We are open Saturday and Sunday!

Call Now to Set Up Tutoring:

(888) 888-0446

Linear Algebra : Linear Equations

Study concepts, example questions & explanations for Linear Algebra

Create An Account

All Linear Algebra Resources

4 Diagnostic Tests 108 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #31 : Reduced Row Echelon Form And Row Operations

True or false:

$\begin{bmatrix} 1& 0& \pi \\ 0& 1& e\\ 0& 0& 0\\ 0& 0& 0 \end{bmatrix}$

is an example of a matrix in row-echelon form.

Possible Answers:

False

True

Correct answer:

True

Explanation:

A matrix is in row-echelon form if and only if it fits three conditions:

1) Any rows comprising only zeroes are at the bottom.

2) Any leading nonzero entries are 1's..

3) Each leading 1 is to the right of the one immediately above.

It can be seen that this matrix fits all three conditions, and is therefore in row-echelon form.

Report an Error

Example Question #31 : Reduced Row Echelon Form And Row Operations

$A = \begin{bmatrix} e & 0 & 0 & 0 \\ 0 & \pi & 0 & 0 \\ 0 & 0 & -\pi & 0 \\ 0 & 0 & 0 & -e \end{bmatrix}$

True or false: is an example of a matrix in row-echelon form.

Possible Answers:

True

False

Correct answer:

False

Explanation:

A matrix is in row-echelon form if and only if it fits three conditions:

1) Any rows comprising only zeroes are at the bottom.

2) Any leading nonzero entries are 1's..

3) 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.

Report an Error

Example Question #31 : Reduced Row Echelon Form And Row Operations

Is the following matrix in reduced row echelon form? Why or why not.

$\begin{bmatrix} 1& 0& 0& 7\\ 0& 1& 0& 5\\ 0& 0& 1& 4 \end{bmatrix}$

Possible Answers:

Yes, the matrix is in reduced row echelon form

No, the matrix has nonzero elements

No, the matrix does not have all zero rows

No, the matrix has non zero lead entries

Correct answer:

Yes, the matrix is in reduced row echelon form

Explanation:

A matrix is in reduced row echelon form if

all nonzero rows are above any all zero rows
the left most nonzero entry in each row (the leading entry) is .
the leading entry is in a column to the right of the leading entry in the column above it
all entries in a column below the leading entry are zero

This matrix meets all four of these criteria.

Report an Error

Example Question #31 : Reduced Row Echelon Form And Row Operations

Is the following matrix in reduced row echelon form? Why or why not.

$\begin{bmatrix} 5& 0& 0& 7\\ 0& 1& 0& 5\\ 0& 0& 1& 4 \end{bmatrix}$

Possible Answers:

Yes, the matrix is in reduced row echelon form

No, the matrix does not have an all zero row

No, the matrix has nonzero elements

No, the leading entry in the first row is not

Correct answer:

No, the leading entry in the first row is not

Explanation:

A matrix is in reduced row echelon form if

all nonzero rows are above any all zero rows
the left most nonzero entry in each row (the leading entry) is .
the leading entry is in a column to the right of the leading entry in the column above it
all entries in a column below the leading entry are zero

For this matrix to be in reduced row echelon form, the leading entry in the first row would need to be , not .

Report an Error

Example Question #31 : Reduced Row Echelon Form And Row Operations

Is the following matrix in reduced row echelon form? Why or why not.

$\begin{bmatrix} 1& 0& 0& 7\\ 0& 0& 0& 0\\ 0& 0& 1& 4 \end{bmatrix}$

Possible Answers:

No, the matrix has nonzero terms

No, there is a nonzero row below an all zero row.

Yes, the matrix is in reduced row echelon form

Yes, the leading entries are all

Correct answer:

No, there is a nonzero row below an all zero row.

Explanation:

A matrix is in reduced row echelon form if

all nonzero rows are above any all zero rows
the left most nonzero entry in each row (the leading entry) is .
the leading entry is in a column to the right of the leading entry in the column above it
all entries in a column below the leading entry are zero

For the matrix to be in reduced row echelon form, the 2nd and 3rd row would beed to be switched.

Report an Error

Example Question #32 : Reduced Row Echelon Form And Row Operations

Is the following matrix in reduced row echelon form? Why or why not.

