Linear Algebra/Cramer's Rule

Cramer's Rule

Let's try to solve the systems of linear equations:
a₁₁x₁+a₁₂x₂+a₁₃x₃+...+a_1nx_n=b₁
a₂₁x₁+a₂₂x₂+a₂₃x₃+...+a_2nx_n=b₂
a₃₁x₁+a₃₂x₂+a₃₃x₃+...+a_3nx_n=b₃
...
a_n1x₁+a_n2x₂+a_n3x₃+...+a_nnx_n=b_n

Consider the matrix

$\left[\begin{array}{ccccc}{a}_{11}& a_{12}& a_{13}& \dots & a_{1n}\\ a_{21}& a_{22}& a_{23}& \dots & a_{2n}\\ a_{31}& a_{23}& a_{33}& \dots & a_{3n}\\ ⋮& ⋮& ⋮& ⋮& ⋮\\ a_{n1}& a_{n3}& a_{n3}& \dots & a_{nn}\end{array}\right]$

to be denoted D.

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

Co(a_1j)a₁₁x₁+Co(a_1j)a₁₂x₂+Co(a_1j)a₁₃x₃+...+Co(a_1j)a_1nx_n+
Co(a_2j)a₂₁x₁+Co(a_2j)a₂₂x₂+Co(a_2j)a₂₃x₃+...+Co(a_2j)a_2nx_n+
Co(a_3j)a₃₁x₁+Co(a_3j)a₃₂x₂+Co(a_3j)a₃₃x₃+...+Co(a_3j)a_3nx_n+
+...+
Co(a_nj)a_n1x₁+Co(a_nj)a_n2x₂+Co(a_nj)a_n3x₃+...+Co(a_nj)a_nnx_n
=
Co(a_1j)b₁+Co(a_2j)b₂+Co(a_3j)b₃+...+Co(a_nj)b_n.

The left side cancels out except for Co(a_1j)a_1jx_j+Co(a_2j)a_2jx_j+Co(a_3j)a_3jx_j+...+Co(a_nj)a_njx_j

which is equal to $x_j\left[\begin{array}{ccccc}{a}_{11}& a_{12}& a_{13}& \dots & a_{1n}\\ a_{21}& a_{22}& a_{23}& \dots & a_{2n}\\ a_{31}& a_{23}& a_{33}& \dots & a_{3n}\\ ⋮& ⋮& ⋮& ⋮& ⋮\\ a_{n1}& a_{n3}& a_{n3}& \dots & a_{nn}\end{array}\right]=D$

amd the right side is equal to

$\left[\begin{array}{ccccccc}{a}_{11}& a_{12}& a_{13}& \dots & b_1& \dots & a_{1n}\\ a_{21}& a_{22}& a_{23}& \dots & b_2& \dots & a_{2n}\\ a_{31}& a_{23}& a_{33}& \dots & b_3& \dots & a_{3n}\\ ⋮& ⋮& ⋮& ⋮& ⋮& ⋮& ⋮\\ a_{n1}& a_{n3}& a_{n3}& \dots & b_n& \dots & a_{nn}\end{array}\right]$, to be denoted D(j), which is the same thing as D but with the jth column replaced by b_k.

Dividing by D gets x_j=$\frac{D_j}{D}$.

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

Example

Consider the system of linear equations below.
$3x_1+2x_2-5x_3=15$
$5x_1+x_3=23$
$x_2+x_3=12$

If we only want the solution for, say, $x_2$, we can apply Cramer's Rule to find that its solution is $\frac{D_2}{D}$, and since we know
$D_2=\left[\begin{array}{ccc}3& 15& -5\\ 5& 23& 1\\ 0& 12& 1\end{array}\right]$,
$x_2=\frac{\det \left[\begin{array}{ccc}3& 15& -5\\ 5& 23& 1\\ 0& 12& 1\end{array}\right]}{\det \left[\begin{array}{ccc}3& 2& -5\\ 5& 0& 1\\ 0& 1& 1\end{array}\right]}=9$.