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Notebook[{

Cell[CellGroupData[{
Cell["Determinants", "Title"],

Cell["Math 338: Linear Algebra", "Subtitle"],

Cell[CellGroupData[{

Cell["The Definition", "Subsection"],

Cell["\<\
There are a number of equivalent ways to define the determinant of a matrix. \
Here are three (there are examples below!):\
\>", "Text"],

Cell[TextData[{
  StyleBox["Theorem/Definition",
    FontWeight->"Bold"],
  ": Let ",
  Cell[BoxData[
      \(TraditionalForm\`A\)]],
  " be a square ",
  Cell[BoxData[
      \(TraditionalForm\`n\ \[Cross]\ n\)]],
  " matrix. The following three quantities are equal, and are called the \
determinant of ",
  Cell[BoxData[
      \(TraditionalForm\`A\)]],
  ".\n\t1)\t",
  Cell[BoxData[
      \(TraditionalForm\`Det[{a}]\  = \ a\)]],
  ", and if ",
  Cell[BoxData[
      \(TraditionalForm\`n > 1\)]],
  " then ",
  Cell[BoxData[
      \(TraditionalForm\`Det[
          A] = \(a\_11\) A\_11 + \(a\_12\) 
            A\_12 + \[CenterEllipsis] + \(a\_\(1  n\)\) A\_\(1  n\)\)]],
  ", where ",
  Cell[BoxData[
      \(TraditionalForm\`A\_\(i\ j\)\)]],
  " is ",
  Cell[BoxData[
      \(TraditionalForm\`\((\(-1\))\)\^\(i + j\)\)]],
  " times the determinant of the matrix obtained by deleting the entire ",
  Cell[BoxData[
      \(TraditionalForm\`i\)]],
  "-th row and ",
  Cell[BoxData[
      \(TraditionalForm\`j\)]],
  "-th column of ",
  Cell[BoxData[
      \(TraditionalForm\`A\)]],
  ".\n\t2)\tIn computing the reduced row echelon form of ",
  Cell[BoxData[
      \(TraditionalForm\`A\)]],
  ", we occasionally use the row operation: \"replace a row by a multiple of \
itself\". If the reduced row echelon form of ",
  Cell[BoxData[
      \(TraditionalForm\`A\)]],
  " is not the identity matrix, then the determinant of ",
  Cell[BoxData[
      \(TraditionalForm\`A\)]],
  " is 0, and otherwise ",
  Cell[BoxData[
      \(TraditionalForm\`Det[A]\)]],
  " is the reciprocal of the product of those multiples (the ones used in \
computing the rref) times ",
  Cell[BoxData[
      \(TraditionalForm\`\((\(-1\))\)\^k\)]],
  ", where ",
  Cell[BoxData[
      \(TraditionalForm\`k\)]],
  " is the number of times adjacent rows were switched.\n\t3)\tA permutation \
is a map \[Sigma] from ",
  Cell[BoxData[
      \(TraditionalForm\`{1, \ 2, \ \[Ellipsis]\ , \ n}\)]],
  " onto {1, 2, \[Ellipsis] , n}. (\"onto\" means that for every ",
  Cell[BoxData[
      \(TraditionalForm\`j \[Element] {1, 2,  ... , n}\)]],
  ", there is some ",
  Cell[BoxData[
      \(TraditionalForm\`i \[Element] {1, 2,  ... , n}\)]],
  " with ",
  Cell[BoxData[
      \(TraditionalForm\`\[Sigma](i) = j\)]],
  "). Write down the list ",
  Cell[BoxData[
      \(TraditionalForm\`\[Sigma](1), \ \[Sigma](2), \  ... , \ \[Sigma](
        n)\)]],
  ", and sort this list into order by swapping adjacent entries: if you used \
an even number set ",
  Cell[BoxData[
      \(TraditionalForm\`sgn(\[Sigma]) = 1\)]],
  ", and otherwise set ",
  Cell[BoxData[
      \(TraditionalForm\`sgn(\[Sigma]) = \(-1\)\)]],
  ". The determinant of ",
  Cell[BoxData[
      \(TraditionalForm\`A\)]],
  " is given by a sum over ",
  StyleBox["all",
    FontSlant->"Italic"],
  " permutations:"
}], "Text"],

