Image of the unit square under linear transformation Let $\mathbf{B} = \begin{pmatrix}1&amp;-1 \\ 3 &amp; -2\end{pmatrix}.$ The
  image of the unit square in $\mathbb{R}^2$ under the linear mapping
  $T_{\mathbf{B}}: \mathbb{R}^2 \to \mathbb{R}^2$ defined by
  $T_\mathbf{B}{\mathbf{x}} = \mathbf{Bx}$ is
  
  (A) a square; (B) a non-square rectangle; (C) a non-rectangular
  parallelogram; (D) a line segment; (E) the point $(1, −2)$; (F) none of
  these.


I think it takes $(0,0) \mapsto (0,0)$, $(1,0) \mapsto (1,3)$, $(0,1) \mapsto (-1,-2)$, and $(1,1) \mapsto (0,1)$. Is this right? I can't figure out what shape it should be.
Here is an image of the unit square image:
If it’s a rectangle adjacent sides will be perpendicular. Take the dot product of adjacent sides. If $0$ it’s a rectangle or a square. If not $0$ it’s a parallelogram. Btw if two adjacent sides point in the same direction it’s a line. If the determinant is zero it’s a point. No matrix is a translation so if determinant is zero, it’s the point $(0,0)$.