Inverse of a matrix

[khavar]

I'm coming along well with understanding the basics of matrices. I need help finding the inverse of the following:
[FONT=&quot][/FONT]Matrix

1-cosA	sinA
sinA	-cosA

My first step would be to find out whether or not the determinant is equal to zero.

Determinant(D)=1-cosA(-cosA)-(sin²A) = -cosA+cos²A-sin²A = -cosA+cos2A

Matrix^-1= (1/D)* adjugate

Therefore, (1/-cosA+cos2A) multiplied by:

-cosA	-sinA
-sinA	1-cosA

I don't know what to do now. Thank you for the help.

 

[girodd]

I'm coming along well with understanding the basics of matrices. I need help finding the inverse of the following:
Matrix

1-cosA	sinA
sinA	-cosA

My first step would be to find out whether or not the determinant is equal to zero.

Determinant(D)=1-cosA(-cosA)-(sin²A) = -cosA+cos²A-sin²A = -cosA+cos2A

Matrix^-1= (1/D)* adjugate

Therefore, (1/-cosA+cos2A) multiplied by:

-cosA	-sinA
-sinA	1-cosA

I don't know what to do now. Thank you for the help.

The determinant will be zero sometimes, depending on what A is. It is more useful to write the determinant as -cos(a) + 1 - 2cos(a)^2, which is a quadratic in cos(a). You can solve using the quadratic formula to determine the values of cos(a) which make the determinant zero and hence the matrix has no inverse.

 

[khavar]

Thank you, Girodd!

So, cos(a)=1 which corresponds with a=0 or 2pi.
cos(a)=-1/2, a=2pi\3

I see now.