A square matrix is called singular if its determinant is 0. So, "\\det (M)=0". Taking into account that "\\det (kM)=k^n\\det (M)" where "n" is a number of rows in "M", we conclude that "\\det (kM)=k^n\\det (M)=k^n\\cdot 0=0" for all "k\\in \\mathbb N." Therefore, "kM" is a singular matrix as well.
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