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According to Shilov's Linear algebra (page no 7) while solving two linear equation,

$a_{11}x_1+a_{12}x_2=b_1...(I) \\ a_{21}x_1+a_{22}x_2=b_2...(II)$

solving in the usual way we get

$x_1=\Large {\frac{b_{1}a_{22}-b_{2}a_{12}}{a_{11}a_{22}-a_{21}a_{12}} }$ and $x_2=\Large{ \frac{b_{2}a_{11}-b_{1}a_{21}}{a_{11}a_{22}-a_{21}a_{12}} }$,

when I solved the two equations (I) and (II), my solutions for $x_1$ matches with the text (as shown above) but for $x_2$ I get following

$\color{Red}{x_2=\Large{ \frac{b_{1}a_{21}-b_{2}a_{11}}{a_{21}a_{12}-a_{11}a_{22}}} }$ {multiplying eq (II) by $a_{11}/a_{21}$ and then subtracting (II) from (I)}

when I used one of the online tools to solve these two equations with values $a_{11}=2, a_{12}=3, b_1=4, a_{21}=5, a_{22}=6, b_2=7$ answer was $x_1=-1, x_2=2$ which matched with the answer from my version of the formula, what is wrong here?

Also, on second thought I feel "my" solution for $x_2$ should be wrong, as the quantity $a_{11}a_{22}$ gets a minus sign which according the "Inversion number" method is NOT correct (the principal diagonal is one of the permutations of $1,2,...,n$ when all the rows/numbers are in ascending order, so the inversion number is zero, $(-1)^0=+1$ )

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The numerator and denominator are both off by negative one. You can convert the formula to it's standard form by multiplying by $1 = \frac{-1}{-1}$. –  EuYu Oct 27 '12 at 4:39
    
+1 for giving context and showing what you tried. –  Noah Snyder Oct 27 '12 at 9:27
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1 Answer

up vote 3 down vote accepted

The two expressions, yours and Shilov's, are equivalent. Multiply top and bottom by $-1$.

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