# Suppose a $3 × 5$ coefficient matrix for a system has three pivot columns. Is the system consistent?

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

• @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. Aug 6, 2022 at 5:06
• You guys are both being sloppy. $b$ is a vector in $\Bbb R^3$. Now think again. Aug 6, 2022 at 5:23
• $Ax=b$ is consistent iff it has a solution iff $b\in C(A)$ iff $b$ is a linear combination of the columns of $A$ . Aug 6, 2022 at 6:25
• @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. Aug 5, 2023 at 17:35

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.

• 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. Aug 6, 2022 at 7:52
• This is precisely what the answer was. When I looked into the next questions in the book, this is what made more sense. Aug 6, 2022 at 10:47

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$$\boxed{•}$$

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