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