Matrices & DeterminantshardFree

Matrices & Determinants: Let Pmatrix Pmatrix

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

Question
Let S={(a,b)  |  A3=A, where A=(1212ab)}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)n(S) is:
A11
B22
C33correct
D00
Solution
Step 1: Form the characteristic equation
λ2−pλ+q=0,p=tr(A)=12+b,q=det⁡(A)=b−a2\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
A2=pA−qIA^2 = pA - qI
A3=pA2−qA=p(pA−qI)−qA=(p2−q)A−pq IA^3 = pA^2 - qA = p(pA - qI) - qA = (p^2 - q)A - pq\,I
 Step 3: Impose the condition A3=AA^3 = A
(p2−q−1)A=pq I(p^2 - q - 1)A = pq\,I
 Since AA is not a scalar matrix (its off-diagonal entry is 12\tfrac{1}{2}), both sides require:
p2−q−1=0andpq=0p^2 - q - 1 = 0 \quad \text{and} \quad pq = 0
 Step 4: Solve for (p,q)(p, q) If q=0q = 0, then p2=1  ⟹  p=±1p^2 = 1 \implies p = \pm 1. If p=0p = 0, then q=−1q = -1.
(p,q)∈{(1,0), (−1,0), (0,−1)}(p, q) \in \{(1, 0),\ (-1, 0),\ (0, -1)\}
 Each pair yields a distinct (a,b)(a, b), so n(S)=3n(S) = 3. Answer: (3)
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