Solve: `|[costheta,sintheta],[costheta,costheta]|=0`

Evaluate- |[costheta,sintheta],[sintheta,costheta]|

Show that costheta [[costheta, sintheta],[-sintheta, cos theta]] + sin theta [[sintheta ,-cos theta],[costheta, sin theta]] =1

costheta[[costheta, sintheta], [-sintheta, costheta]]+sintheta[[sintheta, -costheta], [costheta, sintheta]]=

Evaluate : |{:(costheta,-sintheta),(sintheta,costheta):}|

If A=[[costheta, sintheta], [-sintheta, costheta]] , then

If A=[[costheta, sintheta], [-sintheta, costheta]] , then |A^(-1)|=

The inverse of the matrix [[costheta, -sintheta, 0], [sintheta, costheta, 0], [0, 0, 1]] is

Then inverse of the matrix [[1, 0, 0], [0, costheta, sintheta], [0, sintheta, -costheta]] is

If A=[[costheta,sintheta],[-sintheta,costheta]] and beta=A^4+A then determinant of beta=

Show that every real orthogonal matrix is of any one of the forms {:[(costheta,-sintheta),(sintheta,costheta)]:}or{:[(costheta,sintheta),(sintheta,-costheta)]:}