InterviewSolution
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InterviewSolution
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यदि (IF) ,` A = [(cos theta , sin theta),(-sin theta, cos theta)]` सत्यापित करे कि `(A^(-1)) ^(-1) = A ` । |
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Answer» दिया है , ` A = [(cos theta , sin theta),(-sin theta, cos theta)]` अब `।A।= ।(cos theta , sin theta),(-sin theta, cos theta)।= cos ^(2) theta + sin^(2) theta = 1 ne 0 ` अतः `A^(-1)` का अस्तित्व है । अब `A_(11) = a_(11) "अर्थात " cos theta ` का सहखण्ड (cofacot ) ` = (-1)^(1+1) cos theta = cos theta ` ` A_(12) = a_(12) "अर्थात" sin theta ` सहखण्ड (cofacot ) `= (-1)^(2 +1) (-sin theta) = sin theta ` ` A_(21) = a_(21)"अर्थात" - sin theta ` सहखण्ड (cofacot ) `= 9-1)^(2+1) sin theta = - sin theta ` ` A_(22) = a_(22) "अर्थात" cos theta ` सहखण्ड (cofacot ) `= (-1)^(2+2) cos theta = cos theta ` अब adj `A= [(A_(11) ,A_(12)),(A_(21),A_(22))]"का transpose " = [(A_(11) ,A_(12)),(A_(21),A_(22))]=[(cos theta , - sin theta ),(sin theta , cos theta )]` ` therefore A^(-1) = (adi A)/(|A|) = (adi A)/(1) = [(cos theta , - sin theta ),(sin theta , cos theta )]` ` therefore (A^(-1)) = [(cos theta , -(- sin theta) ),(-sin theta , cos theta )]=[(cos theta , sin theta ),(-sin theta , cos theta )]` ` therefore (A^(-1))^(-1)=(adj (A^(-1)))/(।A^(-1)।) = (adj (A^(-1)))/(1) ` ` = adj(A^(-1))=[(cos theta , sin theta ),(- sin theta , cos theta )]=A ` ltबरgtअतः `(A^(-1))^(-1) = A ` |
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