Elementary Row Operation on Augmented Matrix leads to Equivalent System of Simultaneous Linear Equations/Corollary 2

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Corollary to Elementary Row Operation on Augmented Matrix leads to Equivalent System of Simultaneous Linear Equations

Let $S$ be a system of simultaneous linear equations:

$\displaystyle \forall i \in \set {1, 2, \ldots, m}: \sum_{j \mathop = 1}^n \alpha_{i j} x_j = \beta_i$

Let $\begin {pmatrix} \mathbf A & \mathbf b \end {pmatrix}$ denote the augmented matrix of $S$.

Let $\begin {pmatrix} \mathbf R & \mathbf s \end {pmatrix}$ be a reduced echelon matrix derived from $\begin {pmatrix} \mathbf A & \mathbf b \end {pmatrix}$.


Then $S$ is consistent if and only if the rightmost column of $\begin {pmatrix} \mathbf R & \mathbf s \end {pmatrix}$ does not have a leading coefficient.


Proof

Let $S'$ be the system of simultaneous linear equations:

$\displaystyle \forall i \in \set {1, 2, \ldots, m}: \sum_{j \mathop = 1}^n \rho_{i j} x_j = \sigma_i$

whose augmented matrix is $\begin {pmatrix} \mathbf R & \mathbf s \end {pmatrix}$.

By Corollary 1, $S$ and $S'$ are equivalent.

Hence any every solution to $S'$ is also a solution to $S$.


Consider the structure of $\begin {pmatrix} \mathbf R & \mathbf s \end {pmatrix}$.

Suppose the leading coefficients appear in columns which we name $j_1, j_2, \ldots, j_l$.

Let the remaining columns be named $j_{l + 1}, j_{l + 2}, \ldots, j_n$.

Then we have that $S'$ can be expressed as:

\(\displaystyle x_{j_1} + \sum_{k \mathop = l + 1}^n \rho_{i j_k} x_{j_k}\) \(=\) \(\displaystyle \sigma_1\)
\(\displaystyle x_{j_2} + \sum_{k \mathop = l + 1}^n \rho_{2 j_k} x_{j_k}\) \(=\) \(\displaystyle \sigma_2\)
\(\displaystyle \) \(\cdots\) \(\displaystyle \)
\(\displaystyle x_{j_l} + \sum_{k \mathop = l + 1}^n \rho_{2 j_k} x_{j_k}\) \(=\) \(\displaystyle \sigma_l\)
\(\displaystyle 0\) \(=\) \(\displaystyle \sigma_{l + 1}\)

where $l + 1 \le m$.


We have that $\begin {pmatrix} \mathbf R & \mathbf s \end {pmatrix}$ is in reduced echelon form.

By definition of reduced echelon form, it follows that if $\sigma_{l + 1} \ne 0$ then $\sigma_{l + 1} = 1$.

Then when $\sigma_{l + 1} = 1$, $\sigma_{l + 1}$ is the leading coefficient of row $l + 1$ of $\begin {pmatrix} \mathbf R & \mathbf s \end {pmatrix}$.

This leading coefficient is seen to be in the rightmost column.


Sufficient Condition

Let the rightmost column of $\begin {pmatrix} \mathbf R & \mathbf s \end {pmatrix}$ have no leading coefficient.

That is:

$\sigma_{l + 1} = 0$

Let us set:

$(1): \quad \forall i \in \set {1, 2, \ldots, m}: x_{j_i} = \begin {cases} \sigma i & : i \le l \\0 & : i > l \end {cases}$

Then we have that $S'$ reduces to:

\(\displaystyle x_{j_1}\) \(=\) \(\displaystyle \sigma_1\)
\(\displaystyle x_{j_2}\) \(=\) \(\displaystyle \sigma_2\)
\(\displaystyle \) \(\cdots\) \(\displaystyle \)
\(\displaystyle x_{j_l}\) \(=\) \(\displaystyle \sigma_l\)

and it can be seen by inspection that $(1)$ is a solution to $S$.


Necessary Condition

Let $S$ be consistent.

Aiming for a contradiction, suppose let the rightmost column of $\begin {pmatrix} \mathbf R & \mathbf s \end {pmatrix}$ have no leading coefficient.

That is:

$\sigma_{l + 1} = 1$

Thus row $l + 1$ looks like:

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

which corresponds to the equation:

$0 x_1 + 0 x_2 + \cdots + 0 x_n = 1$

That is:

$0 = 1$

Hence there is no such $\tuple {x_1, x_2, \ldots, x_n}$ that is a solution to $S'$.

As $S$ and $S'$ are equivalent, there can likewise be no solution to $S$.

That is, $S$ is not consistent.

But this contradicts our initial supposition that $S$ is consistent.

Hence, by Proof by Contradiction, it follows that there can exist no leading coefficient in the rightmost column of $\begin {pmatrix} \mathbf R & \mathbf s \end {pmatrix}$

$\blacksquare$


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