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Question: Answered & Verified by Expert
If $\mathrm{D}=\operatorname{diag}\left(\mathrm{d}_{1}, \mathrm{~d}_{2}, \ldots, \mathrm{d}_{\mathrm{n}}\right)$ where $\mathrm{d}_{1} \neq 0$, for $\mathrm{i}=1,2, \ldots, \mathrm{n}$, then $\mathrm{D}^{-1}$ is equal to
MathematicsApplication of DerivativesVITEEEVITEEE 2008
Options:
  • A $\mathrm{D}^{\mathrm{T}}$
  • B $\mathrm{D}$
  • C $\operatorname{Adj}$ (D)
  • D $\operatorname{diag}\left(\mathrm{d}_{1}^{-1}, \mathrm{~d}_{2}^{-1}, \ldots \mathrm{d}_{\mathrm{n}}^{-1}\right)$
Solution:
1807 Upvotes Verified Answer
The correct answer is: $\operatorname{diag}\left(\mathrm{d}_{1}^{-1}, \mathrm{~d}_{2}^{-1}, \ldots \mathrm{d}_{\mathrm{n}}^{-1}\right)$
$D=\operatorname{diag}\left(d_{1}, d_{2}, \ldots, d_{n}\right)$
$\mathrm{D}=\left|\begin{array}{cccccc}\mathrm{d}_{1} & 0 & 0 & - & - & 0 \\ 0 & \mathrm{~d}_{2} & 0 & - & - & 0 \\ 0 & 0 & \mathrm{~d}_{3} & - & - & 0 \\ - & - & - & - & - & - \\ - & - & - & - & - & - \\ 0 & 0 & 0 & 0 & - & 0\end{array}\right|$
$|\mathrm{D}|=\left|\begin{array}{cccccc}\mathrm{d}_{1} & 0 & 0 & - & - & 0 \\ 0 & \mathrm{~d}_{2} & 0 & - & - & 0 \\ 0 & 0 & \mathrm{~d}_{3} & - & - & 0 \\ - & - & - & - & - & - \\ - & - & - & - & - & - \\ 0 & 0 & 0 & 0 & - & 0\end{array}\right|$
$=\mathrm{d}_{1} \mathrm{~d}_{2} \mathrm{~d}_{3} \ldots \ldots \mathrm{d}_{\mathrm{n}}$


$\operatorname{adj}(\mathrm{D})=\frac{1}{\mathrm{~d}_{1} \mathrm{~d}_{2} \mathrm{~d}_{3} \ldots \mathrm{d}_{\mathrm{n}}}$
$\left[\begin{array}{cccccc}\mathrm{d}_{2} \mathrm{~d}_{3} \mathrm{~d}_{4} \cdot \mathrm{dn} & 0 & 0 & - & - & 0 \\ 0 & \mathrm{~d}_{1} \mathrm{~d}_{3} \mathrm{~d}_{4} \ldots \mathrm{dn} & 0 & - & - & 0 \\ 0 & 0 & \mathrm{~d}_{1} \mathrm{~d}_{2} \mathrm{~d}_{4} \ldots \mathrm{dn} & - & - & 0 \\ - & - & - & - & - & - \\ - & - & - & - & - & - \\ 0 & 0 & 0 & 0 & -\mathrm{d}_{1} \mathrm{~d}_{2} \mathrm{~d}_{3} \ldots \mathrm{d}_{\mathrm{n}-1}\end{array}\right]$
$=\left[\begin{array}{cccccc}\frac{1}{d_{1}} & 0 & 0 & - & - & 0 \\ 0 & \frac{1}{d_{2}} & 0 & - & - & 0 \\ 0 & 0 & \frac{1}{d_{3}} & - & - & 0 \\ - & - & - & - & - & - \\ - & - & - & - & - & - \\ 0 & 0 & 0 & 0 & - & \frac{1}{d_{n}}\end{array}\right]$
= $\begin{aligned} &\left[\begin{array}{cccccc}\mathrm{d}_{1}^{-1} & 0 & 0 & - & - & 0 \\ 0 & \mathrm{~d}_{2}^{-1} & 0 & - & - & 0 \\ 0 & 0 & \mathrm{~d}_{3}^{-1} & - & - & 0 \\ - & - & - & - & - & - \\ - & - & - & - & - & - \\ 0 & 0 & 0 & 0 & - & \mathrm{d}_{\mathrm{n}}^{-1}\end{array}\right] \\=& \operatorname{diag}\left(\mathrm{d}_{1}^{-1}, \mathrm{~d}_{2}^{-1}, \ldots, \mathrm{d}_{\mathrm{n}}^{-1}\right) \end{aligned}$

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