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Let $A = \left[\begin{array}{cc}5 & -1 \ 6 & 7\end{array}ight]$ and $B = \left[\begin{array}{cc}2 & 1 \ 3 & 4\end{array}ight]$.First,calculate t

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Let $A = \left[\begin{array}{cc}5 & -1 \ 6 & 7\end{array}ight]$ and $B = \left[\begin{array}{cc}2 & 1 \ 3 & 4\end{array}ight]$.First,calculate the product $AB$:$AB = \left[\begin{array}{cc}5 & -1 \ 6 & 7\end{array}ight] \left[\begin{array}{cc}2 & 1 \ 3 & 4\end{array}ight]$$= \left[\begin{array}{cc}5(2) + (-1)(3) & 5(1) + (-1)(4) \ 6(2) + 7(3) & 6(1) + 7(4)\end{array}ight]$$= \left[\begin{array}{cc}10 - 3 & 5 - 4 \ 12 + 21 & 6 + 28\end{array}ight] = \left[\begin{array}{cc}7 & 1 \ 33 & 34\end{array}ight]$Next,calculate the product $BA$:$BA = \left[\begin{array}{cc}2 & 1 \ 3 & 4\end{array}ight] \left[\begin{array}{cc}5 & -1 \ 6 & 7\end{array}ight]$$= \left[\begin{array}{cc}2(5) + 1(6) & 2(-1) + 1(7) \ 3(5) + 4(6) & 3(-1) + 4(7)\end{array}ight]$$= \left[\begin{array}{cc}10 + 6 & -2 + 7 \ 15 + 24 & -3 + 28\end{array}ight] = \left[\begin{array}{cc}16 & 5 \ 39 & 25\end{array}ight]$Since $\left[\begin{array}{cc}7 & 1 \ 33 & 34\end{array}ight] 
e \left[\begin{array}{cc}16 & 5 \ 39 & 25\end{array}ight]$,it is shown that $AB 
e BA$.

Show that $\left[ {\begin{array}{cc} 5 & -1 \\ 6 & 7 \end{array}} \right] \left[ {\begin{array}{cc} 2 & 1 \\ 3 & 4 \end{array}} \right] \ne \left[ {\begin{array}{cc} 2 & 1 \\ 3 & 4 \end{array}} \right] \left[ {\begin{array}{cc} 5 & -1 \\ 6 & 7 \end{array}} \right]$

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