From a3e17c41fff38c3db3db1c6c71454c2cf751e86c Mon Sep 17 00:00:00 2001 From: Steven Clontz Date: Tue, 19 Aug 2025 15:55:36 +0000 Subject: [PATCH 1/8] refactor RREF discovery to align with @davidaustinm #teamforwardsubstitution --- source/linear-algebra/source/01-LE/02.ptx | 241 ++++++++++++++++------ 1 file changed, 181 insertions(+), 60 deletions(-) diff --git a/source/linear-algebra/source/01-LE/02.ptx b/source/linear-algebra/source/01-LE/02.ptx index baac5e484..f9f369c58 100644 --- a/source/linear-algebra/source/01-LE/02.ptx +++ b/source/linear-algebra/source/01-LE/02.ptx @@ -400,90 +400,122 @@ notation: (Combination of old rows) \rightarrow (New row).

Each of the following linear systems has the same solution set.

- + +

A) - x &\,+\,& 2y &\,+\,& z &\,=\,& 3 + x &\,-\,& 2y &\,-\,& 3z &\,=\,& -12 - -x &\,-\,& y &\,+\,& z &\,=\,& 1 + -2x &\,+\,& 6y &\,+\,& 4z &\,=\,& 28 - 2x &\,+\,& 5y &\,+\,& 3z &\,=\,& 7 + &\,\,& 5y &\,-\,& 4z &\,=\,& 11

B) - 2x &\,+\,& 5y &\,+\,& 3z &\,=\,& 7 + x &\,-\,& 2y &\,-\,& 3z &\,=\,& -12 - -x &\,-\,& y &\,+\,& z &\,=\,& 1 + &\,\,& y &\,-\,& z &\,=\,& 2 - x &\,+\,& 2y &\,+\,& z &\,=\,& 3 + &\,\,& 5y &\,-\,& 4z &\,=\,& 11

+ +

C) - x & & &\,-\,& z &\,=\,& 1 + x &\,\,& &\,\,& &\,=\,& -3 - & & y &\,+\,& 2z &\,=\,& 4 + &\,\,& y &\,\,& &\,=\,& 3 - & & y &\,+\,& z &\,=\,& 1 + &\,\,& &\,\,& z &\,=\,& 1

- -

D) - x &\,+\,& 2y &\,+\,& z &\,=\,& 3 + x &\,-\,& 2y &\,-\,& 3z &\,=\,& -12 - & & y &\,+\,& 2z &\,=\,& 4 + &\,\,& y &\,-\,& z &\,=\,& 2 - 2x &\,+\,& 5y &\,+\,& 3z &\,=\,& 7 + &\,\,& &\,\,& z &\,=\,& 1

+ +

E) - x & & &\,-\,& z &\,=\,& 1 + -2x &\,+\,& 6y &\,+\,& 4z &\,=\,& 28 - & & y &\,+\,& 2z &\,=\,& 4 + x &\,-\,& 2y &\,-\,& 3z &\,=\,& -12 - & & & & z &\,=\,& 3 + &\,\,& 5y &\,-\,& 4z &\,=\,& 11

F) - x &\,+\,& 2y &\,+\,& z &\,=\,& 3 + x &\,-\,& 2y &\,\,& &\,=\,& -9 - & & y &\,+\,& 2z &\,=\,& 4 + &\,\,& y &\,-\,& z &\,=\,& 2 - & & y &\,+\,& z &\,=\,& 1 - + &\,\,& &\,\,& z &\,=\,& 1 +

- + + +

G) + + + x &\,-\,& 2y &\,-\,& 3z &\,=\,& -12 + + + &\,\,& 2y &\,-\,& 2z &\,=\,& 4 + + + &\,\,& 5y &\,-\,& 4z &\,=\,& 11 + + +

+

H) + + + x &\,-\,& 2y &\,\,& &\,=\,& -9 + + + &\,\,& y &\,\,& &\,=\,& 3 + + + &\,\,& &\,\,& z &\,=\,& 1 + + +

+ +

-Sort these six equivalent linear systems from +Sort these eight equivalent linear systems from most complicated to simplest (in your opinion).

