Linear Equations - Linear Algebra
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Solve the following system by reducing its corresponding matrix.


Solve the following system by reducing its corresponding matrix.
The corresponding matrix is

We add
times row one to row two.

This yields the solution

or
and
is a free variable.
The corresponding matrix is
We add times row one to row two.
This yields the solution
or
and
is a free variable.
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Solve the system by reducing its corresponding matrix.


Solve the system by reducing its corresponding matrix.
The corresponding matrix is

We add
times row one to row two.

This yields the solution

or
and
is a free variable.
The corresponding matrix is
We add times row one to row two.
This yields the solution
or
and
is a free variable.
Compare your answer with the correct one above
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Consider the system of linear equations:


Which of the following expresses the solution set in parametric form?
Consider the system of linear equations:
Which of the following expresses the solution set in parametric form?
First, write the matrix of coefficients for the system, seen below:

It is necessary to convert this matrix to reduced row-echelon form as follows:
The first step is usually to get a leading "1" in the first row; however, one is already there.
The next step is to get a "0" in the first position of the second row by multiplying the first row by the additive inverse of the number already there, and adding it to that row. Therefore, perform the operation
:

or

The next step is usually to get a leading "1" in the second row; however, one is already there.
The next step is to get a "0" in the second position of the first row by multiplying the first row by the additive inverse of the number already there, and adding it to that row. Therefore, perform the operation:



The matrix is now in reduced row-echelon form - all leading nonzero elements are "1", and go from upper-left to lower-right, and there are no zero rows. This matrix translates as the linear equations


To state the solution set parametrically, rewrite:


Set
; the parametric form of the solution set is
.
First, write the matrix of coefficients for the system, seen below:
It is necessary to convert this matrix to reduced row-echelon form as follows:
The first step is usually to get a leading "1" in the first row; however, one is already there.
The next step is to get a "0" in the first position of the second row by multiplying the first row by the additive inverse of the number already there, and adding it to that row. Therefore, perform the operation
:
or
The next step is usually to get a leading "1" in the second row; however, one is already there.
The next step is to get a "0" in the second position of the first row by multiplying the first row by the additive inverse of the number already there, and adding it to that row. Therefore, perform the operation:
The matrix is now in reduced row-echelon form - all leading nonzero elements are "1", and go from upper-left to lower-right, and there are no zero rows. This matrix translates as the linear equations
To state the solution set parametrically, rewrite:
Set ; the parametric form of the solution set is
.
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Consider the system of linear equations:



Which of the following gives its complete solution set (in parametric form, if applicable) or its only solution?
Consider the system of linear equations:
Which of the following gives its complete solution set (in parametric form, if applicable) or its only solution?
First, write the matrix of coefficients for the system, seen below:

It is necessary to convert this matrix to reduced row-echelon form as follows:
The first step is usually to get a leading "1" in the first row; however, one is already there.
The next two steps, which can be performed simultaneously, is to get a "0" in the first position of the second and third rows by multiplying the first row by the additive inverse of the number already there, and adding it to that row. Therefore, perform the operations

and



The next step is to get a leading "1" in the second rowby multiplying each entry in that row by the multiplicative inverse of the number already there; therefore, perform the operation



The next two steps, which can be performed simultaneously, is to get a "0" in the second position of the first and third rows by multiplying the first row by the additive inverse of the number already there, and adding it to that row. Therefore, perform the operations

and



The matrix is now in reduced row-echelon form - all leading nonzero elements are "1", and go from upper-left to lower-right, and there are no zero rows. This matrix translates as the linear equations


To state the solution set parametrically, rewrite:


Set
; the parametric form of the solution set is
.
First, write the matrix of coefficients for the system, seen below:
It is necessary to convert this matrix to reduced row-echelon form as follows:
The first step is usually to get a leading "1" in the first row; however, one is already there.
The next two steps, which can be performed simultaneously, is to get a "0" in the first position of the second and third rows by multiplying the first row by the additive inverse of the number already there, and adding it to that row. Therefore, perform the operations
and
The next step is to get a leading "1" in the second rowby multiplying each entry in that row by the multiplicative inverse of the number already there; therefore, perform the operation
The next two steps, which can be performed simultaneously, is to get a "0" in the second position of the first and third rows by multiplying the first row by the additive inverse of the number already there, and adding it to that row. Therefore, perform the operations
and
The matrix is now in reduced row-echelon form - all leading nonzero elements are "1", and go from upper-left to lower-right, and there are no zero rows. This matrix translates as the linear equations
To state the solution set parametrically, rewrite:
Set ; the parametric form of the solution set is
.
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Use row operations to find the inverse of the matrix 
Use row operations to find the inverse of the matrix
add the first row to the second
subtract two times the second row to the first
subtract the last row from the top row
subtract the first row from the last row
subtract two times the last row from the second row
switch the sign in the middle row

The inverse is 
add the first row to the second
subtract two times the second row to the first
subtract the last row from the top row
subtract the first row from the last row
subtract two times the last row from the second row
switch the sign in the middle row
The inverse is
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True or false:
is an example of a matrix in row-echelon form.
True or false: is an example of a matrix in row-echelon form.
A matrix is in row-echelon form if and only if it fits three conditions:
-
Any rows comprising only zeroes are at the bottom.
-
Any leading nonzero entries are 1's..
-
Each leading 1 is to the right of the one immediately above.
All four rows have leading nonzero entries, but none of them are 1's. The matrix violates the conditions of a matrix in row-echelon form.
A matrix is in row-echelon form if and only if it fits three conditions:
-
Any rows comprising only zeroes are at the bottom.
-
Any leading nonzero entries are 1's..
-
Each leading 1 is to the right of the one immediately above.
All four rows have leading nonzero entries, but none of them are 1's. The matrix violates the conditions of a matrix in row-echelon form.
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Find the inverse using row operations

Find the inverse using row operations
To find the inverse, use row operations:
add the third row to the second
subtract the second row from the top
subtract the first row from the second
subtract two times the first row from the bottom row
subtract three times the bottom row from the second row
subtract 2 times the middle row from the bottom row
add the bottom row to the top

The inverse is 
To find the inverse, use row operations:
add the third row to the second
subtract the second row from the top
subtract the first row from the second
subtract two times the first row from the bottom row
subtract three times the bottom row from the second row
subtract 2 times the middle row from the bottom row
add the bottom row to the top
The inverse is
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Find the inverse using row operations: 
Find the inverse using row operations:
subtract two times the second row from the last row
subtract the second row from the first
subtract two times the first row from the second
add the third row to the second
subtract 7 times the second row from the third row, then multiply by -1
add the bottom row to the middle row
add the last row to the top row
subtract two times the second row from the top row

The inverse is

subtract two times the second row from the last row
subtract the second row from the first
subtract two times the first row from the second
add the third row to the second
subtract 7 times the second row from the third row, then multiply by -1
add the bottom row to the middle row
add the last row to the top row
subtract two times the second row from the top row
The inverse is
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