$\begin{bmatrix} 1& 2& 0& 7\\ 0& 1& 0& 5\\ 0& 0& 1& 4 \end{bmatrix}$

Possible Answers:

No, the matrix has nonzero terms

No, there are no nonzero rows

No, there is a nonzero entry in the same column as a leading entry

Yes, the matrix is in reduced row echelon form

Correct answer:

No, there is a nonzero entry in the same column as a leading entry

Explanation:

A matrix is in reduced row echelon form if

all nonzero rows are above any all zero rows
the left most nonzero entry in each row (the leading entry) is .
the leading entry is in a column to the right of the leading entry in the column above it
all entries in a column below the leading entry are zero

For the matrix to be in reduced row echelon form, the in the first row would need to be transformed into a using row operations.

Report an Error

Example Question #31 : Reduced Row Echelon Form And Row Operations

Used reduced row echelon form to solve the following system of equations

x_1+2x_2-2x_3=9

4x_1+3x_2+x_3=0

x_1+2x_2+3x_3=4

Possible Answers:

x_1=-3.8 , x_2=5.4 , x_3=-1

Correct answer:

x_1=-3.8 , x_2=5.4 , x_3=-1

Explanation:

A matrix is in reduced row echelon form if

all nonzero rows are above any all zero rows
the left most nonzero entry in each row (the leading entry) is .
the leading entry is in a column to the right of the leading entry in the column above it
all entries in a column below the leading entry are zero

The first step in solving this system of linear equations is to represent this system as an augmented matrix.

x_1+2x_2-2x_3=9

4x_1+3x_2+x_3=0

x_1+2x_2+3x_3=4

$\begin{bmatrix} 1& 2& -2& 9\\ 4& 3& 1& 0\\ 1& 2& 3& 4 \end{bmatrix}$

We will use row operations to transform this matrix. There are three kinds of row operations that can be performed. Rows can be switched, a row can be multiplied by a constant and rows can be linearly combined ( aR_n+bR_m where and are constants).

The first number in the 1st row is , so no further calculation is needed.

We will make the first number in the 2nd row using the row operation R_2=r_2-4r_1

$\begin{bmatrix} 1& 2& -2& 9\\ 4-4(1)& 3-4(2)& 1-4(-2)& 0-4(9)\\ 1& 2& 3& 4 \end{bmatrix}$ $\Rightarrow$ $\begin{bmatrix} 1& 2& -2& 9\\ 0& -5& 9& -36\\ 1& 2& 3& 4 \end{bmatrix}$

We now make the leading entry of the 2nd row using the row operation R_2=-r_2/5

$\begin{bmatrix} 1& 2& -2& 9\\ 0/-5& -5/-5& 9/-5& -36/-5\\ 1& 2& 3& 4 \end{bmatrix}$ $\Rightarrow$ $\begin{bmatrix} 1& 2& -2& 9\\ 0& 1& -9/5& 36/5\\ 1& 2& 3& 4 \end{bmatrix}$

Next we will make the first number in the 3rd row using the row operation R_3=r_3-r_1

$\begin{bmatrix} 1& 2& -2& 9\\ 0& 1& -9/5& 36/5\\ 1-1& 2-2& 3+2& 4-9 \end{bmatrix}$ $\Rightarrow$ $\begin{bmatrix} 1& 2& -2& 9\\ 0& 1& -9/5& 36/5\\ 0& 0& 5& -5 \end{bmatrix}$

That operation also made the 2nd number in the 3rd row .

We now make the leading entry of the 3rd row using the row operation R_3=r_3/5

$\begin{bmatrix} 1& 2& -2& 9\\ 0& 1& -9/5& 36/5\\ 0/5& 0/5& 5/5& -5/5 \end{bmatrix}$ $\Rightarrow$ $\begin{bmatrix} 1& 2& -2& 9\\ 0& 1& -9/5& 36/5\\ 0& 0& 1& -1 \end{bmatrix}$

The matrix is now in echelon form.