@@ -491,62 +523,151 @@ most complicated to simplest (in your opinion). - +

Here we've written the sorted linear systems from as augmented matrices. - + \left[\begin{array}{ccc|c} - 2 & 5 & 3 & 7 \\ - -1 & -1 & 1 & 1 \\ - 1 & 2 & 1 & 3 -\end{array}\right] & \sim & + -2 & 6 & 4 & 28 \\ + 1 & -2 & -3 & -12 \\ + 0 & 5 & -4 & 11 +\end{array}\right] & \sim \left[\begin{array}{ccc|c} -\markedPivot{1} & 2 & 1 & 3 \\ --1 & -1 & 1 & 1 \\ -2 & 5 & 3 & 7 -\end{array}\right] & \sim & + 1 & -2 & -3 & -12 \\ + -2 & 6 & 4 & 28 \\ + 0 & 5 & -4 & 11 +\end{array}\right] + + +\sim +\left[\begin{array}{ccc|c} + 1 & -2 & -3 & -12 \\ + 0 & 2 & -2 & 4 \\ + 0 & 5 & -4 & 11 +\end{array}\right] & \sim \left[\begin{array}{ccc|c} -\markedPivot{1} & 2 & 1 & 3 \\ -0 & 1 & 2 & 4 \\ -2 & 5 & 3 & 7 -\end{array}\right]\sim + 1 & -2 & -3 & -12 \\ + 0 & 1 & -1 & 2 \\ + 0 & 5 & -4 & 11 +\end{array}\right] -\sim \left[\begin{array}{ccc|c} -\markedPivot{1} & 2 & 1 & 3 \\ -0 & \markedPivot{1} & 2 & 4 \\ -0 & 1 & 1 & 1 -\end{array}\right] & \sim & +\sim +\left[\begin{array}{ccc|c} + 1 & -2 & -3 & -12 \\ + 0 & 1 & -1 & 2 \\ + 0 & 0 & 1 & 1 +\end{array}\right] & \sim \left[\begin{array}{ccc|c} -\markedPivot{1} & 0 & -1 & 1 \\ -0 & \markedPivot{1} & 2 & 4 \\ -0 & 1 & 1 & 1 -\end{array}\right] & \sim & + 1 & -2 & 0 & -9 \\ + 0 & 1 & -1 & 2 \\ + 0 & 0 & 1 & 1 +\end{array}\right] + + +\sim +\left[\begin{array}{ccc|c} + 1 & -2 & 0 & -9 \\ + 0 & 1 & 0 & 3 \\ + 0 & 0 & 1 & 1 +\end{array}\right] & \sim \left[\begin{array}{ccc|c} -\markedPivot{1} & 0 & -1 & 1 \\ -0 & \markedPivot{1} & 2 & 4 \\ -0 & 0 & -1 & -3 \end{array}\right] + 1 & 0 & 0 & -3 \\ + 0 & 1 & 0 & 3 \\ + 0 & 0 & 1 & 1 +\end{array}\right] +

+ + + +

Assign the following row operations to each step used to manipulate each matrix to the next: +

    +
  1. +

    R_1\leftrightarrow R_2

    +
    2. +
  2. +

    R_2+2R_2\to R_2

    +
    3. +
  3. +

    \frac{1}{2}R_2\to R_2

    +
    4. +
  4. +

    R_3-5R_2\to R_3

    +
    5. +
  5. +

    R_1+3R_3\to R_1

    +
    6. +
  6. +

    R_2+1R_3\to R_2

    +
    7. +
  7. +

    R_1+2R_2\to R_1

    +
    8. +

- -

R_3-1R_2\to R_3

-

R_2+1R_1\to R_2

-

R_1\leftrightarrow R_3

- - -

R_3-2R_1\to R_3

-

R_1-2R_3\to R_1

- - + + +

+

    +
  1. +

    R_1\leftrightarrow R_2

    +
    2. +
  2. +

    R_2+2R_2\to R_2

    +
    3. +
  3. +

    \frac{1}{2}R_2\to R_2

    +
    4. +
  4. +

    R_3-5R_2\to R_3

    +
    5. +
  5. +

    R_1+3R_3\to R_1

    +
    6. +
  6. +

    R_2+1R_3\to R_2

    +
    7. +
  7. +

    R_1+2R_2\to R_1

    +
    8. +
+

+ + + + +

+ What is the solution set for the following system? + + + -2x &\,+\,& 6y &\,+\,& 4z &\,=\,& 28 + + + x &\,-\,& 2y &\,-\,& 3z &\,=\,& -12 + + + &\,\,& 5y &\,-\,& 4z &\,=\,& 11 + + +

