Lf A- 2 -1- 3 -2 - then A5-

The correct option is B A A^2=[ 2 amp; 3; 3 amp; 2 ][ 2 amp; 3; 3 amp; 2 ]=[ 1 amp; 0; 0 amp; 1 ] A^3=AI=A ...

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Multiplication of Matrices

lf A=[[ 2 -...

Question

lf $A = [\begin{matrix} 2 & - 1 3 & - 2 \end{matrix}],$ then $A^{5} =$

I

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A

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- A

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A^{2}

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{(\begin{matrix} 2 & - 1 3 & - 2 \end{matrix})}^{n} = (\begin{matrix} 1 & 0 0 & 1 \end{matrix})

n

A = ⎡ ⎢ ⎣ \begin{matrix} 1 & 2 & 1 0 & 1 & - 1 3 & - 1 & 1 \end{matrix} ⎤ ⎥ ⎦

A^{2} - A =

a = cos \frac{2 π}{7} + i sin \frac{2 π}{7}, α = a + a^{2} + a^{4}

β = a^{3} + a^{5} + a^{6}

α, β

a_{1}, a_{2}, a_{3}, a_{4}, a_{5} > 0

\frac{a_{1}}{a_{2} + a_{3}} + \frac{a_{2}}{a_{3} + a_{4}} + \frac{a_{3}}{a_{4} + a_{5}} + \frac{a_{4}}{a_{5} + a_{1}} + \frac{a_{5}}{a_{1} + a_{2}} \geq \frac{5}{2}

A

A^{5} = 0

B = I + A + A^{2} + A^{3} + A^{4}

B

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