线性代数 下面行列式求简单方法?

1个回答

  • (1)

    若b=0, 由行列式=0

    当b≠0时,

    ri-r1, i=2,3,4

    1+a 1 1 1

    -a -a 0 0

    -a 0 b 0

    -a 0 0 -b

    c1-r2+(a/b)c3-(a/b)c4

    a 1 1 1

    0 -a 0 0

    0 0 b 0

    0 0 0 -b

    行列式 = a^2b^2.

    当b=0时,上式也成立.

    (2)

    c1-(1/ai)c(i+1), i=1,2,...,n

    t 1 1 ... 1

    0 a1 0 ... 0

    0 0 a2 ... 0

    .

    0 0 0 ... an

    t = a0-1/a1-1/a2-...-1/an

    行列式 = (a0-1/a1-1/a2-...-1/an)a1a2...an