## Elementary Linear Algebra 7th Edition

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