Cayley-Hamilton Theorem: The Game-Changer for JEE Advanced
Master this powerful theorem to solve complex matrix problems in seconds instead of minutes.
Why Cayley-Hamilton is a JEE Advanced Secret Weapon
The Cayley-Hamilton Theorem transforms complex matrix computations into simple polynomial evaluations. In JEE Advanced, where every second counts, this theorem can save you 3-5 minutes on matrix problems that would otherwise require tedious calculations.
🎯 JEE Advanced Significance
- Appears in 1-2 problems every year in JEE Advanced
- Solves problems that are nearly impossible with conventional methods
- Particularly useful for finding high powers of matrices and matrix inverses
- Reduces computation time by 70-80%
The Cayley-Hamilton Theorem
Every square matrix satisfies its own characteristic equation
Mathematical Statement
If $A$ is an $n \times n$ matrix and $p(\lambda) = \det(A - \lambda I)$ is its characteristic polynomial, then:
Where $p(A)$ means substitute the matrix $A$ into the characteristic polynomial instead of the variable $\lambda$.
Quick Example
For a $2 \times 2$ matrix $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$:
Characteristic equation: $\lambda^2 - (a+d)\lambda + (ad - bc) = 0$
Cayley-Hamilton says: $A^2 - (a+d)A + (ad - bc)I = 0$
Application 1: Finding High Powers of Matrices
Problem Example
If $A = \begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix}$, find $A^5$ using Cayley-Hamilton Theorem.
Step-by-Step Solution
Step 1: Find characteristic equation
$|A - \lambda I| = \begin{vmatrix} 1-\lambda & 2 \\ 2 & 1-\lambda \end{vmatrix} = (1-\lambda)^2 - 4 = \lambda^2 - 2\lambda - 3 = 0$
Step 2: Apply Cayley-Hamilton: $A^2 - 2A - 3I = 0$
So $A^2 = 2A + 3I$
Step 3: Find higher powers recursively:
$A^3 = A \cdot A^2 = A(2A + 3I) = 2A^2 + 3A = 2(2A + 3I) + 3A = 7A + 6I$
$A^4 = A \cdot A^3 = A(7A + 6I) = 7A^2 + 6A = 7(2A + 3I) + 6A = 20A + 21I$
$A^5 = A \cdot A^4 = A(20A + 21I) = 20A^2 + 21A = 20(2A + 3I) + 21A = 61A + 60I$
Step 4: Substitute $A$ and $I$:
$A^5 = 61\begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix} + 60\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 121 & 122 \\ 122 & 121 \end{bmatrix}$
💡 Time-Saving Insight
For $A^n$, express it as $A^n = \alpha A + \beta I$ (for $2\times2$ matrices), then use the characteristic equation to find $\alpha$ and $\beta$.
Application 2: Finding Matrix Inverse
Problem Example
Find $A^{-1}$ for $A = \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix}$ using Cayley-Hamilton Theorem.
Step-by-Step Solution
Step 1: Characteristic equation:
$|A - \lambda I| = \begin{vmatrix} 2-\lambda & 1 \\ 1 & 2-\lambda \end{vmatrix} = (2-\lambda)^2 - 1 = \lambda^2 - 4\lambda + 3 = 0$
Step 2: Cayley-Hamilton: $A^2 - 4A + 3I = 0$
Step 3: Rearrange for inverse:
$A^2 - 4A + 3I = 0 \Rightarrow A^2 - 4A = -3I$
$A(A - 4I) = -3I \Rightarrow A \cdot \frac{4I - A}{3} = I$
Step 4: Therefore: $A^{-1} = \frac{4I - A}{3}$
Step 5: Compute:
$A^{-1} = \frac{1}{3}\left(4\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} - \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix}\right) = \frac{1}{3}\begin{bmatrix} 2 & -1 \\ -1 & 2 \end{bmatrix}$
💡 General Formula
For any $2\times2$ matrix $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$ with characteristic equation $\lambda^2 - S\lambda + D = 0$ where $S = a+d$, $D = ad-bc$:
Application 3: Solving Complex Matrix Equations
Problem Example
If $A = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}$, find $A^4 - 5A^3 + 7A^2 - 2A + I$
Step-by-Step Solution
Step 1: Find characteristic equation:
$|A - \lambda I| = \begin{vmatrix} 3-\lambda & 1 \\ -1 & 2-\lambda \end{vmatrix} = (3-\lambda)(2-\lambda) + 1 = \lambda^2 - 5\lambda + 7 = 0$
Step 2: Cayley-Hamilton: $A^2 - 5A + 7I = 0$
Step 3: Notice the given expression:
$A^4 - 5A^3 + 7A^2 - 2A + I = (A^2 - 5A + 7I)(A^2) - 2A + I$
Step 4: But $A^2 - 5A + 7I = 0$, so:
$= 0 \cdot A^2 - 2A + I = -2A + I$
Step 5: Compute:
$-2\begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} + \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} -5 & -2 \\ 2 & -3 \end{bmatrix}$
💡 Powerful Strategy
For any polynomial expression in $A$, use polynomial division to divide by the characteristic polynomial. The remainder gives you the simplified expression.
📋 Cayley-Hamilton Quick Reference
For 2×2 Matrices
$A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$
Characteristic equation: $\lambda^2 - (a+d)\lambda + (ad-bc) = 0$
Cayley-Hamilton: $A^2 - (a+d)A + (ad-bc)I = 0$
Inverse: $A^{-1} = \frac{(a+d)I - A}{ad-bc}$
For 3×3 Matrices
$A = \begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix}$
Characteristic equation: $\lambda^3 - S_1\lambda^2 + S_2\lambda - S_3 = 0$
Where $S_1 = \text{trace}(A)$, $S_3 = \det(A)$
Cayley-Hamilton: $A^3 - S_1A^2 + S_2A - S_3I = 0$
⚠️ Common Mistakes to Avoid
Mistake 1: Wrong Characteristic Equation
Double-check your determinant calculation. For $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$, it's $\lambda^2 - (a+d)\lambda + (ad-bc) = 0$, not $\lambda^2 - (a+d)\lambda - (ad-bc) = 0$.
Mistake 2: Scalar vs Matrix Operations
Remember that $A^2$ means matrix multiplication $A \times A$, not element-wise squaring. And $0$ in the theorem means the zero matrix, not scalar zero.
Mistake 3: Forgetting the Identity Matrix
In the characteristic polynomial $p(\lambda) = \lambda^2 + \alpha\lambda + \beta$, when substituting $A$, the constant term $\beta$ becomes $\beta I$, not just $\beta$.
🎯 JEE Advanced Practice Problems
1. If $A = \begin{bmatrix} 2 & -1 \\ 1 & 2 \end{bmatrix}$, find $A^6$ using Cayley-Hamilton Theorem.
2. Let $A = \begin{bmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{bmatrix}$. Verify that A satisfies its characteristic equation and hence find $A^{-1}$.
3. If $A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$ and $I$ is the identity matrix, find $A^5 - 5A^4 + 6A^3 + 2A^2 - 3A + I$.
Time yourself: Try to solve each problem in under 4 minutes using Cayley-Hamilton Theorem.
🚀 Exam Strategy for Cayley-Hamilton
When to Use It
- Finding $A^n$ for $n > 3$
- Computing matrix polynomials
- Finding inverse of $2\times2$ and $3\times3$ matrices
- Solving matrix equations
- Proving matrix identities
Speed Tips
- Memorize the $2\times2$ characteristic formula
- Use polynomial division for complex expressions
- For $A^n$, find pattern in coefficients
- Practice mental calculation of determinants
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