    +
  1. \emptyset
    2. +
  2. \left\{\left[\begin{array}{c}-3\\3\\1\end{array}\right]\right\}
    3. +
  3. \left\{ \left[\begin{array}{c} a + 1 \\ -3 \, a - 2 \\ a \end{array}\right] \,\middle|\, a \in\mathbb R \right\}
    4. +
  4. \left\{ \left[\begin{array}{c} a + 3 \, b - 2 \\ a \\ b \end{array}\right] \,\middle|\, a,b \in\mathbb R \right\}
    5. +
+

+ + From e631cfa3bb95747a1691309d308fefc1c12bf7ad Mon Sep 17 00:00:00 2001 From: Steven Clontz Date: Tue, 26 Aug 2025 10:51:42 -0500 Subject: [PATCH 2/8] randomize order of choices --- source/linear-algebra/source/01-LE/02.ptx | 14 +++++++------- 1 file changed, 7 insertions(+), 7 deletions(-) diff --git a/source/linear-algebra/source/01-LE/02.ptx b/source/linear-algebra/source/01-LE/02.ptx index f9f369c58..9197e8a0c 100644 --- a/source/linear-algebra/source/01-LE/02.ptx +++ b/source/linear-algebra/source/01-LE/02.ptx @@ -593,25 +593,25 @@ Assign the following row operations to each step used to manipulate each matrix to the next:
  1. -

    R_1\leftrightarrow R_2

    +

    R_3-5R_2\to R_3

  2. -

    R_2+2R_2\to R_2

    +

    R_1+3R_3\to R_1

  3. -

    \frac{1}{2}R_2\to R_2

    +

    R_1+2R_2\to R_1

  4. -

    R_3-5R_2\to R_3

    +

    R_2+2R_1\to R_2

  5. -

    R_1+3R_3\to R_1

    +

    \frac{1}{2}R_2\to R_2

  6. -

    R_2+1R_3\to R_2

    +

    R_1\leftrightarrow R_2

  7. -

    R_1+2R_2\to R_1

    +

    R_2+1R_3\to R_2

From 7c3c78d122657675bbc373ff13e73f536a9f4db9 Mon Sep 17 00:00:00 2001 From: Steven Clontz Date: Tue, 26 Aug 2025 10:53:35 -0500 Subject: [PATCH 3/8] add RREF with zero row and without zero column --- source/linear-algebra/source/01-LE/02.ptx | 24 +++++++++++++++++------ 1 file changed, 18 insertions(+), 6 deletions(-) diff --git a/source/linear-algebra/source/01-LE/02.ptx b/source/linear-algebra/source/01-LE/02.ptx index 9197e8a0c..25de9f6f3 100644 --- a/source/linear-algebra/source/01-LE/02.ptx +++ b/source/linear-algebra/source/01-LE/02.ptx @@ -680,7 +680,7 @@ A matrix is in reduced row echelon form (RREF)Red
  • Each pivot is to the right of every higher pivot.
    • -
  • Each term that is either above or below a pivot is 0. +
  • Each term that is anywhere above or below a pivot is 0.
  • All zero rows (rows whose terms are all 0) are at the bottom of the matrix.
    • @@ -701,7 +701,7 @@ Recall that a matrix is in reduced row echelon form (RREF
  • Each pivot is to the right of every higher pivot.
    • -
  • Each term that is either above or below a pivot is 0. +
  • Each term that is anywhere above or below a pivot is 0.
  • All zero rows (rows whose terms are all 0) are at the bottom of the matrix.
    • @@ -711,9 +711,21 @@ For each matrix, mark the leading terms, and label it as RREF or not RREF. For the ones not in RREF, determine which rule is violated and how it might be fixed.

