Lab 1, problem 2 for TR: Reduced row echelon form
in Assignments
(10 pts)
This is where we'll report our answers to problem 2 of lab 1. The question itself deals with a randomly generated matrix (personalized for each student) and you are asked to do the following:
- Describe the the problem with prose and your initial typeset matrix.
- Include the code you used to solve the problem as a code block.
- Include the output of your command, also as a code block.
- Describe your interpretation of the output, viewing the matrix as the augmented matrix representation of a linear system.
Comments
The problem is to find the reduced row echelon form of the following matrix using Sage:
$$\left(\begin{matrix}-2 & 5 & 9 & 18 & -8 & 9 \\ -4 & 2 & -6 & -12 & 2 & 9 \\ 9 & 9 & -4 & -8 & 9 & 6 \\ 13 & 7 & 2 & 4 & 7 & -3\end{matrix}\right) $$
Code:
Result:
If this is the augmented matrix of a linear system, the output indicates that the system is underdetermined because there is a row of zeroes. Therefore, there are infinitely many solutions. The free variables are $x_4$ and $x_5$.
Given the following matrix:
$$
\left(\begin{matrix}-8 & -3 & -4 & -7 & 6 & 9 \\
-10 & 3 & -5 & -11 & 1 & 18 \\
-16 & -1 & -8 & 1 & 4 & 7 \\
2 & -6 & 1 & 4 & 5 & -9\end{matrix}\right)
$$
We can reduce it to reduced row echelon form using this code:
Which gives this output:
x1, x2, and x4 are fixed variables and x3 and x5 are free variables.
$$
x1 + 1/2(x3) - 89/450(x5) = -27/50\\
x2 - 77/75(x5) = 17/25\\
x4 - 43/225(x5) = -24/25
$$
I was given the following matrix:
$$
\left(\begin{matrix}-12 & 0 & -3 & -2 & -1 & -6 \\ 14 & 1 & -1 & -2 & 5 & 7 \\ -12 & -7 & -7 & -3 & -4 & -6 \\ 0 & -7 & -4 & -1 & -3 & 0\end{matrix}\right)
$$
I used the following code to reduce it to reduced row echelon form:
The result of running that code gives the following matrix:
The bound variables here are $x_{1}, x_{2}, x_{4}$, and $x_{5}$, with a free variable of $x_{3}$. Additionally since the last row is all zeros, and we are assuming the result to be an augmented matrix, it can be said that the system is undetermined, which means that there are infinite number of solutions.
We have an augmented matrix with four rows and six columns, meaning there are four equations with five terms each and a value equal to each equation in the last (sixth) column:
$$\left(\begin{matrix}-5 & -3 & 2 & 4 & 4 & -8 \\ 9 & -5 & -7 & -14 & -6 & 2 \\ 8 & 8 & -1 & -2 & -8 & 3 \\ 3 & 5 & 1 & 2 & -4 & -5\end{matrix}\right)$$
which is represented in Sage code as follows:
When evaluating with Sage, we get the following answer in reduced row echelon form,
Each row either starts with a one in the leftmost entry or is a zero row, and every leftmost entry in a row is the only nonzero number in its column. The bound variables here (pivots) are $x_1,x_2,$ and $x_3$ with free variables $x_4$ and $x_5$. The corresponding linear system would then be:
$$\begin{array}{rcl}
x_1 & = & \frac{-123}{20} - \frac{-41x_5}{30}\\
x_2 & = & \frac{101}{20} - \frac{7x_5}{30}\\
x_3 & = & \frac{-59}{5} + \frac{16x_5}{15}-2x_4.\\
\end{array}$$
Because the pivots depend on some free variables $x_4$ and $x_5$, this system of equations has infinitely many solutions.
