Question

If A = `[(costheta, theta),(0, costheta)]`, B = `[(sintheta, 0),(0, sintheta)]` then show that A² + B² = I

Answer 1

L.H.S. = A² + B² 

A² = `[(costheta, 0),(0, costheta)][(costheta, 0),(0, costheta)]`

= `[(cos^2theta, 0),(0, cos^2theta)]`

B² = `[(sintheta, 0),(0, sintheta)] [(sintheta, 0),(0, sintheta)]`

= `[(sin^2theta, 0),(0, sin^2theta)]`

A² + B² = `[(cos^2theta, 0),(0, cos^2theta)] + [(sin^2theta, 0),(0, sin^2theta)]`

= `[(sin^2theta + cos^2theta, 0),(0, sin^2theta + cos^2theta)]`

= `[(1, 0),(0, 1)]`

= I

= R.H.S.

Hence proved.