diff --git a/scripts/quickbuild_changes.py b/scripts/quickbuild_changes.py index c6d39f467..050d88247 100644 --- a/scripts/quickbuild_changes.py +++ b/scripts/quickbuild_changes.py @@ -30,7 +30,8 @@ def main(): xml_ids = [] for f in changed_files: if Path("source", b) in f.parents and f.suffix == ".ptx": - root = etree.parse(f).getroot() + parser=etree.XMLParser(recover=True) + root = etree.parse(f,parser=parser).getroot() if root.tag in ["section", "chapter", "preface", "appendix", "frontmatter"]: xml_id = root.get(r"{http://www.w3.org/XML/1998/namespace}id") xml_ids.append({ diff --git a/source/linear-algebra/source/01-LE/01.ptx b/source/linear-algebra/source/01-LE/01.ptx index 89638228e..ecd59c2f9 100644 --- a/source/linear-algebra/source/01-LE/01.ptx +++ b/source/linear-algebra/source/01-LE/01.ptx @@ -342,7 +342,7 @@ is a collection of one or more linear equations. When variables in a large linear system are missing, we prefer to write the system in one of the following standard forms:standardform of a linear system

- +

Original linear system: @@ -386,6 +386,52 @@ is a collection of one or more linear equations.

+ +

+ Original linear system: + + + x_1 + 3x_3 &\,=\,& 3 + + + 3x_1 - 2x_2 + 4x_3 &\,=\,& 0 + + + -x_2 + x_3 &\,=\,& -2 + + +

+ + +

+ Verbose standard form: + + + 1x_1 &\,+\,& 0x_2 &\,+\,& 3x_3 &\,=\,& 3 + + + 3x_1 &\,-\,& 2x_2 &\,+\,& 4x_3 &\,=\,& 0 + + + 0x_1 &\,-\,& 1x_2 &\,+\,& 1x_3 &\,=\,& -2 + + +

+

+ Concise standard form: + + + x_1 & & &\,+\,& 3x_3 &\,=\,& 3 + + + 3x_1 &\,-\,& 2x_2 &\,+\,& 4x_3 &\,=\,& 0 + + + &\,-\,& x_2 &\,+\,& x_3 &\,=\,& -2 + + +

+ @@ -456,7 +502,15 @@ system). Otherwise it is inconsistent.inconsistent linear s
Intersection of three planes at one point - + + + + + +

Three planes are shown to intersect at a single point. An arrow points to the point of intersection at coordinates (1,2,3).

+ + + @@ -468,7 +522,15 @@ system). Otherwise it is inconsistent.inconsistent linear s
Intersection of three planes at a line - + + + + + +

Three planes are shown to intersect along a line of points.

+ + + @@ -480,7 +542,15 @@ system). Otherwise it is inconsistent.inconsistent linear s
Three non-mutually-intersecting planes - + + + + + +

Three planes are shown to intersect at no common point, although each pair of planes intersects along a line of points.

+ + + @@ -704,7 +774,7 @@ Then use these to describe the solution set The only important information in a linear system are its coefficients and constants.

- +

Original linear system: @@ -748,6 +818,52 @@ Then use these to describe the solution set

+ +

+ Original linear system: + + + x_1 + 3x_3 &\,=\,& 3 + + + 3x_1 - 2x_2 + 4x_3 &\,=\,& 0 + + + -x_2 + x_3 &\,=\,& -2 + + +

+ + +

+ Verbose standard form: + + + 1x_1 &\,+\,& 0x_2 &\,+\,& 3x_3 &\,=\,& 3 + + + 3x_1 &\,-\,& 2x_2 &\,+\,& 4x_3 &\,=\,& 0 + + + 0x_1 &\,-\,& 1x_2 &\,+\,& 1x_3 &\,=\,& -2 + + +

+

+ Coefficients/constants: + + + 1 & & 0 &\,\,& 3 &\,|\,& 3 + + + 3 &\, \,& -2 &\,\,& 4 &\,|\,& 0 + + + 0 &\, \,& -1 &\,\,& 1 &\,|\,& -2 + + +

+ diff --git a/source/linear-algebra/source/01-LE/02.ptx b/source/linear-algebra/source/01-LE/02.ptx index 9d4c1d475..36b55ddad 100644 --- a/source/linear-algebra/source/01-LE/02.ptx +++ b/source/linear-algebra/source/01-LE/02.ptx @@ -180,7 +180,7 @@ Swapping two columns, for example:

Add a constant to every term of a row, for example:

- +

\left[\begin{array}{cc|c} 1 & 2 & 4 @@ -197,6 +197,21 @@ Add a constant to every term of a row, for example:

+

+ + \left[\begin{array}{cc|c} 1 & 2 & 4 + \\ 1 & 3 & 5 \end{array}\right] + \sim + \left[\begin{array}{cc|c} 1+3 & 2+3 & 4+3 + \\ 1 & 3 & 5 \end{array}\right] + +

+

+ + x +2y \amp= 4 \amp 4x + 5y \amp= 7 + x + 3y \amp= 5 \amp x +3y \amp= 5 + +

  1. Solutions must be the same.
    2. @@ -211,7 +226,7 @@ Add a constant to every term of a row, for example:

    Multiply a row by a nonzero constant, for example:

    - +

    \left[\begin{array}{cc|c} 1 & 2 & 4 @@ -228,6 +243,21 @@ Multiply a row by a nonzero constant, for example:

    +

    + + \left[\begin{array}{cc|c} 1 & 2 & 4 + \\ 1 & 3 & 5 \end{array}\right] + \sim + \left[\begin{array}{cc|c} 3(1) & 3(2) & 3(4) + \\ 1 & 3 & 5 \end{array}\right] + +

    +

    + + x +2y \amp= 4 \amp 3x + 6y \amp= 12 + x + 3y \amp= 5 \amp x +3y \amp= 5 + +

    1. Solutions must be the same.
      2. @@ -242,7 +272,7 @@ Multiply a row by a nonzero constant, for example:

      Add one row to another row, for example:

      - +

      \left[\begin{array}{cc|c} 1 & 2 & 4 @@ -259,6 +289,21 @@ Add one row to another row, for example:

      +

      + + \left[\begin{array}{cc|c} 1 & 2 & 4 + \\ 1 & 3 & 5 \end{array}\right] + \sim + \left[\begin{array}{cc|c} 1 & 2 & 4 + \\1+1 & 3+2 & 5+4 \end{array}\right] + +

      +

      + + x +2y \amp= 4 \amp \unknown x +\unknown y \amp= \unknown + x + 3y \amp= 5 \amp \unknown x +\unknown y \amp= \unknown + +

      1. Solutions must be the same.
        2. diff --git a/source/linear-algebra/source/02-EV/01.ptx b/source/linear-algebra/source/02-EV/01.ptx index 80f010c7c..4c4b2df93 100644 --- a/source/linear-algebra/source/02-EV/01.ptx +++ b/source/linear-algebra/source/02-EV/01.ptx @@ -125,10 +125,20 @@ we refer to this real number as a scalar.

        Sketch the four Euclidean vectors - 1\left[\begin{array}{c}1\\2\end{array}\right]=\left[\begin{array}{c}1\\2\end{array}\right],\hspace{1em} + 1\left[\begin{array}{c}1\\2\end{array}\right]=\left[\begin{array}{c}1\\2\end{array}\right],\hspace{1em} 3\left[\begin{array}{c}1\\2\end{array}\right]=\left[\begin{array}{c}3\\6\end{array}\right],\hspace{1em} 0\left[\begin{array}{c}1\\2\end{array}\right]=\left[\begin{array}{c}0\\0\end{array}\right],\hspace{1em} -2\left[\begin{array}{c}1\\2\end{array}\right]=\left[\begin{array}{c}-2\\-4\end{array}\right] + + + 1\left[\begin{array}{c}1\\2\end{array}\right]\amp=\left[\begin{array}{c}1\\2\end{array}\right], \amp + 3\left[\begin{array}{c}1\\2\end{array}\right]\amp=\left[\begin{array}{c}3\\6\end{array}\right], + + + 0\left[\begin{array}{c}1\\2\end{array}\right]\amp=\left[\begin{array}{c}0\\0\end{array}\right],\amp + -2\left[\begin{array}{c}1\\2\end{array}\right]\amp\left[\begin{array}{c}-2\\-4\end{array}\right] + + in the same xy plane by drawing an arrow to the (x,y) coordinate associated with each vector.

        @@ -163,22 +173,51 @@ we refer to this real number as a scalar.

