Is there any way possible to know which operation is best to perform on the matrix first, besides using intuition?

Usually with matrices you want to get 1s along the diagonal, so the usual method is to make the upper left most entry 1 by dividing that row by whatever that upper left entry is. So say the first row is 3 7 5 1. you would divide the whole row by 3 and it would become 1 7/3 5/3 1/3. From there you use the first row to make the first column have all 0s under that 1. Then repeat this going along the diagonal.

Step #1 of the last question appears to have the row operation labeled incorrectly. Shouldn't it be R1 + R2 --> R2 ??

Yes, and you're supposed to drag the row operations up/down, so that they match what is done in each step.

Can one also use any of these row operations to add in a new row? This could prove useful in operations where the matrices need to have the same dimensions.

No, row operations can only manipulate existing rows.

Is there only one correct method of using row operations on a specific matrix to solve a system of equations?

As long as you make as many columns that only have a 1 in them it doesn't really matter how you solve it. Making them a diagonal is just mostly used since it's systematic and is easy to keep track of since the answers are all the way on the right, so starting at the left makes it so you can go column by column.

So is the point of this just to solve for x and y at the same time? It seems a little unnecessary to me, since we already know how to solve systems of equations, and just about as quickly, without learning how to read matrices. It just seems like we don't need this . . .

Matrix stuff does get its own uses, this is more like an intro to them to then later use in linear algebra and calculus. Among other places.

If you want to apply the same priciples to columns do you us C1 or C2 instead of R1 or R2 or is there some other letter that you should use?

Technically, yes. On paper you can perform column operations. However, it nullifies the validity of the equations represented in the matrix. In other words, it breaks the equality. Say we have a matrix to represent: 3x + 3y = 15 2x + 2y = 10, where x = 2 and y = 3 Performing the operation 2R1 --> R1 (replace row 1 with 2 times row 1) gives us 4x + 4y+ = 20 = 4x2 + 4x3 = 20, which works But if we did 2C2 --> C2 (replace Column 2 with 2 times column 2), we get 3x + 6y = 15 = 3x2 + 6x3 = 15 = 6 + 18 = 15 2x + 4y = 10 = 2x2 + 4x3 = 10 = 4 + 12 = 10 Ridiculous, ain't it? I was wondering the same thing before I saw you question. It's the same reason you can't operate on one side of an equation without doing the same to the other (after all, it's called an EQUAtion; it has to be EQUAL!) Hope this helped :)

Why is it not legitimate to multiply a row with zero?

Um, remember the multiplication property of 0? What happens when you multiply by 0? Everything becomes 0, so you lose any informaiton given in the original problem.

What's a non zero constant?

Any number that is not zero is a non-zero constant. A variable is not a constant as it is an unknown number.

how did 5 - 13 ended up being 18 on the last challenge? statement says "R1 - R2 -> R1" Maybe a typo?

R1,3 is 5, while R2,3 is -13. When we subtract R1,3-R2,3, we get 5-(-13)=5+13=18. You had 13 mixed up with -13.

7) challenge problem, first row, third column in the matrix: why would the answer be -9? shouldn't 2*(-6) be +12 and then with the 3 added equal 15?

2*(-6) actually equals -12. It is the same as taking (-6) + (-6) which is the same as -6 -6 which equals -12.

Main content

Algebra (all content)

Course: Algebra (all content) > Unit 20

Lesson 3: Elementary matrix row operations

Matrix row operations

Google Classroom

Learn how to perform the matrix elementary row operations. These operations will allow us to solve complicated linear systems with (relatively) little hassle!

Matrix row operations

The following table summarizes the three elementary matrix row operations.

Matrix row operation	Example
Switch any two rows	$\left[\begin{array}{rr}{\blueD2} & {\blueD5} &{ \blueD{3}} \\ \greenD{3} &\greenD {4} &\greenD {6} \end{array}\right]\rightarrow\left[\begin{array}{rr} \greenD{3} & \greenD{4} &\greenD {6}\\\blueD{2} &\blueD {5} &\blueD{ 3} \end{array}\right]\\\\~~\\ \\ {\text{(Interchange row 1 and row 2.)}}$
Multiply a row by a nonzero constant	$\left[\begin{array}{rr}{\maroonD2} & {\maroonD5} &{ \maroonD3} \\ {3} & {4} & {6} \end{array}\right]\rightarrow\left[\begin{array}{rrr}{\goldD3 \cdot \maroonD2} & {\goldD3 \cdot \maroonD5} &{ \goldD3 \cdot \maroonD3} \\ { 3} & { 4} & { 6} \end{array}\right] \\~\\ {\text{(Row 1 becomes 3 times itself.)}}$
Add one row to another	$\left[\begin{array}{rr}{\tealD2} &\tealD5 &{ \tealD{3}} \\ \purpleC{3} &\purpleC {4} &\purpleC {6} \end{array}\right]\rightarrow\left[\begin{array}{rrr} {\tealD2} &\tealD5&{\tealD3}\\\purpleC{3}+\tealD2 & \purpleC{4}+\tealD5 &\purpleC{6} +\tealD3\end{array}\right]\\~~\\ {\text{(Row 2 becomes the sum of rows 2 and 1.)}}$

Matrix row operations can be used to solve systems of equations, but before we look at why, let's practice these skills.

