determinant-volume.xml

<?xml version="1.0" encoding="UTF-8"?>

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<section xml:id="determinants-volumes">
  <title>Determinants and Volumes</title>

  <objectives>
    <ol>
      <li>Understand the relationship between the determinant of a matrix and the volume of a parallelepiped.</li>
      <li>Learn to use determinants to compute volumes of parallelograms and triangles.</li>
      <li>Learn to use determinants to compute the volume of some curvy shapes like ellipses.</li>
      <li><em>Pictures:</em> parallelepiped, the image of a curvy shape under a linear transformation.</li>
      <li><em>Theorem:</em> determinants and volumes.</li>
      <li><em>Vocabulary word:</em> <term>parallelepiped</term>.</li>
    </ol>
  </objectives>

  <introduction>
    <p>
      In this section we give a geometric interpretation of determinants, in terms of <em>volumes.</em>  This will shed light on the reason behind three of the four <xref ref="det-defn-the-defn" text="title">defining properties of the determinant</xref>.  It is also a crucial ingredient in the change-of-variables formula in multivariable calculus.
    </p>
  </introduction>

  <subsection>
    <title>Parallelograms and Paralellepipeds</title>

    <p>
      The determinant computes the volume of the following kind of geometric object.
    </p>

  <definition>
    <idx><h>Parallelepiped</h><h>definition of</h></idx>
    <statement>
      <p>
        The <term>paralellepiped</term> determined by <m>n</m> vectors <m>v_1, v_2,\ldots,v_n</m> in <m>\R^n</m> is the subset
        <me>
          P = \bigl\{a_1x_1 + a_2x_2 + \cdots + a_nx_n \bigm| 0 \leq a_1,a_2,\ldots,a_n\leq 1\bigr\}.
        </me>
      </p>
    </statement>
  </definition>

  <p>
    In other words, a parallelepiped is the set of all linear combinations of <m>n</m> vectors with coefficients in <m>[0,1]</m>.  We can draw parallelepipeds using the parallelogram law for vector addition.
  </p>

  <specialcase>
    <title>The unit cube</title>
    <idx><h>Unit cube</h></idx>
    <p>
      The parallelepiped determined by the standard coordinate vectors <m>e_1,e_2,\ldots,e_n</m> is the unit <m>n</m>-dimensional cube.
      <latex-code mode="bare">
              \usetikzlibrary{math}
      </latex-code>
      <latex-code>
\begin{tikzpicture}[thin border nodes, scale=2]
  \filldraw[fill=seq-orange, fill opacity=.2, very thin]
    (0,0) -- (1,0) -- (1,1) -- (0,1) -- cycle;
  \draw[vector] (0,0) -- node[auto]{$e_2$} (0,1);
  \draw[vector, opacity=.75] (0,0) -- node[auto,swap]{$e_1$} (1,0);
\end{tikzpicture}
\qquad\qquad
\begin{tikzpicture}[thin border nodes, myxyz, scale=2, baseline=-1cm]
  \def\va{(1,0,0)}
  \def\vb{(0,1,0)}
  \def\vc{(0,0,1)}
  \tikzmath{
    coordinate \vab;
    coordinate \vac;
    coordinate \vbc;
    \vab = \va + \vb;
    \vac = \va + \vc;
    \vbc = \vb + \vc;
  }
  \filldraw[fill=seq-orange, opacity=.2, very thin]
    (0,0,0) -- \vb -- (\vbc) -- \vc -- cycle;
  \filldraw[fill=seq-orange, opacity=.2, very thin]
    (0,0,0) -- \vb -- (\vab) -- \va -- cycle;
  \filldraw[fill=seq-orange, opacity=.2, very thin]
    (0,0,0) -- \va -- (\vac) -- \vc -- cycle;
  \filldraw[fill=seq-orange, fill opacity=.2, shift={\vb}, very thin]
    (0,0,0) -- \va -- (\vac) -- \vc -- cycle;
  \filldraw[fill=seq-orange, fill opacity=.2, shift={\va}, very thin]
    (0,0,0) -- \vb -- (\vbc) -- \vc -- cycle;
  \filldraw[fill=seq-orange, fill opacity=.2, shift={\vc}, very thin]
    (0,0,0) -- \vb -- (\vab) -- \va -- cycle;
  \draw[vector, opacity=.4] (0,0,0) -- node[auto] {$e_1$} \va;
  \draw[vector, opacity=.4] (0,0,0) -- node[auto, swap] {$e_2$} \vb;
  \draw[vector, opacity=.4] (0,0,0) -- node[auto,pos=.3] {$e_3$} \vc;
\end{tikzpicture}
      </latex-code>
    </p>
  </specialcase>

  <specialcase>
    <title>Parallelograms</title>
    <idx><h>Parallelepiped</h><h>parallelogram</h></idx>
    <idx><h>Parallelogram</h><see>Parallelepiped</see></idx>
    <p>
      When <m>n = 2</m>, a paralellepiped is just a paralellogram in <m>\R^2</m>.  Note that the edges come in parallel pairs.
      <latex-code><![CDATA[
\begin{tikzpicture}[thin border nodes, scale=1.3]
  \filldraw[fill=seq-orange, fill opacity=.2, very thin]
    (0,0) -- (1,2) -- (3,3) -- (2,1) -- cycle;
  \draw[vector] (0,0) -- node[auto]{$v_2$} (1,2);
  \draw[vector] (0,0) -- node[auto,swap]{$v_1$} (2,1);
  \node at (1.5,1.5) {$P$};
\end{tikzpicture}
      ]]>
      </latex-code>
    </p>
  </specialcase>

  <specialcase>
    <p>
      When <m>n=3</m>, a parallelepiped is a kind of a skewed cube.  Note that the faces come in parallel pairs.
      <latex-code><![CDATA[
\begin{tikzpicture}[thin border nodes, myxyz, scale=2]
  \filldraw[fill=seq-orange, opacity=.2, very thin]
    (0,0,0) -- (-1,0,0) -- (-1,.3,1) -- (0,.3,1) -- cycle;
  \filldraw[fill=seq-orange, opacity=.2, very thin]
    (0,0,0) -- (-1,0,0) -- (-.8,1,0) -- (.2,1,0) -- cycle;
  \filldraw[fill=seq-orange, opacity=.2, shift={(-1,0,0)}, very thin]
    (0,0,0) -- (.2,1,0) -- (.2,1.3,1) -- (0,.3,1) -- cycle;
  \filldraw[fill=seq-orange, fill opacity=.2, very thin]
    (0,0,0) -- (.2,1,0) -- (.2,1.3,1) -- (0,.3,1) -- cycle;
  \filldraw[fill=seq-orange, fill opacity=.2, shift={(.2,1,0)}, very thin]
    (0,0,0) -- (-1,0,0) -- (-1,.3,1) -- (0,.3,1) -- cycle;
  \filldraw[fill=seq-orange, fill opacity=.2, shift={(0,.3,1)}, very thin]
    (0,0,0) -- (-1,0,0) -- (-.8,1,0) -- (.2,1,0) -- cycle;
  \draw[vector] (0,0,0) -- node[auto,swap] {$v_1$} (.2,1,0);
  \draw[vector, opacity=.4] (0,0,0) -- node[auto,swap] {$v_2$} (-1,0,0);
  \draw[vector] (0,0,0) -- node[auto] {$v_3$} (0,.3,1);
  \node at (0,1.2,1.2) {$P$};
\end{tikzpicture}
      ]]>
      </latex-code>
    </p>
  </specialcase>

  <p>
    When does a parallelepiped have zero volume?  This can happen only if the parallelepiped is flat, i.e., it is squashed into a lower dimension.
    <latex-code>
\begin{tikzpicture}[thin border nodes, baseline=0]
  \draw[line width=1mm, seq-orange] (-1,-.5) to["$P$"' black] (2,1);
  \draw[vector, opacity=.75] (0,0) to["$v_1$" opacity=1] (2,1);
  \draw[vector, opacity=.75] (0,0) to["$v_2$"' opacity=1] (-1,-.5);
  \point at (0,0);
\end{tikzpicture}
\qquad
\begin{tikzpicture}[thin border nodes, baseline=0]
  \node[coordinate] (X) at (1,.2) {};
  \node[coordinate] (Y) at (-.7,-1) {};

  \begin{scope}[x=(X), y=(Y), transformxy]
    \draw[help lines, step=.5, opacity=.75] (-2,-1.5) grid (4,1.5);
  \end{scope}