        -

        +

        Sketch the following five Euclidean vectors in the same xy plane. - - 1\left[\begin{array}{c}1\\2\end{array}\right]+ - 0\left[\begin{array}{c}-1\\1\end{array}\right]=\unknown\hspace{2em} - 0\left[\begin{array}{c}1\\2\end{array}\right]+ - 1\left[\begin{array}{c}-1\\1\end{array}\right]=\unknown\hspace{2em} - 1\left[\begin{array}{c}1\\2\end{array}\right]+ - 1\left[\begin{array}{c}-1\\1\end{array}\right]=\unknown - - - -2\left[\begin{array}{c}1\\2\end{array}\right]+ - 1\left[\begin{array}{c}-1\\1\end{array}\right]=\unknown\hspace{2em} - -1\left[\begin{array}{c}1\\2\end{array}\right]+ - -2\left[\begin{array}{c}-1\\1\end{array}\right]=\unknown - + + + 1\left[\begin{array}{c}1\\2\end{array}\right]+ + 0\left[\begin{array}{c}-1\\1\end{array}\right]\amp=\unknown\amp + 0\left[\begin{array}{c}1\\2\end{array}\right]+ + 1\left[\begin{array}{c}-1\\1\end{array}\right]\amp=\unknown\amp + 1\left[\begin{array}{c}1\\2\end{array}\right]+ + 1\left[\begin{array}{c}-1\\1\end{array}\right]\amp=\unknown + + + + + -2\left[\begin{array}{c}1\\2\end{array}\right]+ + 1\left[\begin{array}{c}-1\\1\end{array}\right]\amp=\unknown\amp + -1\left[\begin{array}{c}1\\2\end{array}\right]+ + -2\left[\begin{array}{c}-1\\1\end{array}\right]\amp=\unknown + + +

        +

        + Sketch the following five Euclidean vectors in the same xy plane. + + + 1\left[\begin{array}{c}1\\2\end{array}\right]+ + 0\left[\begin{array}{c}-1\\1\end{array}\right]\amp=\unknown\amp + 0\left[\begin{array}{c}1\\2\end{array}\right]+ + 1\left[\begin{array}{c}-1\\1\end{array}\right]\amp=\unknown + + + + + 1\left[\begin{array}{c}1\\2\end{array}\right]+ + 1\left[\begin{array}{c}-1\\1\end{array}\right]\amp=\unknown \amp + -2\left[\begin{array}{c}1\\2\end{array}\right]+ + 1\left[\begin{array}{c}-1\\1\end{array}\right]\amp=\unknown + . + + + + -1\left[\begin{array}{c}1\\2\end{array}\right]+ + -2\left[\begin{array}{c}-1\\1\end{array}\right]\amp=\unknown + +

        diff --git a/source/linear-algebra/source/02-EV/03.ptx b/source/linear-algebra/source/02-EV/03.ptx index fc272a88e..6ed2eab87 100644 --- a/source/linear-algebra/source/02-EV/03.ptx +++ b/source/linear-algebra/source/02-EV/03.ptx @@ -869,16 +869,34 @@ For all \vec v\in W and c\in \IR (not just specific values),

        Consider these subsets of \IR^3: - - R= + + + R\amp= \setBuilder{ \left[\begin{array}{c}x\\y\\z\end{array}\right]}{y=z+1} - \hspace{2em} - S= + \amp + S\amp= \setBuilder{ \left[\begin{array}{c}x\\y\\z\end{array}\right]}{y=|z|} - \hspace{2em} - T= + \amp + T\amp= \setBuilder{ \left[\begin{array}{c}x\\y\\z\end{array}\right]}{z=xy} - . + + + + + R\amp= + \setBuilder{ \left[\begin{array}{c}x\\y\\z\end{array}\right]}{y=z+1} + \amp + S\amp= + \setBuilder{ \left[\begin{array}{c}x\\y\\z\end{array}\right]}{y=|z|} + + + + + T\amp= + \setBuilder{ \left[\begin{array}{c}x\\y\\z\end{array}\right]}{z=xy} + + + .

        @@ -999,7 +1017,7 @@ that is, (kx)+(ky)=(kx)(ky). This is verified by the following calculatio

        - Given the set of ingredients S=\{\textrm{flour}, \textrm{yeast}, \textrm{salt}, \textrm{water}, \textrm{sugar}, \textrm{milk}\}, how should we think of the subspace \vspan(S)? + Given the set of ingredients S=\{\textrm{flour}, \textrm{yeast}, \textrm{salt}, \textrm{water}, \textrm{sugar}, \textrm{milk}\}, how should we think of the subspace \vspan(S)?

        diff --git a/source/linear-algebra/source/02-EV/06.ptx b/source/linear-algebra/source/02-EV/06.ptx index 567270518..05aaa40ff 100644 --- a/source/linear-algebra/source/02-EV/06.ptx +++ b/source/linear-algebra/source/02-EV/06.ptx @@ -212,21 +212,28 @@ T=\left\{ Even though we found different bases for them, \vspan S and \vspan T are exactly the same subspace of \IR^4, since - - S=\left\{ - \left[\begin{array}{c}2\\3\\0\\1\end{array}\right], - \left[\begin{array}{c}2\\0\\1\\-1\end{array}\right], - \left[\begin{array}{c}2\\-3\\2\\-3\end{array}\right], - \left[\begin{array}{c}1\\5\\-1\\0\end{array}\right] - \right\} - = - \left\{ - \left[\begin{array}{c}2\\0\\1\\-1\end{array}\right], - \left[\begin{array}{c}2\\-3\\2\\-3\end{array}\right], - \left[\begin{array}{c}1\\5\\-1\\0\end{array}\right], - \left[\begin{array}{c}2\\3\\0\\1\end{array}\right] - \right\}=T - . + + + S\amp=\left\{ + \left[\begin{array}{c}2\\3\\0\\1\end{array}\right], + \left[\begin{array}{c}2\\0\\1\\-1\end{array}\right], + \left[\begin{array}{c}2\\-3\\2\\-3\end{array}\right], + \left[\begin{array}{c}1\\5\\-1\\0\end{array}\right] + \right\} + + + \amp= + \left\{ + \left[\begin{array}{c}2\\0\\1\\-1\end{array}\right], + \left[\begin{array}{c}2\\-3\\2\\-3\end{array}\right], + \left[\begin{array}{c}1\\5\\-1\\0\end{array}\right], + \left[\begin{array}{c}2\\3\\0\\1\end{array}\right] + \right\} + + + \amp=T + + .

        Thus a given basis for a subspace need not be unique. @@ -253,7 +260,7 @@ Thus a given basis for a subspace need not be unique. \left[\begin{array}{c}0\\1\\0\end{array}\right], \left[\begin{array}{c}1\\1\\1\end{array}\right] } - \text{ and } + \text{ and } \setList{ \left[\begin{array}{c}1\\0\\-3\end{array}\right], \left[\begin{array}{c}2\\-2\\1\end{array}\right], diff --git a/source/linear-algebra/source/02-EV/07.ptx b/source/linear-algebra/source/02-EV/07.ptx index 3ea8dac4d..76b6eaa86 100644 --- a/source/linear-algebra/source/02-EV/07.ptx +++ b/source/linear-algebra/source/02-EV/07.ptx @@ -250,7 +250,7 @@ Find a basis for this solution space. The coefficients of the free variables in the solution space of a linear system always yield linearly independent vectors that span the solution space.

        -

        +

        Thus if \setBuilder{ diff --git a/source/linear-algebra/source/02-EV/samples/02.ptx b/source/linear-algebra/source/02-EV/samples/02.ptx index 477f25bbb..3f040c076 100644 --- a/source/linear-algebra/source/02-EV/samples/02.ptx +++ b/source/linear-algebra/source/02-EV/samples/02.ptx @@ -14,13 +14,19 @@ The set of vectors spans \mathbb R^4.

        -

        The vector equation +

        The vector equation + + x_1 \left[\begin{array}{c} -1 \\ 1 \\ 0 \\ 2 \end{array}\right] + - x_2 \left[\begin{array}{c} 0 \\ -1 \\ 1 \\ -3 \end{array}\right] + + x_2 \left[\begin{array}{c} 0 \\ -1 \\ 1 \\ -3 \end{array}\right] \amp + x_3 \left[\begin{array}{c} 2 \\ -2 \\ 1 \\ -2 \end{array}\right] + - x_4 \left[\begin{array}{c} -3 \\ -1 \\ 4 \\ -5 \end{array}\right] + + + + x_4 \left[\begin{array}{c} -3 \\ -1 \\ 4 \\ -5 \end{array}\right] \amp + x_5 \left[\begin{array}{c} 7 \\ 8 \\ -15 \\ 28 \end{array}\right] = \vec w - has at least one solution for every \vec w\in\mathbb R^4.