Switch any two rows

Example

Perform the row operation R, start subscript, 1, end subscript, \leftrightarrow, R, start subscript, 2, end subscript on the following matrix.

\left[\begin{array} {rrr} 4 & 8 & 3 \\ 2 & 4 & 5 \\ 7 & 1 & 2 \end{array} \right]

Solution

R, start subscript, start color #11accd, 1, end color #11accd, end subscript, \leftrightarrow, R, start subscript, start color #1fab54, 2, end color #1fab54, end subscript means to interchange row start color #11accd, 1, end color #11accd and row start color #1fab54, 2, end color #1fab54.

So the matrix

\left[\begin{array} {rrr} \blueD4 & \blueD8 & \blueD{3} \\ \greenD2 & \greenD4 & \greenD5 \\ 7 & 1 & 2 \end{array} \right]

becomes

\left[\begin{array} {rrr} \greenD2 & \greenD4 & \greenD5 \\ \blueD4 & \blueD8 & \blueD{3} \\ 7 & 1 & 2 \end{array} \right]

Sometimes you will see the following notation used to indicate this change.

\left[\begin{array} {rrr} 4 & 8 & 3 \\ 2 & 4 & 5 \\ 7 & 1 & 2 \end{array} \right] \xrightarrow{{R_1\leftrightarrow R_2}}\left[\begin{array} {rrr} 2 & 4 & 5 \\ 4 &8 & 3 \\ 7 & 1 & 2 \end{array} \right]

Notice how row 1 replaces row 2 and row 2 replaces row 1. The third row is not changed.

Problem 1

Current

Perform the row operation R, start subscript, 2, end subscript, \leftrightarrow, R, start subscript, 3, end subscript on the following matrix.

\left[\begin{array} {rrr} 7 & 2 & 9 \\ 6 & 4 & 1 \\ 1 & 3 & 12 \end{array} \right]

Multiply a row by a nonzero constant

Example

Perform the row operation 3, R, start subscript, 2, end subscript, right arrow, R, start subscript, 2, end subscript on the following matrix.

\left[\begin{array} {rrr} 6 & 6 & 1 \\ 2 & 3 & 0 \\ 4 & 5 & 9 \end{array} \right]

Solution

start color #ca337c, 3, end color #ca337c, R, start subscript, start color #e07d10, 2, end color #e07d10, end subscript, right arrow, R, start subscript, start color #e07d10, 2, end color #e07d10, end subscript means to replace the start color #e07d10, 2, start text, n, d, end text, end color #e07d10 row with start color #ca337c, 3, end color #ca337c times itself.

\left[\begin{array} {rrr} 6 & 6 & 1 \\ \goldD{2} & \goldD{3} & \goldD{0} \\ 4 & 5 & 9 \end{array} \right]

becomes

\left[\begin{array} {rrr} 6 & 6 & 1 \\ \maroonD{3}\cdot \goldD{2} &\maroonD{3}\cdot \goldD{3} &\maroonD{3}\cdot \goldD{0} \\ 4 & 5 & 9 \end{array} \right] =\left[\begin{array} {rrr} 6 & 6 & 1 \\ 6 & 9 & {0} \\ 4 & 5 & 9 \end{array} \right]

To indicate this matrix row operation, we often see the following:

\left[\begin{array} {rrr} 6 & 6 & 1 \\ 2 & 3 & 0 \\ 4 & 5 & 9 \end{array} \right] \xrightarrow{3R_2\rightarrow R_2}\left[\begin{array} {rrr} 6 & 6 & 1 \\ 6 & 9 & {0} \\ 4 & 5 & 9 \end{array} \right]

Notice here three times the second row replaces the second row. The other rows remain the same.

Problem 3

Current

Perform the row operation 2, R, start subscript, 1, end subscript, right arrow, R, start subscript, 1, end subscript on the following matrix.

