• 2022-05-29
    设\(A\)为\(n\)阶方阵,\(\left| A \right| = 2 \),则\(\left| {\left| A \right|{A^T}} \right|=\)
    A: \({2^{n + 1}} \)
    B: \({2^{n }}\)
    C: \({2^{n - 1}}\)
    D: \(2\)
  • A

    内容

    • 0

      设\( A,B \)均为\( n \)阶方阵,则必有( ) A: \( \left| {A + B} \right| = \left| A \right| + \left| B \right| \) B: \( AB = BA \) C: \( \left| {AB} \right| = \left| {BA} \right| \) D: \( {\left( {A + B} \right)^{ - 1}} = {A^{ - 1}} + {B^{ - 1}} \)

    • 1

      \( {1 \over {1 + x}} \)的麦克劳林公式为( )。 A: \( {1 \over {1 + x}} = 1 + x + { { {x^2}} \over 2} + \cdots + { { {x^n}} \over {n!}} + o\left( { { x^n}} \right) \) B: \( {1 \over {1 + x}} = 1 + x + {x^2} + \cdots + {x^n} + o\left( { { x^n}} \right) \) C: \( {1 \over {1 + x}} = 1 - x + {x^2} - \cdots + {( - 1)^n}{x^n} + o\left( { { x^n}} \right) \) D: \( {1 \over {1 + x}} = 1 - x - { { {x^2}} \over 2}- \cdots - { { {x^n}} \over {n!}} + o\left( { { x^n}} \right) \)

    • 2

      设\( n \) 阶方阵 \( A,B \)相似,则 \( \left| A \right| = \left| B \right| \).

    • 3

      函数$y = \ln x$,则${\left( {\ln x} \right)^{\left( n \right)}} = {\left( { - 1} \right)^{n - 1}}{{\left( {n - 1} \right)!} \over {{x^n}}}$。( )

    • 4

      \( {1 \over {1 + x}} \)的麦克劳林公式为( ). A: \( {1 \over {1 + x}} = 1 + x + { { {x^2}} \over 2} + \cdots + { { {x^n}} \over {n!}} + o\left( { { x^n}} \right) \) B: \( {1 \over {1 + x}} = 1 + x + {x^2} + \cdots + {x^n} + o\left( { { x^n}} \right) \) C: \( {1 \over {1 + x}} = 1 - x + {x^2} - \cdots + {( - 1)^n}{x^n} + o\left( { { x^n}} \right) \)