Adjugate matrix and Cramer's rule/Introduction/Section

Definition

For a square matrix ${}M\in \operatorname {Mat} _{n}(K)$ , we call

M^{\operatorname {adj} }=(b_{ij}){\text{ with }}b_{ij}=(-1)^{i+j}\det M_{ji}

the adjugate matrix of

{}M

, where

{}M_{ji}

arises from

{}M

by deleting the

{}j

-th row and the

{}i

-th column.

Note that in this definition, for the entries of the adjugate, the rows and the columns are swapped.

Theorem

Let ${}K$ be a field, and let ${}M$ denote an ${}n\times n$ -matrix over ${}K$ . Then

{}(M^{\operatorname {adj} })\cdot M=(\det M)E_{n}=M\cdot (M^{\operatorname {adj} })\,

If

{}M

is

invertible, then

{}M^{-1}={\frac {1}{\det M}}\cdot M^{\operatorname {adj} }\,.

Proof

Let ${}M={\left(a_{ij}\right)}_{ij}$ . Let the coefficients of the adjugate matrix be denoted by

{}b_{ik}=(-1)^{i+k}\det M_{ki}\,.

The coefficients of the product ${}(M^{\operatorname {adj} })\cdot M$ are

{}c_{ij}=\sum _{k=1}^{n}b_{ik}a_{kj}=\sum _{k=1}^{n}(-1)^{i+k}a_{kj}\det M_{ki}\,.

In case ${}j=i$ , this is ${}\det M$ , as this sum is the expansion of the determinant with respect to the ${}j$ -th column. So let ${}j\neq i$ , and let ${}N$ denote the matrix that arises from ${}M$ by replacing in ${}M$ the ${}i$ -th column by the ${}j$ -th column. If we expand ${}N$ with respect to the ${}i$ -th column, then we get

{}0=\det N=\sum _{k=1}^{n}(-1)^{i+k}a_{kj}\det M_{ki}=c_{ij}\,.

Therefore, these coefficients are ${}0$ , and the first equation holds.
The second equation is proved similarly, where we use now the expansion of the determinant with respect to the rows.

\Box

The following statement is called Cramer's rule.

Theorem

Let ${}K$ be a field, and let

{\begin{matrix}a_{11}x_{1}+a_{12}x_{2}+\cdots +a_{1n}x_{n}&=&c_{1}\\a_{21}x_{1}+a_{22}x_{2}+\cdots +a_{2n}x_{n}&=&c_{2}\\\vdots &\vdots &\vdots \\a_{n1}x_{1}+a_{n2}x_{2}+\cdots +a_{nn}x_{n}&=&c_{n}\end{matrix}}

be an inhomogeneous linear system over ${}K$ . Suppose that the describing matrix ${}M={\left(a_{ij}\right)}_{ij}$ is invertible. Then the unique solution for ${}x_{j}$ is given by

{}x_{j}={\frac {\det {\begin{pmatrix}a_{11}&\cdots &a_{1,j-1}&c_{1}&a_{1,j+1}&\cdots &a_{1n}\\\vdots &\ddots &\vdots &\vdots &\vdots &\ddots &\vdots \\a_{n1}&\cdots &a_{n,j-1}&c_{n}&a_{n,j+1}&\cdots &a_{nn}\end{pmatrix}}}{\det M}}

.

Proof

For an invertible matrix ${}M$ , the solution of the linear system ${}Mx=c$ can be found by applying ${}M^{-1}$ , that is, ${}x=M^{-1}c$ . Using fact, this means ${}x={\frac {1}{\det M}}(M^{\operatorname {adj} })c$ . For the ${}j$ -th component, this means