\left[\begin{array} {ccc} 2 & 6 & 5 & 1 \\ 7 & 4 & 8 & 0 \end{array} \right]

Add one row to another

Example

Perform the row operation R, start subscript, 1, end subscript, plus, R, start subscript, 2, end subscript, right arrow, R, start subscript, 2, end subscript on the following matrix.

\left[\begin{array} {rrr} 2 & 3 & 4\\ 0 & 8 & 1 \end{array} \right]

Solution

R, start subscript, start color #01a995, 1, end color #01a995, end subscript, plus, R, start subscript, start color #aa87ff, 2, end color #aa87ff, end subscript, right arrow, R, start subscript, 2, end subscript means to replace the 2, start text, n, d, end text row with the sum of the start color #01a995, 1, start text, s, t, end text, end color #01a995 and start color #aa87ff, 2, start text, n, d, end text, end color #aa87ff rows.

\left[\begin{array} {rrr} \tealD2 & \tealD{3} &\tealD{ 4}\\ \purpleC0 & \purpleC8 & \purpleC1 \end{array} \right]

becomes

\left[\begin{array} {lll} \tealD2 &{\tealD3} &{ \tealD4}\\ \tealD2+\purpleC0 & \tealD3+\purpleC8 & \tealD4 +\purpleC1 \end{array} \right]= \left[\begin{array} {rrr} 2 & 3 & 4\\ 2 & 11 & 5 \end{array} \right]

To indicate this matrix row operation, we can write the following:

\left[\begin{array} {rrr} 2 & 3 & 4\\ 0 & 8 & 1 \end{array} \right] \xrightarrow{R_1+R_2\rightarrow R_2} \left[\begin{array} {rrr} 2 & 3 & 4\\ 2 & 11 & 5 \end{array} \right]

Notice how the sum of row 1 and 2 replaces row 2. The other row remains the same.

Problem 5

Current

Perform the row operation R, start subscript, 1, end subscript, plus, R, start subscript, 3, end subscript, right arrow, R, start subscript, 3, end subscript on the following matrix.

\left[\begin{array} {rrr} -1 & 6 & -2 \\ -3 & 5 & 0 \\ 7 & 2 & 1 \end{array} \right]

Challenge problem

Perform the row operation R, start subscript, 1, end subscript, plus, 2, R, start subscript, 3, end subscript, right arrow, R, start subscript, 1, end subscript on the following matrix.

\left[\begin{array} {rrr} -5 & 7 & 3 \\ -2 & -1 & 4 \\ 8 & 8 & -6 \end{array} \right]

Systems of equations and matrix row operations

Recall that in an augmented matrix, each row represents one equation in the system and each column represents a variable or the constant terms.

For example, the system on the left corresponds to the augmented matrix on the right.

System	Matrix
$\begin{aligned} 1x+3y &=5\\2x+5y &=6\end{aligned}$	$\left[\begin{array}{cc:c}1&3&5\\\\2&5&6\end{array}\right]$

When working with augmented matrices, we can perform any of the matrix row operations to create a new augmented matrix that produces an equivalent system of equations. Let's take a look at why.

Switching any two rows

Equivalent Systems	Augmented matrix
$\begin{aligned} \blueD1x+\blueD3y &=\blueD{5} \\\greenD{2}x+\greenD{{5}}y &=\greenD{6} \end{aligned}$	$\left[\begin{array}{cc:c}\blueD1&\blueD3&\blueD5\\\\\greenD2&\greenD5&\greenD6\end{array}\right]$
	\downarrow
$\begin{aligned}\greenD{2}x+\greenD{{5}}y &=\greenD{6}\\ \blueD1x+\blueD3y &=\blueD{5} \end{aligned}$	$\left[\begin{array}{cc:c}\greenD2&\greenD5&\greenD6\\\\\blueD1&\blueD3&\blueD5\end{array}\right]$

The two systems in the above table are equivalent, because the order of the equations doesn't matter. This means that when using an augmented matrix to solve a system, we can interchange any two rows.

Multiply a row by a nonzero constant

We can multiply both sides of an equation by the same nonzero constant to obtain an equivalent equation.

In solving systems of equations, we often do this to eliminate a variable. Because the two equations are equivalent, we see that the two systems are also equivalent.

Equivalent Systems	Augmented matrix
$\begin{aligned} \maroonD1x+\maroonD3y &=\maroonD5 \\2x+5y &=6\end{aligned}$	$\left[\begin{array}{cc:c}\maroonD1 & \maroonD3 &\maroonD5 \\2&5&6\end{array}\right]$
	\downarrow
$\begin{aligned}\goldD{-2}x+(\goldD{-6})y &=\goldD{-10} \\2x+\phantom{(-)}5y &=6\end{aligned}$	$\left[\begin{array}{rr:r}\goldD{-2}&\goldD{-6}& \goldD{-10}\\2&5&6\end{array}\right]$

This means that when using an augmented matrix to solve a system, we can multiply any row by a nonzero constant.

