## Elementary Linear Algebra 7th Edition

(a) $$AB=\left[ \begin {array}{ccc} -8&-2&-5\\ 4&8&17 \\ -20&1&4\end {array} \right].$$ (a) $$BA=\left[ \begin {array}{ccc} 9&5&4\\3&11&-5 \\ -17&-1&-16\end {array} \right].$$
Given $$A=\left[\begin{array}{rrr}{2} & {-1} & {3} \\ {5} & {1} & {-2} \\ {2} & {2} & {3}\end{array}\right], \quad B=\left[\begin{array}{rrr}{0} & {1} & {2} \\ {-4} & {1} & {3} \\ {-4} & {-1} & {-2}\end{array}\right]$$ We have (a) \begin{align*} AB&=\left[\begin{array}{rrr}{2} & {-1} & {3} \\ {5} & {1} & {-2} \\ {2} & {2} & {3}\end{array}\right]\left[\begin{array}{rrr}{0} & {1} & {2} \\ {-4} & {1} & {3} \\ {-4} & {-1} & {-2}\end{array}\right]\\ &=\left[ \begin {array}{ccc} -8&-2&-5\\ 4&8&17 \\ -20&1&4\end {array} \right]. \end{align*} (b) \begin{align*} BA&=\left[\begin{array}{rrr}{0} & {1} & {2} \\ {-4} & {1} & {3} \\ {-4} & {-1} & {-2}\end{array}\right]\left[\begin{array}{rrr}{2} & {-1} & {3} \\ {5} & {1} & {-2} \\ {2} & {2} & {3}\end{array}\right]\\ &=\left[ \begin {array}{ccc} 9&5&4\\3&11&-5 \\ -17&-1&-16\end {array} \right]. \end{align*}