    -

    A=\left[\begin{array}{ccc|c} 1 & 0 & 0 & 3 \\ 0 & 0 & 1 & -1 \\ 0 & 0 & 0 & 0 \end{array}\right]

    -

    B=\left[\begin{array}{ccc|c} 1 & 0 & 4 & 3 \\ 0 & 1 & 0 & -1 \\ 0 & 0 & 1 & 2 \end{array}\right]

    -

    C=\left[\begin{array}{ccc|c} 0 & 0 & 0 & 0 \\ 1 & 2 & 0 & 3 \\ 0 & 0 & 1 & -1 \end{array}\right]

    +

    A=\left[\begin{array}{ccc|c} +1 & -4 & 0 & 3 \\ +0 & 0 & 1 & -1 \\ +0 & 0 & 0 & 0 +\end{array}\right]

    +

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

    +

    C=\left[\begin{array}{ccc|c} +0 & 0 & 0 & 0 \\ +1 & 2 & 0 & 3 \\ +0 & 0 & 1 & -1 +\end{array}\right]

     @@ -729,7 +741,7 @@ Recall that a matrix is in reduced row echelon form (RREF
  • Each pivot is to the right of every higher pivot.
    • -
  • Each term that is either above or below a pivot is 0. +
  • Each term that is anywhere above or below a pivot is 0.
  • All zero rows (rows whose terms are all 0) are at the bottom of the matrix.
    • From 70f4d65fd191f3b550f8cf525abce637367c7aad Mon Sep 17 00:00:00 2001 From: Steven Clontz Date: Tue, 26 Aug 2025 12:21:46 -0500 Subject: [PATCH 4/8] tweak rref def --- source/linear-algebra/source/01-LE/02.ptx | 19 ++++++++++--------- 1 file changed, 10 insertions(+), 9 deletions(-) diff --git a/source/linear-algebra/source/01-LE/02.ptx b/source/linear-algebra/source/01-LE/02.ptx index 25de9f6f3..9186bed8b 100644 --- a/source/linear-algebra/source/01-LE/02.ptx +++ b/source/linear-algebra/source/01-LE/02.ptx @@ -673,14 +673,15 @@ matrix to the next:

    +The leading term of a row is its first nonzero term (if it exists). A matrix is in reduced row echelon form (RREF)Reduced row echelon form if

      -
    1. The leftmost nonzero term of each row is 1. - We call these terms pivots.pivot +
    2. The leading term of each row has value 1. + (We call these terms pivots.)pivot
    3. Each pivot is to the right of every higher pivot.
      4. -
    4. Each term that is anywhere above or below a pivot is 0. +
    5. Every term in the same column as a pivot is zero (besides the pivot itself).
    6. All zero rows (rows whose terms are all 0) are at the bottom of the matrix.
      7. @@ -696,12 +697,12 @@ Every matrix has a unique reduced row echelon form. If A is a matrix, we Recall that a matrix is in reduced row echelon form (RREF) if

        -
      1. The leftmost nonzero term of each row is 1. - We call these terms pivots. +
      2. The leading term of each row has value 1. + (We call these terms pivots.)
      3. Each pivot is to the right of every higher pivot.
        4. -
      4. Each term that is anywhere above or below a pivot is 0. +
      5. Each pivot is the only nonzero term in its column.
      6. All zero rows (rows whose terms are all 0) are at the bottom of the matrix.
        7. @@ -736,12 +737,12 @@ For the ones not in RREF, determine which rule is violated and how it might be f Recall that a matrix is in reduced row echelon form (RREF) if

