## Elementary Linear Algebra 7th Edition

(a) $$AB=\left[\begin{array}{rrr}{0} & {15} \\ {6} & {12} \end{array}\right].$$ (b) $$BA=\left[\begin{array}{rrr}{-2} & {2} \\ {31} & {14} \end{array}\right].$$
Given $$A=\left[\begin{array}{rrr}{1} & {2} \\ {4} & {2} \end{array}\right], \quad \text { and } \quad B=\left[\begin{array}{rrr}{2} & {-1} \\ {-1} & {8} \end{array}\right].$$ We have (a) \begin{align*} AB&=\left[\begin{array}{rrr}{1} & {2} \\ {4} & {2} \end{array}\right] \left[\begin{array}{rrr}{2} & {-1} \\ {-1} & {8} \end{array}\right]\\ &=\left[\begin{array}{rrr}{2-2} & {-1+16} \\ {8-2} & {-4+16} \end{array}\right]\\ &=\left[\begin{array}{rrr}{0} & {15} \\ {6} & {12} \end{array}\right]. \end{align*} (b) \begin{align*} BA&=\left[\begin{array}{rrr}{2} & {-1} \\ {-1} & {8} \end{array}\right]\left[\begin{array}{rrr}{1} & {2} \\ {4} & {2} \end{array}\right] \\ &=\left[\begin{array}{rrr}{2-4} & {4-2} \\ {-1+32} & {-2+16} \end{array}\right]\\ &=\left[\begin{array}{rrr}{-2} & {2} \\ {31} & {14} \end{array}\right]. \end{align*}