Problem:
The following matrix corresponds to a system of linear equations with five variables: $x_1, x_2, ..., x_5$
$$
\left(\begin{matrix}-2 & 1 & -3 & -1 & 8 & 5 \\ 8 & -3 & -9 & 4 & -4 & -8 \\ -10 & 4 & 6 & -5 & 12 & 13 \\ 0 & 3 & -2 & 0 & -2 & 6\end{matrix}\right)
$$
Code:
Output:
Interpretation:
Here we see three pivot columns, which means we have three bound variables $(x_1, x_2, and\, x_3)$ and two free variables $(x_4 \, and \, x_5)$. The last row implies that this system has infinitely many solutions.
I was given the following matrix:
$$\left(\begin{matrix}-5 & 4 & 8 & -2 & -4 & 0 \\ -4 & 1 & 2 & -6 & 5 & -9 \\ 11 & 2 & 4 & 5 & -4 & 2 \\ 6 & 6 & 12 & 3 & -8 & 2\end{matrix}\right)$$
Here's the code to put it in RREF:
The output looks like this! (Added a vertical bat after col[4] to represent augmented matrix)
If we think of this as an augmented matrix, representing a system of equations, then the bottom row would be representative of a free variable, specifically x3 and x5 are our free variables.
$$\begin{align}
&x_{1} = -x_{5}\frac{100}{231}-\frac{172}{232}\\
&x_{2} = x_{5}\frac{257}{231}-2x_{3}+\frac{17}{231}\\
&x_{4} = x_{5}\frac{362}{231}+\frac{464}{231}
\end{align}
$$
Problem 2: Finding the Reduced Row Echelon Form of this system:
$$
\left(\begin{matrix}-9 & -2 & 5 & 8 & 16 & 2 \\ -1 & 8 & 2 & 8 & 16 & -9 \\ -8 & -10 & 3 & 0 & 0 & 11 \\-6 & 4 & 6 & -6 & -12 & 2\end{matrix}\right)
$$
Here is the code used:
This means the unique solution is:
This means there is a free variable in x5.
Here is the matrix a was asked to change into reduced row echelon form:
$$
\left(\begin{matrix} -8 & -10 & 6 & 7 & 1 & -5 \\ -17 & -10 & 6 & 5 & 10 & -5 \\ -8 & -18 & 1 & -1 & -6 & -9 \\ 9 & 0 & 0 & 2 & -9 & 0 \end{matrix}\right) $$
Which gives this matrix:
Since this matrix has 6 columns, if it was the augmented matrix of a linear system then the system would have 5 variables, and as it has 3 non-zero rows it will have 3 fixed variables with the remaining 2 being free variables, and will have an infinite number of solutions in the following form:
$$ x_1 = x_4 +\frac{2}{9}x_5 $$
$$x_2 = -\frac{11}{14}x_4 - \frac{37}{822}x_5 + \frac{1}{2}$$
$$x_3 = -\frac{1}{7}x_4 - \frac{676}{441}x_5 $$
With free variables $x_4$ and $x_5$.
We can use Sage to find the Reduced Row Echelon Form of the augmented matrix
$$
\left(
\begin{array}{cccccc}
-2 & -8 & -1 & -8 & -6 & -5 \\
16 & -8 & 8 & 6 & -5 & 0 \\
8 & 8 & 4 & -8 & -4 & -1 \\
8 & -16 & 4 & 14 & -1 & 1 \\
-12 & 3 & -6 & -5 & 2 & -4 \\
\end{array}
\right).
$$
by typing the following code:
This will return the following matrix output:
Since the augmented matrix has $6$ columns we will have $5$ variables, $x_1$, $x_2$, $x_3$, $x_4$, $x_5$ corresponding to columns $1$ through $5$ respectively. Therefore our reduced row echelon form implies the infinite solutions defined by the system
$$
\begin{align}
x_{1} &= -\frac{6}{49}- \frac{12}{7}x_{5} \\
x_{2} &= \frac{26}{49} -\frac{15}{14}x_{5}\\
x_{3} &= -\frac{11}{98} -\frac{1}{2}x_4-\frac{9}{28}x_{5} \\
\end{align}
$$
with the free variables $x_4$ and $x_5$.
