determinant-definitions-properties.xml

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<section xml:id="determinants-definitions-properties">
  <title>Determinants: Definition</title>

  <objectives>
    <ol>
      <li>Learn the definition of the determinant.</li>
      <li>Learn some ways to eyeball a matrix with zero determinant, and how to compute determinants of upper- and lower-triangular matrices.</li>
      <li>Learn the basic properties of the determinant, and how to apply them.</li>
      <li><em>Recipe:</em> compute the determinant using row and column operations.</li>
      <li><em>Theorems:</em> existence theorem, invertibility property, multiplicativity property, transpose property.</li>
      <li><em>Vocabulary words:</em> <term>diagonal</term>, <term>upper-triangular</term>, <term>lower-triangular</term>, <term>transpose</term>.</li>
      <li><em>Essential vocabulary word:</em> <term>determinant</term>.</li>
    </ol>
  </objectives>

  <introduction>
    <p>
      In this section, we define the determinant, and we present one way to compute it.  Then we discuss some of the many wonderful properties the determinant enjoys.
    </p>

  </introduction>

  <subsection>
    <title>The Definition of the Determinant</title>

    <p>
      The determinant of a square matrix <m>A</m> is a real number <m>\det(A)</m>.  It is defined via its behavior with respect to row operations; this means we can use row reduction to compute it.  We will give a recursive formula for the determinant in <xref ref="determinants-cofactors"/>.  We will also show in this <xref ref="det-defn-magic-props"/> that the determinant is related to invertibility, and in <xref ref="determinants-volumes"/> that it is related to volumes.
    </p>

    <essential xml:id="det-defn-the-defn">
      <idx><h>Determinant</h><h>defining properties of</h></idx>
      <idx><h>Matrix</h><h>determinant of</h><see>Determinant</see></idx>
      <idx><h>Determinant</h><h>identity matrix</h></idx>
      <idx><h>Identity matrix</h><h>determinant of</h></idx>
      <idx><h>Row operations</h><h>and determinants</h></idx>
      <idx><h>Determinant</h><h>and row operations</h></idx>
      <notation><usage>\det(A)</usage><description>The determinant of a matrix</description></notation>
      <statement>
	<p>
	  The <term>determinant</term> is a function
          <me>\det\colon \bigl\{\text{square matrices}\bigr\}\To\R</me>
          satisfying the following 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 the determinant by <m>c</m>.
	    </li>
	    <li>
	      Swapping two rows of a matrix multiplies the determinant by <m>-1</m>.
	    </li>
	    <li>
	      The determinant of the identity matrix <m>I_n</m> is equal to <m>1</m>.
	    </li>
	  </ol>
	</p>
      </statement>
    </essential>

    <p>
      In other words, to every square matrix <m>A</m> we assign a number <m>\det(A)</m> in a way that satisfies the above properties.
    </p>

    <p>
      In each of the first three cases, doing a row operation on a matrix scales the determinant by a <em>nonzero</em> number.  (Multiplying a row by zero is not a row operation.)  Therefore, doing row operations on a square matrix <m>A</m> does not change whether or not the determinant is zero.
    </p>

    <p>
      The main motivation behind using these particular defining properties is geometric:
      see <xref ref="determinants-volumes"/>.  Another motivation for this definition is that it tells us how to compute the determinant: we row reduce and keep track of the changes.
    </p>

    <specialcase>
      <p>
        Let us compute <m>\det\smallmat2114.</m>
        First we row reduce, then we compute the determinant in the opposite order:
          <md>
            <mrow>
              \amp\mat{2 1; 1 4} \amp\strut\det\amp=7
            </mrow>
            <mrow>
  \;\xrightarrow{\hbox to 1.7cm{\hss\tiny$R_1\longleftrightarrow R_2$\hss}}\;\amp
    \mat{1 4; 2 1} \amp \strut\det \amp= -7
            </mrow>
            <mrow>
  \;\xrightarrow{\hbox to 1.7cm{\hss\tiny$R_2 = R_2 - 2R_1$\hss}}\;\amp
    \mat{1 4; 0 -7} \amp \strut\det \amp= -7
            </mrow>
            <mrow>
  \;\xrightarrow{\hbox to 1.7cm{\hss\tiny$R_2 = R_2 \div -7$\hss}}\;\amp
    \mat{1 4; 0 1} \amp \strut\det \amp= 1
            </mrow>
            <mrow>
  \;\xrightarrow{\hbox to 1.7cm{\hss\tiny$R_1 = R_1 - 4R_2$\hss}}\;\amp
    \mat{1 0; 0 1} \amp \strut\det \amp= 1
            </mrow>
          </md>
          The reduced row echelon form of the matrix is the identity matrix <m>I_2</m>, so its determinant is <m>1</m>.  The second-last step in the row reduction was a row replacement, so the second-final matrix also has determinant <m>1</m>.  The previous step in the row reduction was a row scaling by <m>-1/7</m>; since (the determinant of the second matrix times <m>-1/7</m>) is <m>1</m>, the determinant of the second matrix must be <m>-7</m>.  The first step in the row reduction was a row swap, so the determinant of the first matrix is negative the determinant of the second.  Thus, the determinant of the original matrix is <m>7</m>.
      </p>
      <p>
        Note that our answer agrees with this <xref ref="matrix-inv-def-det" text="title">definition</xref> of the determinant.
      </p>
    </specialcase>

    <example>
      <statement>
	<p>Compute <m>\det\mat{1 0; 0 3}.</m></p>
      </statement>
      <solution>
        <p>
	  Let <m>A=\mat{1 0; 0 3}</m>. Since <m>A</m> is obtained from <m>I_2</m> by multiplying the second row by the constant <m>3</m>, we have
          <me>\det(A)=3\det(I_2)=3\cdot 1=3.</me>
	</p>
        <p>
          Note that our answer agrees with this <xref ref="matrix-inv-def-det" text="title">definition</xref> of the determinant.
        </p>
      </solution>
    </example>

    <example>
      <statement>
	<p>
          Compute <m>\det\mat{1 0 0;0 0 1;5 1 0}.</m>
        </p>
      </statement>
      <solution>
        <p>
          First we row reduce, then we compute the determinant in the opposite order:
          <md>
            <mrow>
  \amp\mat{1 0 0;0 0 1;5 1 0} \amp\strut\det\amp=-1
            </mrow>
            <mrow>
  \;\xrightarrow{\hbox to 1.7cm{\hss\tiny$R_2\longleftrightarrow R_3$\hss}}\;\amp
    \mat{1 0 0;5 1 0; 0 0 1} \amp \strut\det \amp= 1
            </mrow>
            <mrow>
  \;\xrightarrow{\hbox to 1.7cm{\hss\tiny$R_2 = R_2 - 5R_1$\hss}}\;\amp
    \mat{1 0 0; 0 1 0; 0 0 1} \amp \strut\det \amp= 1
            </mrow>
          </md>
          The reduced row echelon form is <m>I_3</m>, which has determinant <m>1</m>.
          Working backwards from <m>I_3</m> and using the four <xref ref="det-defn-the-defn">defining properties</xref>, we see that the second matrix also has determinant <m>1</m> (it differs from <m>I_3</m> by a row replacement), and the first matrix has determinant <m>-1</m> (it differs from the second by a row swap).
        </p>
      </solution>
    </example>

    <p>
      Here is the general method for computing determinants using row reduction.
    </p>

    <bluebox xml:id="det-defn-ref-compute" type-name="Recipe">
      <title>Recipe: Computing determinants by row reducing</title>
      <idx><h>Determinant</h><h>computation of</h><h>row reduction</h></idx>
      <idx><h>Row reduction</h><h>computing determinants</h></idx>
	<p>
          Let <m>A</m> be a square matrix.  Suppose that 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) = (-1)^r\cdot
            \frac{\text{(product of the diagonal entries of $B$)}}
            {\text{(product of scaling factors used)}},
	  </me>
          where <m>r</m> is the number of row swaps performed.
	</p>
    </bluebox>

