Editing Cramer's rule (section)

==Other proofs==

===A proof by abstract linear algebra ===

This is a restatement of the proof above in abstract language.

Consider the map <math>\mathbf{x}=(x_1,\ldots, x_n) \mapsto  \frac{1}{\det A} \left(\det (A_1),\ldots, \det(A_n)\right),</math> where <math>A_i</math> is the matrix <math>A</math> with <math>\mathbf{x}</math> substituted in the <math>i</math>th column, as in Cramer's rule. Because of linearity of determinant in every column, this map is linear. Observe that it sends the <math>i</math>th column of <math>A</math> to the <math>i</math>th basis vector <math>\mathbf{e}_i=(0,\ldots, 1, \ldots, 0) </math> (with 1 in the <math>i</math>th place), because determinant of a matrix with a repeated column is 0. So we have a linear map which agrees with the inverse of <math>A</math> on the column space; hence it agrees with <math>A^{-1}</math> on the span of the column space. Since <math>A</math> is invertible, the column vectors span all of <math>\mathbb{R}^n</math>, so our map really is the inverse of <math>A</math>. Cramer's rule follows.

===A short proof===
A short proof of Cramer's rule <ref>{{cite journal | last = Robinson | first = Stephen M. | title = A Short Proof of Cramer's Rule | journal = Mathematics Magazine| volume = 43 | pages = 94–95 | year = 1970| issue = 2 | doi = 10.1080/0025570X.1970.11976018 }}</ref> can be given by noticing that <math>x_1</math> is the determinant of the matrix

:<math>X_1=\begin{bmatrix}
x_1 & 0 & 0 & \cdots & 0\\
x_2 & 1 & 0 & \cdots & 0\\
x_3 & 0 & 1 & \cdots & 0\\
\vdots & \vdots & \vdots & \ddots &\vdots \\
x_n & 0 & 0 & \cdots & 1
\end{bmatrix}</math>

On the other hand, assuming that our original matrix {{mvar|A}} is invertible, this matrix <math>X_1</math> has columns <math>A^{-1}\mathbf{b}, A^{-1}\mathbf{v}_2, \ldots, A^{-1}\mathbf{v}_n </math>, where <math>\mathbf{v}_n</math> is the ''n''-th column of the matrix {{mvar|A}}. Recall that the matrix <math>A_1</math> has columns <math>\mathbf{b}, \mathbf{v}_2, \ldots, \mathbf{v}_n </math>, and therefore <math>X_1=A^{-1}A_1</math>. Hence, by using that the determinant of the product of two matrices is the product of the determinants, we have

:<math> x_1= \det (X_1) = \det (A^{-1}) \det (A_1)= \frac{\det (A_1)}{\det (A)}.</math>

The proof for other <math>x_j</math> is similar.

===Using Geometric Algebra===
{{Main|Comparison of vector algebra and geometric algebra#Matrix related}}