In case of the example B first k columns of A  [i]linearly independent[/i] C start with id[sub]kxk[/sub].[br]making math shorter and formula more comprehensive:[br] [br](6)[br][math]Cij \, := \, \left(\begin{array}{rrr}1&0&c13\\0&1&c23\\\end{array}\right)[/math][br][br](7) B Cij - A [br][math]\to \left(\begin{array}{rrr}0&0&c13 + c23 - 2\\0&0&c13 + 2 \; c23 - 3\\0&0&c13 + 3 \; c23 - 4\\\end{array}\right)[/math][br][br](8) Flatten(Take(Transpose($7), 3, 3))[br][math]\to \left\{ c13 + c23 - 2, c13 + 2 \; c23 - 3, c13 + 3 \; c23 - 4 \right\} [/math][br][br](9) Solve($8)[br][math]\to \left\{ \left\{ c13 = 1, c23 = 1 \right\} \right\} [/math][br][br]