(************** Content-type: application/mathematica **************

                    Mathematica-Compatible Notebook

This notebook can be used with any Mathematica-compatible
application, such as Mathematica, MathReader or Publicon. The data
for the notebook starts with the line containing stars above.

To get the notebook into a Mathematica-compatible application, do
one of the following:

* Save the data starting with the line of stars above into a file
  with a name ending in .nb, then open the file inside the
  application;

* Copy the data starting with the line of stars above to the
  clipboard, then use the Paste menu command inside the application.

Data for notebooks contains only printable 7-bit ASCII and can be
sent directly in email or through ftp in text mode.  Newlines can be
CR, LF or CRLF (Unix, Macintosh or MS-DOS style).

NOTE: If you modify the data for this notebook not in a Mathematica-
compatible application, you must delete the line below containing
the word CacheID, otherwise Mathematica-compatible applications may
try to use invalid cache data.

For more information on notebooks and Mathematica-compatible 
applications, contact Wolfram Research:
  web: http://www.wolfram.com
  email: info@wolfram.com
  phone: +1-217-398-0700 (U.S.)

Notebook reader applications are available free of charge from 
Wolfram Research.
*******************************************************************)

(*CacheID: 232*)


(*NotebookFileLineBreakTest
NotebookFileLineBreakTest*)
(*NotebookOptionsPosition[     21962,        701]*)
(*NotebookOutlinePosition[     22782,        731]*)
(*  CellTagsIndexPosition[     22709,        725]*)
(*WindowFrame->Normal*)



Notebook[{
Cell[TextData[{
  "You are to go through each example carefully and answer any questions \
posed.\n\nThen, select some problems from your text that you would not want \
to do by hand and let ",
  StyleBox["Mathematica",
    FontSlant->"Italic"],
  " do them for you. Be sure to select a variety of problems that cover the \
notion of linear independence of vectors and functions, vectors forming a \
basis and spanning a space, and writing other vectors in the space as linear \
combinations of the spanning set."
}], "Text"],

Cell[CellGroupData[{

Cell["Linear Algebra Concepts", "Title",
  PageWidth->WindowWidth],

Cell[CellGroupData[{

Cell["Matrices and Determinants", "Section",
  PageWidth->WindowWidth],

Cell[TextData[{
  "Many linear algebra commands are built directly into ",
  StyleBox["Mathematica.",
    FontSlant->"Italic"],
  " Selected, more sophisticated commands require the reading in of a ",
  StyleBox["Mathematica",
    FontSlant->"Italic"],
  " Package. "
}], "Text"],

Cell[CellGroupData[{

Cell["Example 1", "Subsection"],

Cell["\<\
To solve a system of equations, we will first enter the coefficient matrix of \
the system and give it a name so that we can refer to it later. \
\>", "Text"],

Cell[BoxData[{
    \(Clear[A]\), "\[IndentingNewLine]", 
    \(A\  = {{1, \ \(-1\), \ 1}, {2, \ \(-3\), \ 4}, \ {\(-2\), \ \(-1\), \ 
          1}}\ \), "\[IndentingNewLine]", 
    \(MatrixForm[A]\)}], "Input"],

Cell[TextData[{
  "You cannot define a matrix in MatrixForm with this command if you wish to \
do manipulations with it. You have to leave it in the original form of ",
  StyleBox["A",
    FontSlant->"Italic"],
  ".  \nNow, you can enter the constants on the right-hand side as a vector \
(equivalent to one row or one column of a matrix). ",
  StyleBox["Mathematica",
    FontSlant->"Italic"],
  " will interpret this as either a row or a column vector, depending on how \
it is called upon."
}], "Text"],

Cell[BoxData[{
    \(Clear[b]\), "\[IndentingNewLine]", 
    \(b = {0, \ \(-2\), \ 7}\)}], "Input"],

Cell[TextData[{
  "You can solve this system of equations several ways. ",
  StyleBox["Mathematica",
    FontSlant->"Italic"],
  " solves it using linear algebra if you call on the ",
  StyleBox["LinearSolve",
    FontWeight->"Bold"],
  " command."
}], "Text",
  PageWidth->PaperWidth],

Cell[BoxData[
    \(LinearSolve[A, b]\)], "Input"],

Cell[TextData[{
  "The Solve command also works. The Solve command does not necessarily call \
upon the same linear algebra techniques as are used in the ",
  StyleBox["LinearSolve",
    FontWeight->"Bold"],
  " command."
}], "Text"],

Cell[BoxData[{
    \(Clear[x, y, z, vars]\), "\[IndentingNewLine]", 
    \(vars = {x, y, z}\), "\[IndentingNewLine]", 
    \(Solve[A . vars \[Equal] b, vars]\)}], "Input"],

