Cayley–Hamilton Theorem and Computing the Inverse

The theorem

For an $n\times n$ matrix $A$, let its characteristic polynomial be

$p(\lambda) = \det(\lambda I - A) = \lambda^{n} + c_{n-1}\lambda^{n-1} + \cdots + c_{1}\lambda + c_{0}.$

The Cayley–Hamilton theorem says $A$ satisfies its own characteristic equation: substituting $A$ for $\lambda$ (and $A^{0}=I$) gives the zero matrix,

$p(A) = A^{n} + c_{n-1}A^{n-1} + \cdots + c_{1}A + c_{0}I = 0.$

The constant term is $c_{0} = p(0) = \det(-A) = (-1)^{n}\det A$.

Why it hands you the inverse

Move the constant term to the other side and factor an $A$ out of every remaining term:

$A\bigl(A^{n-1} + c_{n-1}A^{n-2} + \cdots + c_{1}I\bigr) = -c_{0}I.$

If $A$ is invertible then $c_{0} = (-1)^{n}\det A \ne 0$, so multiply by $A^{-1}/(-c_{0})$:

$A^{-1} = -\frac{1}{c_{0}}\bigl(A^{n-1} + c_{n-1}A^{n-2} + \cdots + c_{1}I\bigr).$

The inverse is just a polynomial in $A$ — you only ever need powers of $A$ up to $A^{n-1}$, never a single cofactor.

Building $p(\lambda)$ for a $3\times 3$ matrix (the konkur case)

Write it from the three invariants of $A$:

$p(\lambda) = \lambda^{3} - (\operatorname{tr} A)\,\lambda^{2} + m_{2}\,\lambda - \det A,$

where $\operatorname{tr} A$ is the sum of the diagonal, $m_{2}$ is the sum of the three principal $2\times 2$ minors (delete row/column $i$ for $i=1,2,3$), and $\det A$ is the determinant. By Cayley–Hamilton,

$A^{3} - (\operatorname{tr} A)A^{2} + m_{2}A - (\det A)I = 0\ \Longrightarrow\ A^{-1} = \frac{1}{\det A}\bigl(A^{2} - (\operatorname{tr} A)A + m_{2}I\bigr).$

Mini-example

Let $A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$. Here $\operatorname{tr} A = 5$ and $\det A = 1\cdot 4 - 2\cdot 3 = -2$, so the characteristic polynomial is $p(\lambda) = \lambda^{2} - 5\lambda - 2$. Cayley–Hamilton gives $A^{2} - 5A - 2I = 0$, hence

$A^{-1} = \frac{1}{-2}\bigl(A - 5I\bigr) = -\frac{1}{2}\begin{bmatrix} 1-5 & 2 \\ 3 & 4-5 \end{bmatrix} = -\frac{1}{2}\begin{bmatrix} -4 & 2 \\ 3 & -1 \end{bmatrix} = \begin{bmatrix} -2 & 1 \\ \tfrac{3}{2} & -\tfrac{1}{2} \end{bmatrix}.$

A quick check: $A\,A^{-1} = I$. (For $2\times 2$ this reproduces the familiar $\tfrac{1}{\det A}\begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$ formula.)

Exam trick: for a $3\times 3$ matrix, compute only the three invariants $\operatorname{tr} A,\ m_{2},\ \det A$, form $A^{2}$ once, and read off $A^{-1} = \tfrac{1}{\det A}\bigl(A^{2} - (\operatorname{tr} A)A + m_{2}I\bigr)$. This is faster and far less error-prone than the $9$ cofactors of the adjugate method — and it instantly catches a singular matrix (if $\det A = 0$, the $c_{0}$ term vanishes and no inverse exists).