If `D=diag(d_1,d_2,d_3,…,d_n)" where "d ne 0" for all " I = 1,2,…,n," then " D^(-1)`is equal to

Options

A. D
B. `diag(d_1^(-1)d_2^(-1),...,d_n^(-1))`
C. In
D. none of these

Answer

To find the inverse of the diagonal matrix \( D = \text{diag}(d_1, d_2, d_3, \ldots, d_n) \) where \( d_i \neq 0 \) for all \( i = 1, 2, \ldots, n \), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Diagonal Matrix**:
The matrix \( D \) is a diagonal matrix, which means all off-diagonal elements are zero. It can be represented 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. **Finding the Inverse of a Diagonal Matrix**:
The inverse of a diagonal matrix can be found by taking the reciprocal of each of the diagonal elements. Thus, the inverse \( D^{-1} \) will be:
\[ D^{-1} = \text{diag}\left(\frac{1}{d_1}, \frac{1}{d_2}, \frac{1}{d_3}, \ldots, \frac{1}{d_n}\right) \]

3. **Writing the Inverse Matrix**:
Therefore, we can write the inverse matrix explicitly 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. **Conclusion**:
Thus, we conclude that the inverse of the diagonal matrix \( D \) is:
\[ D^{-1} = \text{diag}\left(d_1^{-1}, d_2^{-1}, d_3^{-1}, \ldots, d_n^{-1}\right) \]

### Final Answer:
The correct answer is option B: \( D^{-1} = \text{diag}(d_1^{-1}, d_2^{-1}, d_3^{-1}, \ldots, d_n^{-1}) \).

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