Add one row to another

We know that we can add two equal quantities to both sides of an equation to obtain an equivalent equation.

So if A, equals, B and C, equals, D, then A, plus, C, equals, B, plus, D.

We do this often when solving systems of equations. For example, in this system

\begin{aligned}-2x-6y &=-10 \\ {2}x+{{5}}y &={6}\end{aligned}

, we can add the equations to obtain minus, y, equals, minus, 4.

Pairing this new equation with either original equation creates an equivalent system of equations.

[Why do we keep one of the original equations?]

Equivalent Systems	Augmented matrix
$\begin{aligned} -2x-6y &=-10\\2x+5y &=6\end{aligned}$	$\left[\begin{array}{rr:r}-2&-6&-10\\2&5&6\end{array}\right]$
	\downarrow
$\begin{aligned}-2x+(-6)y &=-10\\\purpleC0x+(\purpleC{-1})y &=\purpleC{-4} \end{aligned}$	$\left[\begin{array}{rr:r}-2&-6&-10\\\purpleC0&\purpleC{-1}&\purpleC{-4}\end{array}\right]$

So when using an augmented matrix to solve a system, we can add one row to another.

Concluding challenge problem

A sequence of row operations is performed on the matrix

\left[\begin{array}{rrr}{2} & {2} &{ 10} \\ {-2} & {-3} & {3} \end{array}\right]

. The table below describes the result of each step in the sequence.

Arrange the row operations according to each step.

Original matrix:

\left[\begin{array}{rr:r}2&2&10\\-2 & -3 & 3\end{array}\right]

Notice that the original matrix corresponds to

\begin{aligned} 2x+2y &={10} \\ {-2}x-3y &={ 3} \end{aligned}

, while the final matrix corresponds to

\begin{aligned} x&=18 \\ y&=-13 \end{aligned}

which simply gives the solution.

The system was solved entirely by using augmented matrices and row operations!

Want to join the conversation?

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Isaias B.
Posted 7 years ago. Direct link to Isaias B.'s post “how did 5 - 13 ended up b...”
how did 5 - 13 ended up being 18 on the last challenge?
statement says "R1 - R2 -> R1"

Maybe a typo?
Button navigates to signup pageComment on Isaias B.'s post “how did 5 - 13 ended up b...”
(0 votes)
Answer
- merelephant
  Posted 7 years ago. Direct link to merelephant's post “R1,3 is 5, while R2,3 is ...”
  R1,3 is 5, while R2,3 is -13. When we subtract R1,3-R2,3, we get 5-(-13)=5+13=18. You had 13 mixed up with -13.
  Button navigates to signup page
  (17 votes)
Christopher Hakes
Posted 3 years ago. Direct link to Christopher Hakes's post “Is there any way possible...”
Is there any way possible to know which operation is best to perform on the matrix first, besides using intuition?
Button navigates to signup pageButton navigates to signup page
(19 votes)
Answer
- loumast17
  Posted 3 years ago. Direct link to loumast17's post “Usually with matrices you...”
  Usually with matrices you want to get 1s along the diagonal, so the usual method is to make the upper left most entry 1 by dividing that row by whatever that upper left entry is. So say the first row is 3 7 5 1. you would divide the whole row by 3 and it would become 1 7/3 5/3 1/3. From there you use the first row to make the first column have all 0s under that 1. Then repeat this going along the diagonal.
  Comment on loumast17's post “Usually with matrices you...”
  (17 votes)
George
Posted 6 years ago. Direct link to George's post “If you want to apply the ...”
If you want to apply the same priciples to columns do you us C1 or C2 instead of R1 or R2 or is there some other letter that you should use?
Button navigates to signup pageButton navigates to signup page
(2 votes)
Answer
- Jibril
  Posted 6 years ago. Direct link to Jibril's post “Technically, yes. On pape...”
  Technically, yes. On paper you can perform column operations.
  However, it nullifies the validity of the equations represented in the matrix. In other words, it breaks the equality. Say we have a matrix to represent:
  3x + 3y = 15
  2x + 2y = 10, where x = 2 and y = 3
  
  Performing the operation 2R1 --> R1 (replace row 1 with 2 times row 1) gives us
  4x + 4y+ = 20 = 4x2 + 4x3 = 20, which works
  
  But if we did 2C2 --> C2 (replace Column 2 with 2 times column 2), we get
  
  3x + 6y = 15 = 3x2 + 6x3 = 15 = 6 + 18 = 15
  2x + 4y = 10 = 2x2 + 4x3 = 10 = 4 + 12 = 10
  
  Ridiculous, ain't it? I was wondering the same thing before I saw you question. It's the same reason you can't operate on one side of an equation without doing the same to the other (after all, it's called an EQUAtion; it has to be EQUAL!) Hope this helped :)
  Comment on Jibril's post “Technically, yes. On pape...”
  (12 votes)
atg999
Posted 5 years ago. Direct link to atg999's post “Step #1 of the last quest...”
Step #1 of the last question appears to have the row operation labeled incorrectly.

