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Wize High School Algebra II Textbook (Common Core) > Systems of Linear Equations (Extension Topic)

Reducing a Matrix (Gauss-Jordan Elimination)

Reduced Row Echelon Form (RREF)Previous Section
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Reducing a Matrix

Elementary Row Operations (EROs)

There are three elementary row operations:

1. Swap any two rows: Ri↔Rj\boxed{\quad R_i\leftrightarrow R_j \quad}
Example
R1↔R2R_1\leftrightarrow R_2 : replace Row 1 with Row 2, and vice-versa, in the matrix [4310206]\left[\begin{array}{rr|r} {4}&{3}&{10}\\ {2}&{0}&{6}\\ \end{array}\right].
[4310206]R1↔R2⟶[2064310]\left[\begin{array}{rr|r} \colorOne{4}&\colorOne{3}&\colorOne{10}\\ \colorTwo{2}&\colorTwo{0}&\colorTwo{6}\\ \end{array}\right] \begin{array}{l} \small{R_1\leftrightarrow R_2}\\ \\ \end{array} \quad \longrightarrow \quad \left[\begin{array}{rr|r} \colorTwo{2}&\colorTwo{0}&\colorTwo{6}\\ \colorOne{4}&\colorOne{3}&\colorOne{10}\\ \end{array}\right]


2. Multiply any row by a non-zero constant k∈Rk \in \reals: Ri → k Ri\boxed{\quad R_i \ \rightarrow \ k\ R_i \quad}
Example
R1 → 12R1R_1 \ \rightarrow \ \dfrac{1}{2}R_1 : multiply every entry in Row 1 by 12\dfrac{1}{2} in the matrix [2064310]\left[\begin{array}{rr|r} {2}&{0}&{6} \\ 4&3&10\\ \end{array}\right].
[2064310]12R1⟶[2(12)0(12)6(12)4310]=[1034310]\left[\begin{array}{rr|r} \colorTwo{2}&\colorTwo{0}&\colorTwo{6} \\ 4&3&10\\ \end{array}\right] \begin{array}{l l} \small{\dfrac{1}{2} R_1}\\ \\ \end{array} \quad \longrightarrow \quad \left[\begin{array}{cc|c} \colorTwo{2}\left(\frac{1}{2}\right) & \colorTwo{0}\left(\frac{1}{2}\right) & \colorTwo{6}\left(\frac{1}{2}\right) \\[0.2em] 4&3&10\\ \end{array}\right] = \left[\begin{array}{rr|r} \colorTwo{1}&\colorTwo{0}&\colorTwo{3} \\ 4&3&10\\ \end{array}\right]


3. Add or subtract any multiple of one row to any other row: Ri → Ri±kRj\boxed{\quad R_i \ \rightarrow \ R_i \pm k R_j \quad}
Example
R2 → R2−4R1R_2 \ \rightarrow \ R_2 -4 R_1 : Take Row 2 and subtract 4 times the entries of Row 1 in the matrix [1034310]\left[\begin{array}{rr|r} 1&0&3\\ {4} & {3} & {10} \\ \end{array}\right].
[1034310]R2−4R1⟶[1034−4(1)  3−4(0)  −2−4(3)]=[10303−2]\left[\begin{array}{rr|r} 1&0&3\\ \colorTwo{4} & \colorTwo{3} & \colorTwo{10} \\ \end{array}\right] \begin{array}{l l} \\ \small{R_2 - 4 R_1}\\ \end{array} \quad \longrightarrow \quad \left[\begin{array}{cc|c} \colorOne{1} & \colorOne{0} & \colorOne{3}\\ \colorTwo{4}-4(\colorOne{1}) \ \ & \colorTwo{3}-4(\colorOne{0})\ \ & \colorTwo{-2}-4(\colorOne{3}) \\ \end{array}\right] = \left[\begin{array}{rr|r} 1&0&3\\ \colorTwo{0} & \colorTwo{3} & \colorTwo{-2} \\ \end{array}\right]

Wize Concept
Performing EROs on an augmented matrix preserves solutions.
We say that the new augmented matrix represents an equivalent system.

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Reducing a Matrix (Matrix Reduction)

Goal: Use the process of Gauss-Jordan Elimination to turn a matrix into RREF using EROs! (There are many ways.)

Exam Tip
This is one of the most important concepts of the entire course!
There will be many questions and concepts that use this procedure.

Gauss-Jordan Elimination
  1. At any point if you see a zero row, move it to the bottom of the matrix
  2. Swap rows as necessary to get a leading coefficient as near as possible to the top left
  3. Use EROs on this row to get a leading 1 (divide by the leading coefficient or add/subtract other rows)
  4. Use this leading 1 to make all other entries in that column 0 (adding/subtracting multiples of this row)
  5. Repeat until the matrix is in RREF

Wize Tip
Avoid getting fractions whenever possible for simpler calculations.

