If `D=diag[d_(1),d_(2),…………..,d_(n)]` where `d_(i)!=0AAi=1,2,3,………..n` then `D^(-1)` is equal to

Options

A. Diagonal matrix
B. `I_(n)`
C. `diag(d_(1)^(-1),d_(2)^(-1),…..d_(n)^(-1))`
D. None

Answer

To find the inverse of a diagonal matrix \( D = \text{diag}[d_1, d_2, \ldots, d_n] \) where \( d_i \neq 0 \) for \( i = 1, 2, \ldots, n \), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Structure of the Diagonal Matrix**:
The matrix \( D \) can be expressed as:
\[ D = \begin{pmatrix} d_1 & 0 & 0 & \cdots & 0 \\ 0 & d_2 & 0 & \cdots & 0 \\ 0 & 0 & d_3 & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & d_n \end{pmatrix} \]

2. **Identify the Inverse of a Diagonal Matrix**:
The inverse of a diagonal matrix is also a diagonal matrix where each diagonal element is the reciprocal of the corresponding diagonal element of the original matrix. Thus, we have:
\[ D^{-1} = \text{diag}\left[\frac{1}{d_1}, \frac{1}{d_2}, \ldots, \frac{1}{d_n}\right] \]

3. **Write the Inverse Matrix**:
Therefore, the inverse matrix \( D^{-1} \) can be written as:
\[ D^{-1} = \begin{pmatrix} \frac{1}{d_1} & 0 & 0 & \cdots & 0 \\ 0 & \frac{1}{d_2} & 0 & \cdots & 0 \\ 0 & 0 & \frac{1}{d_3} & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & \frac{1}{d_n} \end{pmatrix} \]

4. **Final Result**:
Thus, the final result for the inverse of the diagonal matrix \( D \) is:
\[ D^{-1} = \text{diag}\left[\frac{1}{d_1}, \frac{1}{d_2}, \ldots, \frac{1}{d_n}\right] \]