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3 and 4 .Determinants and Matrices
hard
यदि समीकरण $x^{2}+x+1=0$ का एक मूल $\alpha$ है तथा आव्यूह $A =\frac{1}{\sqrt{3}}\left[\begin{array}{ccc}1 & 1 & 1 \\ 1 & \alpha & \alpha^{2} \\ 1 & \alpha^{2} & \alpha^{4}\end{array}\right]$ है, तो आव्यूह $A ^{31}$ बराबर है
A
$A^3$
B
$A$
C
$A^2$
D
$I_3$
(JEE MAIN-2020)
Solution
$x^{2}+x+1=0$
$\alpha=\omega$
$\alpha^{2}=\omega^{2}$
$A=\frac{1}{\sqrt{3}}\left[\begin{array}{ccc}{1} & {1} & {1} \\ {1} & {\omega} & {\omega^{2}} \\ {1} & {\omega^{2}} & {\omega}\end{array}\right]$
$A^{2}=\left[\begin{array}{lll}{1} & {0} & {0} \\ {0} & {0} & {1} \\ {0} & {1} & {0}\end{array}\right]$
$\Rightarrow \mathrm{A}^{4}=\mathrm{A}^{2} \cdot \mathrm{A}^{2}=\mathrm{I}_{3}$
$A^{31}=A^{28} . A^{3}=A^{3}$
Standard 12
Mathematics
