Question

If \[ A = \begin{bmatrix} 2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 5 \end{bmatrix}, \] then \( A^{-1} \) is:

• A. \[ A^{-1} = \begin{bmatrix} \frac{1}{2} & 0 & 0 \\ 0 & \frac{1}{3} & 0 \\ 0 & 0 & \frac{1}{5} \end{bmatrix} \]
• B. \[ 30 \begin{bmatrix} \frac{1}{2} & 0 & 0 \\ 0 & \frac{1}{3} & 0 \\ 0 & 0 & \frac{1}{5} \end{bmatrix} \]
• C. \[ \frac{1}{30} \begin{bmatrix} 2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 5 \end{bmatrix} \]
• D. \[ \frac{1}{30} \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \]

Hint:

For a diagonal matrix \( A = {diag}(a_1, a_2, \dots, a_n) \), the inverse is \( A^{-1} = {diag}\left(\frac{1}{a_1}, \frac{1}{a_2}, \dots, \frac{1}{a_n}\right) \), provided \( a_i \neq 0 \).

Answer 1

The Correct Option is A

The inverse of a diagonal matrix is found by inverting each diagonal element. Given

\[ A = \begin{bmatrix} 2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 5 \end{bmatrix}, \]

the diagonal entries are 2, 3, and 5. Consequently:

\[ A^{-1} = \begin{bmatrix} \frac{1}{2} & 0 & 0 \\ 0 & \frac{1}{3} & 0 \\ 0 & 0 & \frac{1}{5} \end{bmatrix}. \]

This corresponds to option (A).

Final Answer: \[ \boxed{(A)} \]