## Elementary Linear Algebra 7th Edition

(a) $$AB=\left[ \begin {array}{cc} 8&2\\1&-14 \\ -4&18\end {array} \right].$$ (b) $BA$ not defined.
Given $$A=\left[\begin{array}{rr}{2} & {1} \\ {-3} & {4} \\ {1} & {6}\end{array}\right], \quad B=\left[\begin{array}{rrr}{0} & {-1} & {0} \\ {4} & {0} & {2} \\ {8} & {-1} & {7}\end{array}\right]$$ We have (a) \begin{align*} AB&=\left[\begin{array}{rr}{2} & {1} \\ {-3} & {4} \\ {1} & {6}\end{array}\right]\left[\begin{array}{rrr}{0} & {-1} & {0} \\ {4} & {0} & {2} \\ {8} & {-1} & {7}\end{array}\right]\\ &=\left[ \begin {array}{cc} 8&2\\1&-14 \\ -4&18\end {array} \right]. \end{align*} (b) $BA$ not defined.