  \def\va{(2,1)}
  \def\vb{(1.5,-.5)}
  \def\vc{(-1.5,0)}

  \coordinate (v1) at ($4/2*(X) - 1/2*(Y)$);
  \coordinate (v2) at ($3/2*(X) + 1*(Y)$);
  \coordinate (v3) at ($-3/2*(X) - 1/2*(Y)$);

  \filldraw[fill=seq-orange, opacity=.2, draw opacity=1, very thin]
    (v1) -- ($(v1) + (v2)$) -- (v2) -- ($(v2) + (v3)$) -- (v3) -- ($(v3) + (v1)$) -- cycle;
  \draw[vector] (0,0) to["$v_1$"] (v1);
  \draw[vector] (0,0) to["$v_2$"] (v2);
  \draw[vector] (0,0) to["$v_3$"'] (v3);
  \node at ($1/2*(v1) + 1/2*(v2)$) {$P$};
  \point at (0,0);
\end{tikzpicture}
    </latex-code>
    This means exactly that <m>\{v_1,v_2,\ldots,v_n\}</m> is <em>linearly dependent,</em> which by this <xref ref="det-defn-dep-det0"/> means that the matrix with rows <m>v_1,v_2,\ldots,v_n</m> has determinant zero.  To summarize:
  </p>

  <note type-name="Key Observation">
    <idx><h>Parallelepiped</h><h>flat</h></idx>
    <p>
      The parallelepiped defined by <m>v_1,v_2,\ldots,v_n</m> has zero volume if and only if the matrix with rows <m>v_1,v_2,\ldots,v_n</m> has zero determinant.
    </p>
  </note>

  </subsection>


  <subsection>
    <title>Determinants and Volumes</title>

    <p>
      The key observation above is only the beginning of the story:
      the volume of a parallelepiped is <em>always</em> a determinant.
    </p>

