Editing Cramer's rule (section)

==Proof==
The proof for Cramer's rule uses the following [[Determinant#Properties_of_the_determinant|properties of the determinants]]: linearity with respect to any given column and the fact that the determinant is zero whenever two columns are equal, which is implied by the property that the sign of the determinant flips if you switch two columns.

Fix the index {{math|''j''}} of a column, and consider that the entries of the other columns have fixed values. This makes the determinant a function of the entries of the {{mvar|j}}th column. Linearity with respect to this column means that this function has the form 
:<math>D_j(a_{1,j}, \ldots, a_{n,j})= C_{1,j}a_{1,j}+\cdots, C_{n,j}a_{n,j},</math>
where the <math>C_{i,j}</math> are coefficients that depend on the entries of {{mvar|A}} that are not in column {{mvar|j}}. So, one has 
:<math>\det(A)=D_j(a_{1,j}, \ldots, a_{n,j})=C_{1,j}a_{1,j}+\cdots, C_{n,j}a_{n,j}</math>
([[Laplace expansion]] provides a formula for computing the <math>C_{i,j}</math> but their expression is not important here.) 

If the function <math>D_j</math> is applied to any ''other'' column {{math|''k''}} of {{mvar|A}}, then the result is the determinant of the matrix obtained from {{mvar|A}} by replacing column {{math|''j''}} by a copy of column {{math|''k''}}, so the resulting determinant is 0 (the case of two equal columns).

Now consider a system of {{mvar|n}} linear equations in {{mvar|n}} unknowns <math>x_1, \ldots,x_n</math>, whose coefficient matrix is {{mvar|A}}, with det(''A'') assumed to be nonzero:

:<math>\begin{matrix}
a_{11}x_1+a_{12}x_2+\cdots+a_{1n}x_n&=&b_1\\
a_{21}x_1+a_{22}x_2+\cdots+a_{2n}x_n&=&b_2\\
&\vdots&\\
a_{n1}x_1+a_{n2}x_2+\cdots+a_{nn}x_n&=&b_n.
\end{matrix}</math>

If one combines these equations by taking {{math|''C''<sub>1,''j''</sub>}} times the first equation, plus {{math|''C''<sub>2,''j''</sub>}} times the second, and so forth until {{math|''C''<sub>''n'',''j''</sub>}} times the last, then for every {{mvar|k}} the resulting coefficient of {{mvar|x<sub>k</sub>}} becomes
:<math>D_j(a_{1,k},\ldots,a_{n,k}).</math>
So, all coefficients become zero, except the coefficient of <math>x_j</math> that becomes <math>\det(A).</math> Similarly, the constant coefficient becomes <math>D_j(b_1,\ldots,b_n),</math> and the resulting equation is thus
:<math>\det(A)x_j=D_j(b_1,\ldots, b_n),</math>
which gives the value of <math>x_j</math> as
:<math>x_j=\frac1{\det(A)}D_j(b_1,\ldots, b_n).</math>

As, by construction, the numerator is the determinant of the matrix obtained from {{mvar|A}} by replacing column {{math|''j''}} by {{math|'''b'''}}, we get the expression of Cramer's rule as a necessary condition for a solution. 

It remains to prove that these values for the unknowns form a solution. Let {{mvar|M}} be the {{math|''n'' × ''n''}} matrix that has the coefficients of <math>D_j</math> as {{mvar|j}}th row, for <math>j=1,\ldots,n</math> (this is the [[adjugate matrix]] for {{mvar|A}}). Expressed in matrix terms, we have thus to prove that 
:<math>\mathbf x = \frac1{\det(A)}M\mathbf b</math>
is a solution; that is, that 
:<math>A\left(\frac1{\det(A)}M\right)\mathbf b=\mathbf b.</math>
For that, it suffices to prove that 
:<math>A\,\left(\frac1{\det(A)}M\right)=I_n,</math>
where <math>I_n</math> is the [[identity matrix]].

The above properties of the functions <math>D_j</math> show that one has {{math|''MA'' {{=}} det(''A'')''I<sub>n</sub>''}}, and therefore,
:<math>\left(\frac1{\det(A)}M\right)\,A=I_n.</math>
This completes the proof, since a [[inverse element |left inverse]] of a square matrix is also a right-inverse (see [[Invertible matrix theorem]]).

For other proofs, see [[#Other proofs|below]].