Wize University Linear Algebra Textbook > Systems of Linear Equations (SLEs) (Linear Systems)

Reducing a Matrix (Gauss-Jordan Elimination)

Reduced Row Echelon Form (RREF)Previous Section

0:00 / 0:00

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}Ri​↔Rj​​
Example
R1↔R2R_1\leftrightarrow R_2R1​↔R2​ : 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][42​30​106​].

[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][42​30​106​]R1​↔R2​​⟶[24​03​610​]

2. Multiply any row by a non-zero constant k∈Rk \in \realsk∈R:   Ri → k Ri\boxed{\quad
R_i \ \rightarrow \ k\ R_i
\quad}Ri​ → k Ri​​
Example
R1 → 12R1R_1 \ \rightarrow \ \dfrac{1}{2}R_1R1​ → 21​R1​ : multiply every entry in Row 1 by 12\dfrac{1}{2}21​ in the matrix [2064310]\left[\begin{array}{rr|r}
{2}&{0}&{6} \\
4&3&10\\
\end{array}\right][24​03​610​].

[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][24​03​610​]21​R1​​⟶[2(21​)4​0(21​)3​6(21​)10​]=[14​03​310​]

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}Ri​ → Ri​±kRj​​
Example
R2 → R2−4R1R_2 \ \rightarrow \ R_2 -4 R_1R2​ → R2​−4R1​ : 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][14​03​310​].

[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][14​03​310​]R2​−4R1​​⟶[14−4(1)  ​03−4(0)  ​3−2−4(3)​]=[10​03​3−2​]

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

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

Wize Tip
Avoid getting fractions whenever possible for simpler calculations.

0:00 / 0:00

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]​210​−26−7​8−711​0−12​​
Perform the following elementary row operations in the given order.
     R1↔R2R_1\leftrightarrow R_2R1​↔R2​

               [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]​210​−26−7​8−711​0−12​​⟶​120​6−2−7​−7811​−102​​

     R2−2R1R_2 -2R_1R2​−2R1​

               [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]​120​6−2−7​−7811​−102​​⟶​100​6−14−7​−72211​−122​​

     −12R2-\dfrac{1}{2} R_2−21​R2​

               [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]​100​6−14−7​−72211​−122​​⟶​100​67−7​−7−1111​−1−12​​

     R3+1R2R_3+1R_2R3​+1R2​

               [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]​100​67−7​−7−1111​−1−12​​⟶​100​670​−7−110​−1−11​​

Is the resulting matrix in REF, RREF, or neither?

The matrix is in REF.

0:00 / 0:00

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]​320​223​−106​05−3​​

First Iteration
Are there any zero rows? No
Swap to get a leading coefficient in the top-left. (No need; already there)
Get a leading 1.

We could do 13R1\dfrac{1}{3}R_131​R1​, but this will introduce fractions.
Instead, let's change Row 1 by doing R1−R2R_1 - R_2R1​−R2​.
[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]​320​223​−106​05−3​​R1​−R2​​⟶​120​023​−106​−55−3​​

Use this new leading 1 to make all other entries in the first column 0.

We want to cancel the 2\colorTwo{2}2  below our first leading 1. We can do this with R2−2R1R_2 \colorTwo{-2}R_1R2​−2R1​.
[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]​120​023​−106​−55−3​​R2​−2R1​​⟶​100​023​−126​−515−3​​

Repeat: move on to the next staircase position →\rightarrow→ row 2
, column 2
PAGE BREAK
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]​100​023​−126​−515−3​​
Are there any zero rows? No
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]​100​023​−126​−515−3​​R2​↔R3​​​100​032​−162​−5−315​​

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]​100​032​−162​−5−315​​31​R2​​⟶​100​012​−122​−5−115​​

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]​100​012​−122​−5−115​​R3​−2R2​​⟶​100​010​−12−2​−5−117​​

