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- #1

- Apr 14, 2013

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We have the matrix $A=\begin{pmatrix}a_1 & b_1 \\ a_2 & b_2\end{pmatrix}$.

We consider the vectors $\vec{v}:=A\vec{e}_1$ and $\vec{w}:=A\vec{e}_2$.

- Justify geometrically, why the area of the parallelogram spanned by $\vec{v}$ and $\vec{w}$ is equal to $\det A$.

- Calculate the determinant of the matrix that we get from $A$ if we swap the two columns. (This is related to whether the orientation of the parallelogram is equal to the orientation of $\mathbb{R}^2$ which is given $\vec{e}_1$ and $\vec{e}_2$.)

- Give the interpretation of the case $\det A=0$. What can we say in this case for the map $f:\mathbb{R}^2\rightarrow \mathbb{R}^2$, $\vec{x}\mapsto A\vec{x}$ ? Is $f$ injectiv?

Could you give me a hint for 1. ?

At 2. we just have to calculate the determinant. The statement of the parentheses is related to the answer of 1., or not?

At 3. if the determinant is equal to $0$, then it follows that the map is injective. But how could we explain that?