Q.If $A=\begin{bmatrix}3 & 1\\ -1 & 2\end{bmatrix}$, show that $A^{2}-5A+7I=0$. Using this equation find $A^{-1}$. OR Express the matrix $A=\begin{bmatrix}3 & 3 & -1\\ -2 & -2 & 1\\ -4 & -5 & 2\end{bmatrix}$ as the sum of a symmetric and a skew symmetric matrix.

The Cayley–Hamilton Theorem

Matrices add, subtract, multiply and scale much like numbers (except $AB\neq BA$ in general), so we can substitute a matrix into a polynomial. For $p(x)=x^2-3x+2$, replacing $x$ by a square matrix $A$ gives

where the constant $2$ becomes $2I$ so it can be added to matrices.

The Cayley–Hamilton theorem makes a striking claim: every square matrix satisfies its own characteristic equation.

The characteristic polynomial

Every $n\times n$ matrix $A$ has a characteristic polynomial

a degree-$n$ polynomial whose roots are the eigenvalues. For a $2\times2$ matrix it is $\lambda^2-(\operatorname{tr}A)\lambda+\det A$. For $A=\begin{pmatrix}1&2\\3&4\end{pmatrix}$ this is $p(\lambda)=\lambda^2-5\lambda-2$.

The statement

For the example, $A^2-5A-2I=\begin{pmatrix}0&0\\0&0\end{pmatrix}$.

Why it is surprising, and a quick check

The polynomial is built from $\det(\lambda I-A)$, yet feeding $A$ back into it annihilates it — and this holds for any $A$, invertible or not. You can verify it on the general $2\times2$ matrix $A=\begin{pmatrix}a&b\\c&d\end{pmatrix}$, where $p(\lambda)=\lambda^2-(a+d)\lambda+(ad-bc)$: a short computation of $A^2-(a+d)A+(ad-bc)I$ gives the zero matrix.

Why it matters

Cayley–Hamilton lets you rewrite any high power $A^k$ (for $k\ge n$) as a combination of $I,A,\dots,A^{n-1}$, which speeds up computing powers, exponentials and inverses. …