Since A is a singular matrix, ∣A∣=0.
∣A∣=1(−1−α)−3(−2−0)−1(2−0)=0
−1−α+6−2=0⇒α=3
The function is f(x)=0∫x(t2+2t+3)dt=3x3+x2+3x
Differentiating with respect to x, we get f′(x)=x2+2x+3=(x+1)2+2>0 for all x.
Thus, f(x) is a strictly increasing function on the interval [1,3].
The maximum value M occurs at x=3:
M=f(3)=333+32+3(3)=9+9+9=27
The minimum value m occurs at x=1:
m=f(1)=313+12+3(1)=313
We need to find 3(M−m):
3(M−m)=3(27−313)=81−13=68
Answer: 68