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

Question

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

Answer 1

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.

Answer 2

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 $C(A) =\Bbb{R}^{\text{no of rows}}$.

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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$\boxed{•}$