Verified Solution • 2025–26 Session moderate

QJEEM21Maf6d4: JEE Main 2021 Mathematics - Class 12 JEEM21Maf6d4 Solution: If the matrix A = ( 0 2 K − 1 ) satisfies , then the value of K …

If the matrix A=(02K−1) satisfies A(A3+3I)=2IA({A^3} + 3I) = 2I , then the value of K is :

Multiple Choice Options

  • A 12{1 \over 2}
  • B −- 12{1 \over 2}
  • C −- 1
  • D 1

Step-by-Step Verified Solution

Given matrix A=[02k−1]

A4+3IA=2I{A^4} + 3IA = 2I

⇒A4=2I−3A\Rightarrow {A^4} = 2I - 3A

Also characteristic equation of A is ∣A−λI∣ =0|A - \lambda I|\, = 0

⇒|0−λ2k−1−λ|=0

⇒λ+λ2−2k=0\Rightarrow \lambda + {\lambda ^2} - 2k = 0

⇒A+A2=2K.I\Rightarrow A + {A^2} = 2K.I

⇒A2=2KI−A\Rightarrow {A^2} = 2KI - A

⇒A4=4K2I+A2−4AK\Rightarrow {A^4} = 4{K^2}I + {A^2} - 4AK

Put A2=2KI−A{A^2} = 2KI - A

and A4=2I−3A{A^4} = 2I - 3A

2I−3A=4K2I+2KI−A−4AK2I - 3A = 4{K^2}I + 2KI - A - 4AK

⇒I(2−2K−4K2)=A(2−4K)\Rightarrow I(2 - 2K - 4{K^2}) = A(2 - 4K)

⇒−2I(2K2+K−1)=2A(1−2K)\Rightarrow - 2I(2{K^2} + K - 1) = 2A(1 - 2K)

⇒−2I(2K−1)(K+1)=2A(1−2K)\Rightarrow - 2I(2K - 1)(K + 1) = 2A(1 - 2K)

⇒(2K−1)(2A)−2I(2K−1)(K+1)=0\Rightarrow (2K - 1)(2A) - 2I(2K - 1)(K + 1) = 0

⇒(2K−1)[2A−2I(K+1)]=0\Rightarrow (2K - 1)[2A - 2I(K + 1)] = 0

⇒K=12\Rightarrow K = {1 \over 2}

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