    <theorem xml:id="det-is-volume">
      <title>Determinants and volumes</title>
      <idx><h>Parallelepiped</h><h>volume of</h></idx>
      <idx><h>Matrix</h><h>parallelepiped determined by</h></idx>
      <idx><h>Determinant</h><h>and volumes</h></idx>
      <idx><h>Volume</h><h>of a parallelepiped</h></idx>
    <statement>
      <p>
        Let <m>v_1,v_2,\ldots,v_n</m> be vectors in <m>\R^n</m>, let <m>P</m> be the parallelepiped determined by these vectors, and let <m>A</m> be the matrix with rows <m>v_1,v_2,\ldots,v_n</m>.  Then the absolute value of the determinant of <m>A</m> is the volume of <m>P</m>:
        <me>
          |\det(A)| = \vol(P).
        </me>
        <notation><usage>\vol(P)</usage><description>Volume of a region</description></notation>
      </p>
    </statement>
    <proof visible="true">
      <p>
        Since the four <xref ref="det-defn-the-defn" text="title">defining properties</xref> characterize the determinant, they also characterize the absolute value of the determinant.  Explicitly, <m>|\det|</m> is a function on square matrices which satisfies these properties:
	<ol>
	  <li>
	    Doing a row replacement on <m>A</m> does not change <m>|\det(A)|</m>.
	  </li>
          <li>
            Scaling a row of <m>A</m> by a scalar <m>c</m> multiplies <m>|\det(A)|</m> by <m>|c|</m>.
	  </li>
	  <li>
	    Swapping two rows of a matrix does not change <m>|\det(A)|</m>.
	  </li>
	  <li>
	    The determinant of the identity matrix <m>I_n</m> is equal to <m>1</m>.
	  </li>
	</ol>
        The absolute value of the determinant is the <em>only</em> such function: indeed, by this <xref ref="det-defn-ref-compute"/>, if you do some number of row operations on <m>A</m> to obtain a matrix <m>B</m> in row echelon form, then
	  <me>
	    |\det(A)| =
            \left|\frac{\text{(product of the diagonal entries of $B$)}}
            {\text{(product of scaling factors used)}}\right|.
	  </me>
      </p>
      <p>
        <notation><usage>\vol(A)</usage><description>Volume of the parallelepiped of a matrix</description></notation>
        For a square matrix <m>A</m>, we abuse notation and let <m>\vol(A)</m> denote the volume of the paralellepiped determined by the rows of <m>A</m>.  Then we can regard <m>\vol</m> as a function from the set of square matrices to the real numbers.  We will show that <m>\vol</m> also satisfies the above four properties.
        <ol>
          <li>
            For simplicity, we consider a row replacement of the form <m>R_n = R_n + cR_i</m>.  The volume of a paralellepiped is the volume of its base, times its height: here the <q>base</q> is the paralellepiped determined by <m>v_1,v_2,\ldots,v_{n-1}</m>, and the <q>height</q> is the perpendicular distance of <m>v_n</m> from the base.
            <latex-code>
\begin{tikzpicture}[thin border nodes, scale=1.3]
  \filldraw[fill=seq-orange, fill opacity=.2, very thin]
    (0,0) -- (1,2) -- (3,3) -- (2,1) -- cycle;
  \draw[line width=1mm, seq-blue] (0,0) to["base"' {pos=.3,sloped}] (2,1);
  \draw[vector] (0,0) -- node[auto]{$v_2$} (1,2);
  \draw[vector, opacity=.75] (0,0) -- node[auto,swap,opacity=1,pos=.7]{$v_1$} (2,1);
  \draw[dashed, very thick, seq-red] (1,2) to["height"] (8/5,4/5);
\end{tikzpicture}
\qquad
\begin{tikzpicture}[thin border nodes, myxyz, scale=2]
  \filldraw[fill=seq-orange, opacity=.2, very thin]
    (0,0,0) -- (-1,0,0) -- (-1,.3,1) -- (0,.3,1) -- cycle;
  \filldraw[fill=seq-blue, opacity=.5, very thin]
    (0,0,0) -- (-1,0,0) -- (-.8,1,0) -- (.2,1,0) -- cycle;
  \node[seq-blue] at (-.4,.5,0) {base};
  \filldraw[fill=seq-orange, opacity=.2, shift={(-1,0,0)}, very thin]
    (0,0,0) -- (.2,1,0) -- (.2,1.3,1) -- (0,.3,1) -- cycle;
  \draw[dashed, very thick, seq-red] (0,.3,1) to["height" pos=.55] (-.2,.2,0);
  \filldraw[fill=seq-orange, fill opacity=.2, very thin]
    (0,0,0) -- (.2,1,0) -- (.2,1.3,1) -- (0,.3,1) -- cycle;
  \filldraw[fill=seq-orange, fill opacity=.2, shift={(.2,1,0)}, very thin]
    (0,0,0) -- (-1,0,0) -- (-1,.3,1) -- (0,.3,1) -- cycle;
  \filldraw[fill=seq-orange, fill opacity=.2, shift={(0,.3,1)}, very thin]
    (0,0,0) -- (-1,0,0) -- (-.8,1,0) -- (.2,1,0) -- cycle;
  \draw[vector] (0,0,0) -- node[auto,swap] {$v_1$} (.2,1,0);
  \draw[vector, opacity=.4] (0,0,0) -- node[auto,pos=.4] {$v_2$} (-1,0,0);
  \draw[vector] (0,0,0) -- node[auto] {$v_3$} (0,.3,1);
\end{tikzpicture}
            </latex-code>
            Translating <m>v_n</m> by a multiple of <m>v_i</m> moves <m>v_n</m> in a direction parallel to the base.  This changes neither the base nor the height!  Thus, <m>\vol(A)</m> is unchanged by row replacements.
            <latex-code mode="bare">
              \usetikzlibrary{math}
            </latex-code>
            <latex-code>
\begin{tikzpicture}[thin border nodes, scale=1.3, baseline=1.3cm]
  \filldraw[fill=seq-orange, fill opacity=.2, very thin]
    (0,0) -- (1,2) -- (3,3) -- (2,1) -- cycle;
  \draw[line width=1mm, seq-blue] (0,0) to["base"' {pos=.3,sloped}] (2,1);
  \draw[vector] (0,0) -- node[auto]{$v_2$} (1,2);
  \draw[vector, opacity=.75] (0,0) -- node[auto,swap,opacity=1,pos=.7]{$v_1$} (2,1);
  \draw[dashed, very thick, seq-red] (1,2) to["height"] (8/5,4/5);
\end{tikzpicture}
$\quad\xrightarrow{\phantom{MMMM}}\quad$
\begin{tikzpicture}[thin border nodes, scale=1.3, baseline=1.3cm]
  \filldraw[fill=seq-orange, fill opacity=.2, very thin]
    (0,0) -- (0,3/2) -- (2,5/2) -- (2,1) -- cycle;
  \draw[line width=1mm, seq-blue] (0,0) to["base"' {pos=.3,sloped}] (2,1);
  \draw[vector, opacity=.5] (0,0) -- node[auto,swap,pos=.8]{$v_2$} (1,2);
  \draw[vector] (0,0) -- node[auto]{$v_2-.5v_1$} (0,3/2);
  \draw[vector, opacity=.75] (0,0) -- node[auto,swap,opacity=1,pos=.7]{$v_1$} (2,1);
  \draw[dashed, very thick, seq-red] (0,3/2) to["height"] (3/5,3/10);
\end{tikzpicture}\\[3mm]
\begin{tikzpicture}[thin border nodes, myxyz, scale=2, baseline=1.3cm]
  \filldraw[fill=seq-orange, opacity=.2, very thin]
    (0,0,0) -- (-1,0,0) -- (-1,.3,1) -- (0,.3,1) -- cycle;
  \filldraw[fill=seq-blue, opacity=.5, very thin]
    (0,0,0) -- (-1,0,0) -- (-.8,1,0) -- (.2,1,0) -- cycle;
  \node[seq-blue] at (-.4,.5,0) {base};
  \filldraw[fill=seq-orange, opacity=.2, shift={(-1,0,0)}, very thin]
    (0,0,0) -- (.2,1,0) -- (.2,1.3,1) -- (0,.3,1) -- cycle;
  \draw[dashed, very thick, seq-red] (0,.3,1) to["height" pos=.55] (-.2,.2,0);
  \filldraw[fill=seq-orange, fill opacity=.2, very thin]
    (0,0,0) -- (.2,1,0) -- (.2,1.3,1) -- (0,.3,1) -- cycle;
  \filldraw[fill=seq-orange, fill opacity=.2, shift={(.2,1,0)}, very thin]
    (0,0,0) -- (-1,0,0) -- (-1,.3,1) -- (0,.3,1) -- cycle;
  \filldraw[fill=seq-orange, fill opacity=.2, shift={(0,.3,1)}, very thin]
    (0,0,0) -- (-1,0,0) -- (-.8,1,0) -- (.2,1,0) -- cycle;
  \draw[vector] (0,0,0) -- node[auto,swap] {$v_1$} (.2,1,0);
  \draw[vector, opacity=.4] (0,0,0) -- node[auto,pos=.4] {$v_2$} (-1,0,0);
  \draw[vector] (0,0,0) -- node[auto] {$v_3$} (0,.3,1);
\end{tikzpicture}
$\quad\xrightarrow{\phantom{MMMM}}\quad$
\begin{tikzpicture}[thin border nodes, myxyz, scale=2, baseline=1.3cm]
  \def\va{(.2,1,0)}
  \def\vb{(-1,0,0)}
  \def\vc{(.1,.8,1)}
  \tikzmath{
    coordinate \vab;
    coordinate \vac;
    coordinate \vbc;
    \vab = \va + \vb;
    \vac = \va + \vc;
    \vbc = \vb + \vc;
  }
  \filldraw[fill=seq-orange, opacity=.2, very thin]
    (0,0,0) -- \vb -- (\vbc) -- \vc -- cycle;
  \filldraw[fill=seq-blue, opacity=.5, very thin]
    (0,0,0) -- \vb -- (\vab) -- \va -- cycle;
  \node[seq-blue] at (-.25,.35,0) {base};
  \filldraw[fill=seq-orange, opacity=.2, shift={\vb}, very thin]
    (0,0,0) -- \va -- (\vac) -- \vc -- cycle;
  \draw[dashed, very thick, seq-red] \vc to["height" pos=.55] ($(-.2,.2,0)+.5*(.2,1,0)$);
  \filldraw[fill=seq-orange, fill opacity=.2, very thin]
    (0,0,0) -- \va -- (\vac) -- \vc -- cycle;
  \filldraw[fill=seq-orange, fill opacity=.2, shift={\va}, very thin]
    (0,0,0) -- \vb -- (\vbc) -- \vc -- cycle;
