## Elementary Linear Algebra 7th Edition

a) $$A+B = \left[\begin{matrix} 7&3\\ 1&9\\ -2&15 \end{matrix}\right]$$ b) $$A-B = \left[\begin{matrix} 5&-5\\ 3&-1\\ -4&-5 \end{matrix}\right]$$ c) $$2A = \left[\begin{matrix} 12&-2\\ 4&8\\ -6&10 \end{matrix}\right]$$ d) $$2A-B = \left[\begin{matrix} 11&-6\\ 5&3\\ -7&0 \end{matrix}\right]$$ e) $$B+\frac{1}{2}A= \left[\begin{matrix} 4&\frac{7}{2}\\ 0&7\\ -\frac{1}{2}&\frac{15}{2} \end{matrix}\right]$$
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 $3$ rows and both have $2$ columns) so we can add them. The result is the matrix $C$ of the same type (3 rows and 2 columns) and its element in the $i$-th row and $j$-th column is simply $A_{ij}+B_{ij}$: $$C=A+B =\left[ \begin{matrix} 6+1&-1+4\\ 2+(-1)&4+5\\ -3+1&5+10 \end{matrix}\right] = \left[ \begin{matrix} 7&3\\ 1&9\\ -2&15 \end{matrix}\right].$$ 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 $A_{ij}-B_{ij}$: $$D=A-B =\left[ \begin{matrix} 6-1&-1-4\\ 2-(-1)&4-5\\ -3-1&5-10 \end{matrix}\right] = \left[ \begin{matrix} 5&-5\\ 3&-1\\ -4&-5 \end{matrix}\right].$$ 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 $A_{ij}$: $$E = 2A =2\left[ \begin{matrix} 6&-1\\ 2&4\\ -3&5 \end{matrix}\right] = \left[ \begin{matrix} 2\times 6&2\times (-1)\\ 2\times 2&2\times 4\\ 2\times (-3)&2\times 5 \end{matrix}\right] = \left[ \begin{matrix} 12&-2\\ 4&8\\ -6&10 \end{matrix}\right].$$ d) $2A$ and $B$ are of the same type so we can do the subtracting and we will use the already calculated $E=2A$: $$F=2A-B = E-B=\left[ \begin{matrix} 12-1&-2-4\\ 4-(-1)&8-5\\ -6-1&10-10 \end{matrix}\right] = \left[ \begin{matrix} 11&-6\\ 5&3\\ -7&0 \end{matrix}\right].$$ e) This is possible since both $B$ and $\frac{1}{2}A$ are of the same type: $$G=B+\frac{1}{2}A = \left[ \begin{matrix} 1&4\\ -1&5\\ 1&10 \end{matrix}\right] + \frac{1}{2}\left[ \begin{matrix} 6&-1\\ 2&4\\ -3&5 \end{matrix}\right] = \left[ \begin{matrix} 1&4\\ -1&5\\ 1&10 \end{matrix}\right]+ \left[ \begin{matrix} 3&-\frac{1}{2}\\ 1&2\\ -\frac{3}{2}&\frac{5}{2} \end{matrix}\right] = \left[ \begin{matrix} 1+3&4-\frac{1}{2}\\ -1+1&5+2\\ 1-\frac{3}{2}&10+\frac{5}{2} \end{matrix}\right] = \left[ \begin{matrix} 4&\frac{7}{2}\\ 0&7\\ -\frac{1}{2}&\frac{15}{2} \end{matrix}\right].$$