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Matrices & Determinants — JEE Maths practice question

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

Question
If A=(abcd)A = \begin{pmatrix}a & b\\ c & d\end{pmatrix} satisfies the equation x2−(a+d)x+k=0x^2 - (a+d)x + k = 0, then:
Ak=bck = bc
Bk=adk = ad
Ck=ad−bck = ad - bccorrect
Dk=a2+b2+c2+d2k = a^2+b^2+c^2+d^2
Solution
Step 1: Apply the Cayley-Hamilton theorem Every square matrix satisfies its own characteristic equation. The characteristic equation of AA is ∣A−λI∣=0|A - \lambda I| = 0:
∣a−λbcd−λ∣=0  ⟹  (a−λ)(d−λ)−bc=0\begin{vmatrix}a-\lambda & b\\ c & d-\lambda\end{vmatrix} = 0 \implies (a-\lambda)(d-\lambda) - bc = 0
  ⟹  λ2−(a+d)λ+(ad−bc)=0\implies \lambda^2 - (a+d)\lambda + (ad-bc) = 0
 Step 2: Compare with the given equation Comparing λ2−(a+d)λ+k=0\lambda^2 - (a+d)\lambda + k = 0 with the characteristic equation:
k=ad−bck = ad - bc
 Answer: (3)
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