  \filldraw[fill=seq-orange, fill opacity=.2, shift={\vc}, very thin]
    (0,0,0) -- \vb -- (\vab) -- \va -- cycle;
  \draw[vector] (0,0,0) -- node[auto,swap] {$v_1$} \va;
  \draw[vector, opacity=.4] (0,0,0) -- node[auto,pos=.8] {$v_2$} \vb;
  \draw[vector] (0,0,0) -- \vc node[above] {$v_3+.5v_1$};
  \draw[vector, opacity=.5] (0,0,0) -- node[auto,pos=.5] {$v_3$} (0,.3,1);
\end{tikzpicture}
            </latex-code>
          </li>
          <li>
            For simplicity, we consider a row scale of the form <m>R_n = cR_n.</m>  This scales the length of <m>v_n</m> by a factor of <m>|c|</m>, which also scales the perpendicular distance of <m>v_n</m> from the base by a factor of <m>|c|</m>.  Thus, <m>\vol(A)</m> is scaled by <m>|c|</m>.
            <latex-code mode="bare">
              \usetikzlibrary{math}
            </latex-code>
            <latex-code>
\begin{tikzpicture}[thin border nodes, scale=1.3, baseline=1.3cm]
  \filldraw[fill=seq-orange, fill opacity=.2, very thin]
    (0,0) -- (1,2) -- (3,3) -- (2,1) -- cycle;
  \draw[line width=1mm, seq-blue] (0,0) to["base"' {pos=.3, sloped}] (2,1);
  \draw[vector] (0,0) -- node[auto]{$v_2$} (1,2);
  \draw[vector, opacity=.75] (0,0) -- node[auto,swap,opacity=1,pos=.7]{$v_1$} (2,1);
  \draw[dashed, very thick, seq-red] (1,2) to["height"] (8/5,4/5);
\end{tikzpicture}
$\quad\xrightarrow{\phantom{MMMM}}\quad$
\begin{tikzpicture}[thin border nodes, scale=1.3, baseline=1.3cm]
  \filldraw[fill=seq-orange, fill opacity=.2, very thin]
    (0,0) -- (.75,1.5) -- (2.75,2.5) -- (2,1) -- cycle;
  \draw[line width=1mm, seq-blue] (0,0) to["base"' {pos=.3, sloped}] (2,1);
  \draw[vector, opacity=.5] (0,0) -- (1,2);
  \draw[vector] (0,0) -- node[auto]{$\frac34v_2$} (.75,1.5);
  \draw[vector, opacity=.75] (0,0) -- node[auto,swap,opacity=1,pos=.7]{$v_1$} (2,1);
  \draw[dashed, very thick, seq-red] (.75,1.5) to["$\frac34$\small height"] ($.75*(8/5,4/5)$);
\end{tikzpicture}\\[3mm]
\begin{tikzpicture}[thin border nodes, myxyz, scale=2, baseline=1.3cm]
  \filldraw[fill=seq-orange, opacity=.2, very thin]
    (0,0,0) -- (-1,0,0) -- (-1,.3,1) -- (0,.3,1) -- cycle;
  \filldraw[fill=seq-blue, opacity=.5, very thin]
    (0,0,0) -- (-1,0,0) -- (-.8,1,0) -- (.2,1,0) -- cycle;
  \node[seq-blue] at (-.4,.5,0) {base};
  \filldraw[fill=seq-orange, opacity=.2, shift={(-1,0,0)}, very thin]
    (0,0,0) -- (.2,1,0) -- (.2,1.3,1) -- (0,.3,1) -- cycle;
  \draw[dashed, very thick, seq-red] (0,.3,1) to["height" pos=.55] (-.2,.2,0);
  \filldraw[fill=seq-orange, fill opacity=.2, very thin]
    (0,0,0) -- (.2,1,0) -- (.2,1.3,1) -- (0,.3,1) -- cycle;
  \filldraw[fill=seq-orange, fill opacity=.2, shift={(.2,1,0)}, very thin]
    (0,0,0) -- (-1,0,0) -- (-1,.3,1) -- (0,.3,1) -- cycle;
  \filldraw[fill=seq-orange, fill opacity=.2, shift={(0,.3,1)}, very thin]
    (0,0,0) -- (-1,0,0) -- (-.8,1,0) -- (.2,1,0) -- cycle;
  \draw[vector] (0,0,0) -- node[auto,swap] {$v_1$} (.2,1,0);
  \draw[vector, opacity=.4] (0,0,0) -- node[auto,pos=.4] {$v_2$} (-1,0,0);
  \draw[vector] (0,0,0) -- node[auto] {$v_3$} (0,.3,1);
\end{tikzpicture}
$\quad\xrightarrow{\phantom{MMMM}}\quad$
\begin{tikzpicture}[thin border nodes, myxyz, scale=2, baseline=1.3cm]
  \def\va{(.2,1,0)}
  \def\vb{(-1,0,0)}
  \def\vc{(0,.4,4/3)}
  \tikzmath{
    coordinate \vab;
    coordinate \vac;
    coordinate \vbc;
    \vab = \va + \vb;
    \vac = \va + \vc;
    \vbc = \vb + \vc;
  }
  \filldraw[fill=seq-orange, opacity=.2, very thin]
    (0,0,0) -- \vb -- (\vbc) -- \vc -- cycle;
  \filldraw[fill=seq-blue, opacity=.5, very thin]
    (0,0,0) -- \vb -- (\vab) -- \va -- cycle;
  \node[seq-blue] at (-.4,.5,0) {base};
  \filldraw[fill=seq-orange, opacity=.2, shift={\vb}, very thin]
    (0,0,0) -- \va -- (\vac) -- \vc -- cycle;
  \draw[dashed, very thick, seq-red] \vc to["$\frac43$\small height" pos=.55] ($4/3*(-.2,.2,0)$);
  \filldraw[fill=seq-orange, fill opacity=.2, very thin]
    (0,0,0) -- \va -- (\vac) -- \vc -- cycle;
  \filldraw[fill=seq-orange, fill opacity=.2, shift={\va}, very thin]
    (0,0,0) -- \vb -- (\vbc) -- \vc -- cycle;
  \filldraw[fill=seq-orange, fill opacity=.2, shift={\vc}, very thin]
    (0,0,0) -- \vb -- (\vab) -- \va -- cycle;
  \draw[vector] (0,0,0) -- node[auto,swap] {$v_1$} \va;
  \draw[vector, opacity=.4] (0,0,0) -- node[auto,pos=.5] {$v_2$} \vb;
  \draw[vector] (0,0,0) -- node[auto,pos=.8] {$\frac 43v_3$} \vc;
  \draw[vector, black!50!white] (0,0,0) -- (0,.3,1);
\end{tikzpicture}
            </latex-code>
          </li>
          <li>
            Swapping two rows of <m>A</m> just reorders the vectors <m>v_1,v_2,\ldots,v_n</m>, hence has no effect on the parallelepiped determined by those vectors.  Thus, <m>\vol(A)</m> is unchanged by row swaps.
            <latex-code>
\begin{tikzpicture}[thin border nodes, scale=1.3, baseline=1.3cm]
  \filldraw[fill=seq-orange, fill opacity=.2, very thin]
    (0,0) -- (1,2) -- (3,3) -- (2,1) -- cycle;
  \draw[vector] (0,0) -- node[auto]{$v_2$} (1,2);
  \draw[vector] (0,0) -- node[auto,swap]{$v_1$} (2,1);
\end{tikzpicture}
$\quad\xrightarrow{\phantom{MMMM}}\quad$
\begin{tikzpicture}[thin border nodes, scale=1.3, baseline=1.3cm]
  \filldraw[fill=seq-orange, fill opacity=.2, very thin]
    (0,0) -- (1,2) -- (3,3) -- (2,1) -- cycle;
  \draw[vector] (0,0) -- node[auto]{$v_1$} (1,2);
  \draw[vector] (0,0) -- node[auto,swap]{$v_2$} (2,1);
\end{tikzpicture}\\[3mm]
\begin{tikzpicture}[thin border nodes, myxyz, scale=2, baseline=1.3cm]
  \filldraw[fill=seq-orange, opacity=.2, very thin]
    (0,0,0) -- (-1,0,0) -- (-1,.3,1) -- (0,.3,1) -- cycle;
  \filldraw[fill=seq-orange, opacity=.2, very thin]
    (0,0,0) -- (-1,0,0) -- (-.8,1,0) -- (.2,1,0) -- cycle;
  \filldraw[fill=seq-orange, opacity=.2, shift={(-1,0,0)}, very thin]
    (0,0,0) -- (.2,1,0) -- (.2,1.3,1) -- (0,.3,1) -- cycle;
  \filldraw[fill=seq-orange, fill opacity=.2, very thin]
    (0,0,0) -- (.2,1,0) -- (.2,1.3,1) -- (0,.3,1) -- cycle;
  \filldraw[fill=seq-orange, fill opacity=.2, shift={(.2,1,0)}, very thin]
    (0,0,0) -- (-1,0,0) -- (-1,.3,1) -- (0,.3,1) -- cycle;
  \filldraw[fill=seq-orange, fill opacity=.2, shift={(0,.3,1)}, very thin]
    (0,0,0) -- (-1,0,0) -- (-.8,1,0) -- (.2,1,0) -- cycle;
  \draw[vector] (0,0,0) -- node[auto,swap] {$v_1$} (.2,1,0);
  \draw[vector, opacity=.4] (0,0,0) -- node[auto] {$v_3$} (-1,0,0);
  \draw[vector] (0,0,0) -- node[auto] {$v_2$} (0,.3,1);
\end{tikzpicture}
$\quad\xrightarrow{\phantom{MMMM}}\quad$
\begin{tikzpicture}[thin border nodes, myxyz, scale=2, baseline=1.3cm]
  \filldraw[fill=seq-orange, opacity=.2, very thin]
    (0,0,0) -- (-1,0,0) -- (-1,.3,1) -- (0,.3,1) -- cycle;
  \filldraw[fill=seq-orange, opacity=.2, very thin]
    (0,0,0) -- (-1,0,0) -- (-.8,1,0) -- (.2,1,0) -- cycle;
  \filldraw[fill=seq-orange, opacity=.2, shift={(-1,0,0)}, very thin]
    (0,0,0) -- (.2,1,0) -- (.2,1.3,1) -- (0,.3,1) -- cycle;
  \filldraw[fill=seq-orange, fill opacity=.2, very thin]
    (0,0,0) -- (.2,1,0) -- (.2,1.3,1) -- (0,.3,1) -- cycle;
  \filldraw[fill=seq-orange, fill opacity=.2, shift={(.2,1,0)}, very thin]
    (0,0,0) -- (-1,0,0) -- (-1,.3,1) -- (0,.3,1) -- cycle;
  \filldraw[fill=seq-orange, fill opacity=.2, shift={(0,.3,1)}, very thin]
    (0,0,0) -- (-1,0,0) -- (-.8,1,0) -- (.2,1,0) -- cycle;
  \draw[vector] (0,0,0) -- node[auto,swap] {$v_1$} (.2,1,0);
  \draw[vector, opacity=.4] (0,0,0) -- node[auto] {$v_2$} (-1,0,0);
  \draw[vector] (0,0,0) -- node[auto] {$v_3$} (0,.3,1);
\end{tikzpicture}
            </latex-code>
          </li>
          <li>
            The rows of the identity matrix <m>I_n</m> are the standard coordinate vectors <m>e_1,e_2,\ldots,e_n</m>.  The associated paralellepiped is the unit cube, which has volume <m>1</m>.  Thus, <m>\vol(I_n) = 1</m>.
          </li>
        </ol>
        Since <m>|\det|</m> is the only function satisfying these properties, we have
        <me>\vol(P) = \vol(A) = |\det(A)|.</me>
        This completes the proof.
      </p>
    </proof>
  </theorem>