Cell[BoxData[
    \(TraditionalForm\`\(\(\t\t\)\(Det[
        A] = \[Sum]\+\[Sigma]\( 
            sgn(\[Sigma])\)\ \(a\_\(1, \[Sigma](1)\)\) \(a\_\(2, \[Sigma](
                2)\)\) \(a\_\(3, \[Sigma](
                3)\)\) \[CenterEllipsis]\ \(\(a\_\(n, \[Sigma](
                  n)\)\)\(.\)\)\)\)\)], "DisplayFormula",
  FormatType->StandardForm],

Cell["\<\
Make sense? If you said \"yes\" then you're not telling the truth. \
\>", "Text"],

Cell[TextData[{
  "Before we start into the examples, allow me to draw your attention to four \
things. First, the spelling of determinant is with an \"a\", not with an \
\"e\". Second, the determinant of a matrix is not a matrix, it is a number. \
It can be negative, positive, zero, and sometimes even involves the square \
root of ",
  Cell[BoxData[
      \(TraditionalForm\`\(-1\)\)]],
  ". It is commonly written ",
  Cell[BoxData[
      \(TraditionalForm\`\(\(|\)\(A\)\(|\)\)\)]],
  " because in some ways it is a generalized absolute value. But it can be \
negative. Third, we have only defined the determinant of square matrices. \
This isn't a matter of convenience, I'm not aware of any reasonable \
definition that extends to non-square matrices. Fourth, the formulas above \
involve quite a bit of multiplication, and will have positive and negative \
numbers even if the matrix entries are all positive; a little bit of \
round-off error can quickly accumulate. In summary, the determinant is a \
single number that tells you something about infinite-precision square \
matrices. "
}], "Text"],

Cell[CellGroupData[{

Cell["Example of first definition", "Subsubsection"],

Cell[TextData[{
  "Here it is again: ",
  Cell[BoxData[
      \(TraditionalForm\`Det[{a}]\  = \ a\)]],
  ", and if ",
  Cell[BoxData[
      \(TraditionalForm\`n > 1\)]],
  " then\n\t ",
  Cell[BoxData[
      \(TraditionalForm\`Det[
          A] = \(a\_11\) A\_11 + \(a\_12\) 
            A\_12 + \[CenterEllipsis] + \(a\_\(1  n\)\) A\_\(1  n\)\)]],
  ", \nwhere ",
  Cell[BoxData[
      \(TraditionalForm\`A\_\(i\ j\)\)]],
  " is ",
  Cell[BoxData[
      \(TraditionalForm\`\((\(-1\))\)\^\(i + j\)\)]],
  " times the determinant of the matrix obtained by deleting the entire ",
  Cell[BoxData[
      \(TraditionalForm\`i\)]],
  "-th row and ",
  Cell[BoxData[
      \(TraditionalForm\`j\)]],
  "-th column of ",
  Cell[BoxData[
      \(TraditionalForm\`A\)]],
  "."
}], "Text"],

Cell[TextData[{
  Cell[BoxData[
      \(TraditionalForm\`Det[{3}]\  = \ 3\)]],
  ", so 1\[Cross]1 matrices are not so interesting."
}], "Text"],

Cell[TextData[{
  "The formula ",
  Cell[BoxData[
      FormBox[
        RowBox[{
          RowBox[{"Det", "[", 
            RowBox[{"(", GridBox[{
                  {"a", "b"},
                  {"c", "d"}
                  }], ")"}], "]"}], "=", \(a\ d\  - \ b\ c\)}], 
        TraditionalForm]]],
  " is the only one worth remembering, but 2\[Cross]2 matrices are too small \
to see what's really going on, even though this is just a special case."
}], "Text"],

Cell[TextData[{
  "Let's consider the matrix ",
  Cell[BoxData[
      FormBox[
        RowBox[{"A", "=", 
          RowBox[{"(", GridBox[{
                {"a", "b", "c"},
                {"d", "e", "f"},
                {"g", "h", "i"}
                }], ")"}]}], TraditionalForm]]],
  ". The definition says that the determinant of ",
  Cell[BoxData[
      \(TraditionalForm\`A\)]],
  " is the sum\n\t",
  Cell[BoxData[
      FormBox[
        RowBox[{
          RowBox[{"a", " ", \(\((\(-1\))\)\^\(1 + 1\)\), 
            RowBox[{"Det", "[", 
              RowBox[{"(", GridBox[{
                    {"e", "f"},
                    {"h", "i"}
                    }], ")"}], "]"}]}], "+", 
          RowBox[{\(\(b(\(-1\))\)\^\(1 + 2\)\), 
            RowBox[{"Det", "[", 
              RowBox[{"(", GridBox[{
                    {"d", "f"},
                    {"g", "i"}
                    }], ")"}], "]"}]}], "+", 
          RowBox[{\(\(c(\(-1\))\)\^\(1 + 3\)\), 
            RowBox[{"Det", "[", 
              RowBox[{"(", GridBox[{
                    {"d", "e"},
                    {"g", "h"}
                    }], ")"}], "]"}]}]}], TraditionalForm]]],
  ".\nDo you see how this comes from the definition?"
}], "Text"],

