[example: Area Preservation Does Not Preserve Angles]
Let $A:\mathbb{R}^2\to\mathbb{R}^2$ be the [linear map](/page/Linear%20Map) determined by $Ae_1=e_1$ and $Ae_2=e_1+e_2$, where $e_1=(1,0)$ and $e_2=(0,1)$. Since the columns of the standard matrix of a linear map are the images of the standard basis vectors, the first column of $A$ is $Ae_1=(1,0)$ and the second column is $Ae_2=e_1+e_2=(1,0)+(0,1)=(1,1)$. Hence
\begin{align*}
A=\begin{pmatrix}1&1\cr 0&1\end{pmatrix}.
\end{align*}
For a $2\times 2$ matrix $\begin{pmatrix}a&b\cr c&d\end{pmatrix}$, the determinant is $ad-bc$. Therefore
\begin{align*}
\det A=(1)(1)-(1)(0)=1-0=1.
\end{align*}
So this shear preserves signed area.
Now compare what $A$ does to the standard perpendicular basis vectors. Their original Euclidean inner product is
\begin{align*}
\langle e_1,e_2\rangle=\langle (1,0),(0,1)\rangle=(1)(0)+(0)(1)=0+0=0.
\end{align*}
The images are
\begin{align*}
Ae_1=e_1=(1,0).
\end{align*}
Also,
\begin{align*}
Ae_2=e_1+e_2=(1,0)+(0,1)=(1+0,0+1)=(1,1).
\end{align*}
Their new inner product is
\begin{align*}
\langle Ae_1,Ae_2\rangle=\langle (1,0),(1,1)\rangle=(1)(1)+(0)(1)=1+0=1.
\end{align*}
Thus two input vectors with inner product $0$ are sent to two output vectors with inner product $1$, so perpendicularity is not preserved.
The same map changes the length of $e_2$. Since $Ae_2=(1,1)$, its squared length is
\begin{align*}
|Ae_2|^2=\langle Ae_2,Ae_2\rangle=\langle (1,1),(1,1)\rangle=(1)(1)+(1)(1)=1+1=2.
\end{align*}
Meanwhile,
\begin{align*}
|e_2|^2=\langle e_2,e_2\rangle=\langle (0,1),(0,1)\rangle=(0)(0)+(1)(1)=0+1=1.
\end{align*}
Since $|Ae_2|^2=2$ and $|e_2|^2=1$, the vector $e_2$ does not keep its length under $A$. Determinant $1$ records signed area preservation here, but it does not force preservation of lengths or angles; the missing condition is preservation of the Euclidean inner product.
[/example]