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Consider the linear transformation $T$: ℝ$^2$ $\rightarrow$ ℝ$^2$, (x,y) $\mapsto$ A (x,y)

for some 2x2 matrix A and let S be the unit square with corners {(0,0),(0,1),(1,1),(1,0)}.

Suppose that A =

\begin{matrix} \sqrt{3} & 1 \\ 1 & \sqrt{3} \\ \end{matrix}

A) Find det. A. (DONE) B) Draw T(S). (DONE) C) Determine the area of T(S). (Recall that the area of a parallelogram with side lengths a and b and angle $\theta$ between them is ab$sin$($\theta$). (How do I find $\theta$?)

Thank you.

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The area is $\det A$ that you already computed. As for two vectors $X, X^\prime$, you have: $$\det(X,X^\prime)=\Vert X \Vert \Vert X^\prime \Vert \sin \theta $$ where $\theta$ is the angle between the two vectors.

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  • $\begingroup$ It works! Thank you. Can you please explain the reasoning behind why it works? I like to have a good conceptual knowledge of why things work. :) $\endgroup$ – The Pointer Sep 19 '16 at 7:50
  • $\begingroup$ First, take the vectors $u=(0,1)$ and $u_\theta=(\cos \theta, \sin \theta)$. The angle between the two vectors is $\theta$ and $\det(u,u_\theta) = \sin \theta$. Then you know that the columns of the matrix $A$ are the images $(T(e_1),T(e_2))$ of the canonical basis. Finally you should remember that if $U$ is positive orthogonal transformation $\det(U.X,U.X^\prime) = \det(X,X^\prime)$. $\endgroup$ – mathcounterexamples.net Sep 19 '16 at 7:59

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