Let A=[acbd] is a singular matrix. Then detA=ad−bc=0. We have that
[acbd]⋅[11]=[13] and thus, a+b=1 and c+d=3. It follows that a=1−b and
d=3−c, and therefore, 0=ad−bc=(1−b)(3−c)−bc=3−3b−c+bc−bc=3−3b−c, and thus c=3−3b .
Let b=0. Then c=3, a=1−b=1, d=3−c=3−3=0.
We conclude that the matrix A=[1300] is singular that maps (1,1)T into (1,3)T.
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