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Example: EROs

Consider the following augmented matrix:
[2−28016−7−10−7112]\left[\begin{array}{rrr|r} 2 & -2 & 8 & 0\\ 1 & 6 & -7 & -1\\ 0 & -7 & 11 & 2 \end{array}\right]
Perform the following elementary row operations in the given order.
R1↔R2R_1\leftrightarrow R_2
[2−28016−7−10−7112]⟶[16−7−12−2800−7112]\left[\begin{array}{rrr|r} 2 & -2 & 8 & 0\\ 1 & 6 & -7 & -1\\ 0 & -7 & 11 & 2 \end{array}\right] \longrightarrow \left[\begin{array}{rrr|r} 1 & 6 & -7 & -1\\ 2 & -2 & 8 & 0\\ 0 & -7 & 11 & 2 \end{array}\right]
R2−2R1R_2 -2R_1
[16−7−12−2800−7112]⟶[16−7−10−142220−7112]\left[\begin{array}{rrr|r} 1 & 6 & -7 & -1\\ 2 & -2 & 8 & 0\\ 0 & -7 & 11 & 2 \end{array}\right] \longrightarrow \left[\begin{array}{rrr|r} 1 & 6 & -7 & -1\\ 0 & -14 & 22 & 2\\ 0 & -7 & 11 & 2 \end{array}\right]
−12R2-\dfrac{1}{2} R_2
[16−7−10−142220−7112]⟶[16−7−107−11−10−7112]\left[\begin{array}{rrr|r} 1 & 6 & -7 & -1\\ 0 & -14 & 22 & 2\\ 0 & -7 & 11 & 2 \end{array}\right] \longrightarrow \left[\begin{array}{rrr|r} 1 & 6 & -7 & -1\\ 0 & 7 & -11 & -1\\ 0 & -7 & 11 & 2 \end{array}\right]
R3+1R2R_3+1R_2
[16−7−107−11−10−7112]⟶[16−7−107−11−10001]\left[\begin{array}{rrr|r} 1 & 6 & -7 & -1\\ 0 & 7 & -11 & -1\\ 0 & -7 & 11 & 2 \end{array}\right] \longrightarrow \left[\begin{array}{rrr|r} 1 & 6 & -7 & -1\\ 0 & 7 & -11 & -1\\ 0 & 0 & 0 & 1 \end{array}\right]
Is the resulting matrix in REF, RREF, or neither?
The matrix is in REF.
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Example: Reducing a Matrix

Reduce the following augmented matrix into reduced row echelon form (RREF):
[32−102205036−3]\left[ \begin{array}{rrr|r} 3 & 2 & -1 &0\\ 2& 2&0&5\\ 0 & 3 & 6&-3 \end{array} \right]

First Iteration

  1. Are there any zero rows?
    No
  2. Swap to get a leading coefficient in the top-left. (No need; already there)
  3. Get a leading 1.
We could do 13R1\dfrac{1}{3}R_1, but this will introduce fractions.
Instead, let's change Row 1 by doing R1−R2R_1 - R_2.
[32−102205036−3]R1−R2⟶[10−1−52205036−3]\left[ \begin{array}{rrr|r} \colorOne{3} & 2 & -1 &0\\ 2& 2&0&5\\ 0 & 3 & 6&-3 \end{array} \right] \begin{array}{l} R_1-R_2\\ \\ \\ \end{array} \longrightarrow \left[ \begin{array}{rrr|r} \colorOne{1}&0&-1&-5\\ 2&2&0&5\\ 0&3&6&-3 \end{array} \right]
  1. Use this new leading 1 to make all other entries in the first column 0.
We want to cancel the 2\colorTwo{2} below our first leading 1. We can do this with R2−2R1R_2 \colorTwo{-2}R_1.
[10−1−52205036−3]R2−2R1⟶[10−1−502215036−3]\left[ \begin{array}{rrr|r} \colorOne{1}&0&-1&-5\\ \colorTwo{2}&2&0&5\\ 0&3&6&-3 \end{array} \right] \begin{array}{l} \\ R_2-2R_1\\ \\ \end{array} \longrightarrow \left[ \begin{array}{rrr|r} \colorOne{1}&0&-1&-5\\ 0&2&2&15\\ 0&3&6&-3 \end{array} \right]
  1. Repeat: move on to the next staircase position →\rightarrow row
    2
    , column
    2
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Second Iteration