Cell["\<\
Yet another alternative is to enter the equations exactly as they first \
appeared, without matrix notation. Note that the equations are enclosed in \
curly braces, since they each are part of what has to be solved.\
\>", "Text"],

Cell[BoxData[
    \(Solve[{x - y + z \[Equal] 0, 
        2  x - 3  y + 4  z \[Equal] \(-2\), \(-2\) x - y + z \[Equal] 7}, {x, 
        y, z}]\)], "Input"],

Cell[TextData[{
  "Yet another way in that you could solve a system of equations would be to \
first enter the system in equation form and let ",
  StyleBox["Mathematica",
    FontSlant->"Italic"],
  " convert the equations to matrix form. To do this, you need to first read \
in a package. It must be read in before attempting to execute the ",
  StyleBox["LinearEquationsToMatrices",
    FontWeight->"Bold"],
  " command."
}], "Text"],

Cell["<< LinearAlgebra`MatrixManipulation` ", "Input",
  CellTags->"S5.51.1"],

Cell[TextData[{
  "Now enter your equations inside the ",
  StyleBox["LinearEquationsToMatrices",
    FontWeight->"Bold"],
  " command. Again, note the double equal sign that must be used."
}], "Text"],

Cell[BoxData[
    \(meq = 
      LinearEquationsToMatrices[{x - y + z \[Equal] 0, 
          2  x - 3  y + 4  z \[Equal] \(-2\), \(-2\) x - y + z \[Equal] 
            7}, {x, y, z}]\)], "Input"],

Cell["\<\
You can use a double bracket [[...]] to pull out parts if lists.\
\>", "Text"],

Cell[BoxData[{
    \(A = meq[\([1]\)]\), "\[IndentingNewLine]", 
    \(b = meq[\([2]\)]\)}], "Input"],

Cell[TextData[{
  "Then use the ",
  StyleBox["LinearSolve",
    FontWeight->"Bold"],
  " or ",
  StyleBox["Solve",
    FontWeight->"Bold"],
  " command as above."
}], "Text"]
}, Closed]],

Cell[CellGroupData[{

Cell["Example 2", "Subsection"],

Cell["See what happens with this system.", "Text"],

Cell[BoxData[{
    \(Clear[A, b]\), "\[IndentingNewLine]", 
    \(\(A = {{1, 1, \(-1\), 2}, {1, 1, 0, 1}, {1, 2, \(-4\), 0}, {2, 1, 2, 
            5}};\)\), "\[IndentingNewLine]", 
    \(MatrixForm[A]\), "\[IndentingNewLine]", 
    \(\(b = {1, 2, 1, 1};\)\), "\[IndentingNewLine]", 
    \(MatrixForm[b]\), "\[IndentingNewLine]", 
    \(LinearSolve[A, b]\)}], "Input"],

Cell["\<\
The blue error message indicates that this system has no solution, so your \
input command is sent back to you.\
\>", "Text"],

Cell["\<\
Let's use Gaussian elimination on this matrix and see what happens. This is \
done using the command RowReduce.\
\>", "Text"],

Cell[BoxData[
    \(RowReduce[A] // MatrixForm\)], "Input"],

Cell["\<\
Let's also check out the determinant of the coefficient matrix.\
\>", "Text"],

Cell[BoxData[
    \(Det[A]\)], "Input"],

Cell[BoxData[
    \(While\ we\ are\ at\ it, \ 
    why\ not\ look\ at\ the\ eigenvalues\ for\ the\ coefficient\ \(\(matrix\)\
\(?\)\)\)], "Input"],

Cell[BoxData[
    \(Eigenvalues[A] // N\)], "Input"],

Cell["Notice that there is an eigenvalue of 0.", "Text"]
}, Closed]],

Cell[CellGroupData[{

Cell["Example 3", "Subsection"],

Cell[TextData[{
  "Note the distinction between ",
  StyleBox["LinearSolve",
    FontWeight->"Bold"],
  " and ",
  StyleBox["Solve",
    FontWeight->"Bold"],
  " in this example."
}], "Text"],

Cell[BoxData[{
    \(Clear[A, b]\), "\[IndentingNewLine]", 
    \(\(A = {{2, 3, \(-1\)}, {\(-1\), \(-1\), 3}, {1, 2, 2}, {0, 1, 
            5}};\)\), "\[IndentingNewLine]", 
    \(MatrixForm[A]\), "\[IndentingNewLine]", 
    \(\(b = {3, 0, 3, 3};\)\), "\[IndentingNewLine]", 
    \(MatrixForm[b]\), "\[IndentingNewLine]", 
    \(LinearSolve[A, b]\)}], "Input"],

