Simultaneous Linear Equations/Examples/Arbitrary System 8

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Example of Simultaneous Linear Equations

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

\(\ds x - y - z\) \(=\) \(\ds 1\)
\(\ds 2 x - y\) \(=\) \(\ds 1\)
\(\ds 2 x + 2 z\) \(=\) \(\ds 1\)


$S$ is inconsistent and so has no solutions.


Proof

We express $S$ in matrix representation:

$\begin {pmatrix} 1 & -1 & -1 \\ 2 & -1 & 0 \\ 2 & 0 & 2 \end {pmatrix} \begin {pmatrix} x \\ y \\ z \end {pmatrix} = \begin {pmatrix} 1 \\ 1 \\ 1 \end {pmatrix}$

and consider the augmented matrix:

$\begin {pmatrix} \mathbf A & \mathbf b \end {pmatrix} = \paren {\begin {array} {ccc|c} 1 & -1 & -1 & 1 \\ 2 & -1 & 0 & 1 \\ 2 & 0 & 2 & 1 \end {array} }$


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 $\begin {pmatrix} \mathbf A_k & \mathbf b_k \end {pmatrix}$.


$e_1 := r_2 \to r_2 - 2 r_1$

$e_2 := r_3 \to r_3 - 2 r_1$

Hence:

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


$e_3 := r_1 \to r_1 + r_2$

$e_4 := r_3 \to r_3 - 2 r_2$

Hence:

$\begin {pmatrix} \mathbf A_4 & \mathbf b_4 \end {pmatrix} = \paren {\begin {array} {ccc|c} 1 & 0 & 1 & 0 \\ 0 & 1 & 2 & -1 \\ 0 & 0 & 0 & 1 \\ \end {array} }$


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