If the matrix A=02k−1 satisfies AA3+3I=2I, then the value of k is :

A=02K−1 A4+3IA=2I⇒A4=2I−3A Also characteristic equation of A is IA−λII=0⇒0−λ2k−1−λ=0⇒λ+λ2−2k=0⇒A+A2=2K.I⇒A2=2KI−A⇒A4=4K2I+A2−4AK Put A2=2KI−A and A4=2I−3A2I−3A=4K2I+2KI−A−4AK⇒I2−2K−4K2=A(2−4K)⇒−2I2K2+K−1=2A(1−2K)⇒−2I(2K−1)(K+1)=2A(1−2K)⇒(2K−1)(2A)−2I(2K−1)(K+1)=0⇒(2K−1)[2A−2I(K+1)]=0⇒K=12

If the matrix A=02k−1 satisfies AA3+3I=2I, then the value of k is :

A=02K−1 A4+3IA=2I⇒A4=2I−3A Also characteristic equation of A is IA−λII=0⇒0−λ2k−1−λ=0⇒λ+λ2−2k=0⇒A+A2=2K.I⇒A2=2KI−A⇒A4=4K2I+A2−4AK Put A2=2KI−A and A4=2I−3A2I−3A=4K2I+2KI−A−4AK⇒I2−2K−4K2=A(2−4K)⇒−2I2K2+K−1=2A(1−2K)⇒−2I(2K−1)(K+1)=2A(1−2K)⇒(2K−1)(2A)−2I(2K−1)(K+1)=0⇒(2K−1)[2A−2I(K+1)]=0⇒K=12

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$If the matrix A = (\begin{array}{cc} 0 & 2 \\ k & - 1 \end{array}) satisfies A (A^{3} + 3 I) = 2 I, then the value of k is :$

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$1$

$- 1$

$\frac{1}{2}$

$- \frac{1}{2}$

answer is B.

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Detailed Solution

$A = [\begin{array}{cc} 0 & 2 \\ K & - 1 \end{array}] \begin{aligned} A^{4} + 3 IA = 2 I \\ \Rightarrow A^{4} = 2 I - 3 A Also characteristic equation of A is \end{aligned} \begin{array}{l} IA - λ II = 0 \\ \Rightarrow |\begin{array}{cc} 0 - λ & 2 \\ k & - 1 - λ \end{array}| = 0 \\ \Rightarrow λ + λ^{2} - 2 k = 0 \\ \Rightarrow A + A^{2} = 2 K . I \\ \Rightarrow A^{2} = 2 KI - A \\ \Rightarrow A^{4} = 4 K^{2} I + A^{2} - 4 AK \\ Put A^{2} = 2 KI - A \\ and A^{4} = 2 I - 3 A \\ 2 I - 3 A = 4 K^{2} I + 2 KI - A - 4 AK \\ \Rightarrow I (2 - 2 K - 4 K^{2}) = A (2 - 4 K) \\ \Rightarrow - 2 I (2 K^{2} + K - 1) = 2 A (1 - 2 K) \\ \Rightarrow - 2 I (2 K - 1) (K + 1) = 2 A (1 - 2 K) \\ \Rightarrow (2 K - 1) (2 A) - 2 I (2 K - 1) (K + 1) = 0 \\ \Rightarrow (2 K - 1) [2 A - 2 I (K + 1)] = 0 \\ \Rightarrow K = \frac{1}{2} \end{array}$