Let us consider the set of equations Ax=bAx=b written out element by element :
a_11 x_1+a_12 x_2 + ... + a_{1n} x_n=b_1
a_21 x_1+a_22 x_2 + ... + a_{2n} x_n=b_2
...
a_{n1} x_1+a_{n2} x_2 + ... + a_{n n} x_n=b_n
If we want to find x_1, say, we can rewrite this in the form
(a_11 x_1-b_1)+a_12 x_2 + ... + a_{1n} x_n=0
(a_21 x_1-b_2)+a_22 x_2 + ... + a_{2n} x_n=0
...
(a_{n1} x_1-b_n)+a_{n2} x_2 + ... + a_{n n} x_n=0
This can be written in the form A prime x prime =0 where
A prime = ((a_{11}x_1 -b_1, a_12,...,a_{1n}),(a_{21}x_1 -b_2, a_22,...,a_{2n}),(...,....,....,...),(a_{n1}x_1 -b_n, a_{n2},...,a_{n n})), qquad x prime = ((1),(x_2),(...),(x_n))
Now the set of equations A prime x prime =0 has a non trivial solution (the first element of x prime is 1 - so clearly it is non-zero), and thus det A prime = 0 (otherwise Aprime would have had an inverse, forcing xprime = (Aprime)^-1 0 to be the null vector.
So
0 = |(a_{11}x_1 -b_1, a_12,...,a_{1n}),(a_{21}x_1 -b_2, a_22,...,a_{2n}),(...,....,....,...),(a_{n1}x_1 -b_n, a_{n2},...,a_{n n})|
= |(a_{11}x_1 , a_12,...,a_{1n}),(a_{21}x_1 , a_22,...,a_{2n}),(...,....,....,...),(a_{n1}x_1 , a_{n2},...,a_{n n})| -|(b_1, a_12,...,a_{1n}),(b_2, a_22,...,a_{2n}),(...,....,....,...),(b_n, a_{n2},...,a_{n n})|
= x_1|(a_{11} , a_12,...,a_{1n}),(a_{21} , a_22,...,a_{2n}),(...,....,....,...),(a_{n1} , a_{n2},...,a_{n n})| -|(b_1, a_12,...,a_{1n}),(b_2, a_22,...,a_{2n}),(...,....,....,...),(b_n, a_{n2},...,a_{n n})|
and so, if det A ne 0 we can write
x_1 = |(b_1, a_12,...,a_{1n}),(b_2, a_22,...,a_{2n}),(...,....,....,...),(b_n, a_{n2},...,a_{n n})| /|(a_{11} , a_12,...,a_{1n}),(a_{21} , a_22,...,a_{2n}),(...,....,....,...),(a_{n1} , a_{n2},...,a_{n n})|
We can similarly derive the expression for all the unknown variables.
Note: in the derivation above we have used two elementary properties of determinants.
