CUET · MATHS · PYQ PAPER 2025
The inverse of the matrix \(A=\left[\begin{array}{ccc}\frac{1}{2} & 0 & 0 \\ 0 & \frac{1}{4} & 0 \\ 0 & 0 & \frac{1}{8}\end{array}\right]\) is
- A \(\frac{1}{64}\left[\begin{array}{lll}2 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 8\end{array}\right]\)
- B \(\frac{1}{64}\left[\begin{array}{lll}\frac{1}{2} & 0 & 0 \\ 0 & \frac{1}{4} & 0 \\ 0 & 0 & \frac{1}{8}\end{array}\right]\)
- C \(\left[\begin{array}{lll}2 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 8\end{array}\right]\)
- D \(\frac{1}{64}\left[\begin{array}{ccc}\frac{1}{32} & 0 & 0 \\ 0 & \frac{1}{16} & 0 \\ 0 & 0 & \frac{1}{8}\end{array}\right]\)
Answer & Solution
Correct Answer
(C) \(\left[\begin{array}{lll}2 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 8\end{array}\right]\)
Step-by-step Solution
Detailed explanation
\(A^{-1} = \left[\begin{array}{ccc}(\frac{1}{2})^{-1} & 0 & 0 \\ 0 & (\frac{1}{4})^{-1} & 0 \\ 0 & 0 & (\frac{1}{8})^{-1}\end{array}\right] = \left[\begin{array}{ccc}2 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 8\end{array}\right]\)
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