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Linear Algebra : Linear Equations

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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:

True

False

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.

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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:

False

True

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.

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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:

No, the matrix does not have all zero rows

Yes, the matrix is in reduced row echelon form

No, the matrix has non zero lead entries

No, the matrix has nonzero elements

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.

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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:

No, the leading entry in the first row is not

Yes, the matrix is in reduced row echelon form

No, the matrix has nonzero elements

No, the matrix does not have an all zero row

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 .

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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:

Yes, the matrix is in reduced row echelon form

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

No, the matrix has nonzero terms

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.

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Example Question #36 : 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, there is a nonzero entry in the same column as a leading entry

Yes, the matrix is in reduced row echelon form

No, there are no nonzero rows

No, the matrix has nonzero terms

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.

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Example Question #37 : 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

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Example Question #38 : 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=5 , x_2=-1 , x_3=6

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

x_1=2 , x_2=5 , x_3=6

x_1=22 , 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

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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=2 , x_2=5 , x_3=12

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

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

x_1=5 , x_2=5 , x_3=7

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

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Example Question #40 : 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=0.5 , x_2=1.5 , x_3=-1

x_1=5 , x_2=3 , x_3=7

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

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

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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 #36 : Reduced Row Echelon Form And Row Operations

Example Question #37 : Reduced Row Echelon Form And Row Operations

Example Question #38 : Reduced Row Echelon Form And Row Operations

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

Example Question #40 : Reduced Row Echelon Form And Row Operations

All Linear Algebra Resources

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