Question

Find a 2×2 singular matrix A that maps
(1, 1)^T into (1, 3)^T

Answer 1

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

$\left[\begin{array}{cc} a & b \\ c & d\end{array}\right] \cdot \left[\begin{array}{c} 1 \\ 1\end{array}\right]=\left[\begin{array}{c} 1 \\ 3\end{array}\right]$ and thus, $a+b=1$ 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 $A=\left[\begin{array}{cc} 1 & 0 \\ 3 & 0\end{array}\right]$ is  singular that maps $(1, 1)^T$ into $(1, 3)^T$.