## Elementary Linear Algebra 7th Edition

$$A^T=\left[\begin{array}{rrr}{1} \\ {2} \\ {-3} \end{array}\right].$$ $$A^TA= \left[\begin{array}{rrr}{1} & {2}&{-3} \\ {2} & {4}&{-6}\\{-3}&{-6}&{9} \end{array}\right].$$ $$AA^T=\left[\begin{array}{rrr}{14} \end{array}\right].$$
Let $A$ be given by $$A=\left[\begin{array}{rrr}{1} & {2} & {-3} \end{array}\right].$$ Now, we have $$A^T=\left[\begin{array}{rrr}{1} \\ {2} \\ {-3} \end{array}\right].$$ $$A^TA=\left[\begin{array}{rrr}{1} \\ {2} \\ {-3} \end{array}\right]\left[\begin{array}{rrr}{1} & {2} & {-3} \end{array}\right]=\left[\begin{array}{rrr}{1} & {2}&{-3} \\ {2} & {4}&{-6}\\{-3}&{-6}&{9} \end{array}\right].$$ $$AA^T=\left[\begin{array}{rrr}{1} & {2} & {-3} \end{array}\right]\left[\begin{array}{rrr}{1} \\ {2} \\ {-3} \end{array}\right]=\left[\begin{array}{rrr}{14} \end{array}\right].$$