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STD 12 - 3 and 4 . determinant and metrices — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 3 and 4 . determinant and metrices4 Marks
MCQ
Let $\alpha$ be a root of the equation $x^{2}+x+1=0$ and the matrix $A=\frac{1}{\sqrt{3}}\left[\begin{array}{ccc}{1} & {1} & {1} \\ {1} & {\alpha} & {\alpha^{2}} \\ {1} & {\alpha^{2}} & {\alpha^{4}}\end{array}\right],$ then the matrix $\mathrm{A}^{31}$ is equal to
  • ✓
    $A^3$
  • B
    $A$
  • C
    $A^2$
  • D
    $I_3$
✓

Answer

Correct option: A.
$A^3$
a
$x^{2}+x+1=0$

$\alpha=\omega$

$\alpha^{2}=\omega^{2}$

$A=\frac{1}{\sqrt{3}}\left[\begin{array}{ccc}{1} & {1} & {1} \\ {1} & {\omega} & {\omega^{2}} \\ {1} & {\omega^{2}} & {\omega}\end{array}\right]$

$A^{2}=\left[\begin{array}{lll}{1} & {0} & {0} \\ {0} & {0} & {1} \\ {0} & {1} & {0}\end{array}\right]$

$\Rightarrow \mathrm{A}^{4}=\mathrm{A}^{2} \cdot \mathrm{A}^{2}=\mathrm{I}_{3}$

$A^{31}=A^{28} . A^{3}=A^{3}$

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