"Let $\alpha$ be a root of the equation x2 + x + 1 = 0 and the matrix A = $${1 \over {\sqrt 3 }}\left[ {\matrix{ 1 & 1 & 1 \cr 1 & \alpha & {{\alpha ^2}} \cr 1 & {{\alpha ^2}} & {{\alpha ^4}} \cr } } \right]$$ then the matrix A31 is equal to"

Question

Accepted Answer

"x² + x + 1 = 0

$\Rightarrow$ x = ${{ - 1 + i\sqrt 3 } \over 2}$ = $\omega$ or ${{ - 1 - i\sqrt 3 } \over 2}$ = ${\omega ^2}$

Let $\alpha$ = $\omega$

$\therefore$ A = $${1 \over {\sqrt 3 }}\left[ {\matrix{ 1 & 1 & 1 \cr 1 & \omega & {{\omega ^2}} \cr 1 & {{\omega ^2}} & {{\omega ^4}} \cr } } \right]$$

A² = $${1 \over 3}\left[ {\matrix{ 3 & 0 & 0 \cr 0 & 0 & 3 \cr 0 & 3 & 0 \cr } } \right]$$ = $$\left[ {\matrix{ 1 & 0 & 0 \cr 0 & 0 & 1 \cr 0 & 1 & 0 \cr } } \right]$$

Now A⁴ = $$\left[ {\matrix{ 1 & 0 & 0 \cr 0 & 1 & 0 \cr 0 & 0 & 1 \cr } } \right]$$ = I

$\therefore$ A³¹ = A²⁸A³ = A³"

Let $\alpha$ be a root of the equation x² + x + 1 = 0 and the
matrix A = $${1 \over {\sqrt 3 }}\left[ {\matrix{ 1 & 1 & 1 \cr 1 & \alpha & {{\alpha ^2}} \cr 1 & {{\alpha ^2}} & {{\alpha ^4}} \cr } } \right]$$

then the matrix A³¹ is equal to

Solution

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Let $\alpha$ be a root of the equation x2 + x + 1 = 0 and the matrix A = $${1 \over {\sqrt 3 }}\left[ {\matrix{ 1 & 1 & 1 \cr 1 & \alpha & {{\alpha ^2}} \cr 1 & {{\alpha ^2}} & {{\alpha ^4}} \cr } } \right]$$ then the matrix A31 is equal to

Solution

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More Matrices and Determinants questions

Drill 25 more like these. Every day. Free.

Let $\alpha$ be a root of the equation x² + x + 1 = 0 and the
matrix A = $${1 \over {\sqrt 3 }}\left[ {\matrix{ 1 & 1 & 1 \cr 1 & \alpha & {{\alpha ^2}} \cr 1 & {{\alpha ^2}} & {{\alpha ^4}} \cr } } \right]$$

then the matrix A³¹ is equal to