Shouldn't it be R1 + R2 --> R2 ??
Button navigates to signup pageButton navigates to signup page
(4 votes)
Answer
- Jerry Nilsson
  Posted 5 years ago. Direct link to Jerry Nilsson's post “Yes, and you're supposed ...”
  Yes, and you're supposed to drag the row operations up/down, so that they match what is done in each step.
  Button navigates to signup page
  (4 votes)
Muhurthana84
Posted 6 years ago. Direct link to Muhurthana84's post “What's a non zero constan...”
What's a non zero constant?
Button navigates to signup pageButton navigates to signup page
(1 vote)
Answer
- Kim Seidel
  Posted 6 years ago. Direct link to Kim Seidel's post “Any number that is not ze...”
  Any number that is not zero is a non-zero constant.
  A variable is not a constant as it is an unknown number.
  Comment on Kim Seidel's post “Any number that is not ze...”
  (8 votes)
Mojos
Posted 3 years ago. Direct link to Mojos's post “Why is it not legitimate ...”
Why is it not legitimate to multiply a row with zero?
Button navigates to signup pageButton navigates to signup page
(2 votes)
Answer
- Kim Seidel
  Posted 3 years ago. Direct link to Kim Seidel's post “Um, remember the multipli...”
  Um, remember the multiplication property of 0? What happens when you multiply by 0? Everything becomes 0, so you lose any informaiton given in the original problem.
  Button navigates to signup page
  (5 votes)
Wovado
Posted 2 years ago. Direct link to Wovado's post “Can one also use any of t...”
Can one also use any of these row operations to add in a new row? This could prove useful in operations where the matrices need to have the same dimensions.
Button navigates to signup pageButton navigates to signup page
(3 votes)
Answer
- KLaudano
  Posted 2 years ago. Direct link to KLaudano's post “No, row operations can on...”
  No, row operations can only manipulate existing rows.
  Button navigates to signup page
  (3 votes)
🎵 periwinkle 🎵
Posted 3 years ago. Direct link to 🎵 periwinkle 🎵's post “Is there only one correct...”
Is there only one correct method of using row
operations on a specific matrix to solve a system of equations?
Button navigates to signup pageButton navigates to signup page
(3 votes)
Answer
- loumast17
  Posted 3 years ago. Direct link to loumast17's post “As long as you make as ma...”
  As long as you make as many columns that only have a 1 in them it doesn't really matter how you solve it. Making them a diagonal is just mostly used since it's systematic and is easy to keep track of since the answers are all the way on the right, so starting at the left makes it so you can go column by column.
  Button navigates to signup page
  (2 votes)
elijahjrlarson
Posted 4 years ago. Direct link to elijahjrlarson's post “So is the point of this j...”
So is the point of this just to solve for x and y at the same time? It seems a little unnecessary to me, since we already know how to solve systems of equations, and just about as quickly, without learning how to read matrices. It just seems like we don't need this . . .
Button navigates to signup pageButton navigates to signup page
(3 votes)
Answer
- loumast17
  Posted 4 years ago. Direct link to loumast17's post “Matrix stuff does get its...”
  Matrix stuff does get its own uses, this is more like an intro to them to then later use in linear algebra and calculus. Among other places.
  Button navigates to signup page
  (2 votes)
Maj Kristin Stuedemann
Posted 7 years ago. Direct link to Maj Kristin Stuedemann's post “7) challenge problem, fir...”
7) challenge problem, first row, third column in the matrix: why would the answer be -9? shouldn't 2*(-6) be +12 and then with the 3 added equal 15?
Button navigates to signup pageComment on Maj Kristin Stuedemann's post “7) challenge problem, fir...”
(0 votes)
Answer
- Sonny Jansson
  Posted 7 years ago. Direct link to Sonny Jansson's post “2*(-6) actually equals -1...”
  2*(-6) actually equals -12. It is the same as taking (-6) + (-6) which is the same as -6 -6 which equals -12.
  Comment on Sonny Jansson's post “2*(-6) actually equals -1...”
  (9 votes)