    <p>
      In other words, the determinant of <m>A</m> is the product of diagonal entries of the row echelon form <m>B</m>, times a factor of <m>\pm1</m> coming from the number of row swaps you made, divided by the product of the scaling factors used in the row reduction.
    </p>

    <remark>
      <p>
        This is an efficient way of computing the determinant of a large matrix, either by hand or by computer.  The computational complexity of row reduction is <m>O(n^3)</m>; by contrast, the cofactor expansion algorithm we will learn in <xref ref="determinants-cofactors"/> has complexity <m>O(n!)\approx O(n^n\sqrt n)</m>, which is much larger.  (Cofactor expansion has other uses.)
      </p>
    </remark>

    <example>
      <statement>
        <p>Compute <m>\det\mat{0 -7 -4; 2 4 6; 3 7 -1}.</m></p>
      </statement>
      <solution>
        <p>
          We row reduce the matrix, keeping track of the number of row swaps and of the scaling factors used.
          <latex-code>
\begin{minipage}{10cm}  % needed for pdf version
\def\rowop#1#2#3{%
  \hbox to 4.2cm{\hss$\xrightarrow{#1}$\;}%
  \hbox to 3cm{#2\hss}%
  \hbox to 3cm{\hskip3mm\begin{minipage}{3.3cm}#3\end{minipage}\hss}%
  }
\leavevmode\rlap{\hbox{$\mat{0 -7 -4; 2 4 6; 3 7 -1}$}}%
\rowop{R_1\longleftrightarrow R_2}{$\mat{2 4 6; 0 -7 -4; 3 7 -1}$}{$r=1$}\\
\rowop{R_1 = R_1 \div 2}{$\mat{1 2 3; 0 -7 -4; 3 7 -1}$}{scaling factors${}=\frac 12$}\\
\rowop{R_3 = R_3 - 3R_1}{$\mat{1 2 3; 0 -7 -4; 0 1 -10}$}{}\\
\rowop{R_2\longleftrightarrow R_3}{$\mat{1 2 3; 0 1 -10; 0 -7 -4}$}{$r=2$}\\
\rowop{R_3 = R_3 + 7R_2}{$\mat{1 2 3; 0 1 -10; 0 0 -74}$}{}
\end{minipage}
          </latex-code>
          We made two row swaps and scaled once by a factor of <m>1/2</m>, so the <xref ref="det-defn-ref-compute"/> says that
          <me>\det\mat{0 -7 -4; 2 4 6; 3 7 -1} = (-1)^2\cdot\frac{1\cdot 1\cdot(-74)}{1/2} = -148.</me>
        </p>
      </solution>
    </example>

    <example>
      <statement>
        <p>Compute <m>\det\mat{1 2 3;2 -1 1;3 0 1}.</m></p>
      </statement>
      <solution>
        <p>
          We row reduce the matrix, keeping track of the number of row swaps and of the scaling factors used.
          <latex-code>
\begin{minipage}{10cm}  % needed for pdf version
\def\rowop#1#2#3#4{%
  \hbox to 4.5cm{\hss$\xrightarrow[#2]{#1}$\;}%
  \hbox to 2.7cm{#3\hss}%
  \hbox to 3cm{\hskip3mm\begin{minipage}{10cm}#4\end{minipage}\hss}%
  }
\leavevmode\rlap{\hbox{$\mat{1 2 3; 2 -1 1; 3 0 1}$}}%
\rowop{R_2=R_2-2R_1}{R_3=R_3-3R_1}{$\mat{1 2 3; 0 -5 -5; 0 -6 -8}$}{}\\
\rowop{R_2=R_2\div-5}{}{$\mat{1 2 3; 0 1 1; 0 -6 -8}$}{scaling factors${}=-\frac 15$}\\
\rowop{R_3=R_3+6R_2}{}{$\mat{1 2 3; 0 1 1; 0 0 -2}$}{}
\end{minipage}
          </latex-code>
          We did not make any row swaps, and we scaled once by a factor of <m>-1/5</m>, so the <xref ref="det-defn-ref-compute"/> says that
          <me>\det\mat{1 2 3; 2 -1 1; 3 0 1} = \frac{1\cdot 1\cdot(-2)}{-1/5} = 10.</me>
        </p>
      </solution>
    </example>

    <specialcase xml:id="det-defn-22-again">
      <title>The determinant of a <m>2\times 2</m> matrix</title>
      <idx><h>Determinant</h><h>of a <m>2\times 2</m> matrix</h></idx>
      <p>
        Let us use the <xref ref="det-defn-ref-compute"/> to compute the determinant of a general <m>2\times 2</m> matrix <m>A = \smallmat abcd</m>.
        <ul>
          <li>
            If <m>a = 0</m>, then
            <me>
              \det\mat{a b; c d} = \det\mat{0 b; c d} = -\det\mat{c d; 0 b} = -bc.
            </me>
          </li>
          <li>
            If <m>a\neq 0</m>, then
            <me>
              \begin{split}
                \det\mat{a b; c d} \amp= a\cdot\det\mat{1 b/a; c d}
                = a\cdot\det\mat{1 b/a; 0 d-c\cdot b/a} \\
                \amp= a\cdot 1\cdot(d-bc/a) = ad-bc.
              \end{split}
            </me>
          </li>
        </ul>
        In either case, we recover the <xref ref="matrix-inv-def-det">formula</xref>:
        <me>\det\mat{a b; c d} = ad-bc.</me>
      </p>
    </specialcase>

    <p>
      If a matrix is already in row echelon form, then you can simply read off the determinant as the product of the diagonal entries.  It turns out this is true for a slightly larger class of matrices called <em>triangular</em>.
    </p>

    <definition>
      <idx><h>Matrix</h><h>diagonal entries of</h></idx>
      <idx><h>Matrix</h><h>upper-triangular</h></idx>
      <idx><h>Matrix</h><h>lower-triangular</h></idx>
      <idx><h>Diagonal</h><h>see Matrix</h></idx>
      <idx><h>Upper-triangular</h><h>see Matrix</h></idx>
      <idx><h>Lower-triangular</h><h>see Matrix</h></idx>
      <statement>
        <p>
          <ul>
            <li>
              The <term>diagonal</term> entries of a  matrix <m>A</m> are the entries <m>a_{11},a_{22},\ldots</m>:
              <latex-code>
            <![CDATA[
\tikzstyle{circle entry} = [draw,rounded corners,thick,blue!50,inner sep=2pt]
  \begin{tikzpicture}
    \matrix[math matrix, nodes={minimum width=1em, minimum height=1em}] (mat1)
      {
        \node[circle entry]{a_{11}}; \& a_{12} \& a_{13} \& a_{14} \\
        a_{21} \& \node[circle entry]{a_{22}}; \& a_{23} \& a_{24} \\
        a_{31} \& a_{32} \& \node[circle entry]{a_{33}}; \& a_{34} \\
    };
    \matrix[math matrix, nodes={minimum width=1em, minimum height=1em}, xshift=4.75cm] (mat2)
      {
        \node[circle entry]{a_{11}}; \& a_{12} \& a_{13} \\
        a_{21} \& \node[circle entry]{a_{22}}; \& a_{23} \\
        a_{31} \& a_{32} \& \node[circle entry]{a_{33}}; \\
        a_{41} \& a_{42} \& a_{43} \\
      };
    \node[circle entry] at (4.75/2, 1.6) {diagonal entries};
  \end{tikzpicture}
            ]]>
              </latex-code>
            </li>
            <li>
              A square matrix is called <term>upper-triangular</term> if its nonzero entries all lie above the diagonal, and it is called <term>lower-triangular</term> if its nonzero entries all lie below the diagonal.  It is called <term>diagonal</term> if all of its nonzero entries lie on the diagonal, i.e., if it is both upper-triangular and lower-triangular.
              <latex-code>
                <![CDATA[
\pgfdeclarelayer{background}
\pgfsetlayers{background,main}
\begin{tikzpicture}

  \node[math matrix, nodes={minimum width=1em,minimum height=1em}] (ut) {
    \star \& \star \& \star \& \star \\
        0 \& \star \& \star \& \star \\
        0 \&     0 \& \star \& \star \\
        0 \&     0 \&     0 \& \star \\
  };