Repeat: move on to the next staircase position →\rightarrow→ row 3
, column 3

PAGE BREAK
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]​100​010​−12−2​−5−117​​
Are there any zero rows? No
Swap to get a leading coefficient in row 3, column 3. (No need; already there)
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]​100​010​−12−2​−5−117​​−21​R3​​⟶​100​010​−121​−5−1−217​​​

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]​100​010​−121​−5−1−217​​​R1​ → R1​+1R3​R2​ → R2​−2R3​​⟶​100​010​001​−227​16−217​​​

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]​3−11​−630​95−1​12−20​​

     R1↔R3R_1\leftrightarrow R_3R1​↔R3​

     R2 → R2+R1R_2 \ \rightarrow\ R_2 + R_1R2​ → R2​+R1​

     R3 → R3−3R1R_3\ \rightarrow \ R_3-3R_1R3​ → R3​−3R1​

     R3 → 12R3R_3 \ \rightarrow\ \dfrac{1}{2} R_3R3​ → 21​R3​

     R3 → R3+R2R_3 \ \rightarrow\ R_3 + R_2R3​ → R3​+R2​

Fill in the missing values of the resulting matrix.

	0	0
0
0	0

I don't know

Given the matrix,
[13−13−6−321−2]\left[\begin{array}{ccc}
1&3&-1\\
3&-6&-3\\
2&1&-2
\end{array}\right]​132​3−61​−1−3−2​​
find the first row of the RREF of this matrix.

A) [100]\begin{bmatrix}
  1&0&0\\
\end{bmatrix}[1​0​0​]

B) [110]\begin{bmatrix}
  1&1&0\\
\end{bmatrix}[1​1​0​]

C) [101]\begin{bmatrix}
  1&0&1\\
\end{bmatrix}[1​0​1​]

D) [10−1]\begin{bmatrix}
  1&0&-1\\
\end{bmatrix}[1​0​−1​]

E) None of the above

I don't know

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}​31−1​102​214​​


0
0	0

I don't know

Extra Practice

Gauss-Jordan Elimination $\tkct{LAT}$

This activity is just here as a convenient reminder / link for the Linear Algebra Toolkit. The solution illustrate a few points about how to use it; however, the actual details of row-reduction should be addressed in the context of the original chapter :)
Use Gauss-Jordan Elimination to find the RREF of the matrix  [0−23  1363−2663  5]\bcb{\begin{bmatrix} 0&-2&3&\ \ 1\\3&6&3&-2\\ 6&6&3&\ \ 5 \end{bmatrix}}​036​−266​333​  1−2  5​​.

Gauss-Jordan Elimination $\tkct{LAT}$

Use Gauss-Jordan Elimination to find the RREF of the matrix  [0−23  1363−2663  5]\bcb{\begin{bmatrix} 0&-2&3&\ \ 1\\3&6&3&-2\\ 6&6&3&\ \ 5 \end{bmatrix}}​036​−266​333​  1−2  5​​.

Given the following augmented matrix of a system of equations, find the reduced row echelon  form:
[1011−23−22−4∣110]\left[\begin{array}{l}
1&0&1\\
1&-2&3\\
-2&2&-4\\
\end{array}\right.\left|\begin{array}{l}1
\\1\\0\end{array}\right]​11−2​0−22​13−4​​110​​

Solving a system of linear equations

Consider the following system of linear equations:
−2x1+3x2+13x3=5−x1+2x2+5x3=14x1−14x2−2x3=14\begin{array}{rcr}
-2x_1+3x_2+13x_3&=&5\\
-x_1+2x_2+5x_3&=&1\\
4x_1-14x_2-2x_3&=&14\\
\end{array}−2x1​+3x2​+13x3​−x1​+2x2​+5x3​4x1​−14x2​−2x3​​===​5114​
where we wish to solve for the triple (x1, x2, x3) of real numbers that satisfy this system.