          -
        1. The leftmost nonzero term of each row is 1. - We call these terms pivots. +
        2. The leading term of each row has value 1. + (We call these terms pivots.)
        3. Each pivot is to the right of every higher pivot.
          4. -
        4. Each term that is anywhere above or below a pivot is 0. +
        5. Each pivot is the only nonzero term in its column.
        6. All zero rows (rows whose terms are all 0) are at the bottom of the matrix.
          7. From 634366cc4e232457384b4341bbbaf545e4eb8b18 Mon Sep 17 00:00:00 2001 From: Steven Clontz Date: Thu, 28 Aug 2025 08:37:18 -0500 Subject: [PATCH 5/8] algorithm tweak --- source/linear-algebra/source/01-LE/02.ptx | 11 +++++++++-- 1 file changed, 9 insertions(+), 2 deletions(-) diff --git a/source/linear-algebra/source/01-LE/02.ptx b/source/linear-algebra/source/01-LE/02.ptx index 9186bed8b..13c737a76 100644 --- a/source/linear-algebra/source/01-LE/02.ptx +++ b/source/linear-algebra/source/01-LE/02.ptx @@ -882,13 +882,20 @@ that does not already have a marked pivot.

        7. -Mark this pivot, then use row operations to change all values above and below the +Mark this pivot, then use row operations to change all values below the marked pivot to 0.

        8. -Repeat these steps until the matrix is in RREF. +Repeat these steps until the lowest non-zero row has a pivot, and all +zero rows are at the bottom. +

          +
          9. +
        9. +

          +Finally, use row operations to change all values above each pivot to zero, +which should result in an RREF matrix.

        From 7b3f22f4d7b1c08424f401e0c21a46595b35d883 Mon Sep 17 00:00:00 2001 From: Steven Clontz Date: Thu, 28 Aug 2025 13:48:17 +0000 Subject: [PATCH 6/8] fix sage cells using octave --- source/linear-algebra/source/02-EV/02.ptx | 18 ++++++------------ 1 file changed, 6 insertions(+), 12 deletions(-) diff --git a/source/linear-algebra/source/02-EV/02.ptx b/source/linear-algebra/source/02-EV/02.ptx index 3ef150b5b..f669cc54e 100644 --- a/source/linear-algebra/source/02-EV/02.ptx +++ b/source/linear-algebra/source/02-EV/02.ptx @@ -193,10 +193,8 @@ since \IR^1=\setBuilder{cx}{c\in\IR}. \vspan\left\{\left[\begin{array}{c}1\\-1\\0\end{array}\right], \left[\begin{array}{c}-2\\0\\1\end{array}\right],\left[\begin{array}{c}-2\\-2\\2\end{array}\right]\right\}.

        - - - - + + @@ -228,10 +226,8 @@ vector can't fail to belong. \vspan\left\{\left[\begin{array}{c}1\\-1\\0\end{array}\right], \left[\begin{array}{c}-2\\0\\1\end{array}\right],\left[\begin{array}{c}-2\\-2\\2\end{array}\right]\right\}.

        - - - - + + @@ -247,10 +243,8 @@ The vector belongs to the span. \vspan\left\{\left[\begin{array}{c}1\\-1\\0\end{array}\right], \left[\begin{array}{c}-2\\0\\1\end{array}\right],\left[\begin{array}{c}-2\\-2\\2\end{array}\right]\right\}.

        - - - - + + From 1bcf10d3fba379c7794352d14aa5429578606776 Mon Sep 17 00:00:00 2001 From: Steven Clontz Date: Thu, 28 Aug 2025 13:59:56 +0000 Subject: [PATCH 7/8] add tech edge case --- source/linear-algebra/source/01-LE/02.ptx | 81 +++++++++++++++++++---- 1 file changed, 67 insertions(+), 14 deletions(-) diff --git a/source/linear-algebra/source/01-LE/02.ptx b/source/linear-algebra/source/01-LE/02.ptx index 13c737a76..ad3f0f442 100644 --- a/source/linear-algebra/source/01-LE/02.ptx +++ b/source/linear-algebra/source/01-LE/02.ptx @@ -894,8 +894,8 @@ zero rows are at the bottom.

      7. -Finally, use row operations to change all values above each pivot to zero, -which should result in an RREF matrix. +Finally, moving right to left, use row operations to change all values above +each pivot to zero, which should result in an RREF matrix.

      @@ -961,7 +961,18 @@ Can \RREF(A) be used to find \RREF(B)?

      Free browser-based technologies for mathematical computation -are available online. +are available online. Let's explore how these can be used to show + +\left[\begin{array}{ccc} +1& 4& 6 \\ +2& 5& 7 +\end{array}\right] +\sim +\left[\begin{array}{ccc} +1& 0& -2/3 \\ +0& 1& 5/3 +\end{array}\right] +

      @@ -972,7 +983,7 @@ Go to .

