proof of Cramer’s rule

Since $\\det(A)\eq 0$ , 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 $A B$ 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.