Cell[TextData[{
  "Get out a piece of paper and a pencil, and use them to compute the \
determinants of the following matrices:\n\t",
  Cell[BoxData[
      FormBox[
        RowBox[{"(", GridBox[{
              {"1", \(-1\), "3"},
              {"1", "0", "0"},
              {"1", "2", "2"}
              }], ")"}], TraditionalForm]]],
  "\n\t",
  Cell[BoxData[
      FormBox[
        RowBox[{"(", GridBox[{
              {"1", "3", \(-1\)},
              {"1", "2", "5"},
              {"0", "0", "0"}
              }], ")"}], TraditionalForm]]],
  "\n\t",
  Cell[BoxData[
      FormBox[
        RowBox[{"(", GridBox[{
              {"1", "0", "0", "0"},
              {"3", \(-2\), "0", "0"},
              {\(\(-1\)/2\), "4", "7", "0"},
              {"1", "1", "1", "3"}
              }], ")"}], TraditionalForm]]],
  "\n\t",
  Cell[BoxData[
      FormBox[
        RowBox[{"(", GridBox[{
              {"5", "2"},
              {"7", "5"}
              }], ")"}], TraditionalForm]]]
}], "Text"],

Cell[TextData[{
  "Here's a theorem:\n",
  StyleBox["Theorem",
    FontWeight->"Bold"],
  ": If an entire row of the square matrix ",
  Cell[BoxData[
      \(TraditionalForm\`A\)]],
  " is 0, then the determinant of ",
  Cell[BoxData[
      \(TraditionalForm\`A\)]],
  " is 0.\n\nCan you write a proof of this? Do so. Hint: first observe that \
the Theorem is true if ",
  Cell[BoxData[
      \(TraditionalForm\`A\)]],
  " is a 1\[Cross]1 matrix. Then, assume that the Theorem is true for every \
",
  Cell[BoxData[
      \(TraditionalForm\`n\[Cross]n\)]],
  " matrix, but that ",
  Cell[BoxData[
      \(TraditionalForm\`A\)]],
  " is an ",
  Cell[BoxData[
      \(TraditionalForm\`\((n + 1)\)\[Cross]\((n + 1)\)\)]],
  " matrix."
}], "Text"],

Cell["\<\
Is the following true? \"If you multiply a row of a matrix by 3, then the \
determinant is also multiplied by 3\". Can you prove it?\
\>", "Text"]
}, Closed]],

Cell[CellGroupData[{

Cell["Example of second definition", "Subsubsection"],

Cell[TextData[{
  "Here is the second definition: in computing the reduced row echelon form \
of ",
  Cell[BoxData[
      \(TraditionalForm\`A\)]],
  ", we occasionally use the row operation: \"replace a row by a multiple of \
itself\". In computing the reduced row echelon form of ",
  Cell[BoxData[
      \(TraditionalForm\`A\)]],
  ", we occasionally use the row operation: \"replace a row by a multiple of \
itself\". If the reduced row echelon form of ",
  Cell[BoxData[
      \(TraditionalForm\`A\)]],
  " is not the identity matrix, then the determinant of ",
  Cell[BoxData[
      \(TraditionalForm\`A\)]],
  " is 0, and otherwise ",
  Cell[BoxData[
      \(TraditionalForm\`Det[A]\)]],
  " is the reciprocal of the product of those multiples (the ones used in \
computing the rref) times ",
  Cell[BoxData[
      \(TraditionalForm\`\((\(-1\))\)\^k\)]],
  ", where ",
  Cell[BoxData[
      \(TraditionalForm\`k\)]],
  " is the number of times adjacent rows were switched."
}], "Text"],

