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Definition of a Determinant

If A = [ 0 ...

Question

If $A = ⎡ ⎢ ⎣ \begin{matrix} 0 & 1 & 0 0 & 0 & 1 p & q & r \end{matrix} ⎤ ⎥ ⎦$ ,then $p I + q A + r A^{2} =$

A

A

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B

A^{2}

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C

A^{3}

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D

A^{4}

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Solution

The correct option is C $A^{3}$
We have
$A^{2} = A A = ⎡ ⎢ ⎣ \begin{matrix} 0 & 1 & 0 0 & 0 & 1 p & q & r \end{matrix} ⎤ ⎥ ⎦ \times ⎡ ⎢ ⎣ \begin{matrix} 0 & 1 & 0 0 & 0 & 1 p & q & r \end{matrix} ⎤ ⎥ ⎦ = ⎡ ⎢ ⎣ \begin{matrix} 0 & 0 & 1 p & q & r p r & p + q r & q + r^{2} \end{matrix} ⎤ ⎥ ⎦$
$A^{3} = A^{2} A = ⎡ ⎢ ⎣ \begin{matrix} 0 & 0 & 1 p & q & r p r & p + q r & q + r^{2} \end{matrix} ⎤ ⎥ ⎦ \times ⎡ ⎢ ⎣ \begin{matrix} 0 & 1 & 0 0 & 0 & 1 p & q & r \end{matrix} ⎤ ⎥ ⎦ = ⎡ ⎢ ⎣ \begin{matrix} p & q & r p r & p + q r & q + r^{2} p q + r^{2} p & p r + q^{2} + q r^{2} & p + 2 q r + r^{3} \end{matrix} ⎤ ⎥ ⎦$
$∴ p I + q I + r A^{2} = p ⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 0 & 1 & 0 0 & 0 & 1 \end{matrix} ⎤ ⎥ ⎦ + q ⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 0 & 1 & 0 0 & 0 & 1 \end{matrix} ⎤ ⎥ ⎦ + r ⎡ ⎢ ⎣ \begin{matrix} 0 & 0 & 1 p & q & r p r & p + q r & q + r^{2} \end{matrix} ⎤ ⎥ ⎦$
$= ⎡ ⎢ ⎣ \begin{matrix} p & 0 & 0 0 & p & 0 0 & 0 & p \end{matrix} ⎤ ⎥ ⎦ + ⎡ ⎢ ⎣ \begin{matrix} q & 0 & 0 0 & q & 0 0 & 0 & q \end{matrix} ⎤ ⎥ ⎦ + ⎡ ⎢ ⎣ \begin{matrix} 0 & 0 & r p r & q r & r^{2} p r^{2} & p r + q r^{2} & q r + r^{3} \end{matrix} ⎤ ⎥ ⎦$
$= ⎡ ⎢ ⎣ \begin{matrix} p + 0 + 0 & 0 + q + 0 & 0 + 0 + r 0 + 0 + p r & p + 0 + q r & 0 + q + r^{2} 0 + p q + p r^{2} & 0 + q^{2} + p r + q r^{2} & p + q r + q r + r^{3} \end{matrix} ⎤ ⎥ ⎦$
$= ⎡ ⎢ ⎣ \begin{matrix} p & q & r p r & p + q r & q + r^{2} p q + r^{2} p & p r + q^{2} + q r^{2} & p + 2 q r + r^{3} \end{matrix} ⎤ ⎥ ⎦ = A^{3}$

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Similar questions

A = ∣ ∣ ∣ ∣ \begin{matrix} 0 & 1 & 0 0 & 0 & 1 p & q & r \end{matrix} ∣ ∣ ∣ ∣

I

is the identity matrix of order

3

, then show that

A^{3} = p I + q A + r A^{2}

.

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_{1}, a_{2}, a_{3}, a_{4}

are in AP, then

\frac{1}{\sqrt{a_{1}} + \sqrt{a_{2}}} + \frac{1}{\sqrt{a_{2}} + \sqrt{a_{3}}} + \frac{1}{\sqrt{a_{3}} + \sqrt{a_{4}}}

\frac{a_{2} a_{3}}{a_{1} a_{4}} = \frac{a_{2} + a_{3}}{a_{1} + a_{4}} = 3 (\frac{a_{2} - a_{3}}{a_{1} - a_{4}})

a_{1}

a_{2}

a_{3}

a_{4}

a_{1}, a_{2}, a_{3}, a_{4}

are terms of an AP, then the relation between them is

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