diff --git a/source/linear-algebra/source/02-EV/04.ptx b/source/linear-algebra/source/02-EV/04.ptx index 42693c03c..49a6a79e1 100644 --- a/source/linear-algebra/source/02-EV/04.ptx +++ b/source/linear-algebra/source/02-EV/04.ptx @@ -150,134 +150,203 @@

Consider the following three vectors in \IR^3: - \vec v_1=\left[\begin{array}{c}-2 \\ 0 \\ 0\end{array}\right], + \vec v_1=\left[\begin{array}{c}-2 \\ 0 \\ 3\end{array}\right], \vec v_2=\left[\begin{array}{c}1 \\ 3 \\ 0\end{array}\right], \text{ and } - \vec v_3=\left[\begin{array}{c}-2 \\ 5 \\ 4\end{array}\right] + \vec v_3=\left[\begin{array}{c}-2 \\ 6 \\ 6\end{array}\right] .

-

- Let \vec w = 3\vec v_1 - \vec v_2 - 5 \vec v_3 = \left[\begin{array}{c}\unknown \\ \unknown \\ \unknown\end{array}\right]. - The set \{\vec v_1,\vec v_2,\vec v_3,\vec w\} is... -

    -

  1. linearly dependent: at least one vector is a linear combination of others

    2. -

  2. linearly independent: no vector is a linear combination of others

    3. -
-

+ +

+ Let \vec v_4 = 3\vec v_1 - \vec v_2 - 2 \vec v_3 = + \left[\begin{array}{c}-3 \\ -15 \\ -3\end{array}\right]. + The set \{\vec v_1,\vec v_2,\vec v_3,\vec v_4\} is... +

    +

  1. linearly dependent: at least one vector is a linear combination of others

    2. +

  2. linearly independent: no vector is a linear combination of others

    3. +
+

+ + +

+ A. We explicitly constructed \vec v_4 as a linear combination of the others. + (It turns out \vec v_3 is a linear combo as well.) +

+ -

- Find \RREF \left[\begin{array}{ccc|c} - \vec v_1 & \vec v_2 & \vec v_3 & \vec w \\ - \end{array}\right]= - \RREF \left[\begin{array}{ccc|c} - -2 & 1 &-2 & \unknown \\ - 0 & 3 & 5 & \unknown \\ - 0 &0 &4 & \unknown - \end{array}\right]= \unknown . -

-

- What does this tell you about solution set for the vector equation - x_1\vec{v}_1+x_2\vec{v}_2+x_3\vec{v}_3 =\vec w? -

    -
  1. -

    - It is inconsistent. -

    -
    2. -
  2. -

    - It is consistent with one solution. -

    -
    3. -
  3. -

    - It is consistent with infinitely many solutions. -

    -
    4. -
-

- + +

+ Find \RREF \left[\begin{array}{ccc|c} + \vec v_1 & \vec v_2 & \vec v_3 & \vec v_4 \\ + \end{array}\right]= + \RREF \left[\begin{array}{ccc|c} + -2 & 1 &-2 & -3 \\ + 0 & 3 & 6 & -15 \\ + 3 &0 &6 & -3 + \end{array}\right]= \unknown . +

+

+ What does this guarantee about the solution set for the vector equation + x_1\vec{v}_1+x_2\vec{v}_2+x_3\vec{v}_3 =\vec v_4? +

    +
  1. +

    + It is inconsistent. +

    +
    2. +
  2. +

    + It is consistent with one solution. +

    +
    3. +
  3. +

    + It is consistent with infinitely many solutions. +

    +
    4. +
  4. +

    + None of these. +

    +
    5. +
+

+ + +

+ C. Since + \RREF \left[\begin{array}{ccc|c} + -2 & 1 &-2 & -3 \\ + 0 & 3 & 6 & -15 \\ + 3 &0 &6 & -3 + \end{array}\right] = + \RREF \left[\begin{array}{ccc|c} + 1 & 0 &2 & -1 \\ + 0 & 1 & 2 & -5 \\ + 0 &0 &0 & 0 + \end{array}\right] + has no contradiction and a free variable, it has infinitely-many solutions. +

+ + -

- Find \RREF \left[\begin{array}{cccc|c} - \vec v_1 & \vec v_2 & \vec v_3 & \vec w & \vec 0\\ - \end{array}\right]= - \RREF \left[\begin{array}{cccc|c} - -2 & 1 &-2 & \unknown & 0\\ - 0 & 3 & 5 & \unknown & 0 \\ - 0 &0 &4 & \unknown & 0 - \end{array}\right]= \unknown . -

-

- What does this tell you about solution set for the vector equation - x_1\vec{v}_1+x_2\vec{v}_2+x_3\vec{v}_3 + x_4\vec w=\vec{0}? -

    -
  1. -

    - It is inconsistent. -

    -
    2. -
  2. -

    - It is consistent with one solution. -

    -
    3. -
  3. -

    - It is consistent with infinitely many solutions. -

    -
    4. -
-

- - +

- Which of the following is the best conclusion obtained when we solved - x_1\vec{v}_1+x_2\vec{v}_2+x_3\vec{v}_3 + x_4\vec w=\vec{0}? + Let \vec w=\left[\begin{array}{c}\unknown \\ \unknown \\ \unknown\end{array}\right] + be some arbitrary vector in \mathbb R^3. + Consider \RREF \left[\begin{array}{cccc} + \vec v_1 & \vec v_2 & \vec v_3 & \vec v_4 \\ + \end{array}\right]= + \RREF \left[\begin{array}{cccc} + -2 & 1 &-2 & -3 \\ + 0 & 3 & 6 & -15 \\ + 3 &0 &6 & -3 + \end{array}\right]= \unknown . +

+

+ What does this guarantee about the solution set for the vector equation + x_1\vec{v}_1+x_2\vec{v}_2+x_3\vec{v}_3 + x_4\vec v_4=\vec w?