Cell[TextData[{
  "Here's a matrix whose determinant you have calculated on a piece of paper: \
",
  Cell[BoxData[
      FormBox[
        RowBox[{"(", GridBox[{
              {"1", \(-1\), "3"},
              {"1", "0", "0"},
              {"1", "2", "2"}
              }], ")"}], TraditionalForm]]],
  ". Here's how the forward portion of Gaussian elimination looks:\n",
  Cell[BoxData[
      FormBox[
        RowBox[{
          RowBox[{"(", GridBox[{
                {"1", \(-1\), "3"},
                {"1", "0", "0"},
                {"1", "2", "2"}
                }], ")"}], "\[Tilde]", 
          RowBox[{"(", GridBox[{
                {"1", \(-1\), "3"},
                {"0", "1", \(-3\)},
                {"1", "2", "2"}
                }], ")"}], "\[Tilde]", 
          RowBox[{"(", GridBox[{
                {"1", \(-1\), "3"},
                {"0", "1", \(-3\)},
                {"0", "3", \(-1\)}
                }], ")"}], "\[Tilde]", 
          RowBox[{"(", GridBox[{
                {"1", \(-1\), "3"},
                {"0", "1", \(-3\)},
                {"0", "0", "8"}
                }], ")"}], "\[Tilde]", 
          RowBox[{"(", GridBox[{
                {"1", \(-1\), "3"},
                {"0", "1", \(-3\)},
                {"0", "0", "1"}
                }], ")"}]}], TraditionalForm]]],
  ".\nFor each \"\[Tilde]\", identify the elementary row operation used. Do \
you see how the definition says that the determinant is 8? Why do we only \
need to do the forward portion of Gaussian elimination?"
}], "Text"],

Cell[TextData[{
  "Get out that paper again, and use this method to compute the determinant \
of\n\t",
  Cell[BoxData[
      FormBox[
        RowBox[{"(", GridBox[{
              {"1", "5"},
              {\(-1\), "7"}
              }], ")"}], TraditionalForm]]],
  "\n\t",
  Cell[BoxData[
      FormBox[
        RowBox[{"(", GridBox[{
              {"1", "3", \(-1\)},
              {"1", "7", "0"},
              {"2", "6", \(-1\)}
              }], ")"}], TraditionalForm]]]
}], "Text"],

Cell["\<\
With this definition, write out a proof that if you multiply a row of the \
matrix by a constant, you multiply the determinant of the matrix by the same \
constant.\
\>", "Text"],

Cell[TextData[{
  "Here's another nice theorem that is easy to understand from this \
definition: Adding a multiple of one row to another does not change the \
determinant. Here's another: The matrix ",
  Cell[BoxData[
      \(TraditionalForm\`A\)]],
  " is invertible if and only if its determinant is not 0."
}], "Text"]
}, Closed]],

Cell[CellGroupData[{

Cell["Example of third definition", "Subsubsection"],

Cell[TextData[{
  "A permutation is a map \[Sigma] from ",
  Cell[BoxData[
      \(TraditionalForm\`{1, \ 2, \ \[Ellipsis]\ , \ n}\)]],
  " onto {1, 2, \[Ellipsis] , n}. (\"onto\" means that for every ",
  Cell[BoxData[
      \(TraditionalForm\`j \[Element] {1, 2,  ... , n}\)]],
  ", there is some ",
  Cell[BoxData[
      \(TraditionalForm\`i \[Element] {1, 2,  ... , n}\)]],
  " with ",
  Cell[BoxData[
      \(TraditionalForm\`\[Sigma](i) = j\)]],
  "). Write down the list ",
  Cell[BoxData[
      \(TraditionalForm\`\[Sigma](1), \ \[Sigma](2), \  ... , \ \[Sigma](
        n)\)]],
  ", and sort this list into order by swapping adjacent entries: if you used \
an even number set ",
  Cell[BoxData[
      \(TraditionalForm\`sgn(\[Sigma]) = 1\)]],
  ", and otherwise set ",
  Cell[BoxData[
      \(TraditionalForm\`sgn(\[Sigma]) = \(-1\)\)]],
  ". The determinant of ",
  Cell[BoxData[
      \(TraditionalForm\`A\)]],
  " is given by a sum over ",
  StyleBox["all",
    FontSlant->"Italic"],
  " permutations:",
  "\n",
  Cell[BoxData[
      \(TraditionalForm\`\(\(\t\t\)\(Det[
          A] = \[Sum]\+\[Sigma]\( 
              sgn(\[Sigma])\)\ \(a\_\(1, \[Sigma](1)\)\) \(a\_\(2, \[Sigma](
                  2)\)\) \(a\_\(3, \[Sigma](
                  3)\)\) \[CenterEllipsis]\ \(\(a\_\(n, \[Sigma](
                    n)\)\)\(.\)\)\)\)\)]]
}], "Text"],

Cell["\<\
Why would anyone use this definition!?! It's the one the book uses, so there \
must be a reason and it is this: this definition is the one that makes the \
most proofs easiest.\
\>", "Text"],

