If 2 columns are identical is there infinite solutions

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Discussion Overview

The discussion centers around the implications of having identical columns in the coefficient matrix of a consistent system of linear equations. Participants explore whether this condition guarantees infinitely many solutions, referencing reduced row echelon form (RREF) and specific examples.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant proposes that if a coefficient matrix has two identical columns, it leads to infinitely many solutions, citing a specific example and its RREF.
  • Another participant confirms the RREF results and translates them into equations, demonstrating that while one variable is fixed, others can take on multiple values, thus indicating infinite solutions.
  • A question is raised about whether the condition of having fewer rows than columns (r
  • One participant expresses confusion regarding the notation used (r and n) and seeks clarification.
  • A later reply clarifies that "r" refers to rows and "n" refers to columns, noting that this notation is common in linear algebra.

Areas of Agreement / Disagreement

Participants generally agree that having identical columns leads to infinitely many solutions, but there is uncertainty regarding the broader implications of the relationship between the number of rows and columns, as well as the role of pivot columns.

Contextual Notes

There is a lack of consensus on the implications of the relationship between the number of rows and columns in terms of guaranteeing infinite solutions, and the discussion includes varying interpretations of the conditions under which these conclusions hold.

karush
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$\tiny{45.4.T40}$
Suppose that the coefficient matrix of a consistent system of linear equations has two columns that are identical. Prove that the system has infinitely many solutions. Refer to the DTSLS Diagram
\item using augmented matrix A for example with c1 and c3 identical
$\left[
\begin{array}{rrr|r}
1&4&1&12\\
2&3&2&14\\
3&2&3&16
\end{array}\right]$
eMH returned the following RREF which show c1 and c3 as pivot columns
this is a violation of RREF
$\text{REFF}(A)=\left[ \begin{array}{rrr|r}
1 & 0 & 1 & 4 \\
0 & 1 & 0 & 2 \\
0 & 0 & 0 & 0
\end{array} \right]$
the matrix was dirived from the the possible set of $x_1=1\ x_2=2\ x_3=3$
a little perplexed as to whar we need to do when one row is all zero's after RREF
also as a result of RREF can this be just an 2x4 augmented matrix, r<n
this is supposed to be a proof which I am not good at
here is the DTSLS Diagram we are supposed to us

03.png
 
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$\left[
\begin{array}{rrr|r}
1&4&1&12\\
2&3&2&14\\
3&2&3&16
\end{array}\right]$
Subtract twice the first row from the second row and three times the first row from the third row:
$\left[\begin{array}{rrr|r} 1&4 & 1 & 12 \\ 0 & -5 & 0 & -10 \\ 0 & -10 & 0 & -20\end{array}\right]$

Divide the second row by -5, subtract 4 times the second row from the first row, and add 10 times the first row to the third row:
$\left[\begin{array}{rrr|r} 1 & 0 & 1 & 4 \\ 0 & 1 & 0 & 2 \\ 0 & 0 & 0 & 0 \end{array}\right]$
Yes that is what "eMH" (whatever that is) gave you. Now, what does it mean?

Rather than referring to a diagram, write it as "x, y, z" equations.

The top row, "1 0 1 | 4" gives the equation x+ z= 4.
The second row, "0 1 0 | 2" gives the equation y= 2.
The third row, "0 0 0 | 0" gives 0= 0 so no equation at all.

y must equal 2 but x and z can be any pair of numbers that add to 4:
x= 0, y= 2, z= 4
x= 1, y= 2, z= 3
x= 2, y= 2, z= 2
x= 3, y= 2, z= 1
x= 4, y= 2, z= 0
x= 5, y= 2, z= -1
etc.

Yes, there are an infinite number of solutions.
 
I have no idea what "r" and "n" mean and they do not appear in your original post.
 
r = rows
n = columns

it is usually a very common notation for linear algebra
 

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