        + + + has at least one solution for every \vec w\in\mathbb R^4.

                 diff --git a/source/linear-algebra/source/02-EV/samples/03.ptx b/source/linear-algebra/source/02-EV/samples/03.ptx index ed94fc0e4..5ebf935ca 100644 --- a/source/linear-algebra/source/02-EV/samples/03.ptx +++ b/source/linear-algebra/source/02-EV/samples/03.ptx @@ -8,10 +8,16 @@

        Consider the following subsets of Euclidean space \mathbb R^4 defined by - U=\left\{ \left[\begin{array}{c} x \\ y \\ z \\ w \end{array}\right] \middle| + + + U\amp=\left\{ \left[\begin{array}{c} x \\ y \\ z \\ w \end{array}\right] \middle| y^{2} - 7 \, z^{2} = x \right\} \hspace{1em} \text{and} - \hspace{1em} W=\left\{ \left[\begin{array}{c} x \\ y \\ z \\ w \end{array}\right] \middle| - -5 \, w - 7 \, x - y = -7 \, z \right\} + + + W\amp=\left\{ \left[\begin{array}{c} x \\ y \\ z \\ w \end{array}\right] \middle| + -5 \, w - 7 \, x - y = -7 \, z \right\} + + Without writing a proof, explain why only one of these subsets is likely to be a subspace.

        diff --git a/source/linear-algebra/source/02-EV/samples/04.ptx b/source/linear-algebra/source/02-EV/samples/04.ptx index a5500cf75..d58cd815a 100644 --- a/source/linear-algebra/source/02-EV/samples/04.ptx +++ b/source/linear-algebra/source/02-EV/samples/04.ptx @@ -55,14 +55,20 @@ The set of vectors is linearly independent.

        -

        The vector equation +

        The vector equation + + x_1\left[\begin{array}{c} 1 \\ -1 \\ 0 \\ 1 \end{array}\right] + - x_2\left[\begin{array}{c} -3 \\ 4 \\ 3 \\ 1 \end{array}\right] + + x_2\left[\begin{array}{c} -3 \\ 4 \\ 3 \\ 1 \end{array}\right] \amp + x_3\left[\begin{array}{c} -5 \\ 5 \\ 1 \\ -2 \end{array}\right] + - x_4\left[\begin{array}{c} -15 \\ 17 \\ 8 \\ -1 \end{array}\right] + - x_5\left[\begin{array}{c} 2 \\ -5 \\ -7 \\ -4 \end{array}\right]= + + + x_4\left[\begin{array}{c} -15 \\ 17 \\ 8 \\ -1 \end{array}\right] \amp + + x_5\left[\begin{array}{c} 2 \\ -5 \\ -7 \\ -4 \end{array}\right] = \left[\begin{array}{c} 0 \\ 0 \\ 0 \\ 0 \end{array}\right] - has exactly one solution: + + + has exactly one solution: \left[\begin{array}{c} 0 \\ 0 \\ 0 \\ 0 \\ 0 \end{array}\right].

        diff --git a/source/linear-algebra/source/03-AT/01.ptx b/source/linear-algebra/source/03-AT/01.ptx index a8af810ba..52c167b86 100644 --- a/source/linear-algebra/source/03-AT/01.ptx +++ b/source/linear-algebra/source/03-AT/01.ptx @@ -600,10 +600,17 @@ space is or is not a subspace: .)

        Consider the following maps of Euclidean vectors P:\mathbb R^3\rightarrow\mathbb R^3 and Q:\mathbb R^3\rightarrow\mathbb R^3 defined by - P\left( \left[\begin{array}{c} x \\ y \\ z \end{array}\right] \right)= - \left[\begin{array}{c} -2 \, x - 3 \, y - 3 \, z \\ 3 \, x + 4 \, y + 4 \, z \\ 3 \, x + 4 \, y + 5 \, z \end{array}\right] - \hspace{1em} \text{and} \hspace{1em} Q\left( \left[\begin{array}{c} x \\ y \\ z \end{array}\right] \right)= - \left[\begin{array}{c} x - 4 \, y + 9 \, z \\ y - 2 \, z \\ 8 \, y^{2} - 3 \, x z \end{array}\right]. + + + P\left( \left[\begin{array}{c} x \\ y \\ z \end{array}\right] \right)\amp= + \left[\begin{array}{c} -2 \, x - 3 \, y - 3 \, z \\ 3 \, x + 4 \, y + 4 \, z \\ 3 \, x + 4 \, y + 5 \, z \end{array}\right] + \hspace{1em} \text{and} + + + Q\left( \left[\begin{array}{c} x \\ y \\ z \end{array}\right] \right)\amp= + \left[\begin{array}{c} x - 4 \, y + 9 \, z \\ y - 2 \, z \\ 8 \, y^{2} - 3 \, x z \end{array}\right]. + + Which do you suspect?

        1. diff --git a/source/linear-algebra/source/03-AT/02.ptx b/source/linear-algebra/source/03-AT/02.ptx index eb4652c48..2c1fec5ba 100644 --- a/source/linear-algebra/source/03-AT/02.ptx +++ b/source/linear-algebra/source/03-AT/02.ptx @@ -78,7 +78,7 @@ and . What is T\left(\left[\begin{array}{c} 3 \\ 0 \\ 0 \end{array}\right]\right)?

          -
            +
            1. \left[\begin{array}{c} 6 \\ 3\end{array}\right]
            2. \left[\begin{array}{c} -9 \\ 6 \end{array}\right] @@ -108,7 +108,7 @@ and . What is T\left(\left[\begin{array}{c} 1 \\ 0 \\ 1 \end{array}\right]\right)?

              -
                +

                1. \left[\begin{array}{c} 2 \\ 1\end{array}\right] @@ -150,7 +150,7 @@ and . What is T\left(\left[\begin{array}{c} -2 \\ 0 \\ -3 \end{array}\right]\right)?

                  -
                    +

                    1. \left[\begin{array}{c} 2 \\ 1\end{array}\right] @@ -192,7 +192,7 @@ and . What piece of information would help you compute T\left(\left[\begin{array}{c}0\\4\\-1\end{array}\right]\right)?

                      -
                        +

                        1. The value of T\left(\left[\begin{array}{c} 0\\4\\0\end{array}\right]\right)= @@ -230,55 +230,63 @@ create a spanning and linearly independent set along with to compute T\left(\left[\begin{array}{c}0\\4\\-1\end{array}\right]\right): + T\left(\left[\begin{array}{c}0\\4\\-1\end{array}\right]\right) - = + \amp= T\left(\left[\begin{array}{c}0\\4\\0\end{array}\right]\right) - T\left(\left[\begin{array}{c}0\\0\\1\end{array}\right]\right) - = + + + \amp= \left[\begin{array}{c} -4 \\ 16\end{array}\right] - \left[\begin{array}{c} -3 \\ 2 \end{array}\right] = \left[\begin{array}{c} -1 \\ 14 \end{array}\right] - - + + T\left(\left[\begin{array}{c}0\\4\\-1\end{array}\right]\right) - = + \amp= 4 T\left(\left[\begin{array}{c}0\\1\\0\end{array}\right]\right) - T\left(\left[\begin{array}{c}0\\0\\1\end{array}\right]\right) - = + + + \amp= 4 \left[\begin{array}{c} -1 \\ 4\end{array}\right] - \left[\begin{array}{c} -3 \\ 2 \end{array}\right] = \left[\begin{array}{c} -1 \\ 14 \end{array}\right] - - + + T\left(\left[\begin{array}{c}0\\4\\-1\end{array}\right]\right) - = + \amp= 4 T\left(\left[\begin{array}{c}1\\1\\1\end{array}\right]\right) -5 T\left(\left[\begin{array}{c}0\\0\\1\end{array}\right]\right) -4 T\left(\left[\begin{array}{c}1\\0\\0\end{array}\right]\right) - - - = + + + \amp= 4 \left[\begin{array}{c} -2 \\ 7\end{array}\right] -5 \left[\begin{array}{c} -3 \\ 2 \end{array}\right] -4 \left[\begin{array}{c} 2 \\ 1 \end{array}\right] - = + + + \amp = \left[\begin{array}{c} -8+15-8 \\ 28-10-4 \end{array}\right] = \left[\begin{array}{c} -1 \\ 14 \end{array}\right] +

