# Free variables for a matrix in REF

• ChiralSuperfields
In summary, the conversation is about understanding free variables and leading variables in matrices, and how to express them in terms of other variables. The variables a, b, c, d, and e are discussed and it is noted that the 2nd and 5th variables are b and e, respectively. The use of notation and the concept of parameters is also mentioned.
ChiralSuperfields
Homework Statement
Relevant Equations
Row operations
For this,

I am not sure what the '2nd and 5th the variables' are. Dose someone please know whether the free variables ##2, 0, 0## from the second column and ##5, 8, \pi##? Or are there only allowed to be one free variable for each column so ##2## and ##5## for the respective columns.

Also for,

Why can b and d be free so are parametric and take on any value in the domain of real numbers? Could a be written as ##a(b,d) = 3b - 7d + 11 b,d ∈ ℝ## in other words a is leading so can take on any value since it is a function of b and d?

Many thanks!

Last edited:
ChiralSuperfields said:
Homework Statement: Please see below
Relevant Equations: Row operations

For this,
View attachment 324911

I am not sure what the '2nd and 5th the variables' are. Dose someone please know whether the free variables ##2, 0, 0## from the second column and ##5, 8, \pi##? Or are there only allowed to be one free variable for each column so ##2## and ##5## for the respective columns.
Note that "dose" and "does" are different words. A "dose" is a specified amount of medicine. The word you want is "does" as in "I do", "you do", "he/she/it does" and so on. I've noticed this error in at least one other thread of yours.

Assuming the variables, which aren't shown, are a, b, c, d, and e, the 2nd and 5th variables would be b and e. The variable b multiplies the vector in the 2nd column. The variable e multiplies the vector in the 5th column.
ChiralSuperfields said:
Also for,
View attachment 324916
Why can b and d be free so are parametric and take on any value in the domain of real numbers? Could a be written as ##a(b,d) = 3b - 7d + 11 b,d ∈ ℝ## in other words a is leading so can take on any value since it is a function of b and d?
I don't know what your notation above is supposed to mean. What you don't seem to understand is that the augmented matrix in your question is just shorthand for this system of equations:

##a + 2b + 3c + 4d + 5e = 6##
##.......... c + 7d + 8e = 9##
##..............d + \pi e = \sqrt 2##

The leading variables here are in the 1st, 3rd, and 4th columns; namely a, c, and d. So each of these can be written in terms of b and e, the 2nd and 5th variables.
ChiralSuperfields said:

berkeman, ChiralSuperfields and FactChecker
Mark44 said:
Note that "dose" and "does" are different words. A "dose" is a specified amount of medicine. The word you want is "does" as in "I do", "you do", "he/she/it does" and so on. I've noticed this error in at least one other thread of yours.
Thank you for your reply @Mark44! Sorry, I get those two words confused. I'll try look at for that in the future.

Mark44 said:
Assuming the variables, which aren't shown, are a, b, c, d, and e, the 2nd and 5th variables would be b and e. The variable b multiplies the vector in the 2nd column.
Ah true I think I see what you mean. I was saying that ##2,0,0## were free variables when they are scalar multiples of variable ##b##. I think the only way to prove that ##b## and ##e## are free (parametric) is to solve the matrix in REF then try to solve for the values for ##b## and ##e##, however, we won't be able to as there is not enough information. My textbook says that we assign a parameter to each free variable, and that Parmeter varies over the real numbers. Do you please know of a better way to explain that?

Mark44 said:
The variable e multiplies the vector in the 5th column.
I don't know what your notation above is supposed to mean. What you don't seem to understand is that the augmented matrix in your question is just shorthand for this system of equations:
Yeah I was just trying use notation to show that a was a function of b and d.
Mark44 said:
##a + 2b + 3c + 4d + 5e = 6##
##.......... c + 7d + 8e = 9##
##..............d + \pi e = \sqrt 2##

The leading variables here are in the 1st, 3rd, and 4th columns; namely a, c, and d. So each of these can be written in terms of b and e, the 2nd and 5th variables.
Many thanks!

ChiralSuperfields said:
I think the only way to prove that b and e are free (parametric) is to solve the matrix in REF then try to solve for the values for b and e, however, we won't be able to as there is not enough information.
You won't be able to solve for a unique solution, but you can get solutions that involve parameters.

The REF (reduced echelon form) matrix you show in post #1 is not completely reduced. An RREF (reduced row echelon form) matrix has 0 entries above and below each leading row entry.

a+2b+3c+4d+5e=6
..........c+7d+8e=9
..............d+πe=##\sqrt 2##

In augmented matrix form, the system above is
##\begin{bmatrix} 1 & 2 & 3 & 4 & 5 &| & 6\\
0 & 0 & 1 & 7 & 8 & | & 9\\
0 & 0 & 0 & 1 & \pi & | & \sqrt 2 \end{bmatrix}##

As an RREF augmented matrix it becomes
##\begin{bmatrix} 1 & 2 & 0 & 0 & r &| & s\\
0 & 0 & 1 & 0 & t & | & u\\
0 & 0 & 0 & 1 & v & | & w \end{bmatrix}##
Note that to save myself some work I've replaced some of the values with the letters r, s, t, u, v, and w.
From the above, the solution can be seen as
a = -2b - re + s
b = b + 0e ------- I.e., b is a free variable that equals itself
c = 0b -te + u
d = 0b - ve + w
e = 0b + 1e ------ e is a free variable that equals itself

The above can be rewritten as a vector equation like so:
##\begin{bmatrix}a\\b\\c\\d\\e\end{bmatrix} = b\begin{bmatrix}-2\\1 \\0 \\0\\0 \end{bmatrix} + e\\
\begin{bmatrix} -r\\0\\ -t \\-v \\1 \end{bmatrix} + \begin{bmatrix} s \\ 0 \\ u \\ w \\ 0 \end{bmatrix}##
b and e (the 2nd and 5th variables) are free, and can be set arbitrarily.

Geometrically, the vector equation above represents a plane in 5-dimensional space.

Last edited:
ChiralSuperfields

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