Additional Practice Problems--Systems of Linear Equations

1. Write the augmented matrix that represents this system of linear equations
x1+x2=7−3x32−x1=7x1+x2+x3=0\begin{array}{rcl}
x_1+x_2&=&7-3x_3\\
2-x_1&=&7\\
x_1+x_2+x_3&=&0
\end{array}x1​+x2​2−x1​x1​+x2​+x3​​===​7−3x3​70​
2. Determine all values of a, ba,\ ba, b and ccc if the augmented matrix [2−1001−3]\left[\begin{array}{rr|r}
2&-1&0\\
0&1&-3
\end{array}\right][20​−11​0−3​] represents the following system of linear equations: ax1=bx2(a+c)x2+3=0\begin{array}{rcl}
ax_1&=&bx_2\\
(a+c)x_2+3&=&0
\end{array}ax1​(a+c)x2​+3​==​bx2​0​

Reducing a 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}​31−1​102​214​​

Reducing a Matrix

Practice: Reducing a Matrix
Given the matrix,
[13−13−6−321−2]\left[\begin{array}{ccc}
1&3&-1\\
3&-6&-3\\
2&1&-2
\end{array}\right]​132​3−61​−1−3−2​​

Reducing a Matrix

Given the matrix,
[13−13−6−321−2]\left[\begin{array}{ccc}
1&3&-1\\
3&-6&-3\\
2&1&-2
\end{array}\right]​132​3−61​−1−3−2​​
find the first row of the RREF of this matrix.

Practice Question: Matrix Reduction

Practice Question: Matrix Reduction
Reduce [6−1812302−64024−1280−1−26400]\left[\begin{array}{rrrr|r}
6 & -18 & 12 &3 &0\\
2 &-6 & 4 &0&2\\
4 & -12 & 8  &0&-1\\
-2&6&4&0&0
 \end{array}\right]​624−2​−18−6−126​12484​3000​02−10​​ into row-reduce echelon form (RREF) and then state the solution.

Let AAA be the row-reduced echelon form of the matrix [120−20132−1]\left[\begin{array}{r}
1&2&0\\
-2&0&1\\
3&2&-1
\end{array}\right]​1−23​202​01−1​​.
Find the first row of AAA.

Which of the following matrices is a result of performing a single elementary row operation on the augmented matrix [2−2051−43100−24]\left[\begin{array}{rrr|r}
2&-2&0&5\\
1&-4&3&1\\
0&0&-2&4
\end{array}\right]​210​−2−40​03−2​514​​?

Solving a system of linear equations

Consider the following system of linear equations:
−2x1−4x2−6x3−19x4=−153x1+10x2−30x3+9x4=24x1+3x2−6x3+4x4=87x1+17x2+45x4=55\begin{array}{rcr}
-2x_1-4x_2-6x_3-19x_4&=&-15\\
3x_1+10x_2-30x_3+9x_4&=&24\\
x_1+3x_2-6x_3+4x_4&=&8\\
7x_1+17x_2+45x_4&=&55
\end{array}−2x1​−4x2​−6x3​−19x4​3x1​+10x2​−30x3​+9x4​x1​+3x2​−6x3​+4x4​7x1​+17x2​+45x4​​====​−1524855​
where we wish to solve for the quadruple (x1, x2, x3, x4) of real numbers that satisfy this system.

Solving a system of linear equations

Consider the following system of linear equations:
−2x1+3x2+13x3=5−x1+2x2+5x3=14x1−14x2−2x3=14\begin{array}{rcr}
-2x_1+3x_2+13x_3&=&5\\
-x_1+2x_2+5x_3&=&1\\
4x_1-14x_2-2x_3&=&14\\
\end{array}−2x1​+3x2​+13x3​−x1​+2x2​+5x3​4x1​−14x2​−2x3​​===​5114​
where we wish to solve for the triple (x1, x2, x3) of real numbers that satisfy this system.