                          @@ -315,19 +323,22 @@ store this information in an m\times n matrix, called the standard For example, let T: \IR^3 \rightarrow \IR^2 be the linear map determined by the following values for T applied to the standard basis of \IR^3. -\scriptsize + + T\left(\vec e_1 \right) -= +\amp= T\left(\left[\begin{array}{c} 1 \\ 0 \\ 0 \end{array}\right] \right) = \left[\begin{array}{c} 2 \\ 1\end{array}\right] -\hspace{2em} + \amp T\left(\vec e_2 \right) -= +\amp= T\left(\left[\begin{array}{c} 0 \\ 1 \\ 0 \end{array}\right] \right) = \left[\begin{array}{c} -1 \\ 4\end{array}\right] -\hspace{2em} + + + T\left(\vec e_3 \right) = T\left(\left[\begin{array}{c} 0 \\ 0 \\ 1 \end{array}\right] \right) @@ -349,21 +360,24 @@ Then the standard matrix corresponding to T is

                          Let T: \IR^4 \rightarrow \IR^3 be the linear transformation given by + T\left(\vec e_1 \right) - = + \amp= \left[\begin{array}{c} 0 \\ 3 \\ -2\end{array}\right] - \hspace{1em} + \amp T\left(\vec e_2 \right) - = + \amp= \left[\begin{array}{c} -3 \\ 0 \\ 1\end{array}\right] - \hspace{1em} + + T\left(\vec e_3 \right) - = + \amp= \left[\begin{array}{c} 4 \\ -2 \\ 1\end{array}\right] - \hspace{1em} + \amp T\left(\vec e_4 \right) - = + \amp= \left[\begin{array}{c} 2 \\ 0 \\ 0\end{array}\right] + Write the standard matrix [T(\vec e_1) \,\cdots\, T(\vec e_n)] for T.

                          diff --git a/source/linear-algebra/source/03-AT/03.ptx b/source/linear-algebra/source/03-AT/03.ptx index f9c1d4a74..864ec9995 100644 --- a/source/linear-algebra/source/03-AT/03.ptx +++ b/source/linear-algebra/source/03-AT/03.ptx @@ -751,14 +751,14 @@ Given a matrix Mcolumn spacerowspace
                        2. the span of the set of all rows is the row space

                          3. - \mbox{Let } M = \left[\begin{array}{ccc|c}1&-1&0\\2&2&4\\-1&0&-1\end{array}\right] + \mbox{Let } M = \left[\begin{array}{ccc}1&-1&0\\2&2&4\\-1&0&-1\end{array}\right]

                          Is \left[\begin{array}{c}2\\1\\3\end{array}\right] in the column space of M? Is it in the row space of M?

                          -
                            +

                            1. Yes. @@ -779,7 +779,7 @@ Given a matrix Mcolumn spacerowspace Is \left[\begin{array}{c}1\\10\\-3\end{array}\right] in the column space of M? Is it in the row space of M?

                              -
                                +

                                1. Yes. @@ -797,11 +797,11 @@ Given a matrix Mcolumn spacerowspace

                                  - \mbox{Let } N = \left[\begin{array}{ccc|c}1&-1&1\\2&2&-3\\-1&0&-1\end{array}\right] + \mbox{Let } N = \left[\begin{array}{ccc}1&-1&1\\2&2&-3\\-1&0&-1\end{array}\right] Are the row space and column space of N both equal to \mathbb{R}^3?

                                  -
                                    +

                                    1. Yes. diff --git a/source/linear-algebra/source/03-AT/06.ptx b/source/linear-algebra/source/03-AT/06.ptx index e7f0ed3d9..5d86e7e48 100644 --- a/source/linear-algebra/source/03-AT/06.ptx +++ b/source/linear-algebra/source/03-AT/06.ptx @@ -186,7 +186,7 @@ Consider the matrix space M_{2,2}=\left\{\left[\begin{array}{cc}

                                      Does the set S span M_{2,2}? -

                                        +

                                        1. No; the matrix \left[\begin{array}{cc}1&3\\2&4\end{array}\right] is not a linear combination of the matrices in S. @@ -214,7 +214,7 @@ Consider the matrix space M_{2,2}=\left\{\left[\begin{array}{cc}

                                          Is the set S linearly independent? -

                                            +

                                            1. No; the matrix \left[\begin{array}{cc}1&0\\0&0\end{array}\right]\in S is a linear combination of the other matrices in S. @@ -271,7 +271,7 @@ Consider the matrix space M_{2,2}=\left\{\left[\begin{array}{cc}

                                              What is the dimension of M_{2,2}? -

                                                +
                                                1. 2
                                                2. 3
                                                3. 4
                                                  4. @@ -284,7 +284,7 @@ Consider the matrix space M_{2,2}=\left\{\left[\begin{array}{cc}

                                                  Which Euclidean space is M_{2,2} isomorphic to? -

                                                    +
                                                    1. \IR^2
                                                    2. \IR^3
                                                    3. \IR^4
                                                      4. @@ -321,7 +321,7 @@ S=\setList{1,y,y^2,y^3,y^4}.

                                                      Does the set S span \P^4? -

                                                        +

                                                        1. No; the polynomial 1+y^2+2y^3 is not a linear combination of the polynomials in S. @@ -349,7 +349,7 @@ S=\setList{1,y,y^2,y^3,y^4}.

                                                          Is the set S linearly independent? -

                                                            +

                                                            1. No; the polynomial y^2 is a linear combination of the other polynomials in S. @@ -382,7 +382,7 @@ S=\setList{1,y,y^2,y^3,y^4}. S=\left\{ 1,y,y^2,y^3,y^4\right\}? -

                                                                +
                                                                1. S is linearly independent
                                                                2. S spans \P^4
                                                                3. S is a basis of \P^4
                                                                  4. @@ -393,7 +393,7 @@ S=\setList{1,y,y^2,y^3,y^4}.

                                                                  What is the dimension of \P^4? -

                                                                    +
                                                                    1. 2
                                                                    2. 3
                                                                    3. 4
                                                                      4. @@ -406,7 +406,7 @@ S=\setList{1,y,y^2,y^3,y^4}.

                                                                      Which Euclidean space is \P^4 isomorphic to? -

                                                                        +
                                                                        1. \IR^2
                                                                        2. \IR^3
                                                                        3. \IR^4
                                                                          4. diff --git a/source/linear-algebra/source/03-AT/samples/01.ptx b/source/linear-algebra/source/03-AT/samples/01.ptx index b20167a47..c58a80523 100644 --- a/source/linear-algebra/source/03-AT/samples/01.ptx +++ b/source/linear-algebra/source/03-AT/samples/01.ptx @@ -7,10 +7,17 @@

                                                                        4. Consider the following maps of Euclidean vectors P:\mathbb R^3\rightarrow\mathbb R^3 and Q:\mathbb R^3\rightarrow\mathbb R^3 defined by - P\left( \left[\begin{array}{c} x \\ y \\ z \end{array}\right] \right)= + + + P\left( \left[\begin{array}{c} x \\ y \\ z \end{array}\right] \right)\amp= \left[\begin{array}{c} 3 \, x - y + z \\ 2 \, x - 2 \, y + 4 \, z \\ -2 \, x - 2 \, y - 3 \, z \end{array}\right] - \hspace{1em} \text{and} \hspace{1em} Q\left( \left[\begin{array}{c} x \\ y \\ z \end{array}\right] \right)= - \left[\begin{array}{c} y - 2 \, z \\ -3 \, x - 4 \, y + 12 \, z \\ 5 \, x y + 3 \, z \end{array}\right]. + \hspace{1em} \text{and} + + + Q\left( \left[\begin{array}{c} x \\ y \\ z \end{array}\right] \right) \amp= + \left[\begin{array}{c} y - 2 \, z \\ -3 \, x - 4 \, y + 12 \, z \\ 5 \, x y + 3 \, z \end{array}\right]. + + Without writing a proof, explain why only one of these maps is likely to be a linear transformation.

