}\right] + \frac{1}{2}\left[ \begin{matrix} 1&-1\\ 2&-1 \end{matrix}\right] = \left[ \begin{matrix} 2&-1\\ -1&8 \end{matrix}\right]+ \left[ \begin{matrix} \frac{1}{2}&-\frac{1}{2}\\ 1&-\frac{1}{2} \end{matrix}\right] = \left[ \begin{matrix} 2+\frac{1}{2}&-1-\frac{1}{2}\\ -1+1&8-\frac{1}{2} \end{matrix}\right] = \left[ \begin{matrix} \frac{5}{2}&-\frac{3}{2}\\ 0&\frac{15}{2} \end{matrix}\right].$$