Matrices & DeterminantshardFree

Matrices & Determinants — JEE Maths practice question

JEE Maths question with a full step-by-step solution.

Question

Let

S = \left\{(a, b)\;\middle|\; A^3 = A,\ \text{where}\ A = \begin{pmatrix} \frac{1}{2} & \frac{1}{2} \\ a & b \end{pmatrix}\right\}

. Then

n(S)

is:

1

2

3

correct

0

Solution

Step 1: Form the characteristic equation

\lambda^2 - p\lambda + q = 0, \quad p = \text{tr}(A) = \frac{1}{2} + b, \quad q = \det(A) = \frac{b-a}{2}

Step 2: Apply the Cayley-Hamilton theorem

A^2 = pA - qI

A^3 = pA^2 - qA = p(pA - qI) - qA = (p^2 - q)A - pq\,I

Step 3: Impose the condition

A^3 = A

(p^2 - q - 1)A = pq\,I

Since

A

is not a scalar matrix (its off-diagonal entry is

\tfrac{1}{2}

), both sides require:

p^2 - q - 1 = 0 \quad \text{and} \quad pq = 0

Step 4: Solve for

(p, q)

q = 0

, then

p^2 = 1 \implies p = \pm 1

. If

p = 0

, then

q = -1

(p, q) \in \{(1, 0),\ (-1, 0),\ (0, -1)\}

Each pair yields a distinct

(a, b)

, so

n(S) = 3

. Answer: (3)

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