  \node[above] at (ut.north) {upper-triangular};

  \begin{pgfonlayer}{background}
  \fill[fill=seq-green!20!white, rounded corners=1.4mm]
    ($(ut-1-1.west)+(-1mm,0)$)
    -- (ut-1-1.north west)
    -- (ut-1-4.north east)
    -- (ut-4-4.south east)
    -- ($(ut-4-4.south)+(0,-1mm)$)
    -- cycle;
  \end{pgfonlayer}

\end{tikzpicture}
\qquad\qquad
\begin{tikzpicture}

  \node[math matrix, nodes={minimum width=1em,minimum height=1em}] (lt) {
    \star \&     0 \&     0 \&     0 \\
    \star \& \star \&     0 \&     0 \\
    \star \& \star \& \star \&     0 \\
    \star \& \star \& \star \& \star \\
  };

  \node[above] at (lt.north) {lower-triangular};

  \begin{pgfonlayer}{background}
  \fill[fill=seq-green!20!white, rounded corners=1.4mm]
    ($(lt-1-1.north)+(0,1mm)$)
    -- (lt-1-1.north west)
    -- (lt-4-1.south west)
    -- (lt-4-4.south east)
    -- ($(lt-4-4.east)+(1mm,0)$)
    -- cycle;
  \end{pgfonlayer}

\end{tikzpicture}
\qquad\qquad
\begin{tikzpicture}

  \node[math matrix, nodes={minimum width=1em,minimum height=1em}] (diag) {
    \star \&     0 \&     0 \&     0 \\
        0 \& \star \&     0 \&     0 \\
        0 \&     0 \& \star \&     0 \\
        0 \&     0 \&     0 \& \star \\
  };

  \node[above] at (diag.north) {diagonal};

  \begin{pgfonlayer}{background}
  \fill[fill=seq-green!20!white, rounded corners=2mm]
    ($(diag-1-1.west)+(-1mm,0)$)
    -- ($(diag-1-1.north)+(0,1mm)$)
    -- ($(diag-4-4.east)+(1mm,0)$)
    -- ($(diag-4-4.south)+(0,-1mm)$) -- cycle;
  \end{pgfonlayer}

\end{tikzpicture}
                ]]>
              </latex-code>
            </li>
          </ul>
        </p>
      </statement>
    </definition>

    <proposition xml:id="defn-det-special-case">
      <idx><h>Matrix</h><h>upper-triangular</h><h>determinant of</h></idx>
      <idx><h>Matrix</h><h>lower-triangular</h><h>determinant of</h></idx>
      <statement>
        <p>
          Let <m>A</m> be an <m>n\times n</m> matrix.
          <ol>
            <li>
              If <m>A</m> has a zero row or column, then <m>\det(A) = 0.</m>
            </li>
            <li>
              If <m>A</m> is upper-triangular or lower-triangular, then <m>\det(A)</m> is the product of its diagonal entries.
            </li>
          </ol>
        </p>
      </statement>
      <proof visible="true">
        <p>
          <ol>
            <li>
              <p>
                Suppose that <m>A</m> has a zero row.  Let <m>B</m> be the matrix obtained by negating the zero row.  Then <m>\det(A) = -\det(B)</m> by the second <xref ref="det-defn-the-defn">defining property</xref>.  But <m>A = B</m>, so <m>\det(A) = \det(B)</m>:
                <me>
                  \mat{1 2 3; 0 0 0; 7 8 9}
                  \;\xrightarrow{R_2 = -R_2}\;
                  \mat{1 2 3; 0 0 0; 7 8 9}.
                </me>
                Putting these together yields <m>\det(A) = -\det(A)</m>, so <m>\det(A)=0</m>.
              </p>
              <p>
                Now suppose that <m>A</m> has a zero column.  Then <m>A</m> is not invertible by the <xref ref="imt-1"/>, so its reduced row echelon form has a zero row.  Since row operations do not change whether the determinant is zero, we conclude <m>\det(A)=0</m>.
              </p>
            </li>
            <li>
              <p>
                First suppose that <m>A</m> is upper-triangular, and that one of the diagonal entries is zero, say <m>a_{ii}=0</m>.  We can perform row operations to clear the entries above the nonzero diagonal entries:
                <me>
                  \mat{a_{11} \star, \star, \star;
                     0 a_{22} \star, \star; 0 0 0 \star; 0 0 0 a_{44}}
                  \;\xrightarrow{\phantom{MMM}}\;
                  \mat{a_{11} 0, \star, 0;
                     0 a_{22} \star, 0; 0 0 0 0; 0 0 0 a_{44}}
                </me>
                In the resulting matrix, the <m>i</m>th row is zero, so <m>\det(A) = 0</m> by the first part.
              </p>
              <p>
                Still assuming that <m>A</m> is upper-triangular, now suppose that all of the diagonal entries of <m>A</m> are nonzero.  Then <m>A</m> can be transformed to the identity matrix by scaling the diagonal entries and then doing row replacements:
                <me>
\begin{split} \amp\mat{a \star, \star; 0 b \star; 0 0 c}
    \;\xrightarrow{%
    \begin{minipage}{1.8cm}%
      \tiny\centering scale by\\$a\inv,b\inv,c\inv$%
    \end{minipage}}\;
    \mat{1 \star, \star; 0 1 \star; 0 0 1}
    \;\xrightarrow{%
    \begin{minipage}{1.6cm}%
      \tiny\centering row\\replacements%
    \end{minipage}}\;
    \mat{1 0 0; 0 1 0; 0 0 1} \\
    \amp
    \hbox to 2.1cm{\hss$\det=abc$\hss} \;\xleftarrow{\hbox to 1.8cm{\hss}}\;
    \hbox to 2.1cm{\hss$\det=1$\hss} \;\xleftarrow{\hbox to 1.6cm{\hss}}\;
    \hbox to 1.8cm{\hss$\det=1$}
\end{split}
                </me>
                Since <m>\det(I_n) = 1</m> and we scaled by the reciprocals of the diagonal entries, this implies <m>\det(A)</m> is the product of the diagonal entries.
              </p>
              <p>
                The same argument works for lower triangular matrices, except that the the row replacements go down instead of up.
              </p>
            </li>
          </ol>
        </p>
      </proof>
    </proposition>

    <example>
      <statement>
        <p>
          Compute the determinants of these matrices:
          <me>
            \mat{1 2 3; 0 4 5; 0 0 6} \qquad
            \mat{-20 0 0; \pi, 0 0; 100 3 -7} \qquad
            \mat{17 -3 4; 0 0 0; 11/2 1 e}.
          </me>
        </p>
      </statement>
      <solution>
        <p>
          The first matrix is upper-triangular, the second is lower-triangular, and the third has a zero row:
          <md>
            <mrow>
              \det\mat{1 2 3; 0 4 5; 0 0 6} \amp= 1 \cdot 4 \cdot 6 = 24
            </mrow>
            <mrow>
              \det\mat{-20 0 0; \pi, 0 0; 100 3 -7} \amp= -20 \cdot 0 \cdot -7 = 0
            </mrow>
            <mrow>
              \det\mat{17 -3 4; 0 0 0; 11/2 1 e} \amp= 0.
            </mrow>
          </md>
        </p>
      </solution>
    </example>

    <p>
      A matrix can always be transformed into row echelon form by a series of  row operations, and a matrix in row echelon form is upper-triangular.  Therefore, we have completely justified the <xref ref="det-defn-ref-compute"/> for computing the determinant.
    </p>

    <p>
      The determinant is characterized by its <xref ref="det-defn-the-defn">defining properties</xref>, since we can compute the determinant of any matrix using row reduction, as in the above <xref ref="det-defn-ref-compute"/>.  However, we have not yet proved the existence of a function satisfying the defining properties!  Row reducing will compute the determinant <em>if it exists</em>, but we cannot use row reduction to prove existence, because we do not yet know that you compute the same number by row reducing in two different ways.
    </p>