With the given system of equations
$$\begin{align}6x_{1} - 4x_{2} - 8x_{3} + 4x_{4} - 4x_{5} &= -2\\
-3x_{1} - 7x_{2} - 2x_{3} - 4x_{4} - x_{5} &= 9\\
5x_{1} + 4x_{2} + 5x_{4} + 3x_{5} &= 7\\
-9x_{1} + 12x_{2} + 3x_{4} 6x_{5} &= 5\\
\end{align}$$
with the matrix
$$\left(\begin{matrix}6 & -4 & -8 & 4 & -4 & 2 \\
-3 & -7 & -2 & -4 & -1 & 5 \\
8 & -1 & 4 & 7 & 2 & 7 \\
2 & 3 & 12 & 3 & 6 & 5
\end{matrix}\right)$$
using the sage code
yielded the resulting matrix
$$\left(\begin{matrix}1 & 0 & 0 & 0 & 0 & \frac{23}{53} \\
0 & 1 & & 0 & 0 & \frac{-57}{53} \\
0 & 0 & 1 & 0 & 0 & \frac{65}{106}\\
0 & 0 & 0 & 1 & 0 & 0\\
\end{matrix}\right)$$
This puts it into what is known as Reduced Row Echelon Form. This matrix gives you a far better idea about what this system actually means. In this situation, $x = \frac{23}{53} , y = \frac{-57}{53} , z = \frac{65}{106}, w = 0$
Problem 2
Lab 1 asked that I use Sage to find the reduced row echelon form of the following matrix:
$$ \left(\begin{array}{cccccc} -12 & -2 & -6 & -7 & -6 & 4 \\ 2 & 1 & -2 & 7 & 1 & 9 \\ 18 & -6 & 3 & -9 & 9 & 5 \\ -16 & 7 & -5 & 16 & -8 & 4 \end{array}\right)$$
This is the Sage code I used:
Its output:
Which can be represented as the following matrix. By the nature of RREF, it is the unique RREF for the original matrix:
$$\left(\begin{array}{cccccc}
1 & 0 & 0 & \dfrac{89}{124} & \dfrac{1}{2} & \dfrac{98}{93} \\
0 & 1 & 0 & \dfrac{187}{62} & 0 & \dfrac{25}{31} \\
0 & 0 & 1 & \dfrac{-79}{62} & 0 & \dfrac{-283}{93} \\
0 & 0 & 0 & 0 & 0 & 0
\end{array}\right)$$
Interpreted as a system of equations, this matrix would represent:
$$\begin{align}-12x_{1} + -2x_{2} -6_{3} - 7x_{4} - 6x_{5} &= 4\\
2x_{1} + 1x_{2} - 2x_{3} - 7x_{4} + 1x_{5} &= 9\\
18x_{1} - 6x_{2} + 3x_{3} - 9x_{4} + 5x_{5}&= 5\\
-16x_{1} +7x_{2} - 5x_{3} + 16x_{4} - 8x_{5} &= 4\end{align}$$
With infinite solutions represented in the form:
$$\begin{align}x_{1} = \dfrac{-89}{124}r_{1} - \dfrac{1}{2}r_{2} + \dfrac{98}{93}\\
x_{2} = \dfrac{-187}{62}r_{1} + \dfrac{25}{31}\\
x_{3} = \dfrac{-79}{62}r_{1} + \dfrac{-283}{93}\\
x_{4} = r_{1}\\
x_{5} = r_{2}\end{align}$$
With the given matrix: $$\left(\begin{matrix}9 & -4 & 18 & -5 & -4 & -1 \\ -8 & 1 & -16 & 0 & -1 & -3 \\ 1 & -3 & 2 & -5 & -5 & -4 \\ -8 & -4 & -16 & -6 & 8 & 6\end{matrix}\right)$$
Which can be interpreted as a system of equations as: $$\begin{align}9x_{1} + -4x_{2} +18_{3} - 5x_{4} - 4x_{5} &= -1\\
-8x_{1} + 1x_{2} - 16x_{3} - 1x_{5} &= -3\\
1x_{1} - 3x_{2} + 2x_{3} - 5x_{4} - 5x_{5}&= -4\\
-8x_{1} -4x_{2} - 16x_{3} - 6x_{4} + 8x_{5} &= 6\end{align}$$
