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Evaluation of a Determinant

If A = diag ...

Question

If A = diag $[d_{1}, d_{2}, d_{3}]$ then $a^{n}$ is equal to

A

diag

[d_{1}^{n - 1}, d_{2}^{n - 1}, d_{3}^{n - 1}]

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B

A

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C

diag

[d_{1}^{n}, d_{2}^{n}, d_{3}^{n}]

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D

none

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Solution

The correct option is D diag $[d_{1}^{n}, d_{2}^{n}, d_{3}^{n}]$
$A = ⎡ ⎢ ⎣ \begin{matrix} d_{1} & 0 & 0 0 & d_{2} & 0 0 & 0 & d_{3} \end{matrix} ⎤ ⎥ ⎦$

then $A^{2} = ⎡ ⎢ ⎣ \begin{matrix} d_{1} & 0 & 0 0 & d_{2} & 0 0 & 0 & d_{3} \end{matrix} ⎤ ⎥ ⎦ ⎡ ⎢ ⎣ \begin{matrix} d_{1} & 0 & 0 0 & d_{2} & 0 0 & 0 & d_{3} \end{matrix} ⎤ ⎥ ⎦$

$= ⎡ ⎢ ⎢ ⎣ \begin{matrix} d_{1}^{2} & 0 & 0 0 & d_{2}^{2} & 0 0 & 0 & d_{3}^{2} \end{matrix} ⎤ ⎥ ⎥ ⎦$

$A^{3} = A^{2} \cdot A = ⎡ ⎢ ⎢ ⎣ \begin{matrix} d_{1}^{3} & 0 & 0 0 & d_{2}^{3} & 0 0 & 0 & d_{3}^{3} \end{matrix} ⎤ ⎥ ⎥ ⎦$

Hence continuing like this

$A^{n} = d i a g [d_{1}^{n}, d_{2}^{n} d_{3}^{n}]$

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Q. Assertion :If

D = d i a g [d_{1}, d_{2}, . . . ., d_{n}]

D^{- 1} = d i a g [d_{1}^{- 1}, d_{2}^{- 1} . . . . ., d_{n}^{- 1}]

D = d i a g [d_{1}, d_{2}, . . . . . . d_{n}],

D^{n} = d i a g [d_{1}^{n}, d_{2}^{n} . . . . . ., d_{n}^{n}] .

d_{1}, d_{2}, d_{3}

are the diameters of the three escribed circles of a triangle, then

d_{1} d_{2} + d_{2} d_{3} + d_{3} d_{1}

\frac{a △^{2}}{r^{2}}

a

1, d_{1}, d_{2}, d_{3}, d_{4}

x^{5} = 1

then the value of expression :

E = \frac{ω - d_{1}}{ω^{2} - d_{1}} \cdot \frac{ω - d_{2}}{ω^{2} - d_{2}} \cdot \frac{ω - d_{3}}{ω^{2} - d_{3}} \cdot \frac{ω - d_{4}}{ω^{2} - d_{4}}

ω

is the cube root of unity]

D_{p} = ∣ ∣ ∣ ∣ ∣ \begin{matrix} p & 15 & 8 p^{2} & 35 & 9 p^{3} & 25 & 10 \end{matrix} ∣ ∣ ∣ ∣ ∣

D_{1} + D_{2} + D_{3} + D_{4} + D_{5}

D_{p}

∣ ∣ ∣ ∣ ∣ \begin{matrix} p & 15 & 8 p^{2} & 55 & 9 p^{3} & 225 & 10 \end{matrix} ∣ ∣ ∣ ∣ ∣

D_{1} + D_{2} + D_{3} + D_{4} + D_{5}

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