    <theorem xml:id="det-defn-unique">
      <title>Existence of the determinant</title>
      <idx><h>Determinant</h><h>existence and uniqueness of</h></idx>
      <statement>
	<p>
	  There exists one and only one function from the set of square matrices to the real numbers, that satisfies the four <xref ref="det-defn-the-defn">defining properties</xref>.
	</p>
      </statement>
    </theorem>

    <p>
      We will prove the existence theorem in <xref ref="determinants-cofactors"/>, by exhibiting a recursive formula for the determinant.
      Again, the real content of the existence theorem is:
    </p>

    <bluebox>
      <p>
        No matter which row operations you do, you will always compute the same value for the determinant.
      </p>
    </bluebox>

  </subsection>

  <subsection xml:id="det-defn-magic-props">
    <title>Magical Properties of the Determinant</title>

    <p>
      In this subsection, we will discuss a number of the amazing properties enjoyed by the determinant: the <xref ref="det-defn-invert-prop">invertibility property</xref>, the <xref ref="det-defn-mult-prop">multiplicativity property</xref>, and the <xref ref="det-defn-trans-prop">transpose property</xref>.
    </p>

    <proposition hide-type="true" xml:id="det-defn-invert-prop">
      <title>Invertibility Property</title>
      <idx><h>Determinant</h><h>invertibility property</h></idx>
      <idx><h>Invertible matrix</h><h>determinant of</h></idx>
      <statement>
        <p>A square matrix is invertible if and only if <m>\det(A)\neq 0</m>.</p>
      </statement>
      <proof>
        <p>
          If <m>A</m> is invertible, then it has a pivot in every row and column by the <xref ref="imt-1"/>, so its reduced row echelon form is the identity matrix.  Since row operations do not change whether the determinant is zero, and since <m>\det(I_n) = 1</m>, this implies <m>\det(A)\neq 0.</m>  Conversely, if <m>A</m> is not invertible, then it is row equivalent to a matrix with a zero row.  Again, row operations do not change whether the determinant is nonzero, so in this case <m>\det(A) = 0.</m>
        </p>
      </proof>
    </proposition>

    <p>
      By the invertibility property, a matrix that does not satisfy any of the properties of the <xref ref="imt-1"/> has zero determinant.
    </p>

    <corollary xml:id="det-defn-dep-det0">
      <idx><h>Linear independence</h><h>and determinants</h></idx>
      <statement>
        <p>
          Let <m>A</m> be a square matrix.  If the rows or columns of <m>A</m> are linearly dependent, then <m>\det(A)=0</m>.
        </p>
      </statement>
      <proof>
        <p>
          If the columns of <m>A</m> are linearly dependent, then <m>A</m> is not invertible by condition 4 of the <xref ref="imt-1"/>.  Suppose now that the rows of <m>A</m> are linearly dependent.  If <m>r_1,r_2,\ldots,r_n</m> are the rows of <m>A</m>, then one of the rows is in the span of the others, so we have an equation like
          <me>r_2 = 3r_1 - r_3 + 2r_4.</me>
          If we perform the following row operations on <m>A</m>:
          <me>
            R_2 = R_2 - 3R_1;\quad
            R_2 = R_2 +  R_3;\quad
            R_2 = R_2 - 2R_4
          </me>
          then the second row of the resulting matrix is zero.  Hence <m>A</m> is not invertible in this case either.
        </p>
        <p>
          Alternatively, if the rows of <m>A</m> are linearly dependent, then one can combine condition 4 of the <xref ref="imt-1"/> and the <xref ref="det-defn-trans-prop">transpose property</xref> below to conclude that <m>\det(A)=0</m>.
        </p>
      </proof>
    </corollary>

    <p>
      In particular, if two rows/columns of <m>A</m> are multiples of each other, then <m>\det(A)=0.</m>  We also recover the fact that a matrix with a row or column of zeros has determinant zero.
    </p>

    <example>
      <p>
	The following matrices all have zero determinant:
	<me>
          \mat{0 2 -1; 0 5 10; 0 -7 3},\quad
          \mat{5 -15 11; 3 -9 2; 2 -6 16},\quad
          \mat{3 1 2 4; 0 0 0 0; 4 2 5 12; -1 3 4 8},\quad
          \mat{\pi, e 11; 3\pi, 3e 33; 12 -7 2}.
        </me>
      </p>
    </example>

    <p>
      The proofs of the <xref ref="det-defn-mult-prop">multiplicativity property</xref> and the <xref ref="det-defn-trans-prop">transpose property</xref> below, as well as the <xref ref="det-cofact-expansion">cofactor expansion theorem</xref> and the <xref ref="det-is-volume">determinants and volumes theorem</xref>, use the following strategy: define another function <m>d\colon\{\text{$n\times n$ matrices}\} \to \R</m>, and prove that <m>d</m> satisfies the same four defining properties as the determinant.  By the <xref ref="det-defn-unique">existence theorem</xref>, <em>the function <m>d</m> is equal to the determinant</em>.  This is an advantage of defining a function via its properties: in order to prove it is equal to another function, one only has to check the defining properties.
    </p>