Next we change the 3rd term in the 2nd row to using the row operation R_2=r_2+(9/5)r_3

$\begin{bmatrix} 1& 2& -2& 9\\ 0+(9/5)*0& 1+(9/5)*0& -9/5+(9/5)*1& 36/5+(9/5)*(-1)\\ 0& 0& 1& -1 \end{bmatrix}$ $\Rightarrow$ $\begin{bmatrix} 1& 2& -2& 9\\ 0& 1& 0& 27/5\\ 0& 0& 1& -1 \end{bmatrix}$

Next we change the 2nd term in the 1st row to using the row operation R_1=r_1-2r_2

$\begin{bmatrix} 1-2*(0)& 2-2*(1)& -2-2*(0)& 9-2*(27/5)\\ 0& 1& 0& 27/5\\ 0& 0& 1& -1 \end{bmatrix}$ $\Rightarrow$

$\begin{bmatrix} 1&0& -2& -9/5\\ 0& 1& 0& 27/5\\ 0& 0& 1& -1 \end{bmatrix}$

Last we change the 3rd term in the 1st row to using the row operation R_1=r_1+2r_3

$\begin{bmatrix} 1+2(0)&0+2(0)& -2+2(1)& -9/5+2(-1)\\ 0& 1& 0& 27/5\\ 0& 0& 1& -1 \end{bmatrix}$ $\Rightarrow$ $\begin{bmatrix} 1&0& 0& -19/5\\ 0& 1& 0& 27/5\\ 0& 0& 1& -1 \end{bmatrix}$

The matrix is now in reduced row echelon form and the last column is the solution to the system of equations.

x_1=-3.8 , x_2=5.4 , x_3=-1

Report an Error

Example Question #31 : Reduced Row Echelon Form And Row Operations

Used reduced row echelon form to solve the following system of equations

x_1+2x_2+2x_3=8

2x_1+6x_2+5x_3=6

x_1+2x_2+3x_3=4

Possible Answers:

x_1=2 , x_2=5 , x_3=6

x_1=5 , x_2=-1 , x_3=6

x_1=22 , x_2=-3 , x_3=-4

x_1=2 , x_2=3 , x_3=-4

Correct answer:

x_1=22 , x_2=-3 , x_3=-4

Explanation:

A matrix is in reduced row echelon form if

all nonzero rows are above any all zero rows
the left most nonzero entry in each row (the leading entry) is .
the leading entry is in a column to the right of the leading entry in the column above it
all entries in a column below the leading entry are zero

The first step in solving this system of linear equations is to represent this system as an augmented matrix

x_1+2x_2+2x_3=8

2x_1+6x_2+5x_3=6

x_1+2x_2+3x_3=4

$\begin{bmatrix} 1& 2& 2& 8\\ 2& 6& 5& 6\\ 1& 2& 3& 4 \end{bmatrix}$

Since the leading entry in the first row is , no further action is needed.

Next we will make the first number in the 2nd row using the row operation R_2=r_2-2r_1

$\begin{bmatrix} 1& 2& 2& 8\\ 2-2(1)& 6-2(2)& 5-2(2)& 6-2(8)\\ 1& 2& 3& 4 \end{bmatrix}$ $\Rightarrow$ $\begin{bmatrix} 1& 2& 2& 8\\ 0& 2& 1& -10\\ 1& 2& 3& 4 \end{bmatrix}$

We now make the leading entry of the 2nd row using the row operation R_2=r_2/2

$\begin{bmatrix} 1& 2& 2& 8\\ 0/2& 2/2& 1/2& -10/2\\ 1& 2& 3& 4 \end{bmatrix}$ $\Rightarrow$ $\begin{bmatrix} 1& 2& 2& 8\\ 0& 1& 1/2& -5\\ 1& 2& 3& 4 \end{bmatrix}$

Next we will make the first number in the 3rd row using the row operation R_3=r_3-r_1

$\begin{bmatrix} 1& 2& 2& 8\\ 0& 1& 1/2& -5\\ 1-1& 2-2& 3-2& 4-8 \end{bmatrix}$ $\Rightarrow$ $\begin{bmatrix} 1& 2& 2& 8\\ 0& 1& 1/2& -5\\ 0& 0& 1& -4 \end{bmatrix}$

The matrix is now in echelon form.

