JEE MAIN - Mathematics (2020 - 3rd September Morning Slot - No. 23)
Let A = $$\left[ {\matrix{
x & 1 \cr
1 & 0 \cr
} } \right]$$, x $$ \in $$ R and A4 = [aij].
If a11 = 109, then a22 is equal to _______ .
If a11 = 109, then a22 is equal to _______ .
Answer
10
Explanation
$${A^2} = \left[ {\matrix{
x & 1 \cr
1 & 0 \cr
} } \right]\left[ {\matrix{
x & 1 \cr
1 & 0 \cr
} } \right] = \left[ {\matrix{
{{x^2} + 1} & x \cr
x & 1 \cr
} } \right]$$
$${A^4} = \left[ {\matrix{ {{x^2} + 1} & x \cr x & 1 \cr } } \right]\left[ {\matrix{ {{x^2} + 1} & x \cr x & 1 \cr } } \right]$$
$$ = \left[ {\matrix{ {{{({x^2} + 1)}^2} + {x^2}} & {x({x^2} + 1) + x} \cr {x({x^2} + 1) + x} & {{x^2} + 1} \cr } } \right]$$
Given $${({x^2} + 1)^2} + {x^2} = 109$$
Let $${x^2} + 1$$ = t
$${t^2} + t - 1 = 109$$
$$ \Rightarrow $$ (t $$ - $$ 10) (t + 11) = 0
$$ \therefore $$ t = 10 = x2 + 1 = a22
$${A^4} = \left[ {\matrix{ {{x^2} + 1} & x \cr x & 1 \cr } } \right]\left[ {\matrix{ {{x^2} + 1} & x \cr x & 1 \cr } } \right]$$
$$ = \left[ {\matrix{ {{{({x^2} + 1)}^2} + {x^2}} & {x({x^2} + 1) + x} \cr {x({x^2} + 1) + x} & {{x^2} + 1} \cr } } \right]$$
Given $${({x^2} + 1)^2} + {x^2} = 109$$
Let $${x^2} + 1$$ = t
$${t^2} + t - 1 = 109$$
$$ \Rightarrow $$ (t $$ - $$ 10) (t + 11) = 0
$$ \therefore $$ t = 10 = x2 + 1 = a22
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