If \( \mathrm{A} \) is a square matrix such that \( |\mathrm{A}|\ne 0 \) and \( \mathrm{A}^2-\mathrm{A}+2\mathrm{I}=0 \) then \( \mathrm{A}^{-1} \) is equal to :(where I is an identity matrix of order as order of matrix A)
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( \displaystyle\frac{1}{2}(\mathrm{I}-\mathrm{A}) \)
Since \( |\mathrm{A}|\ne 0 \), \( \mathrm{A}^{-1} \) exists. Multiplying the given equation by \( \mathrm{A}^{-1} \):
\( \mathrm{A}^{-1}\mathrm{A}^2-\mathrm{A}^{-1}\mathrm{A}+2\mathrm{A}^{-1}=\mathrm{O} \)
\( \mathrm{A}-\mathrm{I}+2\mathrm{A}^{-1}=\mathrm{O} \)
\( \displaystyle \mathrm{A}^{-1}=\frac{1}{2}(\mathrm{I}-\mathrm{A}) \)