# Simultaneous Linear Equations/Examples/Arbitrary System 6/Mistake

## Source Work

1998: Richard Kaye and Robert Wilson: Linear Algebra:

Part $\text I$: Matrices and vector spaces
$1$ Matrices
$1.5$ Row and column operations
Solving linear equations

## Mistake

To solve:

$\begin {array} {rcrcrcr} x & + & y & + & 2 z & = & -1 \\ -x & + & & & z & = & -1 \\ -x & + & y & + & 4 z & = & 3 \\ \end {array}$,

first put the equation in matrix form

$\paren {\begin {array} {rrr} 1 & 1 & 2 \\ -1 & 0 & 1 \\ -1 & 1 & 4 \end {array} } \begin {pmatrix} x \\ y \\ z \end {pmatrix} = \paren {\begin {array} {r} -1 \\ -1 \\ 3 \end {array} }$

and then put the augmented matrix formed from the matrix on the left with the column vector on the right into echelon form:

$\paren {\begin {array} {rrr|r} 1 & 1 & 2 & -1 \\ -1 & 0 & 1 & -1 \\ -1 & 1 & 4 & 3 \end {array} } \to \paren {\begin {array} {rrr|r} 1 & 1 & 2 & -1 \\ 0 & 1 & 3 & -2 \\ 0 & 2 & 6 & -4 \end {array} } \to \paren {\begin {array} {rrr|r} 1 & 1 & 2 & -1 \\ 0 & 1 & 3 & -2 \\ 0 & 0 & 0 & 0 \end {array} }$.

## Correction

That first transformation should be:

$\paren {\begin {array} {rrr|r} 1 & 1 & 2 & -1 \\ -1 & 0 & 1 & -1 \\ -1 & 1 & 4 & 3 \end {array} } \to \paren {\begin {array} {rrr|r} 1 & 1 & 2 & -1 \\ 0 & 1 & 3 & -2 \\ 0 & 2 & 6 & 2 \end {array} }$

$\paren {\begin {array} {rrr|r} 1 & 1 & 2 & -1 \\ 0 & 1 & 3 & -2 \\ 0 & 0 & 0 & 6 \end {array} }$

Hence from the resulting $0 = 6$ it is seen that the initial system of simultaneous linear equations has no solutions.

The original system of simultaneous linear equations can be amended to:

$\begin {array} {rcrcrcr} x & + & y & + & 2 z & = & -1 \\ -x & + & & & z & = & -1 \\ -x & + & y & + & 4 z & = & -3 \\ \end {array}$

from which the solution is derived in Simultaneous Linear Equations: Arbitrary System $6$.

$\blacksquare$