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Composite Function

Let D = dia...

Question

Let $D = d i a g (d_{1}, d_{2}, d_{3}, . . ., d_{n})$ where $d_{i} \neq 0 \forall i$ , then $D^{- 1}$ equals

A

D

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B

d i a g (d_{1}^{n}, d_{2}^{n}, . . ., d_{n}^{n})

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C

I_{n}

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D

d i a g (d_{1}^{- 1}, d_{2}^{- 1}, d_{3}^{- 1}, . . ., d_{n}^{- 1})

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Solution

The correct option is C $d i a g (d_{1}^{- 1}, d_{2}^{- 1}, d_{3}^{- 1}, . . ., d_{n}^{- 1})$
Given $D = ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} d_{1} & 0 & 0 & 0 & . . . & 0 0 & d_{2} & 0 & 0 & . . . & 0 0 & 0 & d_{3} & . . . & . . . & 0 ∣ . . . & . . . & . . . & . . . & . . . & . . . 0 & 0 & 0 & . . . & . . . & d_{n} \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦$
$| D | =$ product of diagonal elements $= d_{1} d_{2} . . . . d_{n}$
Now cofactor of $D_{11} = d_{2} d_{3} . . . d_{n}$
Cofactor of $D_{22} = d_{1} d_{3} . . . . d_{n}$
Cofactor of $D_{i j} = 0 \forall i \neq j$
So, $D^{- 1} = \frac{1}{| D |}$ (adj.of D) $= \frac{1}{d_{1} d_{2} . . . d_{n}} ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} d_{2} d_{3} . . . d_{n} & 0 & 0 & 0 0 & d_{1} d_{3} . . . d_{n} & 0 & 0 . . . & . . . & . . . & . . . . . . & . . . & . . . & . . . 0 & 0 & 0 & d_{1} d_{2} . . . d_{n - 1} \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦$
$= ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} 1 / d_{1} & 0 & 0 & 0 & . . . & 0 0 & 1 / d_{2} & 0 & 0 & . . . & 0 0 & 0 & 1 / d_{3} & 0 & . . . & . . . . . . & . . . & . . . & . . . & . . . & . . . . . . & . . . & . . . & . . . & . . . & . . . 0 & 0 & 0 & 0 & 0 & 1 / d_{n} \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦$
$= d i a g (d_{1}^{- 1}, d_{2}^{- 1}, d_{3}^{- 1}, . . . . ., d_{n}^{- 1})$

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Q. Assertion :If

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

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

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