# Linear Algebra/Cramer's Rule

## Cramer's Rule

Let's try to solve the systems of linear equations:
a11x1+a12x2+a13x3+...+a1nxn=b1
a21x1+a22x2+a23x3+...+a2nxn=b2
a31x1+a32x2+a33x3+...+a3nxn=b3
...
an1x1+an2x2+an3x3+...+annxn=bn

Which is the special case when the number of equations and the number of variables are the same.

Consider the matrix

${\displaystyle {\begin{bmatrix}a_{11}&a_{12}&a_{13}&\ldots &a_{1n}\\a_{21}&a_{22}&a_{23}&\ldots &a_{2n}\\a_{31}&a_{32}&a_{33}&\ldots &a_{3n}\\\vdots &\vdots &\vdots &\vdots &\vdots \\a_{n1}&a_{n2}&a_{n3}&\ldots &a_{nn}\\\end{bmatrix}}}$

to be denoted D.

First, we multiply the nth equation by the cofactor Co(anj) for the jth column, and add it all up. This gets us

Co(a1j)a11x1+Co(a1j)a12x2+Co(a1j)a13x3+...+Co(a1j)a1nxn+
Co(a2j)a21x1+Co(a2j)a22x2+Co(a2j)a23x3+...+Co(a2j)a2nxn+
Co(a3j)a31x1+Co(a3j)a32x2+Co(a3j)a33x3+...+Co(a3j)a3nxn+
+...+
Co(anj)an1x1+Co(anj)an2x2+Co(anj)an3x3+...+Co(anj)annxn
=
Co(a1j)b1+Co(a2j)b2+Co(a3j)b3+...+Co(anj)bn.

The left side cancels out except for Co(a1j)a1jxj+Co(a2j)a2jxj+Co(a3j)a3jxj+...+Co(anj)anjxj

which is equal to ${\displaystyle x_{j}{\begin{bmatrix}a_{11}&a_{12}&a_{13}&\ldots &a_{1n}\\a_{21}&a_{22}&a_{23}&\ldots &a_{2n}\\a_{31}&a_{32}&a_{33}&\ldots &a_{3n}\\\vdots &\vdots &\vdots &\vdots &\vdots \\a_{n1}&a_{n2}&a_{n3}&\ldots &a_{nn}\\\end{bmatrix}}=D}$

and the right side is equal to

${\displaystyle {\begin{bmatrix}a_{11}&a_{12}&a_{13}&\ldots &b_{1}&\ldots &a_{1n}\\a_{21}&a_{22}&a_{23}&\ldots &b_{2}&\ldots &a_{2n}\\a_{31}&a_{32}&a_{33}&\ldots &b_{3}&\ldots &a_{3n}\\\vdots &\vdots &\vdots &\vdots &\vdots &\vdots &\vdots \\a_{n1}&a_{n2}&a_{n3}&\ldots &b_{n}&\ldots &a_{nn}\\\end{bmatrix}}}$, to be denoted D(j), which is the same thing as D but with the jth column replaced by bk.

Dividing by D gets xj=${\displaystyle {\frac {D_{j}}{D}}}$.

This formula is called Cramer's Rule, and this solution exists when D is not equal to 0.

In particular, in the process of finding the solution, we also find that this is the only solution, so this solution is unique.

### Example

Consider the system of linear equations below.
${\displaystyle 3x_{1}+2x_{2}-5x_{3}=15}$
${\displaystyle 5x_{1}+x_{3}=23}$
${\displaystyle x_{2}+x_{3}=12}$

If we only want the solution for, say, ${\displaystyle x_{2}}$, we can apply Cramer's Rule to find that its solution is ${\displaystyle {\frac {D_{2}}{D}}}$, and since we know
${\displaystyle D_{2}={\begin{bmatrix}3&15&-5\\5&23&1\\0&12&1\end{bmatrix}}}$,
${\displaystyle x_{2}={\frac {\det {\begin{bmatrix}3&15&-5\\5&23&1\\0&12&1\end{bmatrix}}}{\det {\begin{bmatrix}3&2&-5\\5&0&1\\0&1&1\end{bmatrix}}}}=9}$.