What is the Cayley-Hamilton Theorem and how is it used in JEE Main?

The Cayley-Hamilton Theorem states that every square matrix satisfies its own characteristic equation. For a 2×2 matrix A with characteristic equation λ²−(trace A)λ+det(A)=0, we get A²−(trace A)·A+(det A)·I=O. This is used in JEE to simplify higher powers of matrices, find inverse, and reduce matrix polynomial expressions efficiently.

What is the formula for determinant of adjoint of a matrix?

For an n×n matrix M, det(adj(M)) = (det(M))^(n−1). For a 2×2 matrix, det(adj(M)) = det(M). For a 3×3 matrix, det(adj(M)) = (det(M))². This formula is crucial in JEE problems involving adjoint matrices and is derived from the identity M·adj(M) = det(M)·I.

How to compute A³ efficiently using Cayley-Hamilton theorem?

If A²=4A−2I from Cayley-Hamilton, multiply both sides by A: A³=4A²−2A=4(4A−2I)−2A=14A−8I. This recursive substitution avoids tedious matrix multiplication and expresses any power of A as a linear combination of A and I for 2×2 matrices.

What is the value of (det(adj(A³−B³)))² in this problem?

The answer is 225. Using Cayley-Hamilton theorem, α=2 and β=2. Then A³=14A−8I and B³=8B−3I. Computing A³−B³ gives a matrix with determinant −15. Since det(adj(M))=det(M) for 2×2 matrices, (det(adj(A³−B³)))²=(−15)²=225.

Why are matrix and determinant PYQs important for JEE Main?

Matrix and determinant questions carry 8-12 marks in JEE Main every session. Topics like Cayley-Hamilton theorem, adjoint properties, and matrix equations are high-frequency areas. Problems involving det(adj(M)) formula appear almost every session and practicing PYQs builds pattern recognition and speed.

For some α, β ∈ ℝ, let A = [α 2; 1 2] and B = [1 1; 1 β] such that A²−4A+2I = B²−3B+I = O. Then (det(adj(A³−B³)))² is equal to? |

For some α, β ∈ ℝ, let A = [α 2; 1 2] and B = [1 1; 1 β] such that A²−4A+2I = B²−3B+I = O. Then (det(adj(A³−B³)))² is equal to? | NeetJeeRankers

Question

For some $\alpha, \beta \in \mathbb{R}$, let

$A =$

$\begin{pmatrix} \alpha & 2 \\ 1 & 2 \end{pmatrix}$

$B =$

$\begin{pmatrix} 1 & 1 \\ 1 & \beta \end{pmatrix}$

$A^2 – 4A + 2I = B^2 – 3B + I = O$

Then $\left(\det\!\left(\text{adj}(A^3 – B^3)\right)\right)^2$ is equal to ______?

Correct Answer

225

$\left(\det\!\left(\text{adj}(A^3 – B^3)\right)\right)^2 = 225$

Step-by-Step Solution

1 Apply Cayley-Hamilton on A

Given: $A^2 – 4A + 2I = O$

This is the characteristic equation of $A$.

For $2 \times 2$ matrix: $\lambda^2 – (\text{tr}\,A)\lambda + \det A = 0$

Comparing coefficients:

$\text{tr}(A) = 4, \quad \det(A) = 2$

Since $\text{tr}(A) = \alpha + 2 = 4$

$\alpha = 2$

Verify: $\det(A) = 2(2) – 2(1) = 2$ ✅

2 Apply Cayley-Hamilton on B

Given: $B^2 – 3B + I = O$

Comparing with characteristic equation:

$\text{tr}(B) = 3, \quad \det(B) = 1$

Since $\text{tr}(B) = 1 + \beta = 3$

$\beta = 2$

Verify: $\det(B) = 1(2) – 1(1) = 1$ ✅

3 Compute $A^3$

From $A^2 = 4A – 2I$, multiply by $A$:

$A^3 = 4A^2 – 2A$

$= 4(4A – 2I) – 2A$

$= 16A – 8I – 2A$

$A^3 = 14A – 8I$

$A = \begin{pmatrix}2&2\\1&2\end{pmatrix}$

$A^3 = 14\begin{pmatrix}2&2\\1&2\end{pmatrix} – 8\begin{pmatrix}1&0\\0&1\end{pmatrix}$

$A^3 = \begin{pmatrix}20&28\\14&20\end{pmatrix}$

4 Compute $B^3$

From $B^2 = 3B – I$, multiply by $B$:

