When we wrote our row operations above we used expressions like \(R_2 = R_2 - 2 \times R_1\). Of course this does not mean that the second row is equal to the second row minus twice the first row. Instead it means that we are replacing the second row with the second row minus twice the first row. This kind of syntax is used frequently in computer programming when we want to change the value of a variable.

\[\begin{aligned}\left(\begin{array}{cc|c} 1 &1& 2\\ 3& 4& 5\\ 4& 5& 9\end{array}\right) \quad\xrightarrow{R_2=R_2-3R_1}\quad & \left(\begin{array}{cc|c} 1 &1& 2\\ \color{red}{0} & 1& -1\\ 4& 5& 9\end{array}\right) \\ {} \quad\xrightarrow{R_3=R_3-4R_1}\quad & \left(\begin{array}{cc|c} 1 &1& 2\\ 0& 1& -1\\ \color{red}{0}& 1& 1\end{array}\right)\end{aligned}\]

This page titled 1.2: Row Reduction is shared under a GNU Free Documentation License 1.3 license and was authored, remixed, and/or curated by Dan Margalit & Joseph Rabinoff via source content that was edited to the style and standards of the LibreTexts platform.

We will solve systems of linear equations algebraically using the elimination method. In other words, we will combine the equations in various ways to try to eliminate as many variables as possible from each equation. There are three valid operations we can perform on our system of equations:

If an augmented matrix is in reduced row echelon form, the corresponding linear system is viewed as solved. We will see below why this is the case, and we will show that any matrix can be put into reduced row echelon form using only row operations.

\( \( \def\d{\displaystyle}\) \( \newcommand{\f}[1]{\mathfrak #1}\) \( \newcommand{\s}[1]{\mathscr #1}\) \(\def\R{\mathbf R}\) \( \def\C{\mathbf C}\) \( \let\mathbb=\mathbf\) \( \let\oldvec=\vec\) \( \let\vec=\spalignvector\) \( \let\mat=\spalignmat\) \( \let\amat=\spalignaumat\) \( \let\hmat=\spalignaugmathalf\) \( \let\syseq=\spalignsys\) \( \let\epsilon=\varepsilon\) \( \def\Span{\operatorname{Span}}\) \( \def\Nul{\operatorname{Nul}}\) \( \def\Col{\operatorname{Col}}\) \( \def\Row{\operatorname{Row}}\) \( \def\rank{\operatorname{rank}}\) \( \def\nullity{\operatorname{nullity}}\) \( \def\Id{\operatorname{Id}}\) \( \def\Tr{\operatorname{Tr}}\) \( \def\adj{\operatorname{adj}}\) \( \def\Re{\operatorname{Re}}\) \( \def\Im{\operatorname{Im}}\) \( \def\dist{\operatorname{dist}}\) \( \def\refl{\operatorname{ref}}\) \( \def\inv{^{-1}}\) \( \let\To=\longrightarrow\) \( \def\rref{\;\xrightarrow{\text{RREF}}\;}\) \( \def\matrow#1{\text{---}\,#1\,\text{---}}\) \( \def\vol{\operatorname{vol}}\) \( \def\sptxt#1{\quad\text{ #1 }\quad}\) \( \let\bar=\overline\) \( \let\hat=\widehat\) \(\( \def\cBParseError: invalid DekiScript (click for details)Callstack: at (Template:MathJazMargalit), /content/body/div/p[4]/span, line 1, column 1 at template() at (Bookshelves/Linear_Algebra/Interactive_Linear_Algebra_(Margalit_and_Rabinoff)/01:_Systems_of_Linear_Equations-_Algebra/1.02:_Row_Reduction), /content/body/p[1]/span, line 1, column 25 \) \(\newcommand{\leading}[1]{\color{red}#1}\) \(\newcommand{\setof}[1]{\left\{#1\right\}}\)

\[\left(\begin{array}{ccc|c} 1 &0& 0& 1\\ 0& 1& 0& -2\\ 0& 0& 1& 3\end{array}\right) \quad\xrightarrow{\text{becomes}}\quad \left\{\begin{array}{rrr} x &=& 1\\ y &=& -2 \\ z &=& 3.\end{array}\right. \nonumber \]

