CUET · MATHS · PYQ PAPER 2023
The matrix \(A = \begin{bmatrix} 2 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 3 \end{bmatrix}\) then \(A^{-1}\) is equal to:
- A \(\begin{bmatrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 2 \end{bmatrix}\)
- B \(\left[\begin{array}{ccc}\frac{1}{2} & 0 & 0 \\ 0 & -\frac{1}{6} & 0 \\ 0 & 0 & \frac{1}{2}\end{array}\right]\)
- C \(\left[\begin{array}{ccc}\frac{1}{2} & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & \frac{1}{3}\end{array}\right]\)
- D \(\begin{bmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & -3 \end{bmatrix}\)
Answer & Solution
Correct Answer
(C) \(\left[\begin{array}{ccc}\frac{1}{2} & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & \frac{1}{3}\end{array}\right]\)
Step-by-step Solution
Detailed explanation
\(A^{-1} = \left[\begin{array}{ccc}\frac{1}{2} & 0 & 0 \\ 0 & \frac{1}{-1} & 0 \\ 0 & 0 & \frac{1}{3}\end{array}\right]\) \(A^{-1} = \left[\begin{array}{ccc}\frac{1}{2} & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & \frac{1}{3}\end{array}\right]\)
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