                                                                          5. @@ -66,14 +73,19 @@ \left[\begin{array}{c}9 \\ -27 \end{array}\right] whereas - - S \left( \left[\begin{array}{c}1 \\1 \end{array}\right]\right) +S\left( \left[\begin{array}{c}2 \\2 \end{array}\right]\right) = + + + S \left( \left[\begin{array}{c}1 \\1 \end{array}\right]\right) +S\left( \left[\begin{array}{c}2 \\2 \end{array}\right]\right) \amp= \left[\begin{array}{c} 1 + 2(1) \\ -3(1)(1) \end{array}\right] + - \left[\begin{array}{c} 2 + 2(2) \\ -3(2)(2) \end{array}\right] = + \left[\begin{array}{c} 2 + 2(2) \\ -3(2)(2) \end{array}\right] + + + \amp= \left[\begin{array}{c} 3 \\ -3 \end{array}\right] + \left[\begin{array}{c} 6 \\-12 \end{array}\right] = \left[\begin{array}{c} 9\\ -15 \end{array} \right]. - + + Since \left[\begin{array}{c} 9 \\ -27\end{array}\right]\neq \left[\begin{array}{c} 9 \\ -15 \end{array}\right] , S fails to preserve addition and thus cannot be a linear transformation.

                                                                          @@ -84,18 +96,26 @@

                                                                          First, let us take two vectors \left[\begin{array}{c}x_1 \\y_1 \end{array}\right], \left[\begin{array}{c}x_2 \\y_2 \end{array}\right] \in \IR^2 and compute - - T \left( \left[\begin{array}{c}x_1 \\y_1 \end{array}\right] + \left[\begin{array}{c}x_2 \\y_2 \end{array}\right]\right) = - T \left( \left[\begin{array}{c}x_1 +x_2 \\y_1 + y_2 \end{array}\right] \right) = + + + T \left( \left[\begin{array}{c}x_1 \\y_1 \end{array}\right] + \left[\begin{array}{c}x_2 \\y_2 \end{array}\right]\right) \amp= + T \left( \left[\begin{array}{c}x_1 +x_2 \\y_1 + y_2 \end{array}\right] \right) + + \amp= \left[\begin{array}{c} -4(x_1+x_2)-5(y_1+y_2) \\ 2(x_1+x_2)-4(y_1+y_2)\end{array} \right] - + + and - - T \left( \left[\begin{array}{c}x_1 \\y_1 \end{array}\right]\right) +T\left( \left[\begin{array}{c}x_2 \\y_2 \end{array}\right]\right) = + + + T \left( \left[\begin{array}{c}x_1 \\y_1 \end{array}\right]\right) +T\left( \left[\begin{array}{c}x_2 \\y_2 \end{array}\right]\right) \amp= \left[\begin{array}{c} -4x_1-5y_1 \\ 2x_1-4y_1 \end{array}\right] + - \left[\begin{array}{c} -4x_2-5y_2 \\ 2x_2-4y_2 \end{array}\right] = + \left[\begin{array}{c} -4x_2-5y_2 \\ 2x_2-4y_2 \end{array}\right] + + \amp= \left[\begin{array}{c} -4x_1-5y_1-4x_2-5y_2 \\ 2x_1-4y_1+2x_2-4y_2\end{array} \right] - + + So we see that T \left( \left[\begin{array}{c}x_1 \\y_1 \end{array}\right] + \left[\begin{array}{c}x_2 \\y_2 \end{array}\right]\right) = T \left( \left[\begin{array}{c}x_1 \\y_1 \end{array}\right]\right) +T\left( \left[\begin{array}{c}x_2 \\y_2 \end{array}\right]\right) , diff --git a/source/linear-algebra/source/03-AT/samples/06.ptx b/source/linear-algebra/source/03-AT/samples/06.ptx index c811f7788..cbb38dcac 100644 --- a/source/linear-algebra/source/03-AT/samples/06.ptx +++ b/source/linear-algebra/source/03-AT/samples/06.ptx @@ -6,7 +6,7 @@

                                                                        5. Given the set -\left\{ x^{3} - 2 \, x^{2} + x + 2 , 2 \, x^{2} - 1 , -x^{3} + 3 \, x^{2} + 3 \, x - 2 , x^{3} - 6 \, x^{2} + 9 \, x + 5 \right\} +{\scriptsize \left\{ x^{3} - 2 \, x^{2} + x + 2 , 2 \, x^{2} - 1 , -x^{3} + 3 \, x^{2} + 3 \, x - 2 , x^{3} - 6 \, x^{2} + 9 \, x + 5 \right\}} write a statement involving the solutions to a polynomial @@ -42,10 +42,24 @@ Explain how to determine which of these statements is true.

                                                                          The set of polynomials \left\{ x^{3} - 2 \, x^{2} + x + 2 , 2 \, x^{2} - 1 , -x^{3} + 3 \, x^{2} + 3 \, x - 2 , x^{3} - 6 \, x^{2} + 9 \, x + 5 \right\} is linearly independent exactly when the polynomial equation - y_1\left( x^{3} - 2 \, x^{2} + x + 2 \right)+y_2\left( 2 \, x^{2} - 1 \right)+y_3\left( -x^{3} + 3 \, x^{2} + 3 \, x - 2 \right)+y_4\left( x^{3} - 6 \, x^{2} + 9 \, x + 5\right)=0 + + + \amp y_1\left( x^{3} - 2 \, x^{2} + x + 2 \right)+y_2\left( 2 \, x^{2} - 1 \right)+ + + + \amp y_3\left( -x^{3} + 3 \, x^{2} + 3 \, x - 2 \right)+y_4\left( x^{3} - 6 \, x^{2} + 9 \, x + 5\right)=0 + + has no nontrivial (i.e. nonzero) solutions. The set is linearly dependent when this equation has a nontrivial (i.e. nonzero) solution.

                                                                          To solve this equation, we distribute and then collect coefficients to obtain - \left(y_1-y_3+y_4\right)x^3+\left(-2y_1+2y_2+3y_3-6y_4\right)x^2+\left(y_1+3y_3+9y_4\right)x+\left(2y_1-y_2-2y_3+5y_4\right)=0. + + + \amp \left(y_1-y_3+y_4\right)x^3+\left(-2y_1+2y_2+3y_3-6y_4\right)x^2+ + + + \amp \left(y_1+3y_3+9y_4\right)x+\left(2y_1-y_2-2y_3+5y_4\right)=0 + + . These polynomials are equal precisely when their coefficients are equal, leading to the system \begin{matrix} y_1 & & &-&y_3 & +&y_4 & = & 0 \\ diff --git a/source/linear-algebra/source/04-MX/01.ptx b/source/linear-algebra/source/04-MX/01.ptx index fe6097bee..ed40756c4 100644 --- a/source/linear-algebra/source/04-MX/01.ptx +++ b/source/linear-algebra/source/04-MX/01.ptx @@ -216,13 +216,18 @@ Find the standard matrix BA of T \circ S.

                                                                          Consider the following three matrices.

                                                                          - - A = \left[\begin{array}{ccc}1&0&-3\\3&2&1\end{array}\right] - \hspace{2em} - B = \left[\begin{array}{ccccc}2&2&1&0&1\\1&1&1&-1&0\\0&0&3&2&1\\-1&5&7&2&1\end{array}\right] - \hspace{2em} - C = \left[\begin{array}{cc}2&2\\0&-1\\3&1\\4&0\end{array}\right] - + + + A \amp= \left[\begin{array}{cc}2&2\\0&-1\\3&1\\4&0\end{array}\right] + \amp + B \amp= \left[\begin{array}{ccccc}2&2&1&0&1\\1&1&1&-1&0\\0&0&3&2&1\\-1&5&7&2&1\end{array}\right] + + + + + C \amp= \left[\begin{array}{ccc}1&0&-3\\3&2&1\end{array}\right] + +

                                                                          @@ -271,20 +276,28 @@ B*A

                                                                          Of the following three matrices, only two may be multiplied. - - A=\left[\begin{array}{cccc} --1 & 3 & -2 & -3 \\ -1 & -4 & 2 & 3 -\end{array}\right] \hspace{1em} B=\left[\begin{array}{ccc} -1 & -6 & -1 \\ -0 & 1 & 0 -\end{array}\right] \hspace{1em} C=\left[\begin{array}{ccc} -1 & -1 & -1 \\ -0 & 1 & -2 \\ --2 & 4 & -1 \\ --2 & 3 & -1 -\end{array}\right] - + + + A\amp=\left[\begin{array}{cccc} + -1 & 3 & -2 & -3 \\ + 1 & -4 & 2 & 3 + \end{array}\right] \amp + B\amp=\left[\begin{array}{ccc} + 1 & -6 & -1 \\ + 0 & 1 & 0 + \end{array}\right] + + + + + C\amp=\left[\begin{array}{ccc} + 1 & -1 & -1 \\ + 0 & 1 & -2 \\ + -2 & 4 & -1 \\ + -2 & 3 & -1 + \end{array}\right] + + Explain which two can be multiplied and why. Then show how to find their product.