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\(\def\R{\mathbf R}\) \( \def\C{\mathbf C}\) \( \let\mathbb=\mathbf\) \( \let\oldvec=\vec\) \( \let\vec=\spalignvector\) \( \let\mat=\spalignmat\) \( \let\amat=\spalignaumat\) \( \let\hmat=\spalignaugmathalf\) \( \let\syseq=\spalignsys\) \( \let\epsilon=\varepsilon\)

\[\left\{\begin{array}{rrrrrrr} x &+& 2y &+& 3z &=& 6 \\ 2x &-& 3y &+& 2z &=& 14 \\ 3x &+& y &-& z &=& -2 \end{array}\right. \quad\xrightarrow{\text{substitute}}\quad \left\{\begin{array}{rrrrrrr} 1 &2&\cdot(-2) &+& 3\cdot 3 &=& 6 \\ 2\cdot 1 &-& 3\cdot(-2) &+& 2\cdot 3 &=& 14 \\ 3\cdot 1 &+& (-2) &-& 3 &=& -2 \end{array}\right. \nonumber\]

This is called an augmented matrix. The word “augmented” refers to the vertical line, which we draw to remind ourselves where the equals sign belongs; a matrix is a grid of numbers without the vertical line. In this notation, our three valid ways of manipulating our equations become row operations:

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\[\begin{aligned} \left(\begin{array}{ccc|c} 1 &2& 3& 6\\ 0& -7& -4& 2\\ 0& -5& -10& -20 \end{array}\right)  \quad\xrightarrow{R_3=R_3\div-5}\quad & \left(\begin{array}{ccc|c} 1 &2& 3& 6\\ 0 &-7& -4& 2\\ 0& \color{red}{1}& 2& 4\end{array}\right) \\ {}\quad\xrightarrow{R_2\longleftrightarrow R_3}\quad & \left(\begin{array}{ccc|c} 1 &2& 3& 6\\ 0& 1& 2& 4 \\ 0& -7& -4& 2\end{array}\right) \\  {}\quad\xrightarrow{R_3 = R_3+7R_2}\quad & \left(\begin{array}{ccc|c} 1 &2& 3& 6\\ 0& 1& 2& 4\\ 0& \color{red}{0} & 10& 30 \end{array}\right) \\ {}\quad\xrightarrow{R_3 = R_3\div 10}\quad & \left(\begin{array}{ccc|c}1 &2& 3& 6\\ 0& 1& 2& 4 \\ 0& 0& \color{red}{1}& 3\end{array}\right)\end{aligned}\]

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An augmented matrix corresponds to an inconsistent system of equations if and only if the last column (i.e., the augmented column) is a pivot column.

\( \def\Span{\operatorname{Span}}\) \( \def\Nul{\operatorname{Nul}}\) \( \def\Col{\operatorname{Col}}\) \( \def\Row{\operatorname{Row}}\) \( \def\rank{\operatorname{rank}}\) \( \def\nullity{\operatorname{nullity}}\) \( \def\Id{\operatorname{Id}}\) \( \def\Tr{\operatorname{Tr}}\) \( \def\adj{\operatorname{adj}}\) \( \def\Re{\operatorname{Re}}\) \( \def\Im{\operatorname{Im}}\) \( \def\dist{\operatorname{dist}}\) \( \def\refl{\operatorname{ref}}\) \( \def\inv{^{-1}}\) \( \let\To=\longrightarrow\) \( \def\rref{\;\xrightarrow{\text{RREF}}\;}\) \( \def\matrow#1{\text{---}\,#1\,\text{---}}\) \( \def\vol{\operatorname{vol}}\) \( \def\sptxt#1{\quad\text{ #1 }\quad}\) \( \let\bar=\overline\) \( \let\hat=\widehat\)

\[\left\{\begin{array}{rrrrrrr} x &+& 2y &+& 3z &=& 6\\ 2x &-& 3y &+& 2z& =& 14\\ 3x &+& y &-& z &=& -2\end{array}\right. \quad\xrightarrow{\text{becomes}}\quad \left(\begin{array}{ccc|c} 1&2&3&6 \\ 2&-3&2&14 \\ 3&1&-1&-2 \end{array}\right).\nonumber\]

Eliminating a variable from an equation means producing a zero to the left of the line in an augmented matrix. First we produce zeros in the first column (i.e. we eliminate \(x\)) by subtracting multiples of the first row.