$B^3 = 3B^2 – B$

$= 3(3B – I) – B$

$= 9B – 3I – B$

$B^3 = 8B – 3I$

$B = \begin{pmatrix}1&1\\1&2\end{pmatrix}$

$B^3 = 8\begin{pmatrix}1&1\\1&2\end{pmatrix} – 3\begin{pmatrix}1&0\\0&1\end{pmatrix}$

$B^3 = \begin{pmatrix}5&8\\8&13\end{pmatrix}$

5 Compute $A^3 – B^3$

$A^3 – B^3 = \begin{pmatrix}20-5&28-8\\14-8&20-13\end{pmatrix}$

$A^3 – B^3 = \begin{pmatrix}15&20\\6&7\end{pmatrix}$

6 Find $\det(A^3 – B^3)$

$\det = 15 \times 7 – 20 \times 6$

$= 105 – 120$

$\det(A^3 – B^3) = -15$

7 Apply $\det(\text{adj})$ Formula

For $n \times n$ matrix $M$:

$\det(\text{adj}(M)) = (\det M)^{n-1}$

Here $n = 2$:

$\det(\text{adj}(A^3 – B^3)) = (\det(A^3-B^3))^1 = -15$

Therefore:

$\left(\det(\text{adj}(A^3 – B^3))\right)^2 = (-15)^2 = \boxed{225}$

Related Theory

🔹 Cayley-Hamilton Theorem

The Cayley-Hamilton theorem is one of the most powerful results in linear algebra and a frequently tested concept in competitive examinations. It states that every square matrix satisfies its own characteristic equation. For a two-by-two matrix M with characteristic polynomial given by lambda squared minus the trace of M times lambda plus the determinant of M equals zero, the theorem guarantees that M squared minus trace M times M plus det M times identity matrix equals the zero matrix. This result is incredibly useful because it allows us to express higher powers of a matrix entirely in terms of M and I, drastically reducing computation time. In this problem, both the given conditions are essentially characteristic equations of matrices A and B respectively. By comparing the coefficients with the standard form, we directly read off the trace and determinant of each matrix, which immediately gives us the unknown parameters alpha and beta.

🔹 Computing Higher Matrix Powers

Once we establish the Cayley-Hamilton relation, computing any power of a matrix becomes a recursive substitution process. From A squared equals 4A minus 2I, multiplying both sides by A gives A cubed equals 4 times A squared minus 2A, and substituting the relation again yields A cubed equals 14A minus 8I. This technique extends to any power and is far more efficient than direct matrix multiplication. For a two-by-two matrix satisfying its characteristic equation, every higher power can be written as a linear combination of M and I with scalar coefficients. This is a standard trick that saves enormous time during competitive exams and students should practice until this substitution becomes automatic.

🔹 Adjoint Matrix and Its Determinant

The adjoint or classical adjugate of a square matrix M is defined as the transpose of its cofactor matrix. For a two-by-two matrix the adjoint is obtained by swapping diagonal elements and negating off-diagonal elements. The most important identity connecting a matrix with its adjoint is M times adj M equals det M times I. Taking determinant on both sides of this identity immediately gives us the formula for determinant of adjoint.

$\det(\text{adj}(M)) = (\det M)^{n-1}$

For n equals 2, this simplifies to det of adj M equals det M. For n equals 3, det of adj M equals det M squared. This formula appears in almost every session of competitive examinations and students must memorize it along with related results like adj of adj M equals det M raised to n minus 2 times M and det of adj of adj M equals det M raised to the power n minus 1 squared.

🔹 Key Formulas to Remember

$M \cdot \text{adj}(M) = \det(M) \cdot I$ — fundamental identity
$\det(\text{adj}(M)) = (\det M)^{n-1}$
$\text{adj}(\text{adj}(M)) = (\det M)^{n-2} \cdot M$
$\text{adj}(AB) = \text{adj}(B) \cdot \text{adj}(A)$ — order reverses
Trace equals sum of eigenvalues, determinant equals product of eigenvalues
$A^3 – B^3 \neq (A-B)(A^2+AB+B^2)$ unless A and B commute

🔹 Why This Problem Matters

This problem beautifully combines multiple core concepts — identifying characteristic equations from matrix polynomial conditions, extracting trace and determinant to find unknowns, computing cube of matrices efficiently through recursive Cayley-Hamilton substitution, and finally applying the adjoint determinant formula correctly. Such multi-concept problems are characteristic of competitive examinations and test deep understanding rather than rote memorization. Students should especially note that matrix factorization identities from scalar algebra do not directly apply to matrices, making individual computation of A cubed and B cubed necessary before subtraction.