Question

011.10 cos α sin α0sin α-cos α

Answer 1

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 ]