Cell[TextData[{
  "It's actually not as bad as it seems. Start with your matrix: take some of \
the entries so that you have one from each row and one from each column, and \
then multiply them together. In fact, do that in every possible way, and add \
all of them together (with some minus signs). One example may help. Let's \
start with the matrix ",
  Cell[BoxData[
      FormBox[
        RowBox[{"A", "=", 
          RowBox[{"(", GridBox[{
                {"a", "b"},
                {"c", "d"}
                }], ")"}]}], TraditionalForm]]],
  ". If you take \"",
  Cell[BoxData[
      \(TraditionalForm\`a, d\)]],
  "\", then you have one thing from each row and from each column. Another \
way to do this is to take \"",
  Cell[BoxData[
      \(TraditionalForm\`b, c\)]],
  "\". The determinant is ",
  Cell[BoxData[
      \(TraditionalForm\`\(\[PlusMinus]a\)\ d\  \[PlusMinus] b\ c\)]],
  ". So how do the signs come out?"
}], "Text"],

Cell[TextData[{
  "Let's get the sign for ",
  Cell[BoxData[
      \(TraditionalForm\`a\ d\)]],
  ". The thing from the first row is in column 1, and the thing from the \
second row is in column 2, so our list is (1,2), which is already in \
increasing order. We need zero swaps to put this list in order, and zero is \
an even number, so this permutation has ",
  Cell[BoxData[
      \(TraditionalForm\`sgn(\[Sigma]) = 1\)]],
  ". Now for the sign before ",
  Cell[BoxData[
      \(TraditionalForm\`b\ c\)]],
  ": the list we get is (2,1), which we need one swap to put in order. One is \
odd, so for this \[Sigma] we have ",
  Cell[BoxData[
      \(TraditionalForm\`sgn(\[Sigma]) = \(-1\)\)]],
  ". Thus, the determinant is ",
  Cell[BoxData[
      \(TraditionalForm\`\(+a\)\ d\  - b\ c\)]],
  "."
}], "Text"],

Cell[TextData[{
  "Explain how to use this definition to compute the determinant of ",
  Cell[BoxData[
      FormBox[
        RowBox[{"(", "\[NegativeThinSpace]", GridBox[{
              {"a", "b", "c"},
              {"d", "e", "f"},
              {"g", "h", "i"}
              }], "\[NegativeThinSpace]", ")"}], TraditionalForm]]],
  "."
}], "Text"],

Cell[TextData[{
  "Prove that switching two adjacent rows multiplies the sign of the \
determinant by ",
  Cell[BoxData[
      \(TraditionalForm\`\(-1\)\)]],
  "."
}], "Text"],

Cell["\<\
Prove that replacing a row by itself plus a multiple of another row does not \
change the determinant.\
\>", "Text"],

Cell["\<\
Prove that multiplying a row by a constant multiplies the determinant by the \
same constant.\
\>", "Text"],

Cell["Prove that this definition implies the first definition.", "Text"],

Cell["Prove that this definition implies the second definition.", "Text"]
}, Closed]]
}, Open  ]],

Cell[CellGroupData[{

Cell[TextData[{
  "Use ",
  StyleBox["Mathematica",
    FontSlant->"Italic"],
  " "
}], "Subsection"],

Cell["\<\
Compute the determinant of the matrix \
{{1,1/2,1/3,1/4},{1/2,1/3,1/4,1/5},{1/3,1/4,1/5,1/6},{1/4,1/5,1/6,1/7}}.\
\>", "Text"],

Cell[TextData[{
  "Find an ",
  Cell[BoxData[
      \(TraditionalForm\`x\)]],
  " so that ",
  Cell[BoxData[
      FormBox[
        RowBox[{"(", "\[NegativeThinSpace]", GridBox[{
              {"1", \(1/2\), \(1/3\), \(1/4\)},
              {\(1/2\), \(1/3\), \(1/4\), \(1/5\)},
              {\(1/3\), \(1/4\), \(1/5\), \(1/6\)},
              {\(1/4\), \(1/5\), \(1/6\), "x"}
              }], "\[NegativeThinSpace]", ")"}], TraditionalForm]]],
  " does not have an inverse. How close is ",
  Cell[BoxData[
      \(TraditionalForm\`x\)]],
  " to ",
  Cell[BoxData[
      \(TraditionalForm\`1/7\)]],
  "?"
}], "Text"]
}, Closed]]
}, Open  ]]
},
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