If A=[−221−3]A=\left[\begin{array}{cc}-2 & 2 \\ 1 & -3\end{array}\right]A=[−212−3], then sinAt\sin A tsinAt isChoose one optionA13[sin(−4t)+2sin(−t)−2sin(−4t)+2sin(−t)−sin(−4t)+sin(−t)2sin(−4t)+sin(−t)]\frac{1}{3}\left[\begin{array}{cc}\sin (-4 t)+2 \sin (-t) & -2 \sin (-4 t)+2 \sin (-t) \\ -\sin (-4 t)+\sin (-t) & 2 \sin (-4 t)+\sin (-t)\end{array}\right]31[sin(−4t)+2sin(−t)−sin(−4t)+sin(−t)−2sin(−4t)+2sin(−t)2sin(−4t)+sin(−t)]B[sin(−2t)sin(2t)sin(t)sin(−3t)]\left[\begin{array}{rr}\sin (-2 t) & \sin (2 t) \\ \sin (t) & \sin (-3 t)\end{array}\right][sin(−2t)sin(t)sin(2t)sin(−3t)]C13[sin(4t)+2sin(t)2sin(−4t)−2sin(−t)−sin(−4t)+sin(t)2sin(4t)+sin(t)]\frac{1}{3}\left[\begin{array}{lc}\sin (4 t)+2 \sin (t) & 2 \sin (-4 t)-2 \sin (-t) \\ -\sin (-4 t)+\sin (t) & 2 \sin (4 t)+\sin (t)\end{array}\right]31[sin(4t)+2sin(t)−sin(−4t)+sin(t)2sin(−4t)−2sin(−t)2sin(4t)+sin(t)]D13[cos(−t)+2cos(t)−2cos(−4t)+2sin(−t)−cos(−4t)+sin(−t)−2cos(−4t)+cos(−t)]\frac{1}{3}\left[\begin{array}{ll}\cos (-t)+2 \cos (t) & -2 \cos (-4 t)+2 \sin (-t) \\ -\cos (-4 t)+\sin (-t) & -2 \cos (-4 t)+\cos (-t)\end{array}\right]31[cos(−t)+2cos(t)−cos(−4t)+sin(−t)−2cos(−4t)+2sin(−t)−2cos(−4t)+cos(−t)]Submit50:50Try againView answerShowHideAnswer: BWorked explanation not available yet.Free practice · Your answers on this page are not saved.