                                                                          diff --git a/source/linear-algebra/source/04-MX/02.ptx b/source/linear-algebra/source/04-MX/02.ptx index be1b00ede..9e4a10736 100644 --- a/source/linear-algebra/source/04-MX/02.ptx +++ b/source/linear-algebra/source/04-MX/02.ptx @@ -57,15 +57,19 @@

                                                                          For any square matrix A, IA=AI=A: - - \left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] - \left[\begin{array}{ccc} 2 & 7 & -1 \\ 0 & 3 & 2 \\ 1 & 1 & -1 \end{array}\right] - = - \left[\begin{array}{ccc} 2 & 7 & -1 \\ 0 & 3 & 2 \\ 1 & 1 & -1 \end{array}\right] - \left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] - = - \left[\begin{array}{ccc} 2 & 7 & -1 \\ 0 & 3 & 2 \\ 1 & 1 & -1 \end{array}\right] - + + + \left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] + \left[\begin{array}{ccc} 2 & 7 & -1 \\ 0 & 3 & 2 \\ 1 & 1 & -1 \end{array}\right] + \amp= + \left[\begin{array}{ccc} 2 & 7 & -1 \\ 0 & 3 & 2 \\ 1 & 1 & -1 \end{array}\right] + \left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] + + + \amp= + \left[\begin{array}{ccc} 2 & 7 & -1 \\ 0 & 3 & 2 \\ 1 & 1 & -1 \end{array}\right] + + .

                                                                          diff --git a/source/linear-algebra/source/04-MX/04.ptx b/source/linear-algebra/source/04-MX/04.ptx index fbb07005c..72f228ff0 100644 --- a/source/linear-algebra/source/04-MX/04.ptx +++ b/source/linear-algebra/source/04-MX/04.ptx @@ -267,7 +267,7 @@ A 0&3&-1\\ 1&2&3\\ \end{array}\right] - &\sim + \amp\sim \left[\begin{array}{ccc} -1&4&5\\ 1&2&3\\ @@ -275,13 +275,15 @@ A \end{array}\right] - &\sim + \amp\sim \left[\begin{array}{ccc} -1+1&4+2&5+3\\ 1&2&3\\ 0&3&-1\\ \end{array}\right] - = + + + \amp= \left[\begin{array}{ccc} 0&6&8\\ 1&2&3\\ diff --git a/source/linear-algebra/source/04-MX/samples/01.ptx b/source/linear-algebra/source/04-MX/samples/01.ptx index 6b7589946..e504e6130 100644 --- a/source/linear-algebra/source/04-MX/samples/01.ptx +++ b/source/linear-algebra/source/04-MX/samples/01.ptx @@ -4,9 +4,18 @@

                                                                          Of the following three matrices, only two may be multiplied. -A=\left[\begin{array}{cccc} -1 & 3 & -2 & -3 \\ 1 & -4 & 2 & 3 \end{array}\right] -\hspace{1em} B=\left[\begin{array}{ccc} 1 & -6 & -1 \\ 0 & 1 & 0 \end{array}\right] -\hspace{1em} C=\left[\begin{array}{ccc} 1 & -1 & -1 \\ 0 & 1 & -2 \\ -2 & 4 & -1 \\ -2 & 3 & -1 \end{array}\right] + + + A=\left[\begin{array}{cccc} -1 & 3 & -2 & -3 \\ 1 & -4 & 2 & 3 \end{array}\right] +\amp \amp \amp B=\left[\begin{array}{ccc} 1 & -6 & -1 \\ 0 & 1 & 0 \end{array}\right] + + + + + C=\left[\begin{array}{ccc} 1 & -1 & -1 \\ 0 & 1 & -2 \\ -2 & 4 & -1 \\ -2 & 3 & -1 \end{array}\right] + + +

                                                                          @@ -29,7 +38,7 @@ Of the following three matrices, only two may be multiplied.

                                                                          Find their product using technology.

                                                                          - + A = [ -1 3 -2 -3 @@ -71,27 +80,35 @@ ans =

                                                                          We may compute each AC\vec e_i to obtain each column of AC: - + + AC\vec e_1 = - A\left[\begin{array}{c}1\\0\\-2\\-2\end{array}\right] = + A\left[\begin{array}{c}1\\0\\-2\\-2\end{array}\right] \amp= 1 \left[\begin{array}{c}-1\\1\end{array}\right] + 0 \left[\begin{array}{c}-1\\1\end{array}\right] - -2 \left[\begin{array}{c}-1\\1\end{array}\right] - 2 \left[\begin{array}{c}-1\\1\end{array}\right] = - \left[\begin{array}{c} 9 \\ -9 \end{array}\right] - - + -2 \left[\begin{array}{c}-1\\1\end{array}\right] - 2 \left[\begin{array}{c}-1\\1\end{array}\right] + + + \amp= \left[\begin{array}{c} 9 \\ -9 \end{array}\right] + + AC\vec e_2 = - A\left[\begin{array}{c}-1\\1\\4\\3\end{array}\right] = + A\left[\begin{array}{c}-1\\1\\4\\3\end{array}\right] \amp= -1 \left[\begin{array}{c}-1\\1\end{array}\right] + 1 \left[\begin{array}{c}-1\\1\end{array}\right] - +4 \left[\begin{array}{c}-1\\1\end{array}\right] + 3 \left[\begin{array}{c}-1\\1\end{array}\right] = + +4 \left[\begin{array}{c}-1\\1\end{array}\right] + 3 \left[\begin{array}{c}-1\\1\end{array}\right] + + \amp = \left[\begin{array}{c} -13 \\ 12 \end{array}\right] - - + + AC\vec e_3 = - A\left[\begin{array}{c}-1\\-2\\-1\\-1\end{array}\right] = + A\left[\begin{array}{c}-1\\-2\\-1\\-1\end{array}\right] \amp = -1 \left[\begin{array}{c}-1\\1\end{array}\right] - 1 \left[\begin{array}{c}-1\\1\end{array}\right] - -2 \left[\begin{array}{c}-1\\1\end{array}\right] - 1 \left[\begin{array}{c}-1\\1\end{array}\right] = + -2 \left[\begin{array}{c}-1\\1\end{array}\right] - 1 \left[\begin{array}{c}-1\\1\end{array}\right] + + \amp = \left[\begin{array}{c} 0 \\ 2 \end{array}\right] - + +

                                                                          diff --git a/source/linear-algebra/source/05-GT/01.ptx b/source/linear-algebra/source/05-GT/01.ptx index ceef0c974..3902a0904 100644 --- a/source/linear-algebra/source/05-GT/01.ptx +++ b/source/linear-algebra/source/05-GT/01.ptx @@ -70,7 +70,7 @@ transforms the unit square. \begin{tikzpicture} -\fill[red!50!white] (0,0) rectangle (1,1);` +\fill[red!50!white] (0,0) rectangle (1,1); \draw[thin,gray,<->] (-4,0)-- (4,0); \draw[thin,gray,<->] (0,-4)-- (0,4); \draw[thick,blue,->] (0,0) -- node[below] {\(A \vec{e}_1= \left[\begin{array}{c}2 \\ 0 \end{array}\right]\)}++ (2,0); @@ -788,11 +788,15 @@ by manipulating columns to reduce each matrix to the identity: Consider the row operation R_1+4R_3\to R_1 applied as follows to show A\sim B:

                                                                          - -A=\left[\begin{array}{cccc}1&2&0&-3\\4&1&3&0\\0&0&-3&-5\\1&1&1&3\end{array}\right] - \sim -\left[\begin{array}{cccc}1+4(0)&2+4(0)&0+4(-3)&-3+4(-5)\\4&1&3&0\\0&0&-3&-5\\1&1&1&3\end{array}\right]=B - + + +A\amp=\left[\begin{array}{cccc}1&2&0&-3\\4&1&3&0\\0&0&-3&-5\\1&1&1&3\end{array}\right] + + + \amp\sim + \left[\begin{array}{cccc}1+4(0)&2+4(0)&0+4(-3)&-3+4(-5)\\4&1&3&0\\0&0&-3&-5\\1&1&1&3\end{array}\right]=B + +

                                                                          Find a matrix R such that B=RA, by applying the same row operation to I=\left[\begin{array}{cccc}1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{array}\right]. diff --git a/source/linear-algebra/source/05-GT/02.ptx b/source/linear-algebra/source/05-GT/02.ptx index 1b65303c7..68b77cc59 100644 --- a/source/linear-algebra/source/05-GT/02.ptx +++ b/source/linear-algebra/source/05-GT/02.ptx @@ -163,27 +163,32 @@ you can cancel both the column and row containing the 1.

