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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 [ cosα0 ] =cosα

A 23 = ( 1 ) 2+3 ( sinα0 ) =sinα

A 31 = ( 1 ) 3+1 [ 00 ] =0

A 32 = ( 1 ) 3+2 [ 1×sinα0×0 ] =sinα

A 33 = ( 1 ) 3+3 [ cosα0 ] =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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