  <p>
    Since <m>\det(A) = \det(A^T)</m> by the <xref ref="det-defn-trans-prop" text="title">transpose property</xref>, the absolute value of <m>\det(A)</m> is also equal to the volume of the paralellepiped determined by the <em>columns</em> of <m>A</m> as well.
  </p>

  <specialcase>
    <title>Length</title>
    <idx><h>Volume</h><h>and length</h></idx>
    <p>
      A <m>1\times 1</m> matrix <m>A</m> is just a number <m>\mat a</m>.  In this case, the parallelepiped <m>P</m> determined by its one row is just the interval <m>[0,a]</m> (or <m>[a,0]</m> if <m>a\lt0</m>).  The <q>volume</q> of a region in <m>\R^1 = \R</m> is just its length, so it is clear in this case that <m>\vol(P) = |a|</m>.
      <latex-code>
\begin{tikzpicture}
  \draw[very thick, draw=seq-orange] (0,0) -- node[above] {$\vol(\textcolor{seq-orange}{P}) = |a|$} (3,0);
  \draw       (0,.15) -- (0,-.15) node[below] {$0$}
              (3,.15) -- (3,-.15) node[below] {$a$};
\end{tikzpicture}
      </latex-code>
    </p>
  </specialcase>

  <specialcase>
    <title>Area</title>
    <idx><h>Parallelepiped</h><h>parallelogram</h><h>area of</h></idx>
    <p>
      When <m>A</m> is a <m>2\times 2</m> matrix, its rows determine a parallelogram in <m>\R^2</m>.  The <q>volume</q> of a region in <m>\R^2</m> is its area, so we obtain a formula for the area of a parallelogram: it is the determinant of the matrix whose rows are the vectors forming two adjacent sides of the parallelogram.
      <latex-code>
\begin{tikzpicture}[thin border nodes, scale=1.3]
  \filldraw[fill=seq-orange, fill opacity=.2, very thin]
    (0,0) -- (1,2) -- (3,3) -- (2,1) -- cycle;
  \draw[vector] (0,0) -- node[auto]{$\vec{a b}$} (1,2);
  \draw[vector] (0,0) -- node[auto,swap,opacity=1]{$\vec{c d}$} (2,1);
  \node at (6, 1.5) {area${}=\left|\det\mat{a b; c d}\right| = |ad-bc|$};
\end{tikzpicture}
      </latex-code>
      It is perhaps surprising that it is possible to compute the area of a parallelogram without trigonometry.  It is a fun geometry problem to prove this formula by hand.  [Hint: first think about the case when the first row of <m>A</m> lies on the <m>x</m>-axis.]
    </p>
  </specialcase>

  <example>
    <statement>
      <p>
        Find the area of the parallelogram with sides <m>(1,3)</m> and <m>(2,-3)</m>.
        <latex-code>
\begin{tikzpicture}[thin border nodes, scale=1]
  \draw[help lines] (-1,-4) grid (4,4);
  \filldraw[fill=seq-orange, fill opacity=.2, very thin]
    (0,0) -- (1,3) -- (3,0) -- (2,-3) -- cycle;
  \draw[vector] (0,0) -- (1,3);
  \draw[vector] (0,0) -- (2,-3);
\end{tikzpicture}
        </latex-code>
      </p>
    </statement>
    <solution>
      <p>
        The area is
        <me>
          \left|\det\mat{1 3; 2 -3}\right| = |-3-6| = 9.
        </me>
      </p>
    </solution>
  </example>

  <example>
    <statement>
      <p>
        Find the area of the parallelogram in the picture.
        <latex-code>
\begin{tikzpicture}[thin border nodes, scale=1]
  \draw[help lines] (-3,-1) grid (2,6);
  \filldraw[fill=seq-orange, fill opacity=.2, very thin]
    (0,0) -- (-2,1) -- (-1,5) -- (1,4) -- cycle;
\end{tikzpicture}
        </latex-code>
      </p>
    </statement>
    <solution>
      <p>
        We choose two adjacent sides to be the rows of a matrix.  We choose the top two:
        <latex-code>
\begin{tikzpicture}[thin border nodes, scale=1]
  \draw[help lines] (-3,-1) grid (2,6);
  \filldraw[fill=seq-orange, fill opacity=.2, very thin]
    (0,0) -- (-2,1) -- (-1,5) -- (1,4) -- cycle;
  \draw[vector] (-1,5) -- node[auto,whitebg,behind path] {$\vec{2 -1}$} (1,4);
  \draw[vector] (-1,5) -- node[auto,swap,whitebg,behind path] {$\vec{-1 -4}$} (-2,1);
\end{tikzpicture}
        </latex-code>
        Note that we do not need to know where the origin is in the picture: vectors are determined by their length and direction, not where they start.  The area is
        <me>
          \left|\det\mat{-1 -4; 2 -1}\right| = |1+8| = 9.
        </me>
      </p>
    </solution>
  </example>

  <example>
    <title>Area of a triangle</title>
    <idx><h>Triangle</h><h>area of</h><see>Volume</see></idx>
    <idx><h>Volume</h><h>of a triangle</h></idx>
    <statement>
      <p>
        Find the area of the triangle with vertices
        <m>(-1,-2), \,(2,-1),\,(1,3).</m>
        <latex-code>
\begin{tikzpicture}[thin border nodes, scale=1]
  \draw[help lines] (-2,-3) grid (3,4);
  \filldraw[fill=seq-orange, fill opacity=.2, very thin]
    (-1,-2) -- (2,-1) -- (1,3) -- cycle;
\end{tikzpicture}
        </latex-code>
      </p>
    </statement>
    <solution>
      <p>
        Doubling a triangle makes a paralellogram.  We choose two of its sides to be the rows of a matrix.
        <latex-code>
\begin{tikzpicture}[thin border nodes, scale=1]
  \draw[help lines] (-2,-3) grid (5,5);
  \filldraw[fill=seq-orange, fill opacity=.2, very thin]
    (-1,-2) -- (2,-1) -- (1,3) -- cycle;
  \filldraw[fill=seq-orange, fill opacity=.1, very thin, dashed]
    (2,-1) -- (4,4) -- (1,3);
  \draw[vector] (-1,-2) -- node[auto,whitebg,swap,behind path] {$\vec{3 1}$} (2,-1);
  \draw[vector] (-1,-2) -- node[auto,whitebg,behind path] {$\vec{2 5}$} (1,3);
\end{tikzpicture}
        </latex-code>
        The area of the parallelogram is
        <me>
          \left|\det\mat{2 5; 3 1}\right| = |2-15| = 13,
        </me>
        so the area of the triangle is <m>13/2</m>.
      </p>
    </solution>
  </example>

  <p>
    You might be wondering: if the absolute value of the determinant is a volume, what is the geometric meaning of the determinant without the absolute value?  The next remark explains that we can think of the determinant as a <em>signed</em> volume.  If you have taken an integral calculus course, you probably computed negative areas under curves; the idea here is similar.
  </p>

