How do we derive Cramer's method?

1 Answer
Feb 26, 2018

Answer:

See the explanation below

Explanation:

Let us consider the set of equations #Ax=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.