Using the code:
The solution received was:
Which can be shown in its RREF form which is unique to the matrix:$$\left(\begin{matrix}1 & 0 & 2 & 0 & -34/31 & -24/31 \\ 0 & 1 & 0 & 0 & -303/31 & -285/31 \\ 0 & 0 & 0 & 1 & 206/31 & 191/31 \\ 0 & 0 & 0 & 0 & 0 & 0\end{matrix}\right)$$
With infinite solutions represented with free variables $r_{1}$ and $r_{2}$ as: $$\begin{align}x_{1}&=-2r_{2} +34/31x_{5} - 24/31\\
x_{2}&=303/31x_{5}-285/31\\
x_{3}&= r_{1}\\
x_{4} &= -206/31x_{5}+191/31\end{align}$$
The problem is to find the reduced row echelon form of the matrix:
$$
\left(\begin{matrix}7 & -5 & -6 & -5 & -2 & -10 \\ -5 & -4 & 0 & -6 & -3 & -8 \\ 8 & 0 & 5 & 1 & 4 & 0 \\ 3 & -4 & 5 & -5 & 1 & -8\end{matrix}\right)
$$
The code I used to solve the problem was:
The output for this was:
The reduced row echelon matrix above provides the solution to the system. The augmented matrix at the start of the problem corresponds to the system:
$$
\begin{align}7x_{1} - 5x_{2} - 6x_{3} - 5x_{4} - 2x_{5} &= -10\\
- 5x_{1} - 4x_{2} - 6x_{4} - 3x_{5} &= -8\\
8x_{1} + 5x_{3} + x_{4} + 4x_{5} &= 0\\
3x_{1} - 4x_{2} + 5x_{3} - 5x_{4} + x_{5} &= -8\end{align}
$$
The reduced row echelon matrix still refers to the system above, but is a more simplified version. This helps the solutions to the system become clearer. Looking at the reduced row echelon matrix the solutions to the system are:
$x_{5} = t$
$x_{4} = z$
$x_{3} =\frac{27}{457}z - \frac{156}{457}t$
$x_{2} = 2 - \frac{593}{457}z - \frac{179}{457}t$
$x_{1} = - \frac{74}{457}z - \frac{131}{457}t$
The solutions indicate that $x_{5}$ and $x_{4}$ are free variables. This means that the system has infinitely many solutions. Any value can be used for $x_{5}$ and $x_{4}$ which can then be used to determine the other unknown variables and one of the infinitely many possible solutions to the system.
I had to find the reduced row echelon form for the following matrix:
$$\left(\begin{matrix}11 & 3 & 1 & 22 & -8 & -7 \\ 4 & 8 & -5 & 8 & 7 & 5 \\ -3 & 6 & -5 & -6 & 2 & 9 \\ 8 & 9 & -4 & 16 & -6 & 2\end{matrix}\right)$$
Here is the Sage code I used:
M = Matrix([[11, 3, 1, 22, -8, -7],[4, 8, -5, 8, 7, 5],[-3, 6, -5, -6, 2, 9],[8, 9, -4, 16, -6, 2],]);M.rref();When the code is run, the output is the following:
[ 1 0 0 2 181/43 -32/43][ 0 1 0 0 -526/43 26/43][ 0 0 1 0 -757/43 -27/43][ 0 0 0 0 0 0]This system would have infinitely many solutions, with 2 free variables in $x_{4}$ and $x_{5}$. $x_{1}, x_{2},$ and $x_{3}$ would all be expressed in terms of $x_{4}$ and $x_{5}$. $x_{4}$ and $x_{5}$ represent any real number.