  <remark>
    <title>Signed volumes</title>
    <idx><h>Volume</h><h>signed</h></idx>
    <p>
      The <xref ref="det-is-volume"/> on determinants and volumes tells us that the <em>absolute value</em> of the determinant is the volume of a paralellepiped.  This raises the question of whether the sign of the determinant has any geometric meaning.
    </p>
    <p>
      A <m>1\times 1</m> matrix <m>A</m> is just a number <m>\mat a</m>.  In this case, the parallelepiped <m>P</m> determined by its one row is just the interval <m>[0,a]</m> if <m>a \geq 0</m>, and it is <m>[a,0]</m> if <m>a\lt0</m>.  In this case, the sign of the determinant determines whether the interval is to the left or the right of the origin.
    </p>
    <p>
      For a <m>2\times 2</m> matrix with rows <m>v_1,v_2</m>, the sign of the determinant determines whether <m>v_2</m> is counterclockwise or clockwise from <m>v_1</m>.  That is, if the counterclockwise angle from <m>v_1</m> to <m>v_2</m> is less than <m>180^\circ</m>, then the determinant is positive; otherwise it is negative (or zero).
      <latex-code>
        <![CDATA[
\usetikzlibrary{angles}
\begin{tikzpicture}[thin border nodes, scale=1.3]
  \filldraw[fill=seq-orange, fill opacity=.2, very thin]
    (0,0) -- (1,2) -- (3,3) -- (2,1) -- cycle;
  \draw[vector] (0,0) -- node[auto]{$v_2$} (1,2);
  \draw[vector] (0,0) -- node[auto,swap]{$v_1$} (2,1);
  \coordinate (o) at (0,0);
  \coordinate (v1) at (2,1);
  \coordinate (v2) at (1,2);
  \pic[draw, ->, angle radius=1cm] {angle=v1--o--v2};
  \node at (1.5,-.7) {$\det\mat{\matrow{v_1}; \matrow{v_2}} > 0$};
\end{tikzpicture}
\qquad
\begin{tikzpicture}[thin border nodes, scale=1.3]
  \filldraw[fill=seq-orange, fill opacity=.2, very thin]
    (0,0) -- (1,2) -- (3,3) -- (2,1) -- cycle;
  \draw[vector] (0,0) -- node[auto]{$v_1$} (1,2);
  \draw[vector] (0,0) -- node[auto,swap]{$v_2$} (2,1);
  \coordinate (o) at (0,0);
  \coordinate (v1) at (2,1);
  \coordinate (v2) at (1,2);
  \pic[draw, <-, angle radius=1cm] {angle=v1--o--v2};
  \node at (1.5,-.7) {$\det\mat{\matrow{v_1}; \matrow{v_2}} < 0$};
  \end{tikzpicture}
        ]]>
      </latex-code>
      For example, if <m>v_1 = {a\choose b}</m>, then the counterclockwise rotation of <m>v_1</m> by <m>90^\circ</m> is <m>v_2 = {-b\choose a}</m> by this <xref ref="linear-trans-rotation-matrix"/>, and
      <me>
        \det\mat{a b; -b a} = a^2 + b^2 > 0.
      </me>
      On the other hand, the <em>clockwise</em> rotation of <m>v_1</m> by <m>90^\circ</m> is <m>b\choose -a</m>, and
      <me>
        \det\mat{a b; b -a} = -a^2 - b^2 \lt 0.
      </me>
    </p>
    <p>
      For a <m>3\times 3</m> matrix with rows <m>v_1,v_2,v_3</m>, the <em>right-hand rule</em> determines the sign of the determinant.  If you point the index finger of your right hand in the direction of <m>v_1</m> and your middle finger in the direction of <m>v_2</m>, then the determinant is positive if your thumb points roughly in the direction of <m>v_3</m>, and it is negative otherwise.
      <latex-code>
\begin{tikzpicture}[thin border nodes, myxyz, scale=2]
  \filldraw[fill=seq-orange, opacity=.2, very thin]
    (0,0,0) -- (-1,0,0) -- (-1,.3,1) -- (0,.3,1) -- cycle;
  \filldraw[fill=seq-orange, opacity=.2, very thin]
    (0,0,0) -- (-1,0,0) -- (-.8,1,0) -- (.2,1,0) -- cycle;
  \filldraw[fill=seq-orange, opacity=.2, shift={(-1,0,0)}, very thin]
    (0,0,0) -- (.2,1,0) -- (.2,1.3,1) -- (0,.3,1) -- cycle;
  \filldraw[fill=seq-orange, fill opacity=.2, very thin]
    (0,0,0) -- (.2,1,0) -- (.2,1.3,1) -- (0,.3,1) -- cycle;
  \filldraw[fill=seq-orange, fill opacity=.2, shift={(.2,1,0)}, very thin]
    (0,0,0) -- (-1,0,0) -- (-1,.3,1) -- (0,.3,1) -- cycle;
  \filldraw[fill=seq-orange, fill opacity=.2, shift={(0,.3,1)}, very thin]
    (0,0,0) -- (-1,0,0) -- (-.8,1,0) -- (.2,1,0) -- cycle;
  \draw[vector] (0,0,0) -- node[auto,swap] {$v_1$} (.2,1,0);
  \draw[vector, opacity=.4] (0,0,0) -- node[auto] {$v_2$} (-1,0,0);
  \draw[vector] (0,0,0) -- node[auto] {$v_3$} (0,.3,1);
  \node[resetxy] at (1,-.75) {$\det\mat{\matrow{v_1};\matrow{v_2};\matrow{v_3}}>0$};
\end{tikzpicture}
\qquad
\begin{tikzpicture}[thin border nodes, myxyz, scale=2]
  \filldraw[fill=seq-orange, opacity=.2, very thin]
    (0,0,0) -- (-1,0,0) -- (-1,.3,1) -- (0,.3,1) -- cycle;
  \filldraw[fill=seq-orange, opacity=.2, very thin]
    (0,0,0) -- (-1,0,0) -- (-.8,1,0) -- (.2,1,0) -- cycle;
  \filldraw[fill=seq-orange, opacity=.2, shift={(-1,0,0)}, very thin]
    (0,0,0) -- (.2,1,0) -- (.2,1.3,1) -- (0,.3,1) -- cycle;
  \filldraw[fill=seq-orange, fill opacity=.2, very thin]
    (0,0,0) -- (.2,1,0) -- (.2,1.3,1) -- (0,.3,1) -- cycle;
  \filldraw[fill=seq-orange, fill opacity=.2, shift={(.2,1,0)}, very thin]
    (0,0,0) -- (-1,0,0) -- (-1,.3,1) -- (0,.3,1) -- cycle;
  \filldraw[fill=seq-orange, fill opacity=.2, shift={(0,.3,1)}, very thin]
    (0,0,0) -- (-1,0,0) -- (-.8,1,0) -- (.2,1,0) -- cycle;
  \draw[vector] (0,0,0) -- node[auto,swap] {$v_2$} (.2,1,0);
  \draw[vector, opacity=.4] (0,0,0) -- node[auto] {$v_1$} (-1,0,0);
  \draw[vector] (0,0,0) -- node[auto] {$v_3$} (0,.3,1);
  \node[resetxy] at (1,-.75) {$\det\mat{\matrow{v_1};\matrow{v_2};\matrow{v_3}}\lt0$};
\end{tikzpicture}
      </latex-code>
    </p>
    <p>
      In higher dimensions, the notion of signed volume is still important, but it is usually <em>defined</em> in terms of the sign of a determinant.
    </p>
  </remark>

  </subsection>


  <subsection>
    <title>Volumes of Regions</title>

    <p>
      Let <m>A</m> be an <m>n\times n</m> matrix with columns <m>v_1,v_2,\ldots,v_n</m>, and let <m>T\colon\R^n\to\R^n</m> be the associated <xref ref="matrix-trans-defn-of" text="title">matrix transformation</xref> <m>T(x)=Ax</m>.  Then <m>T(e_1)=v_1</m> and <m>T(e_2)=v_2</m>, so <m>T</m> takes the unit cube <m>C</m> to the parallelepiped <m>P</m> determined by <m>v_1,v_2,\ldots,v_n</m>:
      <latex-code>
\begin{tikzpicture}[thin border nodes, scale=1.3]
  \begin{scope}[scale=1.7, yshift=.3cm]
  \filldraw[fill=seq-orange, fill opacity=.2, very thin]
    (0,0) -- (1,0) -- (1,1) -- (0,1) -- cycle;
  \node at (.5, .5) {$C$};
  \draw[vector] (0,0) -- node[auto]{$e_2$} (0,1);
  \draw[vector, opacity=.75] (0,0) -- node[auto,swap]{$e_1$} (1,0);
  \coordinate (a) at (1.2, .5);
  \end{scope}

