Evaluate: ∣∣∣x2−x+1x−1x+1x+1∣∣∣
We have, ∣∣∣x2−x+1x−1x+1x+1∣∣∣=∣∣∣x2−2x+2x−10x+1∣∣∣ [∵C1→C1−C2] =(x2−2x+2).(x+1)−(x−1).0=x3−2x2+2x+x2−2x+2=x3−x2+2
Evaluate the determinant (i) ∣∣∣cosθ−sinθsinθcosθ∣∣∣ (ii) ∣∣∣x2−x+1x−1x+1x+1∣∣∣