Cell[TextData[{
  "The ",
  StyleBox["LinearSolve",
    FontWeight->"Bold"],
  " gave us only one set of solutions, but look at what ",
  StyleBox["Solve",
    FontWeight->"Bold"],
  " yields."
}], "Text"],

Cell[BoxData[
    \(Solve[A . {x, y, z} \[Equal] b, {x, y, z}]\)], "Input"],

Cell["\<\
This means that z can can be assigned any value, thus yielding and entire let \
of solutions that fall along a line in 3-space.\
\>", "Text"],

Cell["Let's try RowReduce on this one.", "Text"],

Cell[BoxData[
    \(RowReduce[A] // MatrixForm\)], "Input"],

Cell["\<\
Both this example and the one before yielded a zero row after using Gaussian \
elimination. Explain why the pervious one yielded no solutions and this \
yielded an infinite number of solutions.

We might as well check out the determinant and the eigenvalues for the \
coefficient matrix.\
\>", "Text"],

Cell[BoxData[{
    \(Det[A]\), "\[IndentingNewLine]", 
    \(Eigenvalues[A] // N\)}], "Input"],

Cell["Take note of this pattern you are observing.", "Text"]
}, Closed]],

Cell[CellGroupData[{

Cell["Example 4", "Subsection"],

Cell[BoxData[{
    \(Clear[A, b]\), "\[IndentingNewLine]", 
    \(\(A = {{4, \(-8\), \(-1\), 1, 3}, {5, \(-10\), \(-1\), 2, 
            3}, {3, \(-6\), \(-1\), 1, 2}};\)\), "\[IndentingNewLine]", 
    \(\(b = {0, 0, 0};\)\), "\[IndentingNewLine]", 
    \(vars = {x\_1, x\_2, x\_3, x\_4, x\_5}\), "\[IndentingNewLine]", 
    \(LinearSolve[A, b]\), "\[IndentingNewLine]", 
    \(Solve[A . vars \[Equal] b, vars]\)}], "Input"],

Cell[TextData[{
  "Note that ",
  StyleBox["LinearSolve",
    FontWeight->"Bold"],
  " gave only the trivial solution, while the ",
  StyleBox["Solve",
    FontWeight->"Bold"],
  " command yields formulas an infinite number of solutions, since ",
  Cell[BoxData[
      \(TraditionalForm\`x\_2\)]],
  "and ",
  Cell[BoxData[
      \(TraditionalForm\`x\_5\)]],
  "can take on any values.\n\nAgain look at the determinant and eigenvalues."
}], "Text"],

Cell[BoxData[{
    \(Det[A]\), "\[IndentingNewLine]", 
    \(Eigenvalues[A] // N\)}], "Input"]
}, Closed]],

Cell[CellGroupData[{

Cell["Example 5", "Subsection"],

Cell[TextData[{
  "To find the inverse of a matrix, use the ",
  StyleBox["Inverse",
    FontWeight->"Bold"],
  " command in ",
  StyleBox["Mathematica",
    FontSlant->"Italic"],
  "."
}], "Text"],

Cell[BoxData[{
    \(Clear[A]\), "\[IndentingNewLine]", 
    \(\(A = {{18, \(-24\), 25, 27}, {\(-12\) - 2  \[Pi], 
            14 - \[Pi], \(-17\), \(-27\)}, {15 + 3 \@ 3, \(-18\) - 4 \@ 3, 
            21 + 4 \@ 3, 27}, {\(-10\), 
            12, \(-14\), \(-18\)}};\)\), "\[IndentingNewLine]", 
    \(Inverse[A] // MatrixForm\), "\[IndentingNewLine]", 
    \(\(Inverse[A] // MatrixForm\) // N\)}], "Input"],

Cell["\<\
The first output is in exact form and the second in approximate form. If any \
decimals had been in the original matrix, the computational mode and output \
is automatically decimal.
What if a matrix does not possess an inverse?\
\>", "Text"],

Cell[BoxData[{
    \(\(B = {{1, 1, 0, 0, 0}, {2, 2, 2, 0, 0}, {0, 3, 3, 3, 0}, {0, 0, 4, 4, 
            4}, {0, 0, 0, 5, 5}};\)\), "\[IndentingNewLine]", 
    \(MatrixForm[B]\), "\[IndentingNewLine]", 
    \(Inverse[B]\)}], "Input"],

Cell[BoxData[
    StyleBox[\(Is\ there\ any\ relation\ between\ no\ inverse\ and\ zero\ \
determinant\ and\ a\ zero\ \(\(eigenvalue\)\(?\)\)\),
      FontFamily->"Times New Roman"]], "Text"],

