**Cayley-Hamilton theorem** states that *every square matrix satisfies its own characteristic equation*. This theorem is named after two mathematicians, Arthur **Cayley** & William Rowan **Hamilton**. This theorem provides an alternative way to find the inverse of a matrix.

Let

a_{o}λ^{n} + a_{1}λ^{n-1} + a_{2}λ^{n-2} + ………………. + a_{n-2}λ^{2} + a_{n-1}λ + a_{n} = 0

be the characteristic equation of square matrix P of order n. Then, according to the Cayley-Hamilton theorem, matrix P will satisfy this characteristic equation i.e.

a_{o}P^{n} + a_{1}P^{n-1} + a_{2}P^{n-2} + ………………. + a_{n-2}P^{2} + a_{n-1}P + a_{n} I_{n}= O

(λ is replaced by matrix P in the characteristic equation and a_{n} replaced by a_{n}I_{n}, where I_{n} is the identity matrix of order n and O is the null or zero matrix of order n.

## Inverse of a matrix using Cayley-Hamilton theorem

Let us take an example of 2 x 2 matrix P such that

$P=\begin{bmatrix} 1 & 4\\ 2 & 3 \end{bmatrix}$

The characterisitc equation of matrix P is

|P – λI| = 0

$\Rightarrow \begin{vmatrix} 1-\lambda & 4\\ 2 & 3-\lambda \end{vmatrix}=0 \\ \Rightarrow (1-\lambda)(3-\lambda)-8=0 \\ \Rightarrow \lambda^{2}- 4 \lambda -5 = 0$

According to Cayley-Hamilton theorem, we have

⇒ P^{2} – 4P – 5I = O

$\Rightarrow I=\frac{1}{5}[P^{2}-4P]$

Pre-multiplying by P^{-1}, we get

$P^{-1}=\frac{1}{5}[P-4I] \\ P^{-1}=\frac{1}{5}\left ( \begin{bmatrix} 1 & 4\\ 2 & 3 \end{bmatrix}-4\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix} \right ) \\ P^{-1}=\frac{1}{5}\begin{bmatrix} -3 & 4\\ 2 & -1 \end{bmatrix}$

This is the inverse of matrix P.