The required elementary matrix is
Left multiplication exchanges the two rows. Applying the same exchange twice restores every row, so .
The elementary row operation and its reverse are
Thus replaces row three by row three minus three times row two. The inverse adds those three copies back.
Use multiplicativity of the determinant and the determinant criterion for an invertible matrix:
Neither factor can therefore be zero, so both matrices are nonsingular. The determinant of a triangular matrix is the product of its diagonal entries:
Each product is nonzero, forcing every diagonal factor to be nonzero. These explicitly stated determinant properties justify both conclusions.
First subtract row one from row two, using ; then subtract four times the updated row two from row three, using :
The updated second row is and the final third row is . Hence and . The resulting LU decomposition is
Reversing the order when undoing the elementary operations is essential.
Introduce . Solve by forward substitution and then by back substitution. For these factors, the recurrences are
followed by
The nonzero diagonal entries make both triangular solves well-defined. No matrix inverse is calculated.
With column-vector homogeneous coordinates, the translation matrix is
It maps to . The final coordinate incorporates the constant displacement into matrix multiplication.
Reflection in sends to . Therefore its homogeneous-coordinate matrix is
Its square is the identity, as required for applying the same reflection twice.
Translate the mirror line to zero, reflect there, then translate back. Under the column-vector convention the rightmost operation acts first:
This composition of affine transformations maps to , fixing exactly the points with .

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