Next we change the 3rd term in the 2nd row to using the row operation R_2=r_2-r_3/2

$\begin{bmatrix} 1& 2& 2& 8\\ 0-0/2& 1-0/2& 1/2-1/2& -5+4/2\\ 0& 0& 1& -4 \end{bmatrix}$ $\Rightarrow$ $\begin{bmatrix} 1& 2& 2& 8\\ 0& 1& 0& -3\\ 0& 0& 1& -4 \end{bmatrix}$

Next we change the 2nd term in the 1st row to using the row operation R_1=r_1-2r_2

$\begin{bmatrix} 1-2(0)& 2-2(1)& 2-2(0)& 8-2(-3)\\ 0& 1& 0& -3\\ 0& 0& 1& -4 \end{bmatrix}$ $\Rightarrow$ $\begin{bmatrix} 1& 0& 2& 14\\ 0& 1& 0& -3\\ 0& 0& 1& -4 \end{bmatrix}$

Last we change the 3rd term in the 1st row to using the row operation R_1=r_1-2r_3

$\begin{bmatrix} 1-2(0)& 0-2(0)& 2-2(1)& 14-2(-4)\\ 0& 1& 0& -3\\ 0& 0& 1& -4 \end{bmatrix}$ $\Rightarrow$ $\begin{bmatrix} 1& 0& 0& 22\\ 0& 1& 0& -3\\ 0& 0& 1& -4 \end{bmatrix}$

The matrix is now in reduced row echelon form and the last column is the solution to the system of equations.

x_1=22 , x_2=-3 , x_3=-4

Report an Error

Example Question #31 : Reduced Row Echelon Form And Row Operations

Used reduced row echelon form to solve the following system of equations

2x_1+8x_3=10

5x_1+5x_2+7x_3=24

7x_1+6x_2+12x_3=33

Possible Answers:

x_1=-3 , x_2=5 , x_3=2

x_1=5 , x_2=5 , x_3=7

x_1=2 , x_2=8 , x_3=-10

x_1=2 , x_2=5 , x_3=12

Correct answer:

x_1=-3 , x_2=5 , x_3=2

Explanation:

A matrix is in reduced row echelon form if

all nonzero rows are above any all zero rows
the left most nonzero entry in each row (the leading entry) is .
the leading entry is in a column to the right of the leading entry in the column above it
all entries in a column below the leading entry are zero

The first step in solving this system of linear equations is to represent this system as an augmented matrix.

2x_1+8x_3=10

5x_1+5x_2+7x_3=24

7x_1+6x_2+12x_3=33

$\begin{bmatrix} 2& 0& 8& 10\\ 5& 5& 7& 24\\ 7& 6& 12& 33 \end{bmatrix}$

First we will make the first number in the 1st row using the row operation R_1=r_1/2

$\begin{bmatrix} 2/2& 0/2& 8/2& 10/2\\ 5& 5& 7& 24\\ 7& 6& 12& 33 \end{bmatrix}$ $\Rightarrow$ $\begin{bmatrix} 1& 0& 4& 5\\ 5& 5& 7& 24\\ 7& 6& 12& 33 \end{bmatrix}$

Next we will make the first number in the 2nd row using the row operation R_2=r_2-5r_1

$\begin{bmatrix} 1& 0& 4& 5\\ 5-5(1)& 5-5(0)& 7-5(4)& 24-5(5)\\ 7& 6& 12& 33 \end{bmatrix}$ $\Rightarrow$ $\begin{bmatrix} 1& 0& 4& 5\\ 0& 5& -13& -1\\ 7& 6& 12& 33 \end{bmatrix}$

We now make the leading entry of the 2nd row using the row operation R_2=r_2/5

$\begin{bmatrix} 1& 0& 4& 5\\ 0/5& 5/5& -13/5& -1/5\\ 7& 6& 12& 33 \end{bmatrix}$ $\Rightarrow$ $\begin{bmatrix} 1& 0& 4& 5\\ 0& 1& -13/5& -1/5\\ 7& 6& 12& 33 \end{bmatrix}$

Next we will make the first number in the 3rd row using the row operation R_3=r_3-7r_1

$\begin{bmatrix} 1& 0& 4& 5\\ 0& 1& -13/5& -1/5\\ 7-7(1)& 6-7(0)& 12-7(4)& 33 -7(5)\end{bmatrix}$ $\Rightarrow$ $\begin{bmatrix} 1& 0& 4& 5\\ 0& 1& -13/5& -1/5\\ 0& 6& -16& -2\end{bmatrix}$

Next we will make the 2nd number in the 3rd row using the row operation R_3=r_3-6r_2

$\begin{bmatrix} 1& 0& 4& 5\\ 0& 1& -13/5& -1/5\\ 0-6(0)& 6-6(1)& -16-6(-13/5)& -2-6(-1/5)\end{bmatrix}$ $\Rightarrow$

$\begin{bmatrix} 1& 0& 4& 5\\ 0& 1& -13/5& -1/5\\ 0& 0& -2/5& -4/5\end{bmatrix}$

We now make the leading entry of the 3rd row using the row operation R_3=-5r_3/2

$\begin{bmatrix} 1& 0& 4& 5\\ 0& 1& -13/5& -1/5\\ 0(-5/2)& 0(-5/2)& -2/5(-5/2)& -4/5(-5/2)\end{bmatrix}$ $\Rightarrow$ $\begin{bmatrix} 1& 0& 4& 5\\ 0& 1& -13/5& -1/5\\ 0& 0& 1& 2\end{bmatrix}$

The matrix is now in echelon form.