Cell[BoxData[{
    \(Det[B]\), "\[IndentingNewLine]", 
    \(Eigenvalues[B] // N\)}], "Input"]
}, Closed]],

Cell[CellGroupData[{

Cell["Example 6", "Subsection"],

Cell[TextData[{
  "The transpose of a matrix is the one with rows taking the place of columns \
and vice-versa. It can be found in ",
  StyleBox["Mathematica",
    FontSlant->"Italic"],
  " with the ",
  StyleBox["Transpose",
    FontWeight->"Bold"],
  " command. "
}], "Text"],

Cell[BoxData[{
    \(Clear[A]\), "\[IndentingNewLine]", 
    \(A = {{2, 1, \(-3\), 4, 1}, {\(-2\), 1, 1, 0, 7}, {6, 2, \(-8\), 4, 
          9}, {\(-2\), 0, 2, \(-6\), \(-2\)}, {10, 3, \(-13\), 14, 
          12}}\), "\[IndentingNewLine]", 
    \(Transpose[A] // MatrixForm\), "\[IndentingNewLine]", 
    \(A . Transpose[A] // MatrixForm\), "\[IndentingNewLine]", 
    \(Transpose[A] . A // MatrixForm\)}], "Input"],

Cell[BoxData[{
    \(RowReduce[A] // MatrixForm\), "\[IndentingNewLine]", 
    \(RowReduce[Transpose[A]] // MatrixForm\)}], "Input"],

Cell["\<\
Note that if there is a zero row in A, there will be a zero row in its \
transpose. Compare their eigenvalues.\
\>", "Text"],

Cell[BoxData[{
    \(Eigenvalues[A]\), "\[IndentingNewLine]", 
    \(Eigenvalues[Transpose[A]]\)}], "Input"],

Cell["How about their determinants?", "Text"],

Cell[BoxData[{
    \(Det[A]\), "\[IndentingNewLine]", 
    \(Det[Transpose[A]]\)}], "Input"]
}, Closed]],

Cell[CellGroupData[{

Cell["Example 7", "Subsection"],

Cell["\<\
Let's check all this out for a matrix whose determinant is not zero.\
\>", "Text"],

Cell[BoxData[{
    \(Clear[A]\), "\[IndentingNewLine]", 
    \(\(A = {{4, \(-2\), 16, 27, \(-11\)}, {9, 43, 9, \(-8\), \(-1\)}, {34, 
            20, \(-3\), 0, 21}, {\(-5\), 4, 4, 7, 41}, {0, 
            12, \(-2\), \(-2\), 3}};\)\), "\[IndentingNewLine]", 
    \(MatrixForm[A]\), "\[IndentingNewLine]", 
    \(\(Ainv = Inverse[A];\)\), "\[IndentingNewLine]", 
    \(MatrixForm[Ainv]\), "\[IndentingNewLine]", 
    \(\(Atran = Transpose[A];\)\), "\[IndentingNewLine]", 
    \(MatrixForm[Atran]\)}], "Input"],

Cell["Compare determinants.", "Text"],

Cell[BoxData[{
    \(Det[A]\), "\[IndentingNewLine]", 
    \(Det[Ainv]\), "\[IndentingNewLine]", 
    \(Det[Atran]\)}], "Input"],

Cell["Compare eigenvalues.", "Text"],

Cell[BoxData[{
    \(Print["\<The eigenvalues of A are \>", 
      Eigenvalues[A] // N]\), "\[IndentingNewLine]", 
    \(Print["\<The eigenvalues of Ainverse are \>", 
      Eigenvalues[Ainv] // N]\), "\[IndentingNewLine]", 
    \(Print["\<The eigenvalues of Atranspose are \>", 
      Eigenvalues[Atran] // N]\)}], "Input"]
}, Closed]]
}, Closed]],

Cell[CellGroupData[{

Cell["Vector Spaces", "Section"],

Cell[CellGroupData[{

Cell["Example 8", "Subsection"],

Cell["\<\
One way to determine if a vector is in a space determined by other vectors is \
to see if you can find constants such that that vector can be written as a \
linear combination of the vectors that span the space.\
\>", "Text"],