The given problem is to find the reduced row echelon form of the matrix below and the solution to the corresponding system. The given matrix is:
$$\left(\begin{matrix}-14 & 0 & -10 & 3 & -7 & 6 \\ 8 & 0 & 8 & -8 & 4 & -5 \\ -6 & 0 & -2 & -5 & -3 & 1 \\ 16 & 7 & -3 & -5 & 8 & 1\end{matrix}\right)$$
Using this bit of sage code:
we get an output matrix in row echelon form of:
Looking at the rref of the original matrix, we can tell what the solution to the corresponding system are when looking at it as an augmented matrix. We can write the solution as a system of linear equations where $x_{1}$, $x_{2}$, and $x_{3}$ are all bound variables while $x_{4}$ and $x_{5}$ are free variables.
$$x_{1}=-\frac{7}{4}x_{4}-\frac{1}{2}x_{5} + \frac{1}{16} \\
x_{2}=\frac{165}{28}x_{4}-\frac{33}{112}\\
x_{3}=\frac{11}{4}x_{4}-\frac{11}{16}$$
This means that the system is underdetermined and has an infinite number of solutions
This is my assigned problem:
$$
\left(\begin{matrix}3 & 3 & 6 & 7 & 15 & 17 \\ 9 & 3 & 18 & 5 & 9 & 9 \\ 6 & 0 & 12 & -2 & -6 & -8 \\ -2 & -2 & -4 & -7 & 2 & -2\end{matrix}\right)
$$
Code:
Output:
There are an infinite number of solutions to the matrix provided due to the row of zeroes present on the 4th row. $ x_1 $, $ x_2 $, and $x_4$ are bound variables while $x_3$ and $x_5$ are free variables. The first five columns of each row represent a value that will be multiplied by the corresponding variable to equal the constant at the end.
$ x_1+ 2x_3 - \frac{-19}{7}x_5 = \frac{-8}{3} $
$ x_2 + \frac{138}{7}x_5 = \frac{53}{3} $
$ x_4 + \frac{-36}{7}x_5 = -4 $
These equations can be rewritten to find up to three variables in terms of the other variables/constants, but the remaining two variables will not be defined in a way that will form a unique solution for this system.
The problem is to find the reduced row echelon form of the following matrix:
$$
\left(\begin{matrix}7 & 3 & -6 & -3 & 8 & -3 \\ 2 & 9 & -6 & -7 & 6 & -3 \\ 8 & -8 & 14 & 2 & 7 & 7 \\ 1 & -11 & 20 & 5 & -1 & 10\end{matrix}\right)
$$
Below is the code that I used to find the reduced row echelon form of this matrix:
The outputted reduced row echelon matrix was:
which can be displayed as the following augmented matrix:
$$
\left(\begin{matrix}1 & 0 & 0 & \dfrac{-34}{141} & \dfrac{152}{141} & 0 \\ 0 & 1 & 0 & \dfrac{-367}{423} & \dfrac{239}{423} & 0 \\ 0 & 0 & 1 & \dfrac{-91}{423} & \dfrac{175}{846} & \dfrac{1}{2} \\ 0 & 0 & 0 & 0 & 0 & 0\end{matrix}\right)
$$
The corresponding system can be written in terms of the free variables $x_{4}=t$ and $x_{5}=z$:
$$
\begin{align}
x_{1}&=\dfrac{34}{141}t-\dfrac{152}{141}z\\
x_{2}&=\dfrac{-367}{423}t-\dfrac{239}{423}z+\dfrac{1}{2}\\
x_{3}&=\dfrac{91}{423}t-\dfrac{175}{846}z\\
x_{4}&=t\\
x_{5}&=z
\end{align}
$$
This is the unique RREF for the system. The system is undetermined and has infinitely many solutions.
