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Isn't it possible that the $5$th column be a pivot column too ?

Wouldn't it mean that there is a possibility of an inconsistent solution as the last(fifth) column is a pivot column ?

Like this matrix below ?

$\begin{pmatrix} 1 &0 &0 &0 &0\\ 0 &0 &1 &0 &0\\ 0& 0 &0 &0& 1 \end{pmatrix}$

As we can see, the 5th column is a pivot column, which will cause the solution to be inconsistent, but the answer given says otherwise.

The answer to this question is as follows:

Yes. The system is consistent because with three pivots, there must be a pivot in the third (bottom) row of the coefficient matrix. The reduced echelon form cannot contain a row of the form $\begin{bmatrix}0&0&0&0&0& 1\end{bmatrix}$

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  • $\begingroup$ @SouravGhosh But what if b is (0 0 0 0 0 1), hence the matrix is: \begin{pmatrix} 1 & 0 & 0 & 0 & 0\\ 0 & 1 & 1 & 0 & 0\\ 0 & 0 & 0 & 0 & 1\\ \end{pmatrix} Will the last column will be considered a pivot columns ? But it will cause the solution set to be empty as it is not possible for 0*x = 1. $\endgroup$ Aug 6, 2022 at 5:06
  • $\begingroup$ You guys are both being sloppy. $b$ is a vector in $\Bbb R^3$. Now think again. $\endgroup$ Aug 6, 2022 at 5:23
  • $\begingroup$ $Ax=b$ is consistent iff it has a solution iff $b\in C(A) $ iff $b$ is a linear combination of the columns of $A$ . $\endgroup$ Aug 6, 2022 at 6:25
  • $\begingroup$ @Circuit_Breaker0.7 "the 5th column is a pivot column, which will cause the solution to be inconsistent", how can you say that when you're mentioning that it's a coefficient matrix? In an augmented matrix, if the last column becomes the pivot column then that system is said to be inconsistent. $\endgroup$
    – Shubham
    Aug 5, 2023 at 17:35

2 Answers 2

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It depends on what your matrix represents.

If $A\mathbf x= \mathbf b$

And the matrix above is the augmented matrix $(A\mid\mathbf b)$

If the last row of which is $(0,0,0,0,0 | 1)$

Which means $0x_1 + 0x_2 +0x_3 + 0x_4 = 1$

This is not possible.

However, if the matrix is just $A,$ and $\mathbf b$ is not represented in the discussion then $x_5 = b_3$ and what you have is consistent.

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    $\begingroup$ If you have something to that might help me improve my answer, I would appreciate the feedback, but downvotes with no commetary are not helpful. $\endgroup$
    – user317176
    Aug 6, 2022 at 7:52
  • $\begingroup$ This is precisely what the answer was. When I looked into the next questions in the book, this is what made more sense. $\endgroup$ Aug 6, 2022 at 10:47
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Consider a system of linear equations $$AX=b$$

Then the system is said to be consistent if it has a solution.

The following $2$ conditions are equivalent:

  1. $AX=b$ is consistent.

  2. $b$ is a linear combination of the columns of $A \quad $$[$$C(A)$ : column space of $A$ $]$

$C(A) $ is a linear subspace of $\Bbb{R}^{\text{no of rows}}$

$AX=b$ is solvable for every right hand side $b$ if $C(A) =\Bbb{R}^{\text{no of rows}}$.


Given $A=\begin{pmatrix} 1 &0 &0 &0 &0\\ 0 &0 &1 &0 &0\\ 0& 0 &0 &0& 1 \end{pmatrix}$

$\begin{align}C(A) &=\operatorname{span}\{col_1,col_2,...,col_5\}\\&=\operatorname{span}\{col_1,col_3,col_5\}\\&=\Bbb{R}^3\end{align}$

Hence $\forall b\in \Bbb{R}^3 \implies b\in C(A) $

Hence $AX=b$ is solvable for all $b\in\Bbb{R}^3$ implies any system of linear equations having coefficient matrix $A$ is consistent$\boxed{•}$

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    $\begingroup$ Thanks for the downvote. But can you explain the reason? It will help me do better . $\endgroup$ Aug 6, 2022 at 8:30
  • $\begingroup$ there's an elite club of us who receive them and nothing else.... welcome to the party $\endgroup$ Aug 14, 2022 at 9:31

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