Cell[BoxData[{
    \(\(v1 = {4, \(-2\), 1, 2, 3, 1};\)\), "\[IndentingNewLine]", 
    \(\(v2 = {2, \(-2\), 1, \(-5\), 7, 1};\)\), "\[IndentingNewLine]", 
    \(\(v3 = {\(-17\), 22, \(-1\), 0, 2, 1};\)\), "\[IndentingNewLine]", 
    \(\(v4 = {\(-31\), \(-3\), \(-8\), 1, 1, 0};\)\), "\[IndentingNewLine]", 
    \(\(v5 = {17, 44, 1, \(-22\), 11, 1.9};\)\), "\[IndentingNewLine]", 
    \(\(v6 = {1, 0, 1, 15, \(-3\), 3};\)\), "\[IndentingNewLine]", 
    \(\(newv = {1, 2, \(-2\), 0, \(-1\), 4};\)\), "\[IndentingNewLine]", 
    \(Solve[
      newv \[Equal] c1\ v1 + c2\ v2 + c3\ v3 + c4\ v4 + c5\ v5 + c6\ v6, {c1, 
        c2, c3, c4, c5, c6}]\)}], "Input"],

Cell["\<\
If no constants can be found, the new vector is not in the space determined \
by the other six.  let's see what happens if we try to solve for the \
constants when the newv is the zero vector.\
\>", "Text"],

Cell[BoxData[{
    \(\(newv = {0, 0, 0, 0, 0, 0};\)\), "\[IndentingNewLine]", 
    \(Solve[
      newv \[Equal] c1\ v1 + c2\ v2 + c3\ v3 + c4\ v4 + c5\ v5 + c6\ v6, {c1, 
        c2, c3, c4, c5, c6}]\)}], "Input"],

Cell["\<\
What does all this say about whether or not the vectors are linearly \
independent?

While we are at it,let's look at the matrix formed by these six vectors and \
examine it from different perspectives.\
\>", "Text"],

Cell[BoxData[{
    \(\(M = {v1, v2, v3, v4, v5, v6};\)\), "\[IndentingNewLine]", 
    \(MatrixForm[M]\), "\[IndentingNewLine]", 
    \(\(MT = Transpose[M];\)\), "\[IndentingNewLine]", 
    \(MatrixForm[MT]\)}], "Input"],

Cell["\<\
Note that the first matrix has the vectors as rows and the second has the \
vectors as columns. Now let's look at them in reduced echelon form and their \
determinants and eigenvalues.\
\>", "Text"],

Cell[BoxData[{
    \(RowReduce[M] // MatrixForm\), "\[IndentingNewLine]", 
    \(RowReduce[MT] // MatrixForm\), "\[IndentingNewLine]", 
    \(Det[M]\), "\[IndentingNewLine]", 
    \(Det[MT]\), "\[IndentingNewLine]", 
    \(Eigenvalues[M]\), "\[IndentingNewLine]", 
    \(Eigenvalues[MT]\)}], "Input"],

Cell["What conclusions might you draw?", "Text"]
}, Closed]],

Cell[CellGroupData[{

Cell["Example 9", "Subsection"],

Cell["\<\
Problems with polynomials can be done in the same way,with polynomial \
coefficients replacing the vector components\
\>", "Text"],

Cell[BoxData[{
    \(\(p1 = {1, \(-1\), 0, 3, \(-4\)};\)\), "\[IndentingNewLine]", 
    \(\(p2 = {1, \(-1\), \(-1\), 1, \(-4\)};\)\ \), "\[IndentingNewLine]", 
    \(\(p3 = {0, 1, 1, \(-3\), 3};\)\), "\[IndentingNewLine]", 
    \(\(p4 = {1, 1, \(-2\), 4, \(-8\)};\)\), "\[IndentingNewLine]", 
    \(\(p5 = {5, \(-7\), \(-2\), \(-1\), 9};\)\), "\[IndentingNewLine]", 
    \(\(p6 = {2, \(-7\), 0, 0, 1};\)\), "\[IndentingNewLine]", 
    \(\(newp = {1, 0, 1, 0, 1};\)\), "\[IndentingNewLine]", 
    \(Solve[
      newp \[Equal] c1\ p1 + c2\ p2 + c3\ p3 + c4\ p4 + c5\ p5 + c6\ p6, {c1, 
        c2, c3, c4, c5, c6}]\)}], "Input"],

Cell["\<\
As you probably predicted, there is more than one solution here.
Consider the vectors defined in the problem above. Form a matrix whose rows \
are these vectors, then find the transpose to form a matrix whose columns are \
these vectors and do a row-reduction on this matrix. \
\>", "Text"],