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\[\left(\begin{array}{ccc|c} 1 &2& 3& 6\\ 0& 1& 2& 4\\ 0& 0& 1& 3\end{array}\right) \quad\xrightarrow{\text{becomes}}\quad \left\{\begin{array}{rrrrrrr}  x &+& 2y &+& 3z &=& 6 \\ {}&{}& y &+& 2z &=& 4 \\ {}&{}&{}&{}&z &=& 3\end{array}\right.\nonumber \]

Solving equations by elimination requires writing the variables \(x,y,z\) and the equals sign \(=\) over and over again, merely as placeholders: all that is changing in the equations is the coefficient numbers. We can make our life easier by extracting only the numbers, and putting them in a box:

\[\begin{aligned} \left(\begin{array}{cc|c} 2&10&-1 \\ 3&15&2 \end{array}\right) \quad\xrightarrow{R_1=R_1\div 2}\quad & \left(\begin{array}{cc|c} \color{red}{1}&5&{-\frac{1}{2}} \\ 3&15&2 \end{array}\right) &&\color{blue}{\text{(Step 1b)}} \\ {} \quad\xrightarrow{R_2=R_2-3R_1}\quad & \left(\begin{array}{cc|c} 1&5&{-\frac{1}{2}} \\ \color{red}{0} &0&{\frac{7}{2}} \end{array}\right) &&\color{blue}{\text{(Step 1c)}} \\ {}\quad\xrightarrow{R_2=R_2\times\frac 27}\quad & \left(\begin{array}{cc|c} 1&5&{-\frac{1}{2}} \\ 0&0&\color{red}{1} \end{array}\right) &&\color{blue}{\text{(Step 2b)}} \\ {} \quad\xrightarrow{R_1=R_1+\frac 12R_2}\quad & \left(\begin{array}{cc|c} 1&5&\color{red}{0} \\ 0&0&1\end{array}\right) &&\color{blue}{\text{(Step 2c)}}\end{aligned}\]

\[\left\{\begin{array}{rrrrrrr} x&+&2y&+&3x&=& 6\\ 2x&-&3y&+&2z&=&14 \\ 3x&+&y&-&z&=&-2 \end{array}\right. \quad\xrightarrow{\text{becomes}}\quad \left(\begin{array}{ccc|c}1&2&3&6\\2&-3&2&14\\3&1&-1&-2\end{array}\right).\nonumber \]

\[\left(\begin{array}{ccc}1&0&2 \\ 0&1&-1\end{array}\right)\qquad \left(\begin{array}{cccc}0&1&8&0\end{array}\right) \qquad \left(\begin{array}{cc|c} 1&17&0\\0&0&1\end{array}\right)\qquad\left(\begin{array}{ccc}0&0&0\\0&0&0\end{array}\right).\nonumber\]

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\[\left(\begin{array}{ccc|c}1&0&0&1 \\ 0&1&0&-2\\0&0&1&3\end{array}\right)  \quad\xrightarrow{\text{translates to}}\quad \left\{\begin{array}{rrrrrrr} x&{}&{}&{}&{}&=&1 \\ {}&{}&y&{}&{}&=&-2 \\ {}&{}&{}&{}&z&=&3.\end{array}\right.\nonumber\]

This was made much easier by the fact that the top-left entry is equal to \(1\text{,}\) so we can simply multiply the first row by the number below and subtract. In order to eliminate \(y\) in the same way, we would like to produce a \(1\) in the second column. We could divide the second row by \(-7\text{,}\) but this would produce fractions; instead, let’s divide the third by \(-5\).

What has happened geometrically is that the original blue line has been replaced with the new blue line \(y=1\). We can think of the blue line as rotating, or pivoting, around the solution \((1,1)\). We used the pivot position in the matrix in order to make the blue line pivot like this. This is one possible explanation for the terminology “pivot”.

We swapped the second and third row just to keep things orderly. Now we translate this augmented matrix back into a system of equations:

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It will be very important to know where are the pivots of a matrix after row reducing; this is the reason for the following piece of terminology.

\[\begin{aligned} \left(\begin{array}{ccc|c} 2&1&12&1 \\ 1&2&9&-1\end{array}\right) \quad\xrightarrow{R_1 \longleftrightarrow R_2}\quad & \left(\begin{array}{ccc|c} \color{red}{1} &2&9&-1 \\ 2&1&12&1 \end{array}\right) &&\color{blue}{\text{(Optional)}} \\ {}\quad\xrightarrow{R_2=R_2-2R_1}\quad & \left(\begin{array}{ccc|c} 1&2&9&-1 \\ \color{red}{0}&-3&-6&3 \end{array}\right) &&\color{blue}{\text{(Step 1c)}} \\ {} \quad\xrightarrow{R_2=R_2\div -3}\quad & \left(\begin{array}{ccc|c} 1&2&9&-1 \\ 0&\color{red}{1} &2&-1 \end{array}\right) &&\color{blue}{\text{(Step 2b)}} \\ {}\quad\xrightarrow{R_1=R_1-2R_2}\quad & \left(\begin{array}{ccc|c} 1&\color{red}{0}&5&1 \\ 0&1&2&-1\end{array}\right) &&\color{blue}{\text{(Step 2c)}}\end{aligned}\]

The uniqueness statement is interesting—it means that, no matter how you row reduce, you always get the same matrix in reduced row echelon form.

