Simultaneous Linear Equations/Examples/Arbitrary System 6/Mistake

Source Work

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

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

which leads to:


 * $\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 $6$.