Cayley-Hamilton theorem — verify for a 2×2 matrix

Question

State the Cayley-Hamilton theorem. Verify it for the matrix $A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$ . Use it to find $A^{-1}$ .

(JEE Main 2022, similar pattern)

Solution — Step by Step

Cayley-Hamilton Theorem: Every square matrix satisfies its own characteristic equation. If $p(\lambda) = \det(A - \lambda I) = 0$ is the characteristic equation, then $p(A) = O$ (the zero matrix).

\det(A - \lambda I) = \det \begin{pmatrix} 1-\lambda & 2 \\ 3 & 4-\lambda \end{pmatrix} = (1-\lambda)(4-\lambda) - 6

= \lambda^2 - 5\lambda + 4 - 6 = \lambda^2 - 5\lambda - 2 = 0

By Cayley-Hamilton: $A^2 - 5A - 2I = O$ .

First, compute $A^2$ :

A^2 = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}\begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} = \begin{pmatrix} 7 & 10 \\ 15 & 22 \end{pmatrix}

Now compute $A^2 - 5A - 2I$ :

= \begin{pmatrix} 7 & 10 \\ 15 & 22 \end{pmatrix} - \begin{pmatrix} 5 & 10 \\ 15 & 20 \end{pmatrix} - \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}

We get the zero matrix. Cayley-Hamilton is verified.

From $A^2 - 5A - 2I = O$ , multiply both sides by $A^{-1}$ :

A - 5I - 2A^{-1} = O

A^{-1} = \frac{1}{2}(A - 5I) = \frac{1}{2}\begin{pmatrix} -4 & 2 \\ 3 & -1 \end{pmatrix} = \mathbf{\begin{pmatrix} -2 & 1 \\ 3/2 & -1/2 \end{pmatrix}}

Why This Works

The characteristic equation captures the fundamental algebraic properties of a matrix through its eigenvalues. The Cayley-Hamilton theorem says the matrix itself, when plugged into this polynomial, gives zero. This is a powerful result because it lets us express higher powers of $A$ in terms of lower powers.

For a $2 \times 2$ matrix, the characteristic polynomial is always $\lambda^2 - \text{tr}(A)\lambda + \det(A)$ , where $\text{tr}(A) = a_{11} + a_{22}$ is the trace and $\det(A)$ is the determinant.

Alternative Method

You can find $A^{-1}$ directly using the formula $A^{-1} = \frac{1}{\det A} \text{adj}(A)$ . Here $\det A = 4 - 6 = -2$ and $\text{adj}(A) = \begin{pmatrix} 4 & -2 \\ -3 & 1 \end{pmatrix}$ . This gives $A^{-1} = \begin{pmatrix} -2 & 1 \\ 3/2 & -1/2 \end{pmatrix}$ — same answer. Cayley-Hamilton provides an elegant alternative route.

For JEE, the Cayley-Hamilton approach is especially useful for finding $A^3$ , $A^4$ , or higher powers. From $A^2 = 5A + 2I$ , we get $A^3 = 5A^2 + 2A = 5(5A + 2I) + 2A = 27A + 10I$ , and so on. This avoids messy matrix multiplication.

Common Mistake

When finding $A^{-1}$ from Cayley-Hamilton, students forget that multiplying the equation by $A^{-1}$ requires $A$ to be invertible ( $\det A \neq 0$ ). If $\det A = 0$ , then $A^{-1}$ does not exist and you cannot use this trick. Always check that the constant term in the characteristic equation (which equals $\det A$ ) is non-zero.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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