已知矩阵函数sin(At)=(t+1,e^t;cos(t),2),求矩阵A?

上面一题取P = [1,0,-4;0,1,0;0,3,1], 可算得:
P^(-1)sin(tA)P = [sin(2t),13t·cos(2t),0;0,sin(2t),0;0,0,sin(t)] ①.
注意到sin([a,b;0,a]) = sin(a)cos([0,b;0,0])+cos(a)sin([0,b,0,0])
= [sin(a),b·cos(a);0,sin(a)],
因此①式右端 = sin(t[2,13,0;0,2,0;0,0,1]).
故sin(tA) = sin(tP[2,13,0;0,2,0;0,0,1]P^(-1)),
两边在t = 0处对t求导得:
A = P[2,13,0;0,2,0;0,0,1]P^(-1) = [2,1,4;0,2,0;0,3,1].
下面一题还是有些问题.
可算得等式右端行列式 = (1-t)e^(4t),
而左端行列式 = e^tr(tA) = e^(t·tr(A)),
于是得t·tr(A) = 4t+ln(1-t),
故tr(A)与t有关, A不可能为常矩阵.