Next we change the 3rd term in the 2nd row to using the row operation R_2=r_2+(13/5)r_3

$\begin{bmatrix} 1& 0& 4& 5\\ 0+(13/5)*0& 1+(13/5)*0& -13/5+(13/5)*1& -1/5+(13/5)*2\\ 0& 0& 1& 2\end{bmatrix}$ $\Rightarrow$ $\begin{bmatrix} 1& 0& 4& 5\\ 0& 1&0& 5\\ 0& 0& 1& 2\end{bmatrix}$

Last we change the 3rd term in the 1st row to using the row operation R_1=r_1-4r_3

$\begin{bmatrix} 1-4*(0)& 0-4*(0)& 4-4*(1)& 5-4*(2)\\ 0& 1&0& 5\\ 0& 0& 1& 2\end{bmatrix}$ $\Rightarrow$ $\begin{bmatrix} 1& 0& 0& -3\\ 0& 1&0& 5\\ 0& 0& 1& 2\end{bmatrix}$

The matrix is now in reduced row echelon form and the last column is the solution to the system of equations.

x_1=-3 , x_2=5 , x_3=2

Report an Error

Example Question #31 : Reduced Row Echelon Form And Row Operations

Used reduced row echelon form to solve the following system of equations

7x_1+5x_2+3x_3=8

3x_1-5x_2+2x_3=-8

5x_1+3x_2+7x_3=0

Possible Answers:

x_1=5 , x_2=3 , x_3=7

x_1=7 , x_2=5 , x_3=-3

x_1=0.5 , x_2=1.5 , x_3=-1

x_1=3 , x_2=-5 , x_3=2

Correct answer:

x_1=0.5 , x_2=1.5 , x_3=-1

Explanation:

A matrix is in reduced row echelon form if

all nonzero rows are above any all zero rows
the left most nonzero entry in each row (the leading entry) is .
the leading entry is in a column to the right of the leading entry in the column above it
all entries in a column below the leading entry are zero

The first step in solving this system of linear equations is to represent this system as an augmented matrix.

7x_1+5x_2+3x_3=8

3x_1-5x_2+2x_3=-8

5x_1+3x_2+7x_3=0

$\begin{bmatrix} 7& 5& 3& 8\\ 3& -5& 2& -8\\ 5& 3& 7& 0 \end{bmatrix}$

The leading entry in the 1st row needs to be , so we use the row operation R_1=r_1/7 to change it.

$\begin{bmatrix} 7/7& 5/7& 3/7& 8/7\\ 3& -5& 2& -8\\ 5& 3& 7& 0 \end{bmatrix}$ $\Rightarrow$ $\begin{bmatrix} 1& 5/7& 3/7& 8/7\\ 3& -5& 2& -8\\ 5& 3& 7& 0 \end{bmatrix}$

We will make the first number in the 2nd row using the row operation R_2=r_2-3r_1

$\begin{bmatrix} 1& 5/7& 3/7& 8/7\\ 3-3(1)& -5-3(5/7)& 2-3(3/7)& -8-3(8/7)\\ 5& 3& 7& 0 \end{bmatrix}$ $\Rightarrow$ $\begin{bmatrix} 1& 5/7& 3/7& 8/7\\ 0& -50/7& 5/7& -80/7\\ 5& 3& 7& 0 \end{bmatrix}$

We now make the leading entry of the 2nd row using the row operation R_2=-7r_2/50

$\begin{bmatrix} 1& 5/7& 3/7& 8/7\\ 0*(-7/50)& -50/7*(-7/50)& 5/7*(-7/50)& -80/7*(-7/50)\\ 5& 3& 7& 0 \end{bmatrix}$ $\Rightarrow$ $\begin{bmatrix} 1& 5/7& 3/7& 8/7\\ 0& 1& -1/10& 8/5\\ 5& 3& 7& 0 \end{bmatrix}$