Cell[BoxData[{
    \(\(v1 = {4, \(-2\), 1, 2, 3, 1};\)\), "\[IndentingNewLine]", 
    \(\(v2 = {2, \(-2\), 1, \(-5\), 7, 1};\)\), "\[IndentingNewLine]", 
    \(\(v3 = {\(-17\), 22, \(-1\), 0, 2, 1};\)\), "\[IndentingNewLine]", 
    \(\(v4 = {\(-31\), \(-3\), \(-8\), 1, 1, 0};\)\), "\[IndentingNewLine]", 
    \(\(v5 = {17, 44, 1, \(-22\), 11, 1.9};\)\), "\[IndentingNewLine]", 
    \(\(v6 = {1, 0, 1, 15, \(-3\), 3};\)\), "\[IndentingNewLine]", 
    \(\(mat = {v1, v2, v3, v4, v5, v6};\)\), "\[IndentingNewLine]", 
    \(\(trmat = Transpose[mat];\)\), "\[IndentingNewLine]", 
    \(RowReduce[trmat] // MatrixForm\ \)}], "Input"],

Cell["What does this result tell you?", "Text"]
}, Closed]],

Cell[CellGroupData[{

Cell["Example 10", "Subsection"],

Cell["\<\
The method for showing that matrices form a basis is similar to that for \
showing that vectors form a basis.\
\>", "Text"],

Cell[BoxData[{
    \(\(m1 = {{1, 0}, {\(-3\), 
            33}, {11, \(-2\)}};\)\), "\[IndentingNewLine]", 
    \(\(m2 = {{1, \(-3\)}, {1, 
            13}, {16, \(-2\)}};\)\ \), "\[IndentingNewLine]", 
    \(\(m3 = {{1, 17}, {\(-3\), 
            46}, {32, \(-9\)}};\)\), "\[IndentingNewLine]", 
    \(\(m4 = {{\(-1\), 91}, {6, \(-98\)}, {\(-1\), 
            1}};\)\), "\[IndentingNewLine]", 
    \(\(m5 = {{\(-2\), \(-1\)}, {21, 0}, {0, 
            0}};\)\), "\[IndentingNewLine]", 
    \(\(m6 = {{3, 2}, {5, \(-5\)}, {\(-4\), 0}};\)\), "\[IndentingNewLine]", 
    \(Solve[
      c1\ m1 + c2\ m2 + c3\ m3 + c4\ m4 + c5\ m5 + 
          c6\ m6 \[Equal] {{0, 0}, {0, 0}, {0, 0}}, {c1, c2, c3, c4, c5, 
        c6}]\)}], "Input"],

Cell["What is your conclusion?", "Text"]
}, Closed]],

Cell[CellGroupData[{

Cell["Example 11", "Subsection"],

Cell[TextData[{
  "To find a basis for the nullspace of a matrix, ",
  StyleBox["Mathematica",
    FontSlant->"Italic"],
  " has a ",
  StyleBox["Nullspace",
    FontWeight->"Bold"],
  " command."
}], "Text"],

Cell[BoxData[{
    \(Clear[A]\), "\[IndentingNewLine]", 
    \(\(A = {{\(-2\), 3, \(-1\), 4, \(-2\)}, {3, \(-1\), 2, 6, 8}, {1, 
            7, \(-5\), 1, 4}, {0, 12, \(-5\), 15, 
            8}};\)\), "\[IndentingNewLine]", 
    \(MatrixForm[A]\), "\[IndentingNewLine]", 
    \(nullA = NullSpace[A]\  // MatrixForm\)}], "Input"],

Cell["\<\
What on earth is a null space?  Look the term up in your textbook and find \
out.\
\>", "Text"],

Cell[TextData[{
  "To find a basis for the row vectors, you can use the ",
  StyleBox["RowReduce",
    FontWeight->"Bold"],
  " command."
}], "Text"],

Cell[BoxData[
    \(RowReduce[A]\  // MatrixForm\)], "Input"],

Cell[TextData[{
  "To find a basis for the column vectors, you can use the ",
  StyleBox["RowReduce",
    FontWeight->"Bold"],
  " command on the ",
  StyleBox["Transpose",
    FontWeight->"Bold"],
  " of ",
  StyleBox["A",
    FontSlant->"Italic"],
  "."
}], "Text"],

Cell[BoxData[
    \(RowReduce[Transpose[A]]\  // MatrixForm\)], "Input"]
}, Closed]],

Cell[CellGroupData[{

Cell["Example 12", "Subsection"],

Cell[TextData[{
  "Although there is not a Wronskian command in ",
  StyleBox["Mathematica",
    FontSlant->"Italic"],
  ", it is easy to define using the command for derivatives, ",
  StyleBox["D[f, x]",
    FontWeight->"Bold"],
  " which represents taking the derivative of the function f with respect to \
x. If you want a second derivative, use ",
  StyleBox["D[f, {x,2}]",
    FontWeight->"Bold"],
  ", etc."
}], "Text"],

