Question:

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

(i) AB $\ne$ BA

Difficulty: Easy

Solution:

(i) AB $\ne$ BA

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

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

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

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

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

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

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

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

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