    <proposition hide-type="true" xml:id="det-defn-mult-prop">
      <title>Multiplicativity Property</title>
      <idx><h>Determinant</h><h>multiplicativity property</h></idx>
      <idx><h>Matrix multiplication</h><h>determinant of</h></idx>
      <statement>
        <p>
          If <m>A</m> and <m>B</m> are <m>n\times n</m> matrices, then
          <me>\det(AB) = \det(A)\det(B).</me>
        </p>
      </statement>
      <proof>
        <p>
          In this proof, we need to use the notion of an <term>elementary matrix</term>.  This is a matrix obtained by doing one row operation to the identity matrix.  There are three kinds of elementary matrices: those arising from row replacement, row scaling, and row swaps:
          <me>
            \begin{split}
               \mat{1 0 0;  0 1 0; 0 0 1} \xrightarrow{\parbox{2cm}{\tiny\centering$R_2 = R_2 - 2R_1$}}\amp
               \mat{1 0 0; -2 1 0; 0 0 1} \\
               \mat{1 0 0;  0 1 0; 0 0 1} \xrightarrow{\parbox{2cm}{\tiny\centering$R_1 = 3R_1$}}\amp
               \mat{3 0 0;  0 1 0; 0 0 1} \\
               \mat{1 0 0;  0 1 0; 0 0 1} \xrightarrow{\parbox{2cm}{\tiny\centering$R_1 \longleftrightarrow R_2$}}\amp
               \mat{0 1 0;  1 0 0; 0 0 1}
            \end{split}
          </me>
          The important property of elementary matrices is the following claim.
        </p>
        <p>
          <em>Claim:</em> If <m>E</m> is the elementary matrix for a row operation, then <m>EA</m> is the matrix obtained by performing the same row operation on <m>A</m>.
        </p>
        <p>
          In other words, left-multiplication by an elementary matrix applies a row operation.  For example,
          <me>
            \begin{split}
            \mat{1 0 0; -2 1 0; 0 0 1}
               \mat{a_{11} a_{12} a_{13}; a_{21} a_{22} a_{23}; a_{31} a_{32} a_{33}}
               \amp= \mat[l]{a_{11} a_{12} a_{13};
               a_{21}-2a_{11} a_{22}-2a_{12} a_{23}-2a_{13};
               a_{31} a_{32} a_{33}} \\
            \mat{3 0 0; 0 1 0; 0 0 1}
               \mat{a_{11} a_{12} a_{13}; a_{21} a_{22} a_{23}; a_{31} a_{32} a_{33}}
               \amp= \mat[r]{3a_{11} 3a_{12} 3a_{13};
                          a_{21} a_{22} a_{23};
                          a_{31} a_{32} a_{33}} \\
               \mat{0 1 0;  1 0 0; 0 0 1}
               \mat{a_{11} a_{12} a_{13}; a_{21} a_{22} a_{23}; a_{31} a_{32} a_{33}}
               \amp= \mat{a_{21} a_{22} a_{23}; a_{11} a_{12} a_{13}; a_{31} a_{32} a_{33}}.
            \end{split}
          </me>
          The proof of the Claim is by direct calculation; we leave it to the reader to generalize the above equalities to <m>n\times n</m> matrices.
        </p>
        <p>
          As a consequence of the Claim and the four <xref ref="det-defn-the-defn">defining properties</xref>, we have the following observation.  Let <m>C</m> be any square matrix.
          <ol>
            <li>
              If <m>E</m> is the elementary matrix for a row replacement, then <m>\det(EC) = \det(C).</m>  In other words, <em>left-multiplication by <m>E</m> does not change the determinant.</em>
            </li>
            <li>
              If <m>E</m> is the elementary matrix for a row scale by a factor of <m>c</m>, then <m>\det(EC) = c\det(C).</m>  In other words, <em>left-multiplication by <m>E</m> scales the determinant by a factor of <m>c</m>.</em>
            </li>
            <li>
              If <m>E</m> is the elementary matrix for a row swap, then <m>\det(EC) = -\det(C).</m>  In other words, <em>left-multiplication by <m>E</m> negates the determinant.</em>
            </li>
          </ol>
        </p>
        <p>
          Now we turn to the proof of the multiplicativity property. Suppose to begin that <m>B</m> is not invertible.  Then <m>AB</m> is also not invertible: otherwise, <m>(AB)\inv AB = I_n</m> implies <m>B\inv = (AB)\inv A.</m>  By the <xref ref="det-defn-invert-prop">invertibility property</xref>, both sides of the equation <m>\det(AB) = \det(A)\det(B)</m> are zero.
        </p>
        <p>
          Now assume that <m>B</m> is invertible, so <m>\det(B)\neq 0</m>.  Define a function
          <me>
            d\colon\bigl\{\text{$n\times n$ matrices}\bigr\} \To \R \sptxt{by}
            d(C) = \frac{\det(CB)}{\det(B)}.
          </me>
          We claim that <m>d</m> satisfies the four defining properties of the determinant.
          <ol>
            <li>
              Let <m>C'</m> be the matrix obtained by doing a row replacement on <m>C</m>, and let <m>E</m> be the elementary matrix for this row replacement, so <m>C' = EC</m>.  Since left-multiplication by <m>E</m> does not change the determinant, we have <m>\det(ECB) = \det(CB)</m>, so
              <me>
                d(C') = \frac{\det(C'B)}{\det(B)}
                = \frac{\det(ECB)}{\det(B)}
                = \frac{\det(CB)}{\det(B)}
                = d(C).
              </me>
            </li>
            <li>
              Let <m>C'</m> be the matrix obtained by scaling a row of <m>C</m> by a factor of <m>c</m>, and let <m>E</m> be the elementary matrix for this row replacement, so <m>C' = EC</m>.  Since left-multiplication by <m>E</m> scales the determinant by a factor of <m>c</m>, we have <m>\det(ECB) = c\det(CB)</m>, so
              <me>
                d(C') = \frac{\det(C'B)}{\det(B)}
                = \frac{\det(ECB)}{\det(B)}
                = \frac{c\det(CB)}{\det(B)}
                = c\cdot d(C).
              </me>
            </li>
            <li>
              Let <m>C'</m> be the matrix obtained by swapping two rows of <m>C</m>, and let <m>E</m> be the elementary matrix for this row replacement, so <m>C' = EC</m>.  Since left-multiplication by <m>E</m> negates the determinant, we have <m>\det(ECB) = -\det(CB)</m>, so
              <me>
                d(C') = \frac{\det(C'B)}{\det(B)}
                = \frac{\det(ECB)}{\det(B)}
                = \frac{-\det(CB)}{\det(B)}
                = -d(C).
              </me>
            </li>
            <li>
              We have
              <me>d(I_n) = \frac{\det(I_nB)}{\det(B)} = \frac{\det(B)}{\det(B)} = 1.</me>
            </li>
          </ol>
        </p>
        <p>
          Since <m>d</m> satisfies the four defining properties of the determinant, <em>it is equal to the determinant</em> by the <xref ref="det-defn-unique">existence theorem</xref>.  In other words, for all matrices <m>A</m>, we have
          <me>\det(A) = d(A) = \frac{\det(AB)}{\det(B)}.</me>
          Multiplying through by <m>\det(B)</m> gives <m>\det(A)\det(B)=\det(AB).</m>
        </p>
      </proof>
    </proposition>

    <p>
      Recall that taking a power of a square matrix <m>A</m> means taking products of <m>A</m> with itself:
      <me>A^2 = AA  \qquad A^3 = AAA \qquad \text{etc.}</me>
      If <m>A</m> is invertible, then we define
      <me>A^{-2} = A\inv A\inv \qquad A^{-3} = A\inv A\inv A\inv \qquad \text{etc.}</me>
      For completeness, we set <m>A^0 = I_n</m> if <m>A\neq 0</m>.
    </p>

    <corollary xml:id="det-defn-power-prop">
      <idx><h>Determinant</h><h>and powers of matrices</h></idx>
      <statement>
        <p>
          If <m>A</m> is a square matrix, then
          <me>\det(A^n) = \det(A)^n</me>
          for all <m>n\geq 1</m>.  If <m>A</m> is invertible, then the equation holds for all <m>n\leq 0</m> as well; in particular,
          <me>\det(A\inv) = \frac 1{\det(A)}.</me>
        </p>
      </statement>
      <proof>
        <p>
          Using the <xref ref="det-defn-mult-prop">multiplicativity property</xref>, we compute
          <me>\det(A^2) = \det(AA) = \det(A)\det(A) = \det(A)^2</me>
          and
          <me>\det(A^3) = \det(AAA) = \det(A)\det(AA) = \det(A)\det(A)\det(A) = \det(A)^3;</me>
          the pattern is clear.
        </p>
        <p>
          We have
          <me>1 = \det(I_n) = \det(A A\inv) = \det(A)\det(A\inv)</me>
          by the <xref ref="det-defn-mult-prop">multiplicativity property</xref> and the fourth <xref ref="det-defn-the-defn">defining property</xref>, which shows that <m>\det(A\inv) = \det(A)\inv</m>.  Thus
          <me>\det(A^{-2}) = \det(A\inv A\inv) = \det(A\inv)\det(A\inv) = \det(A\inv)^2 = \det(A)^{-2},</me>
          and so on.
        </p>
      </proof>
    </corollary>

    <example>
      <statement>
        <p>
          Compute <m>\det(A^{100}),</m> where
          <me>A = \mat{4 1; 2 1}.</me>
        </p>
      </statement>
      <solution>
        <p>
          We have <m>\det(A) = 4 - 2 = 2</m>, so
          <me>\det(A^{100}) = \det(A)^{100} = 2^{100}.</me>
          Nowhere did we have to compute the <m>100</m>th power of <m>A</m>!  (We will learn an efficient way to do that in <xref ref="diagonalization"/>.)
        </p>
      </solution>
    </example>

    <p>
      Here is another application of the <xref ref="det-defn-mult-prop">multiplicativity property</xref>.
    </p>

    <corollary>
      <statement>
        <p>
          Let <m>A_1,A_2,\ldots,A_k</m> be <m>n\times n</m> matrices.  Then the product <m>A_1A_2\cdots A_k</m> is invertible if and only if each <m>A_i</m> is invertible.
        </p>
      </statement>
      <proof visible="true">
        <p>
          The determinant of the product is the product of the determinants by the <xref ref="det-defn-mult-prop">multiplicativity property</xref>:
          <me>\det(A_1A_2\cdots A_k) = \det(A_1)\det(A_2)\cdots\det(A_k).</me>
          By the <xref ref="det-defn-invert-prop">invertibility property</xref>, this is nonzero if and only if <m>A_1A_2\cdots A_k</m> is invertible.  On the other hand, <m>\det(A_1)\det(A_2)\cdots\det(A_k)</m> is nonzero if and only if each <m>\det(A_i)\neq0</m>, which means each <m>A_i</m> is invertible.
        </p>
      </proof>
    </corollary>

