Question Number 7 by user1 last updated on 25/Jan/15
![If A= [(2,1,3),(4,1,0) ]and B= [(1,(−1)),(0,2),(5,0) ], verify that (AB)′=B′A′](https://www.tinkutara.com/question/Q7.png)
$$\mathrm{If}\:\:\:{A}=\begin{bmatrix}{\mathrm{2}}&{\mathrm{1}}&{\mathrm{3}}\\{\mathrm{4}}&{\mathrm{1}}&{\mathrm{0}}\end{bmatrix}{and}\:\:{B}=\begin{bmatrix}{\mathrm{1}}&{−\mathrm{1}}\\{\mathrm{0}}&{\mathrm{2}}\\{\mathrm{5}}&{\mathrm{0}}\end{bmatrix}, \\ $$$$\mathrm{verify}\:\mathrm{that}\:\:\left(\mathrm{A}{B}\right)'={B}'{A}' \\ $$
Answered by user1 last updated on 30/Oct/14
![We have AB= [(2,1,3),(4,1,0) ]∙ [(1,(−1)),(0,( 2)),(5,( 0)) ] = [((2∙1+1∙0+3∙5),(2∙(−1)+1∙2+3∙0)),((4∙1+1∙0+0∙5),(4∙(−1)+1∙2+0∙0)) ] = [((17),( 0)),(( 4),(−2)) ] ∴(AB)′= [((17),( 0)),(( 4),(−2)) ]^′ = [((17),( 4)),(( 0),(−2)) ] B′A′= [(1,(−1)),(0,( 2)),(5,( 0)) ]^′ ∙ [(2,1,3),(4,1,0) ]^′ = [(( 1),0,5),((−1),2,0) ] [(2,4),(1,1),(3,0) ] = [(( 1∙2+0∙1+5∙3),( 1∙4+0∙1+5∙0)),(((−1)∙2+2∙1+0∙3),((−1)∙4+2∙1+0∙0)) ] = [((17),( 4)),(( 0),(−2)) ] Hence, (AB)′=B′A′](https://www.tinkutara.com/question/Q8.png)
$$\mathrm{We}\:\mathrm{have}\: \\ $$$${AB}=\begin{bmatrix}{\mathrm{2}}&{\mathrm{1}}&{\mathrm{3}}\\{\mathrm{4}}&{\mathrm{1}}&{\mathrm{0}}\end{bmatrix}\centerdot\begin{bmatrix}{\mathrm{1}}&{−\mathrm{1}}\\{\mathrm{0}}&{\:\:\:\:\mathrm{2}}\\{\mathrm{5}}&{\:\:\:\:\mathrm{0}}\end{bmatrix} \\ $$$$=\begin{bmatrix}{\mathrm{2}\centerdot\mathrm{1}+\mathrm{1}\centerdot\mathrm{0}+\mathrm{3}\centerdot\mathrm{5}}&{\mathrm{2}\centerdot\left(−\mathrm{1}\right)+\mathrm{1}\centerdot\mathrm{2}+\mathrm{3}\centerdot\mathrm{0}}\\{\mathrm{4}\centerdot\mathrm{1}+\mathrm{1}\centerdot\mathrm{0}+\mathrm{0}\centerdot\mathrm{5}}&{\mathrm{4}\centerdot\left(−\mathrm{1}\right)+\mathrm{1}\centerdot\mathrm{2}+\mathrm{0}\centerdot\mathrm{0}}\end{bmatrix} \\ $$$$=\begin{bmatrix}{\mathrm{17}}&{\:\:\:\:\mathrm{0}}\\{\:\:\:\mathrm{4}}&{−\mathrm{2}}\end{bmatrix} \\ $$$$\therefore\left({AB}\right)'=\begin{bmatrix}{\mathrm{17}}&{\:\:\:\:\mathrm{0}}\\{\:\:\mathrm{4}}&{−\mathrm{2}}\end{bmatrix}^{'} =\begin{bmatrix}{\mathrm{17}}&{\:\:\:\:\mathrm{4}}\\{\:\:\:\mathrm{0}}&{−\mathrm{2}}\end{bmatrix} \\ $$$${B}'{A}'=\begin{bmatrix}{\mathrm{1}}&{−\mathrm{1}}\\{\mathrm{0}}&{\:\:\:\:\:\mathrm{2}}\\{\mathrm{5}}&{\:\:\:\:\:\mathrm{0}}\end{bmatrix}^{'} \centerdot\begin{bmatrix}{\mathrm{2}}&{\mathrm{1}}&{\mathrm{3}}\\{\mathrm{4}}&{\mathrm{1}}&{\mathrm{0}}\end{bmatrix}^{'} \\ $$$$=\begin{bmatrix}{\:\:\:\:\:\mathrm{1}}&{\mathrm{0}}&{\mathrm{5}}\\{−\mathrm{1}}&{\mathrm{2}}&{\mathrm{0}}\end{bmatrix}\begin{bmatrix}{\mathrm{2}}&{\mathrm{4}}\\{\mathrm{1}}&{\mathrm{1}}\\{\mathrm{3}}&{\mathrm{0}}\end{bmatrix} \\ $$$$=\begin{bmatrix}{\:\:\:\:\:\:\:\:\:\:\mathrm{1}\centerdot\mathrm{2}+\mathrm{0}\centerdot\mathrm{1}+\mathrm{5}\centerdot\mathrm{3}}&{\:\:\:\:\:\:\:\:\:\:\mathrm{1}\centerdot\mathrm{4}+\mathrm{0}\centerdot\mathrm{1}+\mathrm{5}\centerdot\mathrm{0}}\\{\left(−\mathrm{1}\right)\centerdot\mathrm{2}+\mathrm{2}\centerdot\mathrm{1}+\mathrm{0}\centerdot\mathrm{3}}&{\left(−\mathrm{1}\right)\centerdot\mathrm{4}+\mathrm{2}\centerdot\mathrm{1}+\mathrm{0}\centerdot\mathrm{0}}\end{bmatrix} \\ $$$$=\begin{bmatrix}{\mathrm{17}}&{\:\:\:\:\mathrm{4}}\\{\:\:\:\mathrm{0}}&{−\mathrm{2}}\end{bmatrix} \\ $$$$\mathrm{Hence},\:\:\left({AB}\right)'={B}'{A}' \\ $$