.In the following I deviate from standard matrix element notation because it is hard to differentiate terms of this style.

.⌊a b c⌋ ⌈d e f⌉ g h i⌊X1 ⌋ ⌈X2 ⌉ X3 =⌊ k1⌋ ⌈ k2⌉ k3

.

.We will use Gaussian elimination to get an equation in the form:

ax = k

.

multiply: top row by f and center row by c, subtract center row from top row

⌊ af - cd bf - ce 0⌋ ⌈ d e f⌉ g h i⌊X ⌋ ⌈X 1⌉ X 2 3 =⌊k f - ck ⌋ ⌈ 1 k 2⌉ k2 3

.

.multiply: center row by i and bottom row by f, subtract bottom row from center row

⌊ ⌋ (af - cd) (bf - ce) 0 ⌈ (di- fg) (ei- fh) 0⌉ g h i⌊ ⌋ X1 ⌈X2⌉ X3 =⌊ ⌋ (k1f - ck2) ⌈(k2i- fk3)⌉ k3

.

.multiply: top row by (ei-fh) and center row by (bf-ce) , subtract center row from top row]

.⌊(af - cd)(ei- fh) (bf - ce)(ei- fh) 0⌋ ⌈(di- fg)(bf - ce) (ei- fh)(bf - ce) 0⌉ g h i⌊ X1⌋ ⌈ X2⌉ X3 =⌊ (k1f - ck2)(ei- fh)⌋ ⌈ (k2i- f k3)(bf - ce)⌉ k3

.

.⌊(af - cd)(ei- fh)- (di- fg)(bf - ce) 0 0⌋ ⌈ (di- fg)(bf - ce) (bf - ce)(ei- fh) 0⌉ g h i⌊X ⌋ ⌈X 1⌉ X 2 3 =⌊(k f - ck )(ei- fh) - (k i- fk )(bf - ce)⌋ ⌈ 1 2 (k i- fk )(b2f - ce)2 ⌉ 2 3k 3

.

.Now set up the equivalency of the first term.

.[(af -cd)(ei-fh)-(di-fg)(bf -ce)]X1 = (k1f -ck2)(ei-fh)-(k2i-fk3)(bf -ce)

.

.The terms in parenthesis are 2x2 determinants and for purposes of proof are more easily handled in that notation. The phrase below shows the phrase in this notation.

.[|| |||| || || |||| ||] ||a c||||e f||- ||d f||||b c|| d f h i g i e fX1=[|| |||| || || |||| ||] ||k1 c||||e f|| -||k2 f||||b c|| k2 f h i k3 i e f

.

.Now I will attempt to coerce terms into the form row sum ( element * cofactor )

.

.tNote the following useful relation

.| | ||a c|| |d f|| | ||e f|| |h i| = [| || | | || |] ||a 0||||e f|| ||0 c||||e f || |0 f||h i| +|d 0||h i| I will break all the terms apart into this form.

.

.[| || | | || | | || | | || |] ||a 0||||e f|| ||0 c||||e f || ||b 0||||d f|| ||0 c||||d f|| |0 f||h i|+ |d 0||h i|- |0 f||g i|- |e 0||g i|X1=[| || | | || |] ||k1 c||||e f|| ||b c||||k2 f|| |k2 f||h i|- |e f||k3 i|

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.The left side coefficient is what we are interested in:

.||a 0|| ||0 f||||e f|| ||h i|| + ||0 c|| ||d 0||||e f|| ||h i|| -||b 0|| ||0 f||||d f|| ||g i|| -||0 c|| ||e 0||||d f|| ||g i|| = ||a 0|| ||0 f||||e f|| ||h i|| -||b 0 || ||0 f ||||d f|| ||g i|| + ||0 c|| ||d 0||||e f || ||h i|| -||0 c|| ||e 0||||d f || ||g i||

.

. The a and b terms are in correct form that agre with the standard definition of the determinant with an extra factor of f. This factor of f cancels out since it is also on the other side of the equation. The wrinkle in the rug has been worked to the wall and now we only need to coerce the c term shown below into standard form. Below is the c term:

.

.| | ||0 c|| |d 0|| | ||e f|| |h i| -| | ||0 c|| |e 0|| | ||d f|| |g i| =| | ||0 c|| |d 0|| | ||0 f || |h 0| -| | ||0 c|| |e 0|| | ||0 f|| |g 0| .......you can see this is true by noting the common cdei term that results from the main diagonal term in both products

.

.Note that these are determinants and thus we can move terms around to neaten things up:

|| || ||0 c|| d 0|| || ||e f|| h i -|| || ||0 c|| e 0|| || ||d f|| g i =f *|| || ||0 c|| d 0|| || ||0 1|| h 0 - f *|| || ||0 c|| e 0|| || ||0 1|| g 0 =f * | | ||0 c|| |1 0|| | ||0 d|| |h 0| -f *| | ||0 c|| |1 0|| | ||0 e|| |g 0| =f *| | ||c 0|| |0 1|| | ||d 0|| |0 h| +f *| | ||c 0|| |0 1|| | ||0 e|| |g 0| =f *| | ||c 0|| |0 1|| | ||d e|| |g h|

.

.

.Each term has at least 1 common factor f ....this will occur on the right hand side of the equation also so it will be divided into both sides of the equation

.[||a 0||||e 0|| ||a 0||||0 f|| ||0 c||||0 1|| ||b 0||||d 0|| ||b 0||||0 f|| ||0 c||||0 1||� ||0 1||||0 i||+ ||0 1||||h 0||+ ||d 0||||h 0||- ||0 1||||0 i||- ||0 1||||g 0||- ||e 0||||g 0&#X1=1∕f[||k1 c||||e f|| ||b c||||k2 f||] ||k2 f||||h i||- ||e f||||k3 i||

.

.

.[|| |||| || || |||| || || |||| || || |||| ||] ||a 0||||e f||- ||b 0||||d f||+ ||0 c||||0 1||- ||0 c||||0 1|| 0 1 h i 0 1 g i d 0 h 0 e 0 g 0X1=1∕f[|| |||| || || |||| ||] ||k1 c||||e f||- ||b c||||k2 f|| k2 f h i e f k3 i

.

.a and b are formed up nicely.

.

.form up the c term by moving terms

.[| || | | || | | || | | || |] ||a 0||||e f||- ||b 0||||d f||+ ||0 c||||0 d||- ||0 c||||0 e|| |0 1||h i| |0 1||g i| |1 0||h 0| |1 0||g 0|X1=1∕f[| || | | || |] ||k1 c||||e f||- ||b c||||k2 f|| |k2 f||h i| |e f ||k3 i|

.

.Here we have arrived at the form of the solution for X1 and it is:

.

.[ ||e f|| ||d f|| ||d e||] a||h i||- b||g i||+ c||g h ||X1=1∕f[||k1 c||||e f || ||k2 f||] ||k2 f||||h i||- c||k3 i||=[ ||e f || ||k2 f|| ||k2 e||] k1||h i||- b||k3 i||+ c||k3 h|| .....right side is cramers rule again

.

.this matches the definition of the determinant plus the pesky factor of f which is on both sides of the equation so it cancels.