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AP EAMCET · Maths · Matrices

Let \(A=\left[\begin{array}{cc}1 & 2 \\ -2 & 1\end{array}\right]\) and \(B^{-1}=\left[\begin{array}{ll}1 & 1 \\ 0 & 2\end{array}\right]\). If \(\left(A B^{-1}\right)^{-1}=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]\), then \(2 b+5 c+10 d=\)

  1. A \(0\)
  2. B \(1\)
  3. C \(-1\)
  4. D \(2\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(1\)

Step-by-step Solution

Detailed explanation

\(A B^{-1}=\left[\begin{array}{cc}1 & 2 \\ -2 & 1\end{array}\right]\left[\begin{array}{ll}1 & 1 \\ 0 & 2\end{array}\right]=\left[\begin{array}{cc}1 & 5 \\ -2 & 0\end{array}\right]\) Now, \(\left|A B^{-1}\right|=0-(-10)=10\) and…