Solving a system of linear equations

Consider the following system of linear equations:
−2x1−4x2−6x3−19x4=−153x1+10x2−30x3+9x4=24x1+3x2−6x3+4x4=87x1+17x2+45x4=55\begin{array}{rcr}
-2x_1-4x_2-6x_3-19x_4&=&-15\\
3x_1+10x_2-30x_3+9x_4&=&24\\
x_1+3x_2-6x_3+4x_4&=&8\\
7x_1+17x_2+45x_4&=&55
\end{array}−2x1​−4x2​−6x3​−19x4​3x1​+10x2​−30x3​+9x4​x1​+3x2​−6x3​+4x4​7x1​+17x2​+45x4​​====​−1524855​
where we wish to solve for the quadruple (x1, x2, x3, x4) of real numbers that satisfy this system.

Solving a system of linear equations

Consider the following system of linear equations:
−2x1+3x2+13x3=5−x1+2x2+5x3=14x1−14x2−2x3=14\begin{array}{rcr}
-2x_1+3x_2+13x_3&=&5\\
-x_1+2x_2+5x_3&=&1\\
4x_1-14x_2-2x_3&=&14\\
\end{array}−2x1​+3x2​+13x3​−x1​+2x2​+5x3​4x1​−14x2​−2x3​​===​5114​
where we wish to solve for the triple (x1, x2, x3) of real numbers that satisfy this system.

Solving a system of linear equations

Consider the following system of linear equations:
−2x1−4x2−6x3−19x4=−153x1+10x2−30x3+9x4=24x1+3x2−6x3+4x4=87x1+17x2+45x4=55\begin{array}{rcr}
-2x_1-4x_2-6x_3-19x_4&=&-15\\
3x_1+10x_2-30x_3+9x_4&=&24\\
x_1+3x_2-6x_3+4x_4&=&8\\
7x_1+17x_2+45x_4&=&55
\end{array}−2x1​−4x2​−6x3​−19x4​3x1​+10x2​−30x3​+9x4​x1​+3x2​−6x3​+4x4​7x1​+17x2​+45x4​​====​−1524855​
where we wish to solve for the quadruple (x1,x2,x3,x4)(x_1,x_2,x_3,x_4)(x1​,x2​,x3​,x4​) of real numbers that satisfy this system

Solving a system of linear equations

Consider the following system of linear equations:
−2x1+3x2+13x3=5−x1+2x2+5x3=14x1−14x2−2x3=14\begin{array}{rcr}
-2x_1+3x_2+13x_3&=&5\\
-x_1+2x_2+5x_3&=&1\\
4x_1-14x_2-2x_3&=&14\\
\end{array}−2x1​+3x2​+13x3​−x1​+2x2​+5x3​4x1​−14x2​−2x3​​===​5114​
where we wish to solve for the triple (x1,x2,x3)(x_1,x_2,x_3)(x1​,x2​,x3​) of real numbers that satisfy this system

Augmented Matrix: Gauss-Jordan Elimination

Consider the system of linear equations
x2+4x3=14+2x1−4x4x1−x3−x4+3=02x4−6=0\begin{array}{c}
x_2+4x_3=14+2x_1-4x_4\\
x_1-x_3-x_4+3=0\\
2x_4-6=0
\end{array}x2​+4x3​=14+2x1​−4x4​x1​−x3​−x4​+3=02x4​−6=0​
(a) Write the augmented matrix for the standard form of this system.

Wize University Linear Algebra Textbook > Systems of Linear Equations (SLEs) (Linear Systems)

Reducing a Matrix (Gauss-Jordan Elimination)

Popular Courses

Reducing a Matrix

Elementary Row Operations (EROs)

Reducing a Matrix (Matrix Reduction)

Example: EROs

Example: Reducing a Matrix

First Iteration

Second Iteration

Third Iteration

Practice: Reducing a Matrix

Extra Practice

Gauss-Jordan Elimination $\tkct{LAT}$

Gauss-Jordan Elimination $\tkct{LAT}$

Solving a system of linear equations

Additional Practice Problems--Systems of Linear Equations

Reducing a Matrix

Reducing a Matrix

Reducing a Matrix

Practice Question: Matrix Reduction

Solving a system of linear equations

Solving a system of linear equations

Solving a system of linear equations

Solving a system of linear equations

Solving a system of linear equations

Solving a system of linear equations

Augmented Matrix: Gauss-Jordan Elimination