                                                                          If the 1 is not on the main diagonal, you'll need to use row or column swaps in order to cancel. - + + \det \left[\begin{array}{cccc} 3 & {\color{red} 0} & -1 & 5 \\ -1 & {\color{red} 0} & 1 & 11 \\ {\color{red} 2} & {\color{red} 1} & {\color{red} 4} & {\color{red} 0} \\ 3 & {\color{red} 0} & 0 & 1 - \end{array}\right] = + \end{array}\right] \amp= -\det \left[\begin{array}{cccc} 3 & {\color{red} 0} & -1 & 5 \\ {\color{red} 2} & {\color{red} 1} & {\color{red} 4} & {\color{red} 0} \\ -1 & {\color{red} 0} & 1 & 11 \\ 3 & {\color{red} 0} & 0 & 1 - \end{array}\right] = + \end{array}\right] + + + \amp= -\det \left[\begin{array}{ccc} 3 & -1 & 5 \\ -1 & 1 & 11 \\ 3 & 0 & 1 \end{array}\right] - + +

                                                                          @@ -241,7 +246,7 @@ matrix to a determinant of a smaller matrix. -1 & 4 & 1 & 5 \\ 2 & 8 & 0 & 3 \end{array}\right] - &= + \amp= \det\left[\begin{array}{cccc} 4 & 3 & {\color{red} 0} & 1 \\ 6 & -18 & {\color{red} 0} & -20 \\ @@ -249,7 +254,9 @@ matrix to a determinant of a smaller matrix. {\color{red} 1} & {\color{red} 5} \\ 2 & 8 & {\color{red} 0} & 3 \end{array}\right] - = +                                                                           + + \amp= \det\left[\begin{array}{ccc} 4 & 3 & 1 \\ 6 & -18 & -20 \\ @@ -257,25 +264,29 @@ matrix to a determinant of a smaller matrix. \end{array}\right] - &=\dots= + \amp=\dots= -2\det\left[\begin{array}{ccc} {\color{red} 1} & {\color{red} 3} & {\color{red} 4} \\ {\color{red} 0} & 21 & 43 \\ {\color{red} 0} & -1 & -10 \end{array}\right] - = + + + \amp= -2\det\left[\begin{array}{cc} 21 & 43 \\ -1 & -10 \end{array}\right] - &= \dots= + \amp= \dots= -2\det\left[\begin{array}{cc} -167 & {\color{red}{21}} \\ {\color{red} 0} & {\color{red} 1} \end{array}\right] - = -2\det[-167] - &=-2(-167)\det(I)= + \amp= -2\det[-167] + + + \amp=-2(-167)\det(I)= 334 @@ -317,7 +328,7 @@ in each row or column. This approach is called

                                                                          For example, since - + \color{blue}{ \left[\begin{array}{ccc} 1 & 2 & 4 \end{array}\right] = @@ -326,26 +337,32 @@ For example, since 2\left[\begin{array}{ccc} 0 & 1 & 0 \end{array}\right] + 4\left[\begin{array}{ccc} 0 & 0 & 1 \end{array}\right]} -, +,

                                                                          -\det \left[\begin{array}{ccc} 2 & 3 & 5 \\ -1 & 3 & 5 \\ {\color{blue} 1} & {\color{blue} 2} & {\color{blue} 4} \end{array}\right] &= +\det \left[\begin{array}{ccc} 2 & 3 & 5 \\ -1 & 3 & 5 \\ {\color{blue} 1} & {\color{blue} 2} & {\color{blue} 4} \end{array}\right] +\amp = {\color{blue} 1}\det \left[\begin{array}{ccc} 2 & 3 & 5 \\ -1 & 3 & 5 \\ {\color{blue} 1} & {\color{blue} 0} & {\color{blue} 0} \end{array}\right] + {\color{blue} 2}\det \left[\begin{array}{ccc} 2 & 3 & 5 \\ -1 & 3 & 5 \\ {\color{blue} 0} & {\color{blue} 1} & {\color{blue} 0} \end{array}\right] + + + \amp \phantom{=~} {\color{blue} 4}\det \left[\begin{array}{ccc} 2 & 3 & 5 \\ -1 & 3 & 5 \\ {\color{blue} 0} & {\color{blue} 0} & {\color{blue} 1} \end{array}\right] -&= -1\det \left[\begin{array}{ccc} 5 & 3 & 2 \\ 5 & 3 & -1 \\ 0 & 0 & 1 \end{array}\right] +\amp= -1\det \left[\begin{array}{ccc} 5 & 3 & 2 \\ 5 & 3 & -1 \\ 0 & 0 & 1 \end{array}\right] -2\det \left[\begin{array}{ccc} 2 & 5 & 3 \\ -1 & 5 & 3 \\ 0 & 0 & 1 \end{array}\right] + + + + \amp \phantom{=~} 4\det \left[\begin{array}{ccc} 2 & 3 & 5 \\ -1 & 3 & 5 \\ 0 & 0 & 1 \end{array}\right] -&= -\det \left[\begin{array}{cc} 5 & 3 \\ 5 & 3 \end{array}\right] +\amp= -\det \left[\begin{array}{cc} 5 & 3 \\ 5 & 3 \end{array}\right] -2 \det \left[\begin{array}{cc} 2 & 5 \\ -1 & 5 \end{array}\right] +4 \det \left[\begin{array}{cc} 2 & 3 \\ -1 & 3 \end{array}\right] - + . @@ -364,7 +381,7 @@ There are formulas and algorithms for the determinants of larger matrices, but they can be pretty tedious to use. For example, writing out a formula for a 4\times 4 determinant would require 24 different terms!

                                                                          - + \det\left[\begin{array}{cccc} a_{11} & a_{12} & a_{13} & a_{14} \\ a_{21} & a_{22} & a_{23} & a_{24} \\ diff --git a/source/linear-algebra/source/05-GT/04.ptx b/source/linear-algebra/source/05-GT/04.ptx index b833b903e..a56b03c2b 100644 --- a/source/linear-algebra/source/05-GT/04.ptx +++ b/source/linear-algebra/source/05-GT/04.ptx @@ -17,7 +17,7 @@

                                                                          Which of the following vectors is an eigenvector for A=\left[\begin{array}{cccc} 2 & 4 & -1 & -5 \\ 0 & 0 & -3 & -9 \\ 1 & 1 & 0 & 2 \\ -2 & -2 & 3 & 5 \end{array}\right]?

                                                                          -
                                                                            +

                                                                            1. \left[\begin{matrix}-2\\1\\0\\1\end{matrix}\right] diff --git a/source/linear-algebra/source/05-GT/samples/02.ptx b/source/linear-algebra/source/05-GT/samples/02.ptx index fae2a3c25..02c2868ba 100644 --- a/source/linear-algebra/source/05-GT/samples/02.ptx +++ b/source/linear-algebra/source/05-GT/samples/02.ptx @@ -21,36 +21,56 @@ Show how to compute the determinant of the matrix

                                                                              Here is one possible solution, first applying a single row operation, and then performing Laplace/cofactor expansions to reduce the determinant - to a linear combination of 2\times 2 determinants: -

                                                                              + to a linear combination of 2\times 2 determinants. First, subtracting the second row + from the first and then expanding along the second row, we have -\det - \left[\begin{array}{cccc} - 1 & 3 & 0 & -1 \\ - 1 & 1 & 2 & 4 \\ - 1 & 1 & 1 & 3 \\ - -3 & 1 & 2 & -5 - \end{array}\right] - &= \det \left[\begin{array}{cccc} 1 & 3 & 0 & -1 \\ 0 & 0 & 1 & 1 \\ 1 & 1 & 1 & 3 \\ -3 & 1 & 2 & -5 \end{array}\right] = - (-1) \det \left[\begin{array}{ccc} 1 & 3 & -1 \\ 1 & 1 & 3 \\ -3 & 1 & -5 \end{array}\right] + (1) \det \left[\begin{array}{ccc} 1 & 3 & 0 \\ 1 & 1 & 1 \\ -3 & 1 & 2 \end{array}\right] +\det A + \amp= \det \left[\begin{array}{cccc} 1 & 3 & 0 & -1 \\ 0 & 0 & 1 & 1 \\ 1 & 1 & 1 & 3 \\ -3 & 1 & 2 & -5 \end{array}\right] - &= (-1) \left( (1) \det \left[\begin{array}{cc} 1 & 3 \\ 1 & -5 \end{array}\right] - (1) \det \left[\begin{array}{cc} 3 & -1 \\ 1 & -5 \end{array}\right] + (-3) \det \left[\begin{array}{cc} 3 & -1 \\ 1 & 3 \end{array}\right] \right) + + \amp= + (-1) \det \left[\begin{array}{ccc} 1 & 3 & -1 \\ 1 & 1 & 3 \\ -3 & 1 & -5 \end{array}\right] + (1) \det \left[\begin{array}{ccc} 1 & 3 & 0 \\ 1 & 1 & 1 \\ -3 & 1 & 2 \end{array}\right] + . +

