If `f (theta) = [[cos^(2) theta , cos theta sin theta,-sin theta],[cos theta sin theta , sin^(2) theta , cos theta ],[sin theta ,-cos theta , 0]] ` ,then f (`pi`/ 7) is

Options

A. symmetric
B. skew-symmetric
C. singular
D. non-singular

Answer

D

Explanation

`because abs(f(theta))= abs((Cos^(2) theta , costheta sin theta , - sin theta ),(costheta sin theta, sin^(2)theta,costheta),(sin theta,-costheta,0))`
Om multiplying in `R_(3)` by `costheta` and then take sommon `cos theta`
from `C_(1)`, then
` abs(f(theta))= abs((Cos theta , costheta sin theta , - sin theta ),( sin theta, sin^(2)theta,costheta),(sin theta,-cos^(2)theta,0))`
Applying `R_(2)rarr R_(3) - R_(3)`, we get
` abs(f(theta))= abs((Cos theta , costheta sin theta , - sin theta ),( 0, 1,costheta),(sin theta,-cos^(2)theta,0))=1`
Appluing `C_(2) rarr C_(2) - sin theta C_(1)`, then
` abs(f(theta))= abs((Cos theta ,0 , - sin theta ),( 0, 1,costheta),(sin theta,-1,0))=1`
`therefore f(pi/7)` is non- singlar matrix.