Simultaneous Linear Equations/Examples/Arbitrary System 12

From ProofWiki
Jump to navigation Jump to search

Example of Simultaneous Linear Equations

Let $S$ denote the system of simultaneous linear equations:

\(\ds x + 2 y + 3 z + w\) \(=\) \(\ds -1\)
\(\ds -x + y + x - w\) \(=\) \(\ds 2\)
\(\ds x + 5 y + 7 z + w\) \(=\) \(\ds 1\)


$S$ is inconsistent and so has no solutions.


Proof

We express $S$ in matrix representation:

$\begin {pmatrix} 1 & 2 & 3 & 1 \\ -1 & 1 & 1 & -1 \\ 1 & 5 & 7 & 1 \end {pmatrix} \begin {pmatrix} x \\ y \\ z \\ w \end {pmatrix} = \begin {pmatrix} -1 \\ 2 \\ 1 \end {pmatrix}$

and consider the augmented matrix:

$\begin {pmatrix} \mathbf A & \mathbf b \end {pmatrix} = \paren {\begin {array} {cccc|c} 1 & 2 & 3 & 1 & -1 \\ -1 & 1 & 1 & -1 & 2 \\ 1 & 5 & 7 & 1 & 1 \end {array} }$


In the following, $\sequence {e_n}_{n \mathop \ge 1}$ denotes the sequence of elementary row operations that are to be applied to $\begin {pmatrix} \mathbf A & \mathbf b \end {pmatrix}$.

The matrix that results from having applied $e_1$ to $e_k$ in order is denoted $\begin {pmatrix} \mathbf A_k & \mathbf b_k \end {pmatrix}$.


$e_1 := r_2 \to r_2 + r_1$

$e_2 := r_3 \to r_3 - r_1$

Hence:

$\begin {pmatrix} \mathbf A_2 & \mathbf b_2 \end {pmatrix} = \paren {\begin {array} {cccc|c}
1 & 2 & 3 & 1 & -1 \\
0 & 3 & 4 & 0 &  1 \\
0 & 3 & 4 & 0 &  2

\end {array} }$


$e_3 := r_3 \to r_3 - r_2$

Hence:

$\begin {pmatrix} \mathbf A_3 & \mathbf b_3 \end {pmatrix} = \paren {\begin {array} {cccc|c}
1 & 2 & 3 & 1 & -1 \\
0 & 3 & 4 & 0 &  1 \\
0 & 0 & 0 & 0 &  1

\end {array} }$


The bottom line of this augmented matrix leads to the false statement $0 = 1$.

It follows that this system of simultaneous linear equations is inconsistent.

$\blacksquare$


Sources