Cramer's rule

Let $ C_{ij}$ be the $ (i,j)$-cofactor of $ A$. Then we define the adjugate `` adj$ A$'' of $ A$ by

   adj$\displaystyle ~A
= \begin{bmatrix}
C_{11} & C_{21} & \cdots & C_{n1} \\
C_{12}...
...
\vdots & \vdots & & \vdots \\
C_{1n} & C_{2n} & \cdots & C_{nn}
\end{bmatrix}$

By applying the Laplace expansions and the property (e) of determinants (that is, $ \det [ \mathbf{a}_1 
\ldots \mathbf{a}_k \ldots \mathbf{a}_{k} 
\ldots \mathbf{a}_n]
= 0$), we can find that

$\displaystyle ($adj$\displaystyle ~A) \begin{bmatrix}a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & ...
...dots & 0 \\ \vdots & \vdots & & \vdots \\ 0 & 0 & \cdots & \det A \end{bmatrix}$ (7)

If $ A$ is invertible (that is, $ \det A \neq 0$), the equation above immediately implies that

$\displaystyle A^{-1} = \frac{1}{\det A}$   adj$\displaystyle ~A.$ (8)

By using the equation above we can express the solution to the matrix equation  $ A \mathbf{x} = \mathbf{b}$ by

$\displaystyle \mathbf{x}
= \frac{1}{\det A}
\begin{bmatrix}
C_{11} & C_{21} & \...
...s \\
\det [\,\mathbf{a}_1\,\mathbf{a}_2\, \ldots \,\mathbf{b}\,]
\end{bmatrix}$

Let $ A_i(\mathbf{b})$ be the matrix produced by replacing the $ i$th column by $ \mathbf{b}$. Then the Cramer's rule gives the solution  $ \mathbf{x} = A^{-1} \mathbf{b}$ entry-wise in the following form:

$\displaystyle x_i = \frac{\det A_i(\mathbf{b})}{\det A}$    for $ i=1,\ldots,n$.



Department of Mathematics
Last modified: 2005-10-21