    <example>
      <statement>
        <p>
          For any number <m>n</m> we define
          <me>A_n = \mat{1 n; 1 2}.</me>
          Show that the product
          <me>A_1 A_2 A_3 A_4 A_5</me>
          is not invertible.
        </p>
      </statement>
      <solution>
        <p>
          When <m>n = 2</m>, the matrix <m>A_2</m> is not invertible, because its rows are identical:
          <me>A_2 = \mat{1 2; 1 2}.</me>
          Hence any product involving <m>A_2</m> is not invertible.
        </p>
      </solution>
    </example>

    <p>
      In order to state the transpose property, we need to define the transpose of a matrix.
    </p>

    <definition>
      <idx><h>Matrix</h><h>transpose of</h></idx>
      <idx><h>Transpose</h><see>Matrix</see></idx>
      <notation><usage>A^T</usage><description>Transpose of a matrix</description></notation>
      <statement>
        <p>
          The <term>transpose</term> of an <m>m\times n</m> matrix <m>A</m> is the <m>n\times m</m> matrix <m>A^T</m> whose rows are the columns of <m>A</m>.  In other words, the <m>ij</m> entry of <m>A^T</m> is <m>a_{ji}</m>.
          <latex-code>
            <![CDATA[
  \begin{tikzpicture}[
      every matrix/.append style={nodes={
          minimum width=1.5em, minimum height=1.5em},
          row sep=.3em, column sep=.3em}
      ]
    \matrix[math matrix, label={[yshift=1mm]above:$A$}] (aij)
      {
        a_{11} \& a_{12} \& a_{13} \& a_{14} \\
        a_{21} \& a_{22} \& a_{23} \& a_{24} \\
        a_{31} \& a_{32} \& a_{33} \& a_{34} \\
      };
    \matrix[math matrix, right=2.4cm of aij,
        label={[yshift=1mm]above:$A^T$}] (aijT)
      {
        a_{11} \& a_{21} \& a_{31} \\
        a_{12} \& a_{22} \& a_{32} \\
        a_{13} \& a_{23} \& a_{33} \\
        a_{14} \& a_{24} \& a_{34} \\
      };
    \draw[->, thick, shorten=6mm] (aij.east) -- (aijT.west);
    \draw[green!50!black, opacity=.5, shorten >=-8mm]
      (aij-1-1.north west) -- (aij-3-3.south east)
      coordinate[pos=1.3, below left=3mm] (left)
      coordinate[pos=1.3, above right=3mm] (right)
      node[pos=1.2, below right, opaque] {\small flip};
    \draw[<->, green!50!black] (left) to[bend left] (right);
    \draw[green!50!black, opacity=.5, shorten >=-8mm]
      (aijT-1-1.north west) -- (aijT-3-3.south east);
  \end{tikzpicture}
                ]]>
          </latex-code>
        </p>
      </statement>
    </definition>

    <p>
      Like inversion, transposition reverses the order of matrix multiplication.
    </p>

    <fact xml:id="det-defn-prod-trans">
      <idx><h>Matrix</h><h>transpose of</h><h>and products</h></idx>
      <statement>
        <p>
          Let <m>A</m> be an <m>m\times n</m> matrix, and let <m>B</m> be an <m>n\times p</m> matrix.  Then
          <me>(AB)^T = B^TA^T.</me>
        </p>
      </statement>
      <proof>
        <p>
          First suppose that <m>A</m> is a row vector an <m>B</m> is a column vector, i.e., <m>m = p = 1</m>.  Then
          <me>
            \begin{split}
            AB \amp= \mat{a_1 a_2 \cdots, a_n}\vec{b_1 b_2 \vdots, b_n}
            = a_1b_1 + a_2b_2 + \cdots + a_nb_n \\
            \amp= \mat{b_1 b_2 \cdots, b_n}\vec{a_1 a_2 \vdots, a_n}
            = B^TA^T.
            \end{split}
          </me>
        </p>
        <p>
          Now we use the row-column rule for matrix multiplication.  Let <m>r_1,r_2,\ldots,r_m</m> be the rows of <m>A</m>, and let <m>c_1,c_2,\ldots,c_p</m> be the columns of <m>B</m>, so
          <me>
            AB =
            \mat[c]{ \matrow{r_1}; \matrow{r_2}; \vdots ; \matrow{r_m}}
            \mat{| | ,, |; c_1 c_2 \cdots, c_p; | | ,, |}
            = \mat{ r_1c_1 r_1c_2 \cdots, r_1c_p;
              r_2c_1 r_2c_2 \cdots, r_2c_p;
              \vdots, \vdots, , \vdots;
              r_mc_1 r_mc_2 \cdots, r_mc_p}.
          </me>
          By the case we handled above, we have <m>r_ic_j = c_j^Tr_i^T</m>. Then
          <me>
            \begin{split}
            (AB)^T
            \amp= \mat{ r_1c_1 r_2c_1 \cdots, r_mc_1;
              r_1c_2 r_2c_2 \cdots, r_mc_2;
              \vdots, \vdots, , \vdots;
              r_1c_p r_2c_p \cdots, r_mc_p} \\
            \amp= \mat{ c_1^Tr_1^T c_1^Tr_2^T \cdots, c_1^Tr_m^T;
              c_2^Tr_1^T c_2^Tr_2^T \cdots, c_2^Tr_m^T;
              \vdots, \vdots, , \vdots;
              c_p^Tr_1^T c_p^Tr_2^T \cdots, c_p^Tr_m^T} \\
            \amp= \mat[c]{ \matrow{c_1^T}; \matrow{c_2^T}; \vdots ; \matrow{c_p^T}}
            \mat{| | ,, |; r_1^T r_2^T \cdots, r_m^T; | | ,, |}
            = B^TA^T.
            \end{split}
          </me>
          <em></em>               <!-- fix QED sign -->
        </p>
      </proof>
    </fact>

    <proposition hide-type="true" xml:id="det-defn-trans-prop">
      <title>Transpose Property</title>
      <idx><h>Determinant</h><h>transpose property</h></idx>
      <idx><h>Matrix</h><h>transpose of</h><h>determinant of</h></idx>
      <statement>
        <p>
          For any square matrix <m>A</m>, we have
          <me>\det(A) = \det(A^T).</me>
        </p>
      </statement>
      <proof>
        <p>
          We follow the same strategy as in the proof of the <xref ref="det-defn-mult-prop">multiplicativity property</xref>: namely, we define
          <me>d(A) = \det(A^T), </me>
          and we show that <m>d</m> satisfies the four defining properties of the determinant.  Again we use elementary matrices, also introduced in the proof of the <xref ref="det-defn-mult-prop">multiplicativity property</xref>.
          <ol>
            <li>
              Let <m>C'</m> be the matrix obtained by doing a row replacement on <m>C</m>, and let <m>E</m> be the elementary matrix for this row replacement, so <m>C' = EC</m>.  The elementary matrix for a row replacement is either upper-triangular or lower-triangular, with ones on the diagonal:
              <me>
                R_1 = R_1 + 3R_3:\;
                \mat{1 0 3; 0 1 0 ; 0 0 1}
                \qquad
                R_3 = R_3 + 3R_1:\;
                \mat{1 0 0; 0 1 0 ; 3 0 1}.
              </me>
              It follows that <m>E^T</m> is also either upper-triangular or lower-triangular, with ones on the diagonal, so <m>\det(E^T) = 1</m> by this <xref ref="defn-det-special-case"/>.   By the <xref ref="det-defn-prod-trans"/> and the <xref ref="det-defn-mult-prop">multiplicativity property</xref>,
              <me>
                \begin{split}
                d(C') \amp= \det((C')^T) = \det((EC)^T) = \det(C^TE^T) \\
                \amp= \det(C^T)\det(E^T) = \det(C^T) = d(C).
                \end{split}
              </me>
            </li>
            <li>
              Let <m>C'</m> be the matrix obtained by scaling a row of <m>C</m> by a factor of <m>c</m>, and let <m>E</m> be the elementary matrix for this row replacement, so <m>C' = EC</m>.  Then <m>E</m> is a diagonal matrix:
              <me>
                R_2 = cR_2:\;\mat{1 0 0; 0 c 0; 0 0 1}.
              </me>
              Thus <m>\det(E^T) = c</m>.  By the <xref ref="det-defn-prod-trans"/> and the <xref ref="det-defn-mult-prop">multiplicativity property</xref>,
              <me>
                \begin{split}
                d(C') \amp= \det((C')^T) = \det((EC)^T) = \det(C^TE^T) \\
                \amp= \det(C^T)\det(E^T) = c\det(C^T) = c\cdot d(C).
                \end{split}
              </me>
            </li>
            <li>
              Let <m>C'</m> be the matrix obtained by swapping two rows of <m>C</m>, and let <m>E</m> be the elementary matrix for this row replacement, so <m>C' = EC</m>.  The <m>E</m> is equal to its own transpose:
              <me>
                R_1\longleftrightarrow R_2:\;
                \mat{0 1 0; 1 0 0; 0 0 1} = \mat{0 1 0; 1 0 0; 0 0 1}^T.
              </me>
              Since <m>E</m> (hence <m>E^T</m>) is obtained by performing one row swap on the identity matrix, we have <m>\det(E^T) = -1</m>. By the <xref ref="det-defn-prod-trans"/> and the <xref ref="det-defn-mult-prop">multiplicativity property</xref>,
              <me>
                \begin{split}
                d(C') \amp= \det((C')^T) = \det((EC)^T) = \det(C^TE^T) \\
                \amp= \det(C^T)\det(E^T) = -\det(C^T) = - d(C).
                \end{split}
              </me>
            </li>
            <li>
              Since <m>I_n^T = I_n,</m> we have
              <me>d(I_n) = \det(I_n^T) = det(I_n) = 1.</me>
            </li>
          </ol>
        </p>
        <p>
          Since <m>d</m> satisfies the four defining properties of the determinant, <em>it is equal to the determinant</em> by the <xref ref="det-defn-unique">existence theorem</xref>.  In other words, for all matrices <m>A</m>, we have
          <me>\det(A) = d(A) = \det(A^T).</me>
        </p>
      </proof>
    </proposition>