  \draw[->, thick] (a) to[bend left]
      node[above=2pt] {$T$}
      node[below=4pt] {$\mat{| |; v_1 v_2; | |}$}
      ($(a) + (2.8cm,0)$);

  \begin{scope}[xshift=5cm]
  \filldraw[fill=seq-orange, fill opacity=.2, very thin]
    (0,0) -- (1,2) -- (3,3) -- (2,1) -- cycle;
  \draw[vector] (0,0) -- node[auto]{$v_2$} (1,2);
  \draw[vector] (0,0) -- node[auto,swap]{$v_1$} (2,1);
  \node at (1.5, 1.5) {$P$};
  \end{scope}
\end{tikzpicture}
      </latex-code>
      Since the unit cube has volume <m>1</m> and its image has volume <m>|\det(A)|</m>, the transformation <m>T</m> scaled the volume of the cube by a factor of <m>|\det(A)|</m>.  To rephrase:
    </p>

    <bluebox>
      <p>
        If <m>A</m> is an <m>n\times n</m> matrix with corresponding matrix transformation <m>T\colon\R^n\to\R^n</m>, and if <m>C</m> is the unit cube in <m>\R^n</m>, then the volume of <m>T(C)</m> is <m>|\det(A)|.</m>
      </p>
    </bluebox>

    <p>
      <notation><usage>T(S)</usage><description>The image of a region under a transformation</description></notation>
      The notation <m>T(S)</m> means the image of the region <m>S</m> under the transformation <m>T</m>.  In <xref ref="set-builder-notation" text="title">set builder notation</xref>, this is the subset
      <me>T(S) = \bigl\{T(x)\mid x \text{ in } S\bigr\}. </me>
    </p>

    <p>
      In fact, <m>T</m> scales the volume of <em>any</em> region in <m>\R^n</m> by the same factor, even for curvy regions.
    </p>

    <theorem xml:id="det-vol-regions">
      <idx><h>Volume</h><h>of a region</h></idx>
      <idx><h>Matrix transformation</h><h>and volumes</h></idx>
      <idx><h>Linear transformation</h><h>and volumes</h><see>Matrix transformation</see></idx>
      <statement>
        <p>
          Let <m>A</m> be an <m>n\times n</m> matrix, and let <m>T\colon\R^n\to\R^n</m> be the associated matrix transformation <m>T(x)=Ax</m>.  If <m>S</m> is any region in <m>\R^n</m>, then
          <me>
	    \vol(T(S))=|\det(A)|\cdot\vol(S).
          </me>
        </p>
      </statement>
      <proof visible="true">
        <p>
          Let <m>C</m> be the unit cube, let <m>v_1,v_2,\ldots,v_n</m> be the columns of <m>A</m>, and let <m>P</m> be the paralellepiped determined by these vectors, so <m>T(C) = P</m> and <m>\vol(P) = |\det(A)|</m>.  For <m>\epsilon > 0</m> we let <m>\epsilon C</m> be the cube with side lengths <m>\epsilon</m>, i.e., the paralellepiped determined by the vectors <m>\epsilon e_1,\epsilon e_2,\ldots,\epsilon e_n</m>, and we define <m>\epsilon P</m> similarly.  By the second <xref ref="linear-trans-defn" text="title">defining property</xref>, <m>T</m> takes <m>\epsilon C</m> to <m>\epsilon P</m>.  The volume of <m>\epsilon C</m> is <m>\epsilon^n</m> (we scaled each of the <m>n</m> standard vectors by a factor of <m>\epsilon</m>) and the volume of <m>\epsilon P</m> is <m>\epsilon^n|\det(A)|</m> (for the same reason), so we have shown that <m>T</m> scales the volume of <m>\epsilon C</m> by <m>|\det(A)|</m>.
          <latex-code>
\begin{tikzpicture}[thin border nodes, scale=1.3]
  \begin{scope}[scale=1.7, yshift=.3cm]
  \filldraw[very thin, black!50!white, fill=seq-orange!10!white]
    (0,0) -- (1,0) -- (1,1) -- (0,1) -- cycle;
  \filldraw[fill=seq-orange, fill opacity=.2, very thin]
    (0,0) -- (.6,0) -- (.6,.6) -- (0,.6) -- cycle;
  \draw[vector, black!50!white] (0, 0) -- (1, 0);
  \draw[vector, black!50!white] (0, 0) -- (0, 1);
  \draw[vector] (0,0) -- node[auto]{$\epsilon e_2$} (0,.6);
  \draw[vector, opacity=.75] (0,0) -- node[auto,swap]{$\epsilon e_1$} (.6,0);
  \node at (.3, .3) {$\epsilon C$};
  \coordinate (a) at (1.2, .5);
  \end{scope}

  \node at (1.7/2, -.5) {$\vol(\epsilon C) = \epsilon^n$};

  \draw[->, thick] (a) to[bend left]
      node[above=2pt] {$T$}
      node[below=4pt] {$\mat{| |; v_1 v_2; | |}$}
      ($(a) + (2.8cm,0)$);

  \begin{scope}[xshift=5cm]
  \filldraw[black!50!white, fill=seq-orange!10!white, very thin]
    (0,0) -- (1,2) -- (3,3) -- (2,1) -- cycle;
  \filldraw[fill=seq-orange, fill opacity=.2, very thin]
    (0,0) -- (.6,1.2) -- (1.8,1.8) -- (1.2,.6) -- cycle;
  \draw[vector, black!50!white] (0, 0) -- (1,2);
  \draw[vector, black!50!white] (0, 0) -- (2,1);
  \draw[vector] (0,0) -- node[auto]{$\epsilon v_2$} (.6,1.2);
  \draw[vector] (0,0) -- node[auto,swap]{$\epsilon v_1$} (1.2,.6);
  \node at (.9,.9) {$\epsilon P$};
  \end{scope}

  \node at (6.3, -.5) {$\vol(\epsilon P) = \epsilon^n|\det(A)|$};
\end{tikzpicture}
          </latex-code>
        </p>
        <p>
          By the first <xref ref="linear-trans-defn" text="title">defining property</xref>, the image of a translate of <m>\epsilon C</m> is a translate of <m>\epsilon P</m>:
          <me>T(x + \epsilon C) = T(x) + \epsilon T(C) = T(x) + \epsilon P.</me>
          Since a translation does not change volumes, this proves that <m>T</m> scales the volume of a translate of <m>\epsilon C</m> by <m>|\det(A)|</m>.
        </p>
        <p>
          At this point, we need to use techniques from multivariable calculus, so we only give an idea of the rest of the proof.  Any region <m>S</m> can be approximated by a collection of very small cubes of the form <m>x + \epsilon C</m>.  The image <m>T(S)</m> is then approximated by the image of this collection of cubes, which is a collection of very small paralellepipeds of the form <m>T(x) + \epsilon P</m>.
          <latex-code>
\def\curvy{(-37.3333, 2.6667)
     .. controls (-37.3333, -5.3333) and (-26.6667, -10.6667) .. (-16, -24)
     .. controls (-5.3333, -37.3333) and (5.3333, -58.6667) .. (18.6667, -48)
     .. controls (32, -37.3333) and (48, 5.3333) .. (45.3333, 18.6667)
     .. controls (42.6667, 32) and (21.3333, 16) .. (8, 13.3333)
     .. controls (-5.3333, 10.6667) and (-10.6667, 21.3333) .. (-18.6667, 21.3333)
     .. controls (-26.6667, 21.3333) and (-37.3333, 10.6667) .. cycle}

\begin{tikzpicture}[scale=1, thin border nodes]
  \draw[help lines, black!25] (-2,-2) rectangle (2,2);
  \begin{scope}[fill=seq-orange, fill opacity=.2, thin,
      x=1.7*.7bp, y=1.7*.7bp, yshift=.4cm, xshift=-.2cm]
    \expandafter\filldraw\curvy;
    \expandafter\clip\curvy;
    \fill[seq-red, opacity=.5, shift={(2*4,-4*4)}] (0,0) rectangle (4,4);
    \coordinate (a) at (2*4,-4*4);
    \draw[very thin, black!70!white, step=4] (-60,-60) grid (60,60);
  \end{scope}

