TeachingLinear Algebra – Determinants – Geometric applications
Theorem: If \(A\) is a \(2\times2\) matrix, then \(|\det(A)|\) is the area of the parallelogram spanned by the columns of \(A\).
Proof: See Lay §3.3.
Theorem: If \(A\) is a \(3\times3\) matrix, then \(|\det(A)|\) is the volume of the parallelepiped spanned by the columns of \(A\).
Proof: See Lay §3.3.
Examples:
Consider the points \(P=(1,-2)\), \(Q=(2,5)\) and \(R=(-1,2)\). Then what is the area of the triangle \(PQR\)?
Translate the triangle such that the point \(P\) coincides with the origin. Then the area of the triangle \(PQR\) equals half the area of the parallelogram spanned by the vectors
\[\vec{PQ}=\mathbf{q}-\mathbf{p}=\begin{pmatrix}2\\5\end{pmatrix}-\begin{pmatrix}1\\-2\end{pmatrix}=\begin{pmatrix}1\\7\end{pmatrix} \quad\text{and}\quad\vec{PR}=\mathbf{r}-\mathbf{p}=\begin{pmatrix}-1\\2\end{pmatrix}-\begin{pmatrix}1\\-2\end{pmatrix}=\begin{pmatrix}-2\\4\end{pmatrix}.\]Hence:
\[\text{area}(PQR)=\frac{1}{2}\cdot\left|\det\begin{pmatrix}1&-2\\7&4\end{pmatrix}\right|=\frac{1}{2}\cdot|4+14|=9.\]Consider the points \(A=(-1,1)\), \(B=(3,-1)\), \(C=(4,3)\), \(D=(2,5)\) and \(E=(0,4)\). What is the area of the pentagon \(ABCDE\)?
Note that the area of the pentagon \(ABCDE\) equals the sum of the areas of the triangles \(ABC\), \(ACD\) and \(ADE\). In order to find these we translate the pentagon \(ABCDE\) such that the vertex \(A\) coincides with the origin. For the area of the triangles we use the vectors
\[\mathbf{b}-\mathbf{a}=\begin{pmatrix}3\\-1\end{pmatrix}-\begin{pmatrix}-1\\1\end{pmatrix}=\begin{pmatrix}4\\-2\end{pmatrix}, \quad\mathbf{c}-\mathbf{a}=\begin{pmatrix}4\\3\end{pmatrix}-\begin{pmatrix}-1\\1\end{pmatrix}=\begin{pmatrix}5\\2\end{pmatrix},\] \[\mathbf{d}-\mathbf{a}=\begin{pmatrix}2\\5\end{pmatrix}-\begin{pmatrix}-1\\1\end{pmatrix}=\begin{pmatrix}3\\4\end{pmatrix} \quad\text{and}\quad\mathbf{e}-\mathbf{a}=\begin{pmatrix}0\\4\end{pmatrix}-\begin{pmatrix}-1\\1\end{pmatrix}=\begin{pmatrix}1\\3\end{pmatrix}.\]So we find:
\[\text{area}(ABCDE)=\frac{1}{2}\cdot\left(\left|\det\begin{pmatrix}4&5\\-2&2\end{pmatrix}\right|+\left|\det\begin{pmatrix}5&3\\2&5\end{pmatrix}\right| +\left|\det\begin{pmatrix}3&1\\4&3\end{pmatrix}\right|\right)=\frac{1}{2}\cdot(18+14+5)=\frac{37}{2}.\]The volume of the parallelepiped spanned by the vectors \(\mathbf{a}=\begin{pmatrix}1\\-1\\1\end{pmatrix}\), \(\mathbf{b}=\begin{pmatrix}-1\\3\\1\end{pmatrix}\) and \(\mathbf{c}=\begin{pmatrix}2\\0\\3\end{pmatrix}\) is equal to \(\left|\det\begin{pmatrix}1&-1&2\\-1&3&0\\1&1&3\end{pmatrix}\right|\).
We obtain:
\[\begin{vmatrix}1&-1&2\\-1&3&0\\1&1&3\end{vmatrix}=\begin{vmatrix}1&-1&2\\0&2&2\\0&2&1\end{vmatrix} =\begin{vmatrix}2&2\\2&1\end{vmatrix}=2-4=-2.\]So the volume of the parallellepiped equals \(2\).
Last modified on March 1, 2021
