# proof of Cramer’s rule

Since $\det(A)\neq 0$, by properties of the determinant we know that $A$ is invertible.

We claim that this implies that the equation $Ax=b$ has a unique solution. Note that $A^{-1}b$ is a solution since $A(A^{-1}b)=(AA^{-1})b=b$, so we know that a solution exists.

Let $s$ be an arbitrary solution to the equation, so $As=b$. But then $s=(A^{-1}A)s=A^{-1}(As)=A^{-1}b$, so we see that $A^{-1}b$ is the only solution.

For each integer $i$, $1\leq i\leq n$, let $a_{i}$ denote the $i$th column of $A$, let $e_{i}$ denote the $i$th column of the identity matrix $I_{n}$, and let $X_{i}$ denote the matrix obtained from $I_{n}$ by replacing column $i$ with the column vector $x$.

We know that for any matrices $A,B$ that the $k$th column of the product $AB$ is simply the product of $A$ and the $k$th column of $B$. Also observe that $Ae_{k}=a_{k}$ for $k=1,\ldots,n$. Thus, by multiplication, we have:

 $\begin{array}[]{lll}AX_{i}&=&A(e_{1},\ldots,e_{i-1},x,e_{i+1},\ldots,e_{n})\\ &=&(Ae_{1},\ldots,Ae_{i-1},Ax,Ae_{i+1},\ldots,Ae_{n})\\ &=&(a_{1},\ldots,a_{i-1},b,a_{i+1},\ldots,a_{n})\\ &=&M_{i}\end{array}$

Since $X_{i}$ is $I_{n}$ with column $i$ replaced with $x$, computing the determinant of $X_{i}$ with cofactor expansion gives:

 $\det(X_{i})=(-1)^{(i+i)}x_{i}\det(I_{n-1})=1\cdot x_{i}\cdot 1=x_{i}$

Thus by the multiplicative property of the determinant,

 $\det(M_{i})=\det(AX_{i})=\det(A)\det(X_{i})=\det(A)x_{i}$

and so $x_{i}=\frac{\det(M_{i})}{\det(A)}$ as required.

Title proof of Cramer’s rule ProofOfCramersRule 2013-03-22 13:03:24 2013-03-22 13:03:24 rmilson (146) rmilson (146) 11 rmilson (146) Proof msc 15A15