If A, B then prove that AB is not equal to BA

Question

Question:

If A = $\left[ \begin{matrix} 0 & 1 \ 2 & -3 \ \end{matrix} ight]$, B = $\left[ \begin{matrix} -3 & 4 \ 5 & -2 \ \end{matrix} ight]$, then prove that

(i) AB $ e$ BA

Accepted Answer

Solution:

(i) AB $ e$ BA

AB = $\left[ \begin{matrix} 0 & 1 \ 2 & -3 \ \end{matrix} ight]\left[ \begin{matrix} -3 & 4 \ 5 & -2 \ \end{matrix} ight]$

= $\left[ \begin{matrix} 0 imes \left( -3 ight)+1~ imes 5 & 0 imes 4+1~ imes \left( -2 ight) \ 2 imes \left( -3 ight)+\left( -3 ight)~ imes 5 & 2 imes 4+\left( -3 ight)~ imes \left( -2 ight) \ \end{matrix} ight]$

= $\left[ \begin{matrix} 0+5 & 0-2 \ -6-15 & 8+6 \ \end{matrix} ight]$

= $\left[ \begin{matrix} 5 & -2 \ -21 & 14 \ \end{matrix} ight] \quad\quad\quad\quad\quad ~...(1)$

BA = $\left[ \begin{matrix} -3 & 4 \ 5 & -2 \ \end{matrix} ight]\left[ \begin{matrix} 0 & 1 \ 2 & -3 \ \end{matrix} ight]$

= $\left[ \begin{matrix} \left( -3 ight) imes 0+4~ imes 2 & \left( -3 ight) imes 1+4~ imes \left( -3 ight) \ 5 imes 0+\left( -2 ight) imes 2 & 5 imes 1+\left( -2 ight) imes \left( -3 ight) \ \end{matrix} ight]$

= $\left[ \begin{matrix} 0+8 & -3-12 \ 0-4 & 5+6 \ \end{matrix} ight]$

= $\left[ \begin{matrix} 8 & -15 \ -4 & 11 \ \end{matrix} ight]\quad\quad\quad\quad\quad ~...(2)$

From $(1)$ and $(2)$, it is clear that AB $ e$ BA