Question

# 011.10 cos α sin α0sin α-cos α

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Solution

## The given matrix is, A=[ 1 0 0 0 cosα sinα 0 sinα −cosα ] The inverse of a matrix exists only if it satisfies the condition of | A |≠0. So, the determinant of A is, | A |=1( − cos 2 α− sin 2 α ) =−1 Since, | A |≠0, so inverse of the matrix A exists. The formula to calculate the inverse of the matrix is, A −1 = adjA | A | (1) The cofactors of each element of the matrix is, A 11 = ( −1 ) 1+1 [ − cos 2 α− sin 2 α ] =−1 A 12 = ( −1 ) 1+2 [ 0( −cosα )−0sinα ] =0 A 13 = ( −1 ) 1+3 [ ( 0 )sinα−( 0 )cosα ] =0 A 21 = ( −1 ) 2+1 [ 0( −cosα )−( 0 )sinα ] =0 A 22 = ( −1 ) 2+2 [ −cosα−0 ] =−cosα A 23 = ( −1 ) 2+3 ( sinα−0 ) =−sinα A 31 = ( −1 ) 3+1 [ 0−0 ] =0 A 32 = ( −1 ) 3+2 [ 1×sinα−0×0 ] =−sinα A 33 = ( −1 ) 3+3 [ cosα−0 ] =cosα The adjoint of A will be, adjA=[ A 11 A 21 A 31 A 12 A 22 A 32 A 13 A 23 A 33 ] =[ −1 0 0 0 −cosα −sinα 0 −sinα cosα ] Substitute [ −1 0 0 0 −cosα −sinα 0 −sinα cosα ] for adjA and −1 for | A | in equation (1), A −1 =−1[ −1 0 0 0 −cosα −sinα 0 −sinα cosα ] =[ 1 0 0 0 cosα sinα 0 sinα −cosα ]

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