Next we will make the first number in the 3rd row using the row operation R_3=r_3-5r_1

$\begin{bmatrix} 1& 5/7& 3/7& 8/7\\ 0& 1& -1/10& 8/5\\ 5-5(1)& 3-5(5/7)& 7-5(3/7)& 0-5(8/7) \end{bmatrix}$ $\Rightarrow$ $\begin{bmatrix} 1& 5/7& 3/7& 8/7\\ 0& 1& -1/10& 8/5\\ 0& -4/7& 34/7& -40/7 \end{bmatrix}$

Next we make the 2nd number in the 3rd row using the row operation R_3=r_3+4r_2/7

$\begin{bmatrix} 1& 5/7& 3/7& 8/7\\ 0& 1& -1/10& 8/5\\ 0+(4/7)(0)& -4/7+(4/7)(1)& 34/7+(4/7)(-1/10)& -40/7+(4/7)(8/5) \end{bmatrix}$ $\Rightarrow$

$\begin{bmatrix} 1& 5/7& 3/7& 8/7\\ 0& 1& -1/10& 8/5\\ 0& 0& 4.8& -4.8 \end{bmatrix}$

We now make the leading entry of the 3rd row using the row operation R_3=r_3/4.8

$\begin{bmatrix} 1& 5/7& 3/7& 8/7\\ 0& 1& -1/10& 8/5\\ 0& 0& 4.8/4.8& -4.8/4.8 \end{bmatrix}$ $\Rightarrow$ $\begin{bmatrix} 1& 5/7& 3/7& 8/7\\ 0& 1& -1/10& 8/5\\ 0& 0& 1& -1 \end{bmatrix}$

The matrix is now in echelon form.

Next we change the 3rd term in the 2nd row to using the row operation R_2=r_2+(1/10)r_3

$\begin{bmatrix} 1& 5/7& 3/7& 8/7\\ 0& 1& -1/10+(1/10)*1& 8/5+(1/10)*(-1)\\ 0& 0& 1& -1 \end{bmatrix}$ $\Rightarrow$ $\begin{bmatrix} 1& 5/7& 3/7& 8/7\\ 0& 1& 0& 1.5\\ 0& 0& 1& -1 \end{bmatrix}$

Next we change the 2nd term in the 1st row to using the row operation R_1=r_1-5r_2/7

$\begin{bmatrix} 1-(5/7)*0& 5/7-(5/7)*1& 3/7-(5/7)*0& 8/7-(5/7)*1.5\\ 0& 1& 0& 1.5\\ 0& 0& 1& -1 \end{bmatrix}$ $\Rightarrow$

$\begin{bmatrix} 1& 0& 3/7& 1/14\\ 0& 1& 0& 1.5\\ 0& 0& 1& -1 \end{bmatrix}$

Last we change the 3rd term in the 1st row to using the row operation R_1=r_1-3r_3/7

$\begin{bmatrix} 1& 0& 3/7-(3/7)*1& 1/14-(3/7)*(-1)\\ 0& 1& 0& 1.5\\ 0& 0& 1& -1 \end{bmatrix}$ $\Rightarrow$ $\begin{bmatrix} 1& 0& 0& 7/14\\ 0& 1& 0& 1.5\\ 0& 0& 1& -1 \end{bmatrix}$

The matrix is now in reduced row echelon form and the last column is the solution to the system of equations.

x_1=0.5 , x_2=1.5 , x_3=-1

Report an Error

View Linear Algebra Tutors

Dalton
Certified Tutor

University of Massachusetts Amherst, Doctor of Philosophy, Mathematics.

View Linear Algebra Tutors

Vicky
Certified Tutor

Clemson University, Doctor of Science, Civil Engineering.

View Linear Algebra Tutors

Christopher
Certified Tutor

University of Michigan-Ann Arbor, Bachelor of Science, Computer Science.

