Let A= (x110 ) ,x ∈Rand A^4= (a_ ij ). If a_11=109, then a_22 is equal to

Question

Accepted Answer

Correct answer: 10. 10 JEE Main 2020 A^2= (x110 ) (x110 )=(x^2+1xx1) A^4=(x^2+1xx1)(x^2+1xx1) ((x^2+1)^2+x^2x(x^2+1)+xx(x^2+1)+xx^2+1) Given (x^2+1)+x^2=109 Let x^2+1=t t^2+t-1=109 (t-10)(t+11)=0 t=10=x^2+1=a_22

$Let A = (x 110), x \in R and A^{4} = (a_{ij}) . If a_{11} = 109, then a_{22} is equal to$

Solution

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