I was asked to put the following matrix in reduced row echelon form:
$$
\left(\begin{matrix}-2 & -8 & 3 & -1 & -14 & -16 \\
-2 & -1 & -3 & -3 & -9 & -2 \\
1 & -2 & -3 & -8 & -9 & -4 \\
3 & 6 & -6 & -7 & 5 & 12\\
\end{matrix}\right)
$$
So I put the following code into Sage:
In return, Sage gave me this solution:
Because the last row is all zeroes, we know that the corresponding linear system has an infinite number of solutions, because any value would make the statement "0=0" true.
the assigned matrix for me was
$$
\left(\begin{matrix}6 & -1 & -1 & -9 & -10 & -5\\
5 & 4 & 7 & 5 & 8 & 4\\
-1 & 6 & -4 & 6 & -8 & -4\\
5 & 5 & -5 & -3 & -18 & -9\\
\end{matrix}\right)
$$
the code that I used was:
the result of such being:
$$\left(\begin{matrix}1 & 0 & 0 & -454/395 & -524/395 & -262/395\\
0 & 1 & 0 & 524/395 & -46/395 & -23/395\\
0 & 0 & 1 & 307/395 & 852/395 & 426/395\\
0 & 0 & 0 & 0 & 0\\
\end{matrix}\right)
$$
this reveals that the first row the x1 value is paramatrized by r1 and r2, as is x2 and x3. this leaves the last row as a free variable. and due to the paramatrization of the variables by real numbers r1 and r2 the result shows infinitely many solutions. ( i am marking x4 and x5 as r1 and r2 respectively)
@Student22
So, you've got a little formatting issue - just use include a backslash in front of your LaTeX commands. For example, you should have a
\fracinstead offrac. More importantly, your interpretation is incomplete.@Student19, @Student02, @Student14, @Student15, @Student04
Can you all say a bit more about the structure of the general solution?
My matrix was
$$\left(\begin{matrix}2 & 8 & 1 & 5 & -4 & -4 \\14 & -5 & 7 & -5 & -5 & -7 \\-16 & 3 & -8 & -4 & 3 & 5 \\ -12 & 13 & -6 & 10 & 1 & 3\end{matrix}\right)$$
I used the input
to put into row Echelon form:
Writing the system in terms of its free variables gives us:
$$x1=-(1/2)t+(105/242)z-(68/121)\\
x2=(83/121)z-(81/121)\\
x3=t\\
x4=(-57/121)z+(60/121)\\
x5=z$$
Which has infinite solutions
The matrix was:
$$\left(\begin{matrix}9 & 5 & -8 & 5 & 10 & -6 \\
-7 & 4 & -3 & -2 & -4 & -8 \\
9 & 5 & -3 & -6 & -12 & 3 \\
-16 & -1 & 5 & -7 & -14 & -2\\
\end{matrix}\right)$$
The code that I used to find the RREF was:
The output was:
This matrix has infinitely many solutions because the final row is all zeros.
For this problem I was asked to use Sage to find the reduced row echelon form of the following matrix:
\begin{pmatrix}
14 & 12 & -5 & 7 & -11 & 6 \\
-9 & -14 & 2 & -2 & -8 & -7 \\
5 & -4 & -1 & 6 & -9 & -2 \\
-9 & -16 & 4 & -1 & 2 & -8\\
\end{pmatrix}
The code I used was:
The output I got was:
There will be 5 variables: x1,x2,x3,x4, and x5 which will correspond to the columns 1 through 5 from this matrix. Thus, the reduced row echelon form of our matrix has infinitely many solutions defined by the following system:
\begin{align}
x_1&= 31/35x_4 + 2/5x_5\\
x_2&= -29/70x_4+6/5x_5\\
x_3&=3/35x_4+31/5x_5 \\
\end{align}
with x4 and x5 being the free variables.