Cell[BoxData[{
    \(functions = {\[ExponentialE]\^\(3  x\), 
        x\ \[ExponentialE]\^\(3  x\), \[ExponentialE]\^\(3  x\)\ Cos[
            2  x], \[ExponentialE]\^\(3  x\)\ Sin[2  x], x\ Cos[x], 
        x\ Sin[x]}\), "\[IndentingNewLine]", 
    \(\(wronmatrix = {functions, D[functions, x], D[functions, {x, 2}], 
          D[functions, {x, 3}], D[functions, {x, 4}], 
          D[functions, {x, 5}]};\)\), "\[IndentingNewLine]", 
    \(MatrixForm[wronmatrix]\), "\[IndentingNewLine]", 
    \(wronskian = Det[wronmatrix] // Simplify\), "\[IndentingNewLine]", 
    \(wronskian /. \(\(x\)\(\[Rule]\)\(0\)\(\ \)\)\)}], "Input"],

Cell["\<\
The Simplify command is very helpful here. The last line of code asks for the \
value of the Wronskian when x is 0.\
\>", "Text"]
}, Closed]]
}, Closed]]
}, Open  ]]
},
FrontEndVersion->"4.1 for Microsoft Windows",
ScreenRectangle->{{0, 1024}, {0, 695}},
WindowSize->{497, 537},
WindowMargins->{{0, Automatic}, {Automatic, 0}},
PrintingCopies->1,
PrintingPageRange->{Automatic, Automatic}
]

(*******************************************************************
Cached data follows.  If you edit this Notebook file directly, not
using Mathematica, you must remove the line containing CacheID at
the top of  the file.  The cache data will then be recreated when
you save this file from within Mathematica.
*******************************************************************)

(*CellTagsOutline
CellTagsIndex->{
  "S5.51.1"->{
    Cell[5342, 164, 77, 1, 30, "Input",
      CellTags->"S5.51.1"]}
  }
*)

(*CellTagsIndex
CellTagsIndex->{
  {"S5.51.1", 22615, 718}
  }
*)

(*NotebookFileOutline
Notebook[{
Cell[1705, 50, 525, 10, 128, "Text"],

Cell[CellGroupData[{
Cell[2255, 64, 66, 1, 115, "Title"],

Cell[CellGroupData[{
Cell[2346, 69, 70, 1, 59, "Section"],
Cell[2419, 72, 279, 8, 52, "Text"],

Cell[CellGroupData[{
Cell[2723, 84, 31, 0, 47, "Subsection"],
Cell[2757, 86, 168, 3, 52, "Text"],
Cell[2928, 91, 210, 4, 70, "Input"],
Cell[3141, 97, 505, 11, 109, "Text"],
Cell[3649, 110, 99, 2, 50, "Input"],
Cell[3751, 114, 285, 9, 52, "Text"],
Cell[4039, 125, 50, 1, 30, "Input"],
Cell[4092, 128, 233, 6, 52, "Text"],
Cell[4328, 136, 171, 3, 70, "Input"],
Cell[4502, 141, 239, 4, 71, "Text"],
Cell[4744, 147, 156, 3, 50, "Input"],
Cell[4903, 152, 436, 10, 90, "Text"],
Cell[5342, 164, 77, 1, 30, "Input",
  CellTags->"S5.51.1"],
Cell[5422, 167, 201, 5, 52, "Text"],
Cell[5626, 174, 195, 4, 70, "Input"],
Cell[5824, 180, 88, 2, 33, "Text"],
Cell[5915, 184, 101, 2, 50, "Input"],
Cell[6019, 188, 175, 8, 33, "Text"]
}, Closed]],

Cell[CellGroupData[{
Cell[6231, 201, 31, 0, 31, "Subsection"],
Cell[6265, 203, 50, 0, 33, "Text"],
Cell[6318, 205, 368, 7, 150, "Input"],
Cell[6689, 214, 135, 3, 52, "Text"],
Cell[6827, 219, 135, 3, 52, "Text"],
Cell[6965, 224, 59, 1, 30, "Input"],
Cell[7027, 227, 87, 2, 33, "Text"],
Cell[7117, 231, 39, 1, 30, "Input"],
Cell[7159, 234, 146, 3, 70, "Input"],
Cell[7308, 239, 52, 1, 30, "Input"],
Cell[7363, 242, 56, 0, 33, "Text"]
}, Closed]],

Cell[CellGroupData[{
Cell[7456, 247, 31, 0, 31, "Subsection"],
Cell[7490, 249, 191, 8, 33, "Text"],
Cell[7684, 259, 361, 7, 130, "Input"],
Cell[8048, 268, 205, 8, 33, "Text"],
Cell[8256, 278, 75, 1, 30, "Input"],
Cell[8334, 281, 151, 3, 52, "Text"],
Cell[8488, 286, 48, 0, 33, "Text"],
Cell[8539, 288, 59, 1, 30, "Input"],
Cell[8601, 291, 311, 7, 128, "Text"],
Cell[8915, 300, 94, 2, 50, "Input"],
Cell[9012, 304, 60, 0, 33, "Text"]
}, Closed]],

