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−λ$$II=0$$⇒$$0$$−λ$$2k$$−$$1$$−λ$$=0$$⇒λ$$+$$λ$$2$$−$$2k=0$$⇒$$A+A2=2K$$.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)[$$2A$$−$$2I(K+1)$$]$$=0$$⇒$$K=12$$

Q.

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

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By Expert Faculty of Sri Chaitanya

$1$

$\frac{1}{2}$

$- 1$

$- \frac{1}{2}$

answer is B.

(Detailed Solution Below)

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Detailed Solution

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

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