If A is a non zero column matrix of order m?1 and B is a non zero row matrix of order 1?n then the rank of AB is

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$$\begin{gathered} \mathrm{Although} \mathrm{rank} \mathrm{is} \mathrm{a} \mathrm{topic} \mathrm{of} \mathrm{Engineering} \mathrm{level} \mathrm{Maths} \mathrm{but} \mathrm{you} \mathrm{can} \mathrm{find} \\ \mathrm{solution} \mathrm{here}. \\ \mathrm{order} \mathrm{of} \mathrm{A} \mathrm{is} \mathrm{m}\times 1 \mathrm{and} \mathrm{that} \mathrm{of} \mathrm{B} \mathrm{is} 1\times \mathrm{n} \\ \mathrm{Hence} \mathrm{order} \mathrm{of} \mathrm{AB} \mathrm{is} \mathrm{m}\times \mathrm{n} \\ \mathrm{Now} \mathrm{as} \mathrm{both} \mathrm{matrix} \mathrm{have} \mathrm{non}-\mathrm{zero} \mathrm{entries}, \mathrm{hence} \mathrm{each} \mathrm{element} \mathrm{in} \\ \mathrm{AB} \mathrm{will} \mathrm{be} \mathrm{non}-\mathrm{zero} \mathrm{as} \mathrm{well}. \\ \mathrm{Now} \\ \mathrm{Rank} \mathrm{is} \mathrm{a} \mathrm{number} \mathrm{associated} \mathrm{with} \mathrm{a} \mathrm{square} \mathrm{matrix} \mathrm{which} \mathrm{is} \mathrm{the} \mathrm{highest} \mathrm{order} \mathrm{of} \mathrm{non}-\mathrm{singular} \mathrm{sub} \mathrm{matrix}. \\ \mathrm{Hence} \mathrm{for} \mathrm{rank} \mathrm{to} \mathrm{exist} \mathrm{we} \mathrm{must} \mathrm{have} \mathrm{m}=\mathrm{n} \\ \mathrm{In} \mathrm{case} \mathrm{m}=\mathrm{n}, \mathrm{then} \mathrm{rank} \mathrm{will} \mathrm{be} \mathrm{m} \mathrm{as} \mathrm{will} \mathrm{be} \mathrm{highest} \mathrm{order} \mathrm{of} \mathrm{non}-\mathrm{singular} \mathrm{sub} \mathrm{matrix}. \end{gathered}$$

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