    <p>
      The <xref ref="det-defn-trans-prop">transpose property</xref> is very useful.
      For concreteness, we note that <m>\det(A)=\det(A^T)</m> means, for instance, that
      <me>\det\mat{1 2 3; 4 5 6; 7 8 9} = \det\mat{1 4 7; 2 5 8; 3 6 9}.</me>
      This implies that the determinant has the curious feature that it also behaves well with respect to <em>column</em> operations.  Indeed, a column operation on <m>A</m> is the same as a row operation on <m>A^T</m>, and <m>\det(A) = \det(A^T)</m>.
    </p>

    <corollary xml:id="defn-det-col-ops">
      <idx><h>Determinant</h><h>and column operations</h></idx>
      <statement>
        <p>
          The determinant satisfies the following properties with respect to column operations:
	  <ol>
	    <li>
	      Doing a column replacement on <m>A</m> does not change <m>\det(A)</m>.
	    </li>
            <li>
              Scaling a column of <m>A</m> by a scalar <m>c</m> multiplies the determinant by <m>c</m>.
	    </li>
	    <li>
	      Swapping two columns of a matrix multiplies the determinant by <m>-1</m>.
	    </li>
	  </ol>
        </p>
      </statement>
    </corollary>

    <p>
      The previous corollary makes it easier to compute the determinant: one is allowed to do row <em>and</em> column operations when simplifying the matrix.  (Of course, one still has to keep track of how the row and column operations change the determinant.)
    </p>

    <example>
      <statement>
        <p>Compute <m>\det\mat{2 7 4; 3 1 3; 4 0 1}.</m></p>
      </statement>
      <solution>
        <p>
          It takes fewer column operations than row operations to make this matrix upper-triangular:
          <latex-code>
\begin{minipage}{8cm}
\def\rowop#1#2#3{%
  \hbox to 4cm{\hss$\xrightarrow{#1}$\;}%
  \hbox to 3cm{#2\hss}%
  \hbox to 3cm{\hskip3mm\begin{minipage}{3.3cm}#3\end{minipage}\hss}%
  }
\leavevmode\rlap{\hbox{$\mat{2 7 4; 3 1 3; 4 0 1}$}}%
\rowop{C_1=C_1-4C_3}{$\mat{-14 7 4; -9 1 3; 0 0 1}$}{}\\
\rowop{C_1=C_1+9C_2}{$\mat{49 7 4; 0 1 3; 0 0 1}$}{}
\end{minipage}
          </latex-code>
          We performed two column replacements, which does not change the determinant; therefore,
          <me>\det\mat{2 7 4; 3 1 3; 4 0 1} = \det\mat{49 7 4; 0 1 3; 0 0 1} = 49.</me>
        </p>
      </solution>
    </example>

    <paragraphs visible="false">
      <title>Multilinearity</title>

      <p>
        The following observation is useful for theoretical purposes.
      </p>

    <p>
      We can think of <m>\det</m> as a function of the rows of a matrix:
      <me>\det(v_1,v_2,\ldots,v_n)
      = \det\mat{\matrow{v_1}; \matrow{v_2}; \vdots; \matrow{v_n}}.</me>
    </p>

    <proposition hide-type="true" xml:id="det-defn-linear-prop">
      <title>Multilinearity Property</title>
      <idx><h>Determinant</h><h>multilinearity property</h></idx>
      <statement>
        <p>
          Let <m>i</m> be a whole number between <m>1</m> and <m>n</m>, and fix <m>n-1</m> vectors <m>v_1,v_2,\ldots,v_{i-1},v_{i+1},\ldots,v_n</m> in <m>\R^n</m>.  Then the transformation <m>T\colon\R^n\to\R</m> defined by
          <me>T(x) = \det(v_1,v_2,\ldots,v_{i-1},x,v_{i+1},\ldots,v_n)</me>
          is <em>linear</em>.
        </p>
      </statement>
      <proof>
        <p>
          First assume that <m>i=1</m>, so
          <me>T(x) = \det(x,v_2,\ldots,v_n).</me>
          We have to show that <m>T</m> satisfies the <xref ref="linear-trans-defn">defining properties</xref>.
          <ul>
            <li>
              By the first <xref ref="det-defn-the-defn">defining property</xref>, scaling any row of a matrix by a number <m>c</m> scales the determinant by a factor of <m>c</m>.  This implies that <m>T</m> satisfies the second property, i.e., that
              <me>T(cx) = \det(cx,v_2,\ldots,v_n) = c\det(x,v_2,\ldots,v_n) = cT(x).</me>
            </li>
            <li>
              <p>
                We claim that <m>T(v+w) = T(v) + T(w)</m>.
                If <m>w</m> is in <m>\Span\{v,v_2,\ldots,v_n\}</m>, then
                <me>w = cv + c_2v_2 + \cdots + c_nv_n</me>
                for some scalars <m>c,c_2,\ldots,c_n</m>.  Let <m>A</m> be the matrix with rows <m>v+w,v_2,\ldots,v_n</m>, so <m>T(v+w) = \det(A).</m>  By performing the row operations
                <me>
                  R_1 = R_1 - c_2R_2;\quad R_1 = R_1 - c_3R_3;\quad\ldots\quad
                  R_1 = R_1 - c_nR_n,
                </me>
                the first row of the matrix <m>A</m> becomes
                <me>v+w-(c_2v_2+\cdots+c_nv_n) = v + cv = (1+c)v.</me>
                Therefore,
                <me>
                  \begin{split}
                  T(v+w)
                  = \det(A) \amp= \det((1+c)v,v_2,\ldots,v_n) \\
                  \amp= (1+c)\det(v,v_2,\ldots,v_n) \\
                  \amp= T(v) + cT(v) = T(v) + T(cv).
                  \end{split}
                </me>
                Doing the opposite row operations
                <me>
                  R_1 = R_1 + c_2R_2;\quad R_1 = R_1 + c_3R_3;\quad\ldots\quad
                  R_1 = R_1 + c_nR_n
                </me>
                to the matrix with rows <m>cv,v_2,\ldots,v_n</m> shows that
                <me>
                  \begin{split}
                  T(cv) \amp= \det(cv,v_2,\ldots,v_n) \\
                  \amp= \det(cv+c_2v_2+\cdots+c_nv_n,v_2,\ldots,v_n) \\
                  \amp= \det(w,v_2,\ldots,v_n) = T(w),
                  \end{split}
                </me>
                which finishes the proof of the first property in this case.
              </p>
              <p>
                Now suppose that <m>w</m> is not in <m>\Span\{v,v_2,\ldots,v_n\}</m>.  This implies that <m>\{v,v_2,\ldots,v_n\}</m> is linearly <em>dependent</em> (otherwise it would form a basis for <m>\R^n</m>), so <m>T(v)</m> = 0.  If <m>v</m> is not in <m>\Span\{v_2,\ldots,v_n\}</m>, then <m>\{v_2,\ldots,v_n\}</m> is linearly dependent by the <xref ref="linindep-increasing-span">increasing span criterion</xref>, so <m>T(x) = 0</m> for all <m>x</m>, as the matrix with rows <m>x,v_2,\ldots,v_n</m> is not invertible.  Hence we may assume <m>v</m> is in <m>\Span\{v_2,\ldots,v_n\}</m>.   By the above argument with the roles of <m>v</m> and <m>w</m> reversed, we have <m>T(v+w) = T(v)+T(w).</m>
              </p>
            </li>
          </ul>
        </p>
        <p>
          For <m>i\neq 1</m>, we note that
          <me>
            \begin{split}
            T(x) \amp= \det(v_1,v_2,\ldots,v_{i-1},x,v_{i+1},\ldots,v_n) \\
            \amp= -\det(x,v_2,\ldots,v_{i-1},v_1,v_{i+1},\ldots,v_n).
            \end{split}
          </me>
          By the previously handled case, we know that <m>-T</m> is linear:
          <me>-T(cx) = -cT(x) \qquad -T(v+w) = -T(v) - T(w).</me>
          Multiplying both sides by <m>-1</m>, we see that <m>T</m> is linear.
        </p>
      </proof>
    </proposition>

