线性代数求解。如图第四题。

解：，此题可以考虑采用递推算法，设原行列式为Dn，则
Dn=xD(n-1)-yzD(n-2)································（1），且：D1=x，D2=x^2-yz，可得D3=x^3-2xyz
由（1）可得：Dn-([x+√(x^2-4yz)]/2)D(n-1)=([x-√(x^2-4yz)]/2)*{D(n-1)-([x+√(x^2-4yz)]/2)*D(n-2)}··········································································（3）
令bn=Dn-([x+√(x^2-4yz)]/2)D(n-1)，则b2={x^2-2yz-x√(x^2-4yz)}/2
bn=([x-√(x^2-4yz)]/2)*b(n-1)，解得bn
Dn-([x-√(x^2-4yz)]/2)D(n-1)=([x+√(x^2-4yz)]/2)*{D(n-1)-([x-√(x^2-4yz)]/2)*D(n-2)}··········································································（4）
令cn=Dn-([x-√(x^2-4yz)]/2)D(n-1)，则c2={x^2-2yz+x√(x^2-4yz)}/2
cn=([x+√(x^2-4yz)]/2)*c(n-1)，解得cn
由bn，cn可得：
Dn= {[{[x+√(x^2-4yz)]/2}^(n-1)]*{x^2-2yz+x√(x^2-4yz)}/2 - [{[x+√(x^2-4yz)]/2}^(n-1)]*{x^2-2yz+x√(x^2-4yz)}/2}/√(x^2-4yz)