      In the dropdown on the right, you can select a number of different languages. -Select "Octave" for the Matlab-compatible syntax used by this text. +Select "Octave" for the open-source Matlab-compatible syntax used by this text.

      @@ -987,30 +998,72 @@ Now try using whitespace to write out the matrix and compute \RREF instea

      -A = [1 3 2 +A = [1 4 6 2 5 7] rref(A)       - - - - - - + +

      +Try using format rat to format the numbers as rational +fractions in the output, rather than decimal approximations. +

      + + format rat A = [ - 1 4 6 - 2 5 7 +1 4 6 +2 5 7 ] rref(A) - + + +       + + + + + +

      +Because the algorithm used to find RREF matrices as quickly as possible +in Octave is different than the one we would use by hand, sometimes +the output is slightly different than expected. +

      +       + +

      +Use Octave to compute +\RREF \left[\begin{array}{cccc} +-6& 8& -9& 22 \\ +4& -3& 5& -5 \\ +-9& 7& -8& 19 \\ +4& -3& 3& -9 +\end{array}\right] + +both with and without the format rat command. +

      +       + +

      +What number do you think the * (with format rat) +and 1.38778e-17 (without format rat) outputs you +found represent? +

        +
      1. 1.38
        2. +
      2. 1.38^{-17}
        3. +
      3. 0
        4. +
      4. \infty
        5. +
      +

      +       +       + From b9fa6e5294d9f57894561917446b2e7d60e8eee2 Mon Sep 17 00:00:00 2001 From: Steven Clontz Date: Thu, 28 Aug 2025 14:17:06 +0000 Subject: [PATCH 8/8] finish LE2 draft --- source/linear-algebra/source/01-LE/02.ptx | 155 ++++++++++++++-------- 1 file changed, 103 insertions(+), 52 deletions(-) diff --git a/source/linear-algebra/source/01-LE/02.ptx b/source/linear-algebra/source/01-LE/02.ptx index ad3f0f442..c860bb47e 100644 --- a/source/linear-algebra/source/01-LE/02.ptx +++ b/source/linear-algebra/source/01-LE/02.ptx @@ -906,61 +906,15 @@ This will keep you from undoing your progress towards an RREF matrix.

      - - -

      -Complete the following RREF calculation (multiple row operations may be needed -for certain steps): - - A= - \left[\begin{array}{cccc}2 & 3 & 2 & 3 \\ -2 & 1 & 6 & 1 \\ -1 & -3 & -4 & 1 \end{array}\right] - \sim \left[\begin{array}{cccc}\markedPivot{1} & \unknown & \unknown & \unknown \\ -2 & 1 & 6 & 1 \\ -1 & -3 & -4 & 1 \end{array}\right] - \sim \left[\begin{array}{cccc}\markedPivot{1} & \unknown & \unknown & \unknown \\ 0 & \unknown & \unknown & \unknown \\ 0 & \unknown & \unknown & \unknown \end{array}\right] - - - \sim \left[\begin{array}{cccc}\markedPivot{1} & \unknown & \unknown & \unknown \\ 0 & \markedPivot{1} & \unknown & \unknown \\ 0 & \unknown & \unknown & \unknown \end{array}\right] - \sim \left[\begin{array}{cccc}\markedPivot{1} & 0 & \unknown & \unknown \\ 0 & \markedPivot{1} & \unknown & \unknown \\ 0 & 0 & \unknown & \unknown \end{array}\right] - \sim \cdots - \sim \left[\begin{array}{cccc}\markedPivot{1} & 0 & -2 & 0 \\ 0 & \markedPivot{1} & 2 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right] - -

      -       -       - - - -

      -Consider the matrix -A=\left[\begin{array}{cccc} 2 & 4 & 2 & -4 \\ -2 & -4 & 1 & 1 \\ 3 & 6 & -1 & -4 \end{array}\right]. -

