Q.

 If the matrix A=02k1 satisfies AA3+3I=2I, then the value of k is : 

Moderate

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By Expert Faculty of Sri Chaitanya

a

1

b

12

c

-1

d

-12

answer is B.

(Detailed Solution Below)

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Detailed Solution

A=02K1 A4+3IA=2IA4=2I3A  Also characteristic equation of A is  IAλII=00λ2k1λ=0λ+λ22k=0A+A2=2K.IA2=2KIAA4=4K2I+A24AK Put A2=2KIA and A4=2I3A2I3A=4K2I+2KIA4AKI22K4K2=A(24K)2I2K2+K1=2A(12K)2I(2K1)(K+1)=2A(12K)(2K1)(2A)2I(2K1)(K+1)=0(2K1)[2A2I(K+1)]=0K=12

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