  \draw[thin] (-1.2, -1.2) node[below] {$x + \epsilon C$} -- (a);

  \node at (0,1.5) {$S$};
  %\point at (0,0);

  \draw[->, thick] (2.1,0) to[bend left, "$T$" above=1mm]
    (8-2.1,0);

  \begin{scope}[xshift=8cm]
    \draw[help lines, black!25] (-2,-2) rectangle (2,2);
    \begin{scope}[cm={2,1,1,2,(0,.5)}, x=.7bp, y=.7bp,
        fill=seq-orange, fill opacity=.2, thin]
      \expandafter\filldraw\curvy;
      \expandafter\clip\curvy;
      \fill[seq-red, opacity=.5, shift={(2*4,-4*4)}] (0,0) rectangle (4,4);
      \draw[very thin, black!70!white, step=4] (-60,-60) grid (60,60);
    \coordinate (b) at (3*4,-4*4);
  \end{scope}

  \draw[thin] (.9,-1.3) node[below] {$T(x) + \epsilon P$} -- (b);

  \node at (-.7,1.5) {$T(S)$};
  % \point at (0,0);
  \end{scope}
\end{tikzpicture}
          </latex-code>
          The volume of <m>S</m> is closely approximated by the sum of the volumes of the cubes; in fact, as <m>\epsilon</m> goes to zero, the limit of this sum is precisely <m>\vol(S)</m>.  Likewise, the volume of <m>T(S)</m> is equal to the sum of the volumes of the paralellepipeds, take in the limit as <m>\epsilon\to 0</m>.  The key point is that <em>the volume of each cube is scaled by <m>|\det(A)|</m>.</em>  Therefore, the sum of the volumes of the paralellepipeds is <m>|\det(A)|</m> times the sum of the volumes of the cubes.  This proves that <m>\vol(T(S)) = |\det(A)|\vol(S).</m>
        </p>
      </proof>
    </theorem>

    <example>
      <statement>
        <p>
          Let <m>S</m> be a half-circle of radius <m>1</m>, let
          <me>A = \mat{1 2; 2 1},</me>
          and define <m>T\colon\R^2\to\R^2</m> by <m>T(x)=Ax</m>.  What is the area of <m>T(S)</m>?
          <latex-code>
\begin{tikzpicture}
  \begin{scope}[yshift=-.3cm]
    \draw[grid lines] (-1.5,-.5) grid (1.5, 1.5);
    \filldraw[fill=seq-orange, fill opacity=.2, thin]
      (1,0) arc[start angle=0, end angle=180, radius=1] -- cycle;
    \node at (0,.5) {$S$};
  \end{scope}

  \draw[->, thick] (2.1,0) to[bend left, "$T$" above=1mm]
    node[below=1mm]{$\mat{1 2; 2 1}$} (7-2.1,0);

  \begin{scope}[xshift=7cm]
    \draw[grid lines] (-1.5,-2.5) grid (2.5, 2.5);
    \begin{scope}[cm={1,2,2,1,(0,0)}]
      \filldraw[fill=seq-orange, fill opacity=.2, thin]
        (1,0) arc[start angle=0, end angle=180, radius=1] -- cycle;
      \node at (0,.5) {$T(S)$};
    \end{scope}
  \end{scope}

\end{tikzpicture}
          </latex-code>
        </p>
      </statement>
      <solution>
        <p>
          The area of the unit circle is <m>\pi</m>, so the area of <m>S</m> is <m>\pi/2</m>.  The transformation <m>T</m> scales areas by a factor of <m>|\det(A)| = |1 - 4| = 3</m>, so
          <me>\vol(T(S)) = 3\vol(S) = \frac{3\pi}2.</me>
        </p>
      </solution>
    </example>

    <example>
      <title>Area of an ellipse</title>
      <idx><h>Volume</h><h>of an ellipse</h></idx>
      <idx><h>Ellipse</h><h>area of</h><see>Volume</see></idx>
      <statement>
        <p>
          Find the area of the interior <m>E</m> of the ellipse defined by the equation
          <me>\left(\frac{2x-y}{2}\right)^2 + \left(\frac{y+3x}{3}\right)^2 = 1.</me>
        </p>
      </statement>
      <solution>
        <p>
          This ellipse is obtained from the unit circle <m>X^2+Y^2=1</m> by the linear change of coordinates
          <me>
            \begin{split}
            X \amp= \frac{2x-y}2 \\
            Y \amp= \frac{y+3x}3.
            \end{split}
          </me>
          In other words, if we define a linear transformation <m>T\colon\R^2\to\R^2</m> by
          <me>
            T\vec{x y} = \vec{(2x-y)/2 (y+3x)/3},
          </me>
          then <m>T{x\choose y}</m> lies on the unit circle <m>C</m> whenever <m>x\choose y</m> lies on <m>E</m>.
          <latex-code>
\begin{tikzpicture}
  \begin{scope}[xshift=7cm]
    \draw[grid lines] (-1.5,-1.5) grid (1.5, 1.5);
    \filldraw[fill=seq-orange, fill opacity=.2, thin]
      (0,0) circle[radius=1];
    \node at (0,0) {$C$};
  \end{scope}

  \draw[->, thick] (2.1,0) to[bend left, "$T$" above=1mm]
    node[below=2mm]{$\mat{1 -1/2; 1 1/3}$} (7-2.1,0);

  \begin{scope}[xshift=0cm]
    \draw[grid lines] (-1.5,-2.5) grid (1.5, 2.5);
    \begin{scope}[cm={2/5,-6/5,3/5,6/5,(0,0)}]
      \filldraw[fill=seq-orange, fill opacity=.2, thin]
        (0,0) circle[radius=1];
      \node at (0,0) {$E$};
    \end{scope}
  \end{scope}

\end{tikzpicture}
          </latex-code>
          We compute the standard matrix <m>A</m> for <m>T</m> by evaluating on the standard coordinate vectors:
          <me>
            T\vec{1 0} = \vec{1 1} \qquad T\vec{0 1} = \vec{-1/2 1/3}
            \quad\implies\quad A = \mat{1 -1/2; 1 1/3}.
          </me>
          Therefore, <m>T</m> scales areas by a factor of <m>|\det(A)| = |\frac13+\frac12| = \frac56.</m>  The area of the unit circle is <m>\pi</m>, so
          <me>\pi = \vol(C) = \vol(T(E)) = |\det(A)|\cdot\vol(E) = \frac56\vol(E),</me>
          and thus the area of <m>E</m> is <m>6\pi/5</m>.
        </p>
      </solution>
    </example>

    <remark>
      <title>Multiplicativity of <m>|\det|\,</m></title>
      <idx><h>Determinant</h><h>multiplicativity property</h><h>and volumes</h></idx>
    <p>
      The above <xref ref="det-vol-regions"/> also gives a geometric reason for multiplicativity of the (absolute value of the) determinant.  Indeed, let <m>A</m> and <m>B</m> be <m>n\times n</m> matrices, and let <m>T,U\colon\R^n\to\R^n</m> be the corresponding matrix transformations.  If <m>C</m> is the unit cube, then
      <me>
        \begin{split}
        \vol\bigl(T\circ U(C)\bigr) = \vol\bigl(T(U(C))\bigr)
        \amp= |\det(A)|\vol(U(C)) \\
        \amp= |\det(A)|\cdot|\det(B)|\vol(C) \\
        \amp= |\det(A)|\cdot|\det(B)|.
        \end{split}
      </me>
      On the other hand, the matrix for the composition <m>T\circ U</m> is the product <m>AB</m>, so
      <me>
        \vol\bigl(T\circ U(C)\bigr) = |\det(AB)|\vol(C) = |\det(AB)|.
      </me>
      Thus <m>|\det(AB)| = |\det(A)|\cdot|\det(B)|</m>.
    </p>
    </remark>

  </subsection>

</section>