                                                                              +

                                                                              Next we compute + - &\phantom{==} (1) \left( (1) \det \left[\begin{array}{cc} 1 & 1 \\ 1 & 2 \end{array}\right] - (3) \det \left[\begin{array}{cc} 1 & 1 \\ -3 & 2 \end{array}\right] \right) + \det \left[\begin{array}{ccc} 1 & 3 & -1 \\ 1 & 1 & 3 \\ -3 & 1 & -5 \end{array}\right] \amp= +(1) \det \left[\begin{array}{cc} 1 & 3 \\ 1 & -5 \end{array}\right] - (1) \det \left[\begin{array}{cc} 3 & -1 \\ 1 & -5 \end{array}\right] + - - % &= (-1)\left( (1)(-8)-(1)(-14)+(-3)(10) \right) + (1) \left( (1)(1)-(3)(5) \right) + \amp \phantom{=}~ +(-3) \det \left[\begin{array}{cc} 3 & -1 \\ 1 & 3 \end{array}\right] - &= (-1) \left( -8+14-30 \right) + (1) \left(1-15 \right) + \amp= (1)(-8)-(1)(-14)+(-3)(10) - &=10 + \amp=-24 - + + and + + + \det \left[\begin{array}{ccc} 1 & 3 & 0 \\ 1 & 1 & 1 \\ -3 & 1 & 2 \end{array}\right] + \amp = + (1) \det \left[\begin{array}{cc} 1 & 1 \\ 1 & 2 \end{array}\right] - (3) \det \left[\begin{array}{cc} 1 & 1 \\ -3 & 2 \end{array}\right] + + + \amp = 1-15 + + + \amp=-14 + + , + so + + + \det A =-1(-24)+(1)(-14)=10 + + . +

                                                                              Here is another possible solution, using row and column operations to first reduce the determinant to a 3\times 3 matrix and then applying a formula: @@ -64,19 +84,24 @@ the determinant to a 3\times 3 matrix and then applying a formula: 1 & 1 & 1 & 3 \\ -3 & 1 & 2 & -5 \end{array}\right] -&= -\det \left[\begin{array}{cccc} 1 & 3 & 0 & -1 \\ 0 & 0 & 1 & 1 \\ 1 & 1 & 1 & 3 \\ -3 & 1 & 2 & -5 \end{array}\right] = +\amp= +\det \left[\begin{array}{cccc} 1 & 3 & 0 & -1 \\ 0 & 0 & 1 & 1 \\ 1 & 1 & 1 & 3 \\ -3 & 1 & 2 & -5 \end{array}\right] + + \amp= \det \left[\begin{array}{cccc} 1 & 3 & 0 & -1 \\ 0 & 0 & 1 & 0 \\ 1 & 1 & 1 & 2 \\ -3 & 1 & 2 & -7 \end{array}\right] -&=-\det \left[\begin{array}{cccc} 1 & 3 & 0 & -1 \\ 1 & 1 & 1 & 2 \\ 0 & 0 & 1 & 0 \\ -3 & 1 & 2 & -7 \end{array}\right] = +\amp=-\det \left[\begin{array}{cccc} 1 & 3 & 0 & -1 \\ 1 & 1 & 1 & 2 \\ 0 & 0 & 1 & 0 \\ -3 & 1 & 2 & -7 \end{array}\right] + + + \amp= -\det \left[\begin{array}{ccc} 1 & 3 & -1 \\ 1 & 1 & 2 \\ -3 & 1 & -7 \end{array}\right] - &=-((-7-18-1)-(3+2-21)) + \amp=-((-7-18-1)-(3+2-21)) - &=10 + \amp=10 diff --git a/source/linear-algebra/source/applications/geology.ptx b/source/linear-algebra/source/applications/geology.ptx index 0e2bccdfe..17c11ac1a 100644 --- a/source/linear-algebra/source/applications/geology.ptx +++ b/source/linear-algebra/source/applications/geology.ptx @@ -36,11 +36,17 @@ and the 5 phases: Geologists already know (or can easily deduce) that -\vec{p}_1 &= 3\vec{c}_1 + \vec{c}_2 + 2 \vec{c}_3 & \vec{p}_2 &= \vec{c}_1 +\vec{c}_2 + \vec{c}_3 & -\vec{p}_3 &= \vec{c}_1 + 0\vec{c}_2 + \vec{c}_3 +\vec{p}_1 \amp= 3\vec{c}_1 + \vec{c}_2 + 2 \vec{c}_3 \amp +\vec{p}_2 \amp= \vec{c}_1 +\vec{c}_2 + \vec{c}_3 -\vec{p}_4 &= \vec{c}_1 +\vec{c}_2 + 2\vec{c}_3 & \vec{p}_5 &= 2\vec{c}_1 + \vec{c}_2 + 2 \vec{c}_3 +\vec{p}_3 \amp= \vec{c}_1 + 0\vec{c}_2 + \vec{c}_3 \amp +\vec{p}_4 \amp= \vec{c}_1 +\vec{c}_2 + 2\vec{c}_3 + + + + + \vec{p}_5 &= 2\vec{c}_1 + \vec{c}_2 + 2 \vec{c}_3 since, for example: @@ -72,11 +78,21 @@ Determine if the set of phases is linearly dependent or linearly independent.

                                                                              Geologists are interested in knowing all the possible chemical reactions among the 5 phases: - \vec{p}_1 = \mathrm{Ca_3MgSi_2O_8} = \left[\begin{array}{c} 3 \\ 1 \\ 2 \end{array}\right] \hspace{1em} -\vec{p}_2 = \mathrm{CaMgSiO_4} = \left[\begin{array}{c} 1 \\ 1 \\ 1 \end{array}\right] \hspace{1em} -\vec{p}_3 = \mathrm{CaSiO_3} = \left[\begin{array}{c} 1 \\ 0 \\ 1 \end{array}\right] -\vec{p}_4 = \mathrm{CaMgSi_2O_6} = \left[\begin{array}{c} 1 \\ 1 \\ 2 \end{array}\right] \hspace{1em} -\vec{p}_5 = \mathrm{Ca_2MgSi_2O_7} = \left[\begin{array}{c} 2 \\ 1 \\ 2 \end{array}\right]. + + + \vec{p}_1 \amp = \mathrm{Ca_3MgSi_2O_8} = \left[\begin{array}{c} 3 \\ 1 \\ 2 \end{array}\right] \amp +\vec{p}_2 \amp = \mathrm{CaMgSiO_4} = \left[\begin{array}{c} 1 \\ 1 \\ 1 \end{array}\right] + + +\vec{p}_3 \amp = \mathrm{CaSiO_3} = \left[\begin{array}{c} 1 \\ 0 \\ 1 \end{array}\right] \amp +\vec{p}_4 \amp = \mathrm{CaMgSi_2O_6} = \left[\begin{array}{c} 1 \\ 1 \\ 2 \end{array}\right] + + + + +\vec{p}_5 = \mathrm{Ca_2MgSi_2O_7} = \left[\begin{array}{c} 2 \\ 1 \\ 2 \end{array}\right]. + + That is, they want to find numbers x_1,x_2,x_3,x_4,x_5 such that x_1\vec{p}_1+x_2\vec{p}_2+x_3\vec{p}_3+x_4\vec{p}_4+x_5\vec{p}_5 = 0. diff --git a/source/linear-algebra/source/applications/pagerank.ptx b/source/linear-algebra/source/applications/pagerank.ptx index 24ebccf87..05f007c96 100644 --- a/source/linear-algebra/source/applications/pagerank.ptx +++ b/source/linear-algebra/source/applications/pagerank.ptx @@ -120,7 +120,7 @@ A webpage distributes its importance equally among all the pages it links to Consider this small network with only three pages. Let x_1, x_2, x_3 be the importance of the three pages respectively.

                                                                              - + \begin{tikzpicture} \begin{scope}[every node/.style={circle,thick,draw}] @@ -168,7 +168,7 @@ This corresponds to the page rank system:
                                                                              - + \begin{tikzpicture} \begin{scope}[every node/.style={circle,thick,draw}]