计算行列式的值:
Dn+1=∣∣∣∣∣∣∣∣∣∣∣anan−1⋮a1(a−1)n(a−1)n−1⋮a−11…………(a−n)n(a−n)n−1⋮a−n1∣∣∣∣∣∣∣∣∣∣∣
解:将行列式上下翻转,得
Dn+1=(−1)2n(n+1)∣∣∣∣∣∣∣∣∣∣∣1a⋮an−1an1a−1⋮(a−1)n−1(a−1)n…………1a−n⋮(a−n)n−1(a−nn∣∣∣∣∣∣∣∣∣∣∣
此行列式为范德蒙德行列式.故
Dn+1=(−1)2n(n+1)n+1≥i≥j≥1∏[(a−i+1)−(a−j+1)]=(−1)2n(n+1)n+1≥i≥j≥1∏[−(i−j)]=(−1)2n(n+1)×(−1)2n+n−1+⋯+1×n+1≥i≥j≥1∏(i−j)=n+1≥i≥j≥1∏(i−j).