Prove that inverse of every square matrix if exist, is unique.

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$$\begin{gathered} \mathrm{Consider} \mathrm{a} \mathrm{square} \mathrm{matrix} \mathrm{A} \\ \mathrm{Let} \mathrm{if} \mathrm{possible}, \mathrm{there} \mathrm{exist} \mathrm{two} \mathrm{inverses} \mathrm{B} \mathrm{and} \mathrm{C}\hspace{0.17em}\mathrm{of} \mathrm{A}. \\ \mathrm{Then} \mathrm{it} \mathrm{implies} \mathrm{that}, \\ \mathrm{AB}=\mathrm{BA}=\mathrm{I} .....\left(1\right) \\ \mathrm{and} \mathrm{AC}=\mathrm{CA}=\mathrm{I} .....\left(2\right) \\ \mathrm{Where} \mathrm{I} \mathrm{is} \mathrm{the} \mathrm{identity} \mathrm{matrix} \mathrm{of} \mathrm{the} \mathrm{same} \mathrm{order} \\ \mathrm{as} \mathrm{A} \mathrm{and} \mathrm{B} \\ \mathrm{Note} \mathrm{that} \mathrm{any} \mathrm{matrix} \mathrm{multiplied} \mathrm{by} \mathrm{identity} \mathrm{matrix} \mathrm{is} \mathrm{the} \\ \mathrm{matrix} \mathrm{itself}. \mathrm{This} \mathrm{implies} \mathrm{that}, \\ \mathrm{B}=\mathrm{BI} \\ =\mathrm{B}\left(\mathrm{AC}\right) \left(\mathrm{I}=\mathrm{AC} \mathrm{from} \left(2\right) \right) \\ =\left(\mathrm{BA}\right)\mathrm{C} \left(\mathrm{By} \mathrm{associativity} \mathrm{of} \mathrm{matrix} \right) \\ =\left(\mathrm{I}\right)\mathrm{C} \left(\mathrm{BA}=\mathrm{I} \mathrm{from} \left(1\right) \right) \\ =\mathrm{C} \\ \mathrm{This} \mathrm{proves} \mathrm{that} \mathrm{B}=\mathrm{C} \\ \mathbf{Therefore}\mathbf{ }\mathbf{it}\mathbf{ }\mathbf{is}\mathbf{ }\mathbf{proved}\mathbf{ }\mathbf{that}\mathbf{ }\mathbf{the}\mathbf{ }\mathbf{inverse}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{any}\mathbf{ }\mathbf{square}\mathbf{ }\mathbf{matrix}\mathbf{ }\mathbf{is}\mathbf{ }\mathbf{unique}\mathbf{.} \end{gathered}$$

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