    <p>
      For example, we have
      <me>
  \det\mat{\matrow{\makebox[\widthof{$av+bw$}]{$v_1$}};
    \matrow{av+bw}; \matrow{\makebox[\widthof{$av+bw$}]{$v_3$}}}
        = a\det\mat{\matrow{v_1}; \matrow{\makebox[\widthof{$v_1$}]{$v$}}; \matrow{v_3}}
        + b\det\mat{\matrow{v_1}; \matrow{\makebox[\widthof{$v_1$}]{$w$}}; \matrow{v_3}}
      </me>
      By the <xref ref="det-defn-trans-prop">transpose property</xref>, the determinant is also multilinear in the <em>columns</em> of a matrix:
      <me>
        \det\mat{| | |; v_1 av+bw v_3; | | |}
        = a\det\mat{| | |; v_1 v v_3; | | |} + b\det\mat{| | |; v_1 w v_3; | | |}.
      </me>
    </p>

    <remark xml:id="det-defn-use-multi">
      <title>Alternative defining properties</title>
      <idx><h>Determinant</h><h>alternative defining properties of</h></idx>
      <p>
        In more theoretical treatments of the topic, where row reduction plays a secondary role, the defining properties of the determinant are often taken to be:
        <ol>
          <li>
            The determinant <m>\det(A)</m> is multilinear in the rows of <m>A</m>.
          </li>
          <li>
            If <m>A</m> has two identical rows, then <m>\det(A) = 0</m>.
          </li>
          <li>
            The determinant of the identity matrix is equal to one.
          </li>
        </ol>
        We have already shown that our four <xref ref="det-defn-the-defn">defining properties</xref> imply these three.  Conversely, we will prove that these three alternative properties imply our four, so that both sets of properties are equivalent.
      </p>
      <p>
        Defining property <m>2</m> is just the second <xref ref="linear-trans-defn">defining property</xref>.  Suppose that the rows of <m>A</m> are <m>v_1,v_2,\ldots,v_n</m>.  If we perform the row replacement <m>R_i = R_i + cR_j</m> on <m>A</m>, then the rows of our new matrix are <m>v_1,v_2,\ldots,v_{i-1},v_i+cv_j,v_{i+1},\ldots,v_n</m>, so by linearity in the <m>i</m>th row,
        <me>
          \begin{split}
          \det(\amp v_1,v_2,\ldots,v_{i-1},v_i+cv_j,v_{i+1},\ldots,v_n) \\
          \amp= \det(v_1,v_2,\ldots,v_{i-1},v_i,v_{i+1},\ldots,v_n) +
          c\det(v_1,v_2,\ldots,v_{i-1},v_j,v_{i+1},\ldots,v_n) \\
          \amp= \det(v_1,v_2,\ldots,v_{i-1},v_i,v_{i+1},\ldots,v_n)
          = \det(A),
          \end{split}
        </me>
        where <m>\det(v_1,v_2,\ldots,v_{i-1},v_j,v_{i+1},\ldots,v_n)=0</m> because <m>v_j</m> is repeated.  Thus, the alternative defining properties imply our first two defining properties.  For the third, suppose that we want to swap row <m>i</m> with row <m>j</m>.  Using the second alternative defining property and multilinearity in the <m>i</m>th and <m>j</m>th rows, we have
        <me>
          \begin{split}
          0 \amp= \det(v_1,\ldots,v_i+v_j,\ldots,v_i+v_j,\ldots,v_n) \\
          \amp= \det(v_1,\ldots,v_i,\ldots,v_i+v_j,\ldots,v_n)
          + \det(v_1,\ldots,v_j,\ldots,v_i+v_j,\ldots,v_n) \\
          \amp= \det(v_1,\ldots,v_i,\ldots,v_i,\ldots,v_n)
          + \det(v_1,\ldots,v_i,\ldots,v_j,\ldots,v_n) \\
          \amp\qquad+\det(v_1,\ldots,v_j,\ldots,v_i,\ldots,v_n)
          + \det(v_1,\ldots,v_j,\ldots,v_j,\ldots,v_n) \\
          \amp= \det(v_1,\ldots,v_i,\ldots,v_j,\ldots,v_n)
          + \det(v_1,\ldots,v_j,\ldots,v_i,\ldots,v_n),
          \end{split}
        </me>
        as desired.
      </p>
    </remark>

    <example>
      <p>
        We have
        <me>
          \vec{-1 2 3} = -\vec{1 0 0} + 2\vec{0 1 0} + 3\vec{0 0 1}.
        </me>
        Therefore,
        <me>
          \begin{split}
          \det\amp\mat{-1 7 2; 2 -3 2; 3 1 1}
          = -\det\mat{1 7 2; 0 -3 2; 0 1 1} \\
          \amp+ 2\det\mat{0 7 2; 1 -3 2; 0 1 1}
          + 3\det\mat{0 7 2; 0 -3 2; 1 1 1}.
          \end{split}
        </me>
        This is the basic idea behind cofactor expansions in <xref ref="determinants-cofactors"/>.
      </p>
    </example>

  </paragraphs>

    <bluebox>
      <title>Summary: Magical Properties of the Determinant</title>
      <idx><h>Determinant</h><h>properties of</h></idx>
      <p>
        <ol>
          <li>
            There is one and only one function <m>\det\colon\{n\times n\text{ matrices}\}\to\R</m> satisfying the four <xref ref="det-defn-the-defn">defining properties</xref>.
          </li>
          <li>
            The determinant of an upper-triangular or lower-triangular matrix is the product of the diagonal entries.
          </li>
          <li>
            A square matrix is invertible if and only if <m>\det(A)\neq 0</m>; in this case,
            <me>\det(A\inv) = \frac 1{\det(A)}.</me>
          </li>
          <li>
            If <m>A</m> and <m>B</m> are <m>n\times n</m> matrices, then
            <me>\det(AB) = \det(A)\det(B).</me>
          </li>
          <li>
            For any square matrix <m>A</m>, we have
            <me>\det(A^T) = \det(A).</me>
          </li>
          <li>
            The determinant can be computed by performing row and/or column operations.
          </li>
        </ol>
      </p>
    </bluebox>

  </subsection>

</section>