Cell[CellGroupData[{
Cell[9109, 309, 31, 0, 31, "Subsection"],
Cell[9143, 311, 424, 7, 150, "Input"],
Cell[9570, 320, 448, 14, 90, "Text"],
Cell[10021, 336, 94, 2, 50, "Input"]
}, Closed]],

Cell[CellGroupData[{
Cell[10152, 343, 31, 0, 31, "Subsection"],
Cell[10186, 345, 197, 8, 33, "Text"],
Cell[10386, 355, 408, 7, 135, "Input"],
Cell[10797, 364, 252, 5, 90, "Text"],
Cell[11052, 371, 233, 4, 90, "Input"],
Cell[11288, 377, 190, 3, 31, "Text"],
Cell[11481, 382, 94, 2, 50, "Input"]
}, Closed]],

Cell[CellGroupData[{
Cell[11612, 389, 31, 0, 31, "Subsection"],
Cell[11646, 391, 277, 9, 52, "Text"],
Cell[11926, 402, 415, 7, 130, "Input"],
Cell[12344, 411, 132, 2, 50, "Input"],
Cell[12479, 415, 134, 3, 52, "Text"],
Cell[12616, 420, 108, 2, 50, "Input"],
Cell[12727, 424, 45, 0, 33, "Text"],
Cell[12775, 426, 92, 2, 50, "Input"]
}, Closed]],

Cell[CellGroupData[{
Cell[12904, 433, 31, 0, 31, "Subsection"],
Cell[12938, 435, 92, 2, 33, "Text"],
Cell[13033, 439, 509, 9, 190, "Input"],
Cell[13545, 450, 37, 0, 33, "Text"],
Cell[13585, 452, 128, 3, 70, "Input"],
Cell[13716, 457, 36, 0, 33, "Text"],
Cell[13755, 459, 324, 6, 110, "Input"]
}, Closed]]
}, Closed]],

Cell[CellGroupData[{
Cell[14128, 471, 32, 0, 39, "Section"],

Cell[CellGroupData[{
Cell[14185, 475, 31, 0, 47, "Subsection"],
Cell[14219, 477, 235, 4, 71, "Text"],
Cell[14457, 483, 655, 10, 190, "Input"],
Cell[15115, 495, 216, 4, 71, "Text"],
Cell[15334, 501, 213, 4, 70, "Input"],
Cell[15550, 507, 226, 6, 90, "Text"],
Cell[15779, 515, 219, 4, 90, "Input"],
Cell[16001, 521, 208, 4, 71, "Text"],
Cell[16212, 527, 300, 6, 130, "Input"],
Cell[16515, 535, 48, 0, 33, "Text"]
}, Closed]],

Cell[CellGroupData[{
Cell[16600, 540, 31, 0, 31, "Subsection"],
Cell[16634, 542, 140, 3, 52, "Text"],
Cell[16777, 547, 626, 10, 190, "Input"],
Cell[17406, 559, 300, 5, 90, "Text"],
Cell[17709, 566, 629, 9, 190, "Input"],
Cell[18341, 577, 47, 0, 33, "Text"]
}, Closed]],

Cell[CellGroupData[{
Cell[18425, 582, 32, 0, 31, "Subsection"],
Cell[18460, 584, 133, 3, 52, "Text"],
Cell[18596, 589, 730, 15, 170, "Input"],
Cell[19329, 606, 40, 0, 33, "Text"]
}, Closed]],

Cell[CellGroupData[{
Cell[19406, 611, 32, 0, 31, "Subsection"],
Cell[19441, 613, 208, 8, 33, "Text"],
Cell[19652, 623, 329, 6, 110, "Input"],
Cell[19984, 631, 105, 3, 33, "Text"],
Cell[20092, 636, 149, 5, 33, "Text"],
Cell[20244, 643, 61, 1, 30, "Input"],
Cell[20308, 646, 267, 11, 52, "Text"],
Cell[20578, 659, 72, 1, 30, "Input"]
}, Closed]],

Cell[CellGroupData[{
Cell[20687, 665, 32, 0, 31, "Subsection"],
Cell[20722, 667, 425, 12, 71, "Text"],
Cell[21150, 681, 630, 10, 172, "Input"],
Cell[21783, 693, 139, 3, 52, "Text"]
}, Closed]]
}, Closed]]
}, Open  ]]
}
]
*)



(*******************************************************************
End of Mathematica Notebook file.
*******************************************************************)