MHT CET · Maths · Matrices
Inverse of the matrix \(\left[\begin{array}{cc}0.8 & -0.6 \\ 0.6 & 0.8\end{array}\right]\) is
- A \(\left[\begin{array}{cc}0.8 & 0.6 \\ -0.6 & 0.8\end{array}\right]\)
- B \(\left[\begin{array}{ll}-0.8 & 0.6 \\ -0.6 & 0.8\end{array}\right]\)
- C \(\left[\begin{array}{cc}-0.8 & -0.6 \\ 0.6 & 0.8\end{array}\right]\)
- D \(\left[\begin{array}{cc}8 & -6 \\ 6 & 8\end{array}\right]\)
Answer & Solution
Correct Answer
(A) \(\left[\begin{array}{cc}0.8 & 0.6 \\ -0.6 & 0.8\end{array}\right]\)
Step-by-step Solution
Detailed explanation
Let \(A=\left[\begin{array}{cc}0.8 & -0.6 \\ 0.6 & 0.8\end{array}\right]\)
\(\therefore\) \(| A |=\left|\begin{array}{cc}0.8 & -0.6 \\ 0.6 & 0.8\end{array}\right|\)
\(=0.64+0.36\)
\(=1 \neq 0\)
\(\therefore\) \(A^{-1}=\left[\begin{array}{cc}0.8 & 0.6 \\ -0.6 & 0.8\end{array}\right]\)
\(\therefore\) \(| A |=\left|\begin{array}{cc}0.8 & -0.6 \\ 0.6 & 0.8\end{array}\right|\)
\(=0.64+0.36\)
\(=1 \neq 0\)
\(\therefore\) \(A^{-1}=\left[\begin{array}{cc}0.8 & 0.6 \\ -0.6 & 0.8\end{array}\right]\)
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