  1. -A pivot column in the augmented matrix \RREF \left[\begin{array}{cccc|c} - \vec v_1 & \vec v_2 & \vec v_3 & \vec w & \vec 0 \\ - \end{array}\right] guarantees the linear independence -of \{\vec v_1,\vec v_2,\vec v_3,\vec w\} -by preventing contradictions. + It is inconsistent.

  2. -A pivot column in the coefficient matrix \RREF \left[\begin{array}{cccc} - \vec v_1 & \vec v_2 & \vec v_3 & \vec w \\ - \end{array}\right] guarantees the linear independence -of \{\vec v_1,\vec v_2,\vec v_3,\vec w\} -by preventing contradictions. + It is consistent with exactly one solution.

  3. -A non-pivot column in the augmented matrix \RREF \left[\begin{array}{cccc|c} - \vec v_1 & \vec v_2 & \vec v_3 & \vec w & \vec 0 \\ - \end{array}\right] guarantees the linear dependence -of \{\vec v_1,\vec v_2,\vec v_3,\vec w\} -by describing a linear combination of one vector in terms of the others. + It is not consistent with exactly one solution.

  4. -A non-pivot column in the coefficient matrix \RREF \left[\begin{array}{cccc} - \vec v_1 & \vec v_2 & \vec v_3 & \vec w \\ - \end{array}\right] guarantees the linear dependence -of \{\vec v_1,\vec v_2,\vec v_3,\vec w\} -by describing a linear combination of one vector in terms of the others. + It is consistent with infinitely many solutions.

- - + + +

+ D. Due to the row of zeros, it might be inconsistent. But if it is consistent, + the free variable column guarantees it will have infinitely-many solutions. +

+ + + + +

+ Which of the following describes any solution to the equation + x_1\vec{v}_1+x_2\vec{v}_2+x_3\vec{v}_3 + x_4\vec v_4=\vec{w} + for any linearly dependent set \{\vec v_1,\vec v_2,\vec v_3,\vec v_4\} + and some vector \vec w\in \mathbb R^n? +

    +
  1. +

    +No such solution can exist. +

    +
    2. +
  2. +

    +It cannot be unique. +

    +
    3. +
  3. +

    +It must be unique. +

    +
    4. +
+

+ + +

+ B. If we assume the solution exists, the previous answer shows us that it cannot be the only one, + due to the free variable column. +

+ + + + +

+ Which of the following describes any solution to the equation + x_1\vec{v}_1+x_2\vec{v}_2+x_3\vec{v}_3 + x_4\vec v_4=\vec{w} + for any linearly independent set \{\vec v_1,\vec v_2,\vec v_3,\vec v_4\} + and some vector \vec w\in \mathbb R^n? +

    +
  1. +

    +No such solution can exist. +

    +
    2. +
  2. +

    +It cannot be unique. +

    +
    3. +
  3. +

    +It must be unique. +

    +
    4. +
+

+ + +

+ C. If we assume the solution exists, it must be the only one, + due to the lack of a free variable column. +

+ + + @@ -286,18 +355,15 @@ by describing a linear combination of one vector in terms of the others.

For any vector space, the set \{\vec v_1,\dots\vec v_n\} is linearly dependent if and only - if the vector equation x_1\vec v_1+ x_2 \vec v_2+\dots+x_n\vec v_n=\vec{0} is consistent with - infinitely many solutions. + if the vector equation x_1\vec v_1+ x_2 \vec v_2+\dots+x_n\vec v_n=\vec w cannot have + a unique solution for any \vec w.

- Likewise, the set of vectors + On the other hand, the set of vectors \{\vec v_1,\dots\vec v_n\} is linearly independent - if and only the vector equation x_1\vec v_1+ x_2 \vec v_2 + \cdots + x_n\vec v_n = \vec{0} - has exactly one solution: \left[\begin{array}{c} - x_1 \\ \vdots \\ x_n - \end{array}\right]=\left[\begin{array}{c} - 0 \\ \vdots \\ 0 - \end{array}\right]. + if and only if any solution to the vector equation + x_1\vec v_1+ x_2 \vec v_2 + \cdots + x_n\vec v_n = \vec w + must be unique for any \vec w.

@@ -322,8 +388,8 @@ by describing a linear combination of one vector in terms of the others. \left[\begin{array}{c}4\\3\\0\\1\end{array}\right] \right\} - is linearly dependent (the part that shows its linear system has - infinitely many solutions). + is linearly dependent (the part that shows its linear system cannot have + a unique solution).