In the previous Subsection The Elimination Method we saw how to translate a system of linear equations into an augmented matrix. We want to find an algorithm for “solving” such an augmented matrix. First we must decide what it means for an augmented matrix to be “solved”.

\[\left(\begin{array}{ccc|c}\color{red}{0}&-7&-4&2 \\ 2&\color{red}{4}&6&12 \\ 3&1&\color{red}{-1}&-2\end{array}\right).\nonumber\]

\[\left\{\begin{array}{rrrrr} x &+& y &=& 2\\ 3x &+& 4y &=& 5\\ 4x &+& 5y &=& 9\end{array}\right. \quad\xrightarrow{\text{augmented matrix}}\quad \left(\begin{array}{cc|c} 1 &1& 2\\ 3 &4& 5\\ 4& 5& 9\end{array}\right) \nonumber\]

\[\left(\begin{array}{ccccc} \color{red}{1} &0&\star &0&\star \\ 0&\color{red}{1} &\star &0 &\star \\ 0&0&0&\color{red}{1}&\star \\ 0&0&0&0&0\end{array}\right)  \qquad \begin{aligned} \star &= \text{any number} \\ \color{red}1 &= \text{pivot} \end{aligned} \nonumber\]

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\(\( \def\cBParseError: invalid DekiScript (click for details)Callstack: at (Template:MathJazMargalit), /content/body/div/p[4]/span, line 1, column 1 at template() at (Bookshelves/Linear_Algebra/Interactive_Linear_Algebra_(Margalit_and_Rabinoff)/01:_Systems_of_Linear_Equations-_Algebra/1.02:_Row_Reduction), /content/body/p[1]/span, line 1, column 25 \)

We immediately see that \(z=3\text{,}\) which implies \(y = 4-2\cdot 3 = -2\) and \(x = 6 - 2(-2) - 3\cdot 3 = 1.\) See Example \(\PageIndex{3}\).

\[\left(\begin{array}{cc}2&1\\0&1\end{array}\right)\qquad \left(\begin{array}{ccc|c} 2&7&1&4 \\ 0&0&2&1 \\ 0&0&0&3\end{array}\right)\qquad \left(\begin{array}{ccc}1&17&0\\0&1&1\end{array}\right) \qquad \left(\begin{array}{ccc}2&1&3\\0&0&0\end{array}\right).\nonumber\]

\[\left(\begin{array}{ccccc} \color{red}{\boxed{\star}} &\star &\star &\star &\star \\ 0&\color{red}{\boxed{\star}} & \star &\star &\star \\ 0&0&0&\color{red}{\boxed{\star}} &\star \\ 0&0&0&0&0 \end{array}\right) \qquad \begin{aligned} \star &= \text{any number} \\ \color{red}\boxed\star &= \text{any nonzero number} \end{aligned} \nonumber \]

We can visualize this system as a pair of lines in \(\mathbb{R}^2\) (red and blue, respectively, in the picture below) that intersect at the point \((1,1)\). If we subtract the first equation from the second, we obtain the equation \(2y=2\text{,}\) or \(y=1\). This results in the system of equations:

\[\left(\begin{array}{cccc|c} 1&0&\star &\star &\color{red}{0} \\ 0&1&\star &\star &\color{red}{0}\\ 0&0&0&0&\color{red}{1}\end{array}\right)\nonumber\]

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\[\begin{aligned} \left(\begin{array}{cc|c} 1&-1&0 \\ 1&1&2\end{array}\right) \quad\xrightarrow{R_2=R_2-R_1}\quad & \left(\begin{array}{cc|c} 1&-1&0 \\ 0&2&2\end{array}\right) \\ {} \quad\xrightarrow{R_2=\frac{1}{2}R_2}\quad & \left(\begin{array}{cc|c} 1&-1&0\\0&1&1\end{array}\right)\end{aligned}\]

When deciding if an augmented matrix is in (reduced) row echelon form, there is nothing special about the augmented column(s). Just ignore the vertical line.

