If 2 columns are identical is there infinite solutions

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The discussion establishes that a consistent system of linear equations with two identical columns in its coefficient matrix has infinitely many solutions. Using the augmented matrix example, the reduced row echelon form (RREF) reveals that one row becomes entirely zero, indicating a free variable. The presence of free variables, alongside the condition where the number of rows (r) is less than the number of columns (n), confirms the existence of infinite solutions. The DTSLS Diagram is referenced for visual understanding, but the discussion emphasizes the algebraic interpretation of the equations derived from the RREF.

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karush
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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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