Q.Show that for the matrix $A = \begin{bmatrix} 1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3 \end{bmatrix}$, $A^3 - 6A^2 + 5A + 11I = 0$. Hence, find $A^{-1}$.

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.

One caution: the theorem is not proved by literally putting $\lambda=A$ into the scalar determinant — the honest proof uses the adjugate. For a first meeting, learn the statement and practise verifying it on small matrices.

The Cayley-Hamilton Theorem is not part of the core CBSE Class 12 Matrices syllabus but is an important topic for JEE Advanced and other engineering entrance exams that build on the NCERT Class 12 Mathematics curriculum on matrices and determinants. "Cayley Hamilton theorem proof and application to find inverse" is a frequently searched topic among students preparing for these higher-level linear algebra questions.

The Inverse Matrix Method

Many problems reduce to a system of linear equations, for example

The inverse matrix method solves such a system by writing it as a single matrix equation and then undoing the coefficient matrix with its inverse — the matrix analogue of dividing.

Writing the system as $AX=B$

Collect the coefficients, the unknowns and the constants:

so the whole system becomes $AX=B$.

The idea

For numbers, $ax=b$ gives $x=a^{-1}b$ provided $a\neq0$. The same works for matrices: if $A$ is invertible, multiply $AX=B$ on the left by $A^{-1}$:

Multiplying on the left matters — matrix products do not commute, so $BA^{-1}$ would be wrong.

When it works

The inverse $A^{-1}$ exists only when $\det A\neq0$, so:

• $\det A\neq0$: the system is consistent with the unique solution $X=A^{-1}B$.
• $\det A=0$: no inverse; the system is either inconsistent (no solution) or has infinitely many — handle it by another method.

Worked steps

For the system above, $\det A=(2)(-1)-(3)(1)=-5\neq0$, and

Then

so $x=1,\ y=2$.

The inverse matrix method (X = A⁻¹B) is one of two standard techniques — alongside Cramer's rule — for solving simultaneous linear equations in the NCERT Class 12 Determinants chapter, and it's a guaranteed board-exam and JEE Main topic. Searches for 'solving system of equations using inverse matrix method class 12' or 'inverse matrix method important questions' almost always lead back to exactly this AX = B setup.