Answer to Question #151365 in Linear Algebra for Ashweta Padhan

Question #151365
Find a 2×2 singular matrix A that maps
(1, 1)^T into (1, 3)^T
1
Expert's answer
2020-12-21T17:32:51-0500

Let A=[abcd]A=\left[\begin{array}{cc} a & b \\ c & d\end{array}\right] is a singular matrix. Then detA=adbc=0\det A=ad-bc=0. We have that


[abcd][11]=[13]\left[\begin{array}{cc} a & b \\ c & d\end{array}\right] \cdot \left[\begin{array}{c} 1 \\ 1\end{array}\right]=\left[\begin{array}{c} 1 \\ 3\end{array}\right] and thus, a+b=1a+b=1 and c+d=3c+d=3. It follows that a=1ba=1-b and


d=3cd=3-c, and therefore, 0=adbc=(1b)(3c)bc=33bc+bcbc=33bc0=ad-bc=(1-b)(3-c)-bc=3-3b-c+bc-bc=3-3b-c, and thus c=33bc=3-3b .


Let b=0b=0. Then c=3, a=1b=1, d=3c=33=0.c=3,\ a=1-b=1,\ d=3-c=3-3=0.


We conclude that the matrix A=[1030]A=\left[\begin{array}{cc} 1 & 0 \\ 3 & 0\end{array}\right] is  singular that maps (1,1)T(1, 1)^T into (1,3)T(1, 3)^T.



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