All Linear Algebra Resources

4 Diagnostic Tests 108 Practice Tests Question of the Day Flashcards Learn by Concept

Popular Subjects

Statistics Tutors in Chicago, Biology Tutors in Houston, Chemistry Tutors in Seattle, Physics Tutors in Houston, MCAT Tutors in Houston, Computer Science Tutors in Dallas Fort Worth, SSAT Tutors in Los Angeles, MCAT Tutors in Philadelphia, ISEE Tutors in Washington DC, French Tutors in Atlanta

Popular Courses & Classes

MCAT Courses & Classes in Denver, Spanish Courses & Classes in Boston, GRE Courses & Classes in Los Angeles, SAT Courses & Classes in San Diego, MCAT Courses & Classes in Washington DC, LSAT Courses & Classes in Dallas Fort Worth, SSAT Courses & Classes in Boston, ACT Courses & Classes in Boston, GRE Courses & Classes in San Francisco-Bay Area, ISEE Courses & Classes in Washington DC

Popular Test Prep

ISEE Test Prep in San Diego, SAT Test Prep in San Francisco-Bay Area, GRE Test Prep in Philadelphia, ISEE Test Prep in Atlanta, LSAT Test Prep in Washington DC, MCAT Test Prep in Phoenix, SAT Test Prep in Chicago, MCAT Test Prep in Miami, GRE Test Prep in Seattle, SAT Test Prep in Los Angeles

DMCA Complaint

If you believe that content available by means of the Website (as defined in our Terms of Service) infringes one or more of your copyrights, please notify us by providing a written notice (“Infringement Notice”) containing the information described below to the designated agent listed below. If Varsity Tutors takes action in response to an Infringement Notice, it will make a good faith attempt to contact the party that made such content available by means of the most recent email address, if any, provided by such party to Varsity Tutors.

Your Infringement Notice may be forwarded to the party that made the content available or to third parties such as ChillingEffects.org.

Please be advised that you will be liable for damages (including costs and attorneys’ fees) if you materially misrepresent that a product or activity is infringing your copyrights. Thus, if you are not sure content located on or linked-to by the Website infringes your copyright, you should consider first contacting an attorney.

Please follow these steps to file a notice:

You must include the following:

A physical or electronic signature of the copyright owner or a person authorized to act on their behalf; An identification of the copyright claimed to have been infringed; A description of the nature and exact location of the content that you claim to infringe your copyright, in \ sufficient detail to permit Varsity Tutors to find and positively identify that content; for example we require a link to the specific question (not just the name of the question) that contains the content and a description of which specific portion of the question – an image, a link, the text, etc – your complaint refers to; Your name, address, telephone number and email address; and A statement by you: (a) that you believe in good faith that the use of the content that you claim to infringe your copyright is not authorized by law, or by the copyright owner or such owner’s agent; (b) that all of the information contained in your Infringement Notice is accurate, and (c) under penalty of perjury, that you are either the copyright owner or a person authorized to act on their behalf.

Send your complaint to our designated agent at:

Charles Cohn Varsity Tutors LLC
101 S. Hanley Rd, Suite 300
St. Louis, MO 63105

Or fill out the form below:

Do not fill in this field

Contact Information

* Full legal name

Company name

* Phone number

* Email address

Address

* Address1

Address2

* City

* State

* Zipcode

* Country

Complaint Details

* URL of copyrighted work (where your original material is located)

* Please describe the copyrighted work so that it may be easily identified

I am the owner, or an agent authorized to act on behalf of the owner of an exclusive right that is allegedly infringed.

I have a good faith belief that the use of the material in the manner complained of is not authorized by the copyright owner, its agent, or the law.

This notification is accurate.

I acknowledge that there may be adverse legal consequences for making false or bad faith allegations of copyright infringement by using this process.

* Signature (your digital signature is legally binding)

HIGH SCHOOL

GRADUATE SCHOOL

K-8

Search 50+ Tests

math tutoring

science tutoring

foreign languages

elementary tutoring

other

Search 350+ Subjects

Linear Algebra : Linear Equations

Study concepts, example questions & explanations for Linear Algebra

All Linear Algebra Resources

Example Questions

Example Question #31 : Reduced Row Echelon Form And Row Operations

Example Question #31 : Reduced Row Echelon Form And Row Operations

Example Question #31 : Reduced Row Echelon Form And Row Operations

Example Question #31 : Reduced Row Echelon Form And Row Operations

Example Question #31 : Reduced Row Echelon Form And Row Operations

Example Question #32 : Reduced Row Echelon Form And Row Operations

Example Question #31 : Reduced Row Echelon Form And Row Operations

Example Question #31 : Reduced Row Echelon Form And Row Operations

Example Question #31 : Reduced Row Echelon Form And Row Operations

Example Question #31 : Reduced Row Echelon Form And Row Operations

All Linear Algebra Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: