Simultaneous Linear Equations/Examples/Arbitrary System 7

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Example of Simultaneous Linear Equations

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

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


$S$ has as its solution set:

\(\ds 2 x + z\) \(=\) \(\ds 3\)
\(\ds 4 y + z\) \(=\) \(\ds 1\)

where $z$ can be any number.


Proof

We express $S$ in matrix representation:

$\begin {pmatrix} 1 & 2 & 1 \\ -1 & 2 & 0 \\ 5 & -2 & 2 \end {pmatrix} \begin {pmatrix} x \\ y \\ z \end {pmatrix} = \begin {pmatrix} 2 \\ -1 \\ 7 \end {pmatrix}$

and consider the augmented matrix:

$\begin {pmatrix} \mathbf A & \mathbf b \end {pmatrix} = \paren {\begin {array} {ccc|c} 1 & 2 & 1 & 2 \\ -1 & 2 & 0 & -1 \\ 5 & -2 & 2 & 7 \end {array} }$


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


$e_1 := r_2 \to r_2 + r_1$

$e_2 := r_3 \to r_3 - 5 r_1$

Hence:

$\begin {pmatrix} \mathbf A_2 & \mathbf b_2 \end {pmatrix} = \paren {\begin {array} {ccc|c}

1 & 2 & 1 & 2 \\ 0 & 4 & 1 & 1 \\ 0 & -12 & -3 & -3 \\ \end {array} }$


$e_3 := r_3 \to r_3 + 3 r_2$

Hence:

$\begin {pmatrix} \mathbf A_3 & \mathbf b_3 \end {pmatrix} = \paren {\begin {array} {ccc|c}

1 & 2 & 1 & 2 \\ 0 & 4 & 1 & 1 \\ 0 & 0 & 0 & 0 \\ \end {array} }$


$e_4 := r_2 \to \dfrac 1 4 r_2$

Hence:

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

1 & 2 & 1 & 2 \\ 0 & 1 & 1/4 & 1/4 \\ 0 & 0 & 0 & 0 \\ \end {array} }$


$e_5 := r_1 \to r_1 - 2 r_2$

Hence:

$\begin {pmatrix} \mathbf A_5 & \mathbf b_5 \end {pmatrix} = \paren {\begin {array} {ccc|c}

1 & 0 & 1/2 & 3/2 \\ 0 & 1 & 1/4 & 1/4 \\ 

0 & 0 &  0 &  0 \\

\end {array} }$


$e_6 := r_1 \to 2 r_1$

$e_7 := r_5 \to 4 r_2$

Hence:

$\begin {pmatrix} \mathbf A_7 & \mathbf b_7 \end {pmatrix} = \paren {\begin {array} {ccc|c}

2 & 0 & 1 & 3 \\ 0 & 4 & 1 & 1 \\ 0 & 0 & 0 & 0 \\ \end {array} }$


Thus $z$ is not restricted, and can be anything, leaving us with:

\(\ds 2 x + z\) \(=\) \(\ds 3\)
\(\ds 4 y + z\) \(=\) \(\ds 1\)

as the solution set:

$\blacksquare$


Sources

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