# Solve this following

Question:

Show that $\mathrm{AB}=\mathrm{BA}$ in each of the following cases:

$A=\left[\begin{array}{rrr}1 & 3 & -1 \\ 2 & 2 & -1 \\ 3 & 0 & -1\end{array}\right]$ and $B=\left[\begin{array}{rrr}-2 & 3 & -1 \\ -1 & 2 & -1 \\ -6 & 9 & -4\end{array}\right]$

Solution:

$\left[\begin{array}{lll}a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33}\end{array}\right] \times\left[\begin{array}{lll}b_{11} & b_{12} & b_{13} \\ b_{21} & b_{22} & b_{23} \\ b_{31} & b_{32} & b_{33}\end{array}\right]$

$=\left[\begin{array}{lll}a_{11} b_{11}+a_{12} b_{21}+a_{13} b_{31} & a_{11} b_{12}+a_{12} b_{22}+a_{13} b_{32} & a_{11} b_{13}+a_{12} b_{23}+a_{13} b_{33} \\ a_{21} b_{11}+a_{22} b_{21}+a_{23} b_{31} & a_{21} b_{12}+a_{22} b_{22}+a_{23} b_{32} & a_{21} b_{13}+a_{22} b_{23}+a_{23} b_{33} \\ a_{31} b_{11}+a_{32} b_{21}+a_{33} b_{31} & a_{31} b_{12}+a_{32} b_{22}+a_{33} b_{32} & a_{31} b_{13}+a_{32} b_{23}+a_{33} b_{33}\end{array}\right]$