यदि (IF) ,` A = [(cos theta , sin theta),(-sin theta, cos theta)]` सत्यापित करे कि `(A^(-1)) ^(-1) = A ` ।

Explanation

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 `