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\[\left(\begin{array}{ccc|c} 2&7&1&4\\0&0&2&1\\0&0&1&3 \end{array}\right)\qquad\left(\begin{array}{cc|c}0&17&0\\0&2&1\end{array}\right)\qquad\left(\begin{array}{cc}2&1\\2&1\end{array}\right) \qquad \left(\begin{array}{c}0\\1\\0\\0\end{array}\right).\nonumber\]

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\[\begin{aligned} \left\{\begin{array}{rrrrrrr} x &+& 2y &+& 3z &=& 6\\ 2x &-& 3y &+& 2z &=& 14\\ 3x &+& y &-& z &=& -2\end{array}\right.  \quad\xrightarrow{\text{2nd ${}={}$ 2nd$-2\times$1st}}\quad &  \left\{\begin{array}{rrrrrrr}x &+& 2y &+& 3z &=& 6\\ {}&{}& -7y &-& 4z &=& 2\\ 3x &+& y &-& z &=& -2 \end{array}\right. \\ {} \quad\xrightarrow{\text{3rd ${}={}$ 3rd$-3\times$1st}}\quad& \left\{\begin{array}{rrrrrrr} x &+& 2y &+& 3z &=& 6 \\ {}&{}& -7y &-& 4z &=& 2\\ {}&{}& -5y &-& 10z &=& -20 \end{array}\right. \\ {} \quad\xrightarrow{\text{2nd $\longleftrightarrow$ 3rd}}\quad & \left\{\begin{array}{rrrrrrr} x &+& 2y &+& 3z &=& 6\\ {}&{}& -5y &-& 10z &=& -20 \\ {}&{}& -7y &-& 4z &=& 2\end{array}\right. \\ {} \quad\xrightarrow{\text{divide 2nd by $-5$}}\quad & \left\{\begin{array}{rrrrrrr} x &+& 2y &+& 3z &=& 6 \\ {}&{}& y &+& 2z &=& 4\\ {}&{}& -7y &-& 4z &=& 2 \end{array}\right. \\ {} \quad\xrightarrow{\text{3rd ${}={}$ 3rd$+7\times$2nd}}\quad & \left\{\begin{array}{rrrrrrr} x &+& 2y &+& 3z &=& 6\\ {}&{}&y &+& 2z &=& 4 \\ {}&{}&{}&{}&10z &=& 30 \end{array}\right.\end{aligned}\]

\[\left(\begin{array}{cc|c} 1 &1& 2\\ 0& 1& -1\\ 0 &1& 1\end{array}\right) \quad\xrightarrow{R_3=R_3-R_2}\quad \left(\begin{array}{cc|c} 1 &1& 2\\ 0& 1& -1\\ 0& \color{red}{0}& 2\end{array}\right) \nonumber \]

\[\left(\begin{array}{ccc|c} 1 &2& 3& 6\\ 0& 1& 2& 4 \\ 0& 0& 10& 30 \end{array}\right) \quad\xrightarrow{\text{becomes}}\quad \left\{\begin{array}{rrrrrrr} x &+& 2y &+& 3z &=& 6 \\ {}&{}& y &+& 2z& =& 4 \\ {}&{}&{}&{}& 10z &=& 30. \end{array}\right. \nonumber\]

Our original system has the same solution set as this system. But this system has no solutions: there are no values of \(x,y\) making the third equation true! We conclude that our original equation was inconsistent.

We will give an algorithm, called row reduction or Gaussian elimination, which demonstrates that every matrix is row equivalent to at least one matrix in reduced row echelon form.

At this point we’ve eliminated both \(x\) and \(y\) from the third equation, and we can solve \(10z=30\) to get \(z=3\). Substituting for \(z\) in the second equation gives \(y+2\cdot3=4\text{,}\) or \(y=-2\). Substituting for \(y\) and \(z\) in the first equation gives \(x + 2\cdot(-2) + 3\cdot3 = 6\text{,}\) or \(x=3\). Thus the only solution is \((x,y,z)=(1,-2,3)\).

\[\begin{aligned} \left(\begin{array}{ccc|c} 1 &2 &3& 6\\ 2& -3& 2& 14\\ 3& 1& -1& -2\end{array}\right) & \quad\xrightarrow{R_2=R_2-2R_1}\quad \left(\begin{array}{ccc|c} 1 &2 &3& 6\\ \color{red}{0}& -7& -4& 2\\ 3& 1& -1& -2\end{array}\right) \\ & \quad\xrightarrow{R_3=R_3-3R_1}\quad \left(\begin{array}{ccc|c} 1 &2& 3& 6\\ 0& -7& -4& 2\\ \color{red}{0}& -5& -10& -20\end{array}\right) \end{aligned}\]