[10−1−502215036−3]\left[ \begin{array}{rrr|r} 1&0&-1&-5\\ 0&2&2&15\\ 0&3&6&-3 \end{array} \right]
  1. Are there any zero rows?
    No
  2. Swap to get a leading coefficient in row 2, column 2.
Thinking ahead, let's swap Row 2 and Row 3 since we can easily divide Row 3 to get a leading 1.
[10−1−502215036−3]→R2↔R3[10−1−5036−302215]\left[ \begin{array}{rrr|r} 1&0&-1&-5\\ 0&2&2&15\\ 0&3&6&-3 \end{array} \right] \xrightarrow{\normalsize R_2\leftrightarrow R_3} \left[ \begin{array}{rrr|r} 1&0&-1&-5\\ 0&3&6&-3\\ 0&2&2&15\\ \end{array} \right]
  1. Get a leading 1.
[10−1−5036−302215]13R2⟶[10−1−5012−102215]\left[ \begin{array}{rrr|r} 1&0&-1&-5\\ 0&\colorOne{3}&6&-3\\ 0&2&2&15\\ \end{array} \right] \begin{array}{l} \\ \dfrac{1}{3}R_2\\ \\ \end{array} \longrightarrow \left[ \begin{array}{rrr|r} 1&0&-1&-5\\ 0&\colorOne{1}&2&-1\\ 0&2&2&15\\ \end{array} \right]
  1. Use this new leading 1 to make all other entries in the second column 0.
[10−1−5012−102215]R3−2R2⟶[10−1−5012−100−217]\left[ \begin{array}{rrr|r} 1&0&-1&-5\\ 0&\colorOne{1}&2&-1\\ 0&\colorTwo{2}&2&15\\ \end{array} \right] \begin{array}{l} \\ \\ R_3\colorTwo{-2}R_2\\ \end{array} \longrightarrow \left[ \begin{array}{rrr|r} 1&0&-1&-5\\ 0&\colorOne{1}&2&-1\\ 0&0&-2&17\\ \end{array} \right]
  1. Repeat: move on to the next staircase position →\rightarrow row
    3
    , column
    3

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Third Iteration

[10−1−5012−100−217]\left[ \begin{array}{rrr|r} 1&0&-1&-5\\ 0&1&2&-1\\ 0&0&-2&17\\ \end{array} \right]
  1. Are there any zero rows?
    No
  2. Swap to get a leading coefficient in row 3, column 3. (No need; already there)
  3. Get a leading 1.
Now we have no choice but to create fractions.
[10−1−5012−100−217]−12R3⟶[10−1−5012−1001−172]\left[ \begin{array}{rrr|r} 1&0&-1&-5\\[0.6em] 0&1&2&-1\\[0.6em] 0&0&\colorOne{-2}&17\\ \end{array} \right] \begin{array}{l} \\[0.6em] \\[0.6em] -\dfrac{1}{2}R_3\\ \end{array} \longrightarrow \left[ \begin{array}{rrr|r} 1&0&-1&-5\\[0.6em] 0&1&2&-1\\[0.6em] 0&0&\colorOne{1}&-\dfrac{17}{2}\\ \end{array} \right]
  1. Use this new leading 1 to make all other entries in the third column 0.
[10−1−5012−1001−172]R1 → R1+1R3R2 → R2−2R3⟶[100−27201016001−172]\left[ \begin{array}{rrr|r} 1&0&\colorFour{-1}&-5\\[1em] 0&1&\colorTwo{2}&-1\\[1em] 0&0&\colorOne{1}&-\dfrac{17}{2}\\ \end{array} \right] \begin{array}{l} R_1\ \rightarrow \ R_1\colorFour{+1}R_3\\[1em] R_2\ \rightarrow \ R_2\colorTwo{-2}R_3\\[1.8em] \\ \end{array} \longrightarrow \left[ \begin{array}{rrr|r} 1&0&0&-\dfrac{27}{2}\\[1em] 0&1&0&16\\[1em] 0&0&1&-\dfrac{17}{2}\\ \end{array} \right]

This is now in RREF.
Perform the indicated elementary row operations, in the given order, on the following augmented matrix:
[3−6912−135−210−10]\left[\begin{array}{rrr|r} 3&-6&9&12\\ -1&3&5&-2\\ 1&0&-1&0 \end{array}\right]

R1↔R3R_1\leftrightarrow R_3

R2 → R2+R1R_2 \ \rightarrow\ R_2 + R_1

R3 → R3−3R1R_3\ \rightarrow \ R_3-3R_1

R3 → 12R3R_3 \ \rightarrow\ \dfrac{1}{2} R_3

R3 → R3+R2R_3 \ \rightarrow\ R_3 + R_2

Fill in the missing values of the resulting matrix.

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Given the matrix,
[13−13−6−321−2]\left[\begin{array}{ccc} 1&3&-1\\ 3&-6&-3\\ 2&1&-2 \end{array}\right]
find the first row of the RREF of this matrix.

Practice: Reducing a Matrix

Find the reduced row echelon form (RREF) of the matrix:
[312101−124]\begin{bmatrix} 3&1&2\\ 1&0&1\\ -1&2&4 \end{bmatrix}

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