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Section 4.1 Determinants and their Geometry

Subsection 4.1.1 Linear Distortions

Investigation 4.1.1. Distorted Area.

Suppose we form a parallelogram with the vectors \(\vec{v}=\left\lt3,1\right\gt\) and \(\vec{w}=\left\lt1,2\right\gt\text{.}\) To find the area we need to multiply the length of the base by the height, or (from Figure 4.1.1)

\begin{equation*} Area = ||\vec{v}||\, ||\vec{w}-proj_{\vec{v}}\vec{w}||. \end{equation*}

\begin{align*} \vec{w}-proj_{\vec{v}}\vec{w}\amp=\left\lt1,2\right\gt-\frac{5}{10}\left\lt3,1\right\gt\\ \amp=\left\lt-\frac{1}{2},\frac{3}{2}\right\gt \end{align*}
\begin{align*} ||\vec{v}||\, ||\vec{w}-proj_{\vec{v}}\vec{w}||\amp= \left(\sqrt{10}\right)\left(\frac{\sqrt{10}}{2}\right)\\ \amp = 5 \end{align*}
Figure 4.1.1. Area of a Parallelogram
So, the area of the parallelogram is 5. Interestingly if we define the transformation

\begin{equation*} T\left(\vec{x}\right)= \left( \begin{array}{rr} 3 \amp 1\\ 1 \amp 2 \end{array} \right)\vec{x}, \end{equation*}

which carries the square of area 1 made by the standard basis to this parallelogram, then

\begin{equation*} a_{11}\, a_{22}-a_{12}\, a_{21}=3\cdot2-1\cdot1=5. \end{equation*}

This is because given a matrix

\begin{equation*} A= \left( \begin{array}{rr} a_{11} \amp a_{12}\\ a_{21} \amp a_{22} \end{array} \right), \end{equation*}

the value

\begin{equation*} det\, A=a_{11}\, a_{22}-a_{12}\, a_{21} \end{equation*}

called the determinant of \(A\) measures how much \(A\) distorts \(\mathbb{R}^2\) if we think of it as a transformation.

Investigation 4.1.2. Distorted Volume.

Suppose we have a parallelepiped as in the figure, for simplicity assume the base is a rectangle. The the volume will be area of the base times the height. The area of the base is easy, \(Area=||\vec{v}||\, ||\vec{w}||\text{.}\) To get the height we need to use projections, in particular:

\begin{equation*} ||\vec{u}-proj_{\vec{v}}\vec{u}-proj_{\vec{w}}\vec{u}||. \end{equation*}
Figure 4.1.2. Volume of a Parallelepiped
\begin{align*} ||\vec{u}-proj_{\vec{v}}\vec{u}-proj_{\vec{w}}\vec{u}||\amp=\left|\left|\vec{u}-\frac{u_1v_1+u_2v_2+u_3v_3}{||\vec{v}||^2}\vec{v}-\frac{u_1w_1+u_2w_2+u_3w_3}{||\vec{w}||^2}\vec{v}\right|\right|\\ \amp=\left|\left|\left\lt u_1,u_2,u_3\right\gt-\frac{u_1v_1+u_2v_2+u_3v_3}{||\vec{v}||^2}\left\lt v_1,v_2,v_3\right\gt-\frac{u_1w_1+u_2w_2+u_3w_3}{||\vec{w}||^2}\left\lt w_1,w_2,w_3\right\gt\right|\right|\\ \amp \vdots\\ \mbox{Then A}\ \amp \mbox{Miracle Occurs}\\ \amp \vdots\\ \amp = \frac{|v_1(u_2w_3-w_2u_3)-v_2(u_1w_3-w_1u_3)+v_3(u_1w_2-w_1u_2)|}{||\vec{v}||\, ||\vec{w}||} \end{align*}

So, that the magnitude will be

\begin{equation*} |v_1(u_2w_3-w_2u_3)-v_2(u_1w_3-w_1u_3)+v_3(u_1w_2-w_1u_2)|, \end{equation*}

the expression inside the absolute value is the determinant of the matrix

\begin{equation*} \left( \begin{array}{lll} v_1 \amp u_1 \amp w_1\\ v_2 \amp u_2 \amp w_2\\ v_3 \amp u_3 \amp w_3\\ \end{array} \right) \end{equation*}

which transforms the vectors of the standard basis into the vectors \(\vec{v}\text{,}\) \(\vec{w}\text{,}\) and \(\vec{u}\text{.}\)

Subsection 4.1.2 Calculating Determinants (Laplace's Formula)

Investigation 4.1.3. Two-By-Two.

Consider the matrix...

Investigation 4.1.4. Three-By-Three.

Consider the matrix...

Investigation 4.1.5. N-By-N.

Consider the matrix...

Properties of Determinants Summarized.