      -

      -Compute \RREF(A). -

      -       -       - - - -

      -Consider the non-augmented and augmented matrices -A=\left[\begin{array}{cccc} 2 & 4 & 2 & -4 \\ -2 & -4 & 1 & 1 \\ 3 & 6 & -1 & -4 \end{array}\right]\hspace{2em} -B=\left[\begin{array}{ccc|c} 2 & 4 & 2 & -4 \\ -2 & -4 & 1 & 1 \\ 3 & 6 & -1 & -4 \end{array}\right]. -

      -

      -Can \RREF(A) be used to find \RREF(B)? -

        -
      1. Yes, \RREF(A) and \RREF(B) are exactly the same.
        2. -
      2. Yes, \RREF(A) may be slightly modified to find \RREF(B).
        3. -
      3. No, a new calculation is required.
        4. -
      -

      -       -       -

      -Free browser-based technologies for mathematical computation +Since Gauss-Jordan elimination is an algorithm that could be +performed the same way every time, computers can be programmed to +do this calculation for us. +

      +

      +Free browser-based technologies for computing RREF are available online. Let's explore how these can be used to show \left[\begin{array}{ccc} @@ -1064,6 +1018,103 @@ found represent? + + +

      +Complete the following RREF calculation (multiple row operations may be needed +for certain steps): + + +\left[\begin{array}{cccc} +2 & 3 & 2 & 3 \\ +-2 & 1 & 6 & 1 \\ +-1 & -3 & -4 & 1 +\end{array}\right] + &\sim +\left[\begin{array}{cccc} +\unknown & \unknown & \unknown & \unknown \\ +-2 & 1 & 6 & 1 \\ +2 & 3 & 2 & 3 \\ +\end{array}\right] + + \sim +\left[\begin{array}{cccc} +\markedPivot{1} & \unknown & \unknown & \unknown \\ +-2 & 1 & 6 & 1 \\ +2 & 3 & 2 & 3 \\ +\end{array}\right] + &\sim +\left[\begin{array}{cccc} +\markedPivot{1} & \unknown & \unknown & \unknown \\ +0 & \unknown& \unknown& \unknown \\ +0 & \unknown& \unknown& \unknown \\ +\end{array}\right] + + \sim +\hspace{3em}\cdots\hspace{3em} + &\sim +\left[\begin{array}{cccc} +\markedPivot{1} & 3 & 4 & -1 \\ +0 & \markedPivot{1} & 2 & -1 \\ +0 & 0 & 0 & \markedPivot{1} \\ +\end{array}\right] + + \sim +\left[\begin{array}{cccc} +\markedPivot{1} & 3 & 4 & \unknown \\ +0 & \markedPivot{1} & 2 & \unknown \\ +0 & 0 & 0 & \markedPivot{1} \\ +\end{array}\right] + &\sim +\left[\begin{array}{cccc} +\markedPivot{1} & \unknown & \unknown & \unknown \\ +0 & \markedPivot{1} & 2 & \unknown \\ +0 & 0 & 0 & \markedPivot{1} \\ +\end{array}\right] + + +

      +       + + +

      + Check your work using technology. +

      +       +       + + + + +

      +Consider the matrix +A=\left[\begin{array}{cccc} 2 & 4 & 2 & -4 \\ -2 & -4 & 1 & 1 \\ 3 & 6 & -1 & -4 \end{array}\right]. +

      +

      +Compute \RREF(A) using technology, then show how to compute +this by hand. +

      +       +       + + + +

      +Consider the non-augmented and augmented matrices +A=\left[\begin{array}{cccc} 2 & 4 & 2 & -4 \\ -2 & -4 & 1 & 1 \\ 3 & 6 & -1 & -4 \end{array}\right]\hspace{2em} +B=\left[\begin{array}{ccc|c} 2 & 4 & 2 & -4 \\ -2 & -4 & 1 & 1 \\ 3 & 6 & -1 & -4 \end{array}\right]. +

      +

      +Can \RREF(A) be used to find \RREF(B)? +

        +
      1. Yes, \RREF(A) and \RREF(B) are exactly the same.
        2. +
      2. Yes, \RREF(A) may be slightly modified to find \RREF(B).
        3. +
      3. No, a new calculation is required.
        4. +
      +

      +       +       +