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If `A(theta)=[(sin theta, i cos theta),(i cos theta, sin theta)]`, then which of the following is not true ?A. `A(theta)^(-t)=A(pi-theta)`B. `A(theta)+A(pi+theta)` is a null matrixC. `A(theta)` is invertible for all `theta in R`D. `A(theta)^(-1)=A(-theta)` |
Answer» Correct Answer - A::B::C We have, `|A(theta)|=1` Hence, A is invertiable. `A(pi+theta)=A(pi + theta)=[(sin(pi+theta),i cos (pi+theta)),(i cos (pi+theta),sin (pi+theta))]` `=[(-sin theta,-i co theta),(-i cos theta,-sin theta)]=-A(theta)` adj `(A(theta))=[(sin theta,-i cos theta),(-i cos theta,sin theta)]` `implies A(theta)^(-1) =[(sin theta,-i cos theta),(-i cos theta,sin theta)]=A(pi-theta)` |
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