高等工程数学(张韵华,汪琥庭,宋立功)—— 第一篇:线性代数

第一篇:线性代数

https://zhuanlan.zhihu.com/p/80690520 公式链接1

第一章:矩阵和向量​

第1题和第5题类型点题了

  1. A , B A, B A,B 是 3 阶方阵.已知 ∣ A ∣ = − 1 , ∣ B ∣ = 3 , |A|=-1,|B|=3, A=1,B=3, ∣ 2 A A 0 − B ∣ = 24 \left|\begin{array}{cr}2 A & A \\ 0 & -B\end{array}\right|= 24 2A0AB=24

    = > ∣ 2 A ∣ ∗ ∣ − B ∣ = 2 3 ∗ ∣ A ∣ ∗ ( − 1 ) 3 ∣ B ∣ = 8 ∗ ( − 1 ) ∗ ( − 1 ) 3 ∗ 3 = 24 => |2A|*|-B|=2^3*|A|*(-1)^3|B|=8*(-1)*(-1)^3*3=24 =>2AB=23A(1)3B=8(1)(1)33=24

  2. 证明:不存在n阶实方阵 A , B A,B A,B满足 A B − B A = I AB-BA=I ABBA=I

    n阶实方阵知道 n ≠ 0 n \neq 0 n=0,

    如果 A B − B A = I , 则 n = t r ( I ) = t r ( A B − B A ) = t r ( A B ) − t r ( B A ) = 0 AB-BA=I,则n=tr(I)=tr(AB-BA)=tr(AB)-tr(BA)=0 ABBA=I,n=tr(I)=tr(ABBA)=tr(AB)tr(BA)=0,矛盾。

  3. A , B A, B A,B n n n 阶方阵, λ ∈ R . \lambda \in \mathbf{R} . λR. 证明 :
    (1) ( λ A ) ∗ = λ n − 1 A ∗ ; (\lambda \boldsymbol{A})^{*}=\lambda^{n-1} \boldsymbol{A}^{*} ; (λA)=λn1A;
    (2) det ⁡ ( A ∗ ) = ( det ⁡ ( A ) ) n − 1 \operatorname{det}\left(A^{*}\right)=(\operatorname{det}(A))^{n-1} det(A)=(det(A))n1

    (1)伴随矩阵由代数余子式构成,A乘以 λ \lambda λ以后,代数余子式有公共系数 λ n − 1 \lambda^{n-1} λn1,故得证。

    (2) A ∗ A ∗ = ∣ A ∣ I n A*A^{*}=|A|I_n AA=AIn,求行列式,得到 ∣ A ∣ ∗ ∣ A ∗ ∣ = ∣ A ∣ n |A|*|A^{*}|=|A|^{n} AA=An,化简后得证。

  4. A A A n n n 阶方阵. 证明:
    (1) 当 A A A 可逆时, ( A − 1 ) T = ( A T ) − 1 , ( A − 1 ) ∗ = ( A ∗ ) − 1 ; \left(A^{-1}\right)^{\mathrm{T}}=\left(A^{\mathrm{T}}\right)^{-1},\left(A^{-1}\right)^{*}=\left(A^{*}\right)^{-1} ; (A1)T=(AT)1,(A1)=(A)1;
    (2) ( A ∗ ) T = ( A T ) ∗ \left(\boldsymbol{A}^{*}\right)^{\mathrm{T}}=\left(\boldsymbol{A}^{\mathrm{T}}\right)^{*} (A)T=(AT)

    (1) A ⊤ ⋅ ( A − 1 ) ⊤ = ( A ⋅ A − 1 ) ⊤ = E ⊤ = E A^{\top} \cdot\left(A^{-1}\right)^{\top}=\left(A \cdot A^{-1}\right)^{\top}=E^{\top}=E A(A1)=(AA1)=E=E
    A T ⋅ ( A ⊤ ) − 1 = E A^{T} \cdot\left(A^{\top}\right)^{-1}=E AT(A)1=E
    ( A − 1 ) ⊤ = ( A ⊤ ) − 1 \left(A^{-1}\right)^{\top}=\left(A^{\top}\right)^{-1} (A1)=(A)1

    A ⋅ A ∗ = ∣ A ∣ E A \cdot A^{*}=|A| E AA=AE
    A − 1 ⋅ ( A − 1 ) ∗ = ∣ A − 1 ∣ ⇒ ( A − 1 ) ∗ = A ⋅ ∣ A − 1 ∣ A^{-1} \cdot\left(A^{-1}\right)^{*}=\left|A^{-1}\right| \quad \Rightarrow\left(A^{-1}\right)^{*}=A \cdot\left|A^{-1}\right| A1(A1)=A1(A1)=AA1
    A ∗ = A − 1 ∣ A ∣ ( A ∗ ) − 1 = A ∣ A ∣ − 1 = A ∣ A − 1 ∣ A^{*}=A^{-1}|A| \quad\quad\left(A^{*}\right)^{-1}=A|A|^{-1}=A\left|A^{-1}\right| A=A1A(A)1=AA1=AA1

    得证;

    (2) A ∗ = ∣ A ∣ A − 1 ⇒ ( A ∗ ) ⊤ = ∣ A ∣ ⋅ ( A − 1 ) ⊤ = ∣ A T ∣ ⋅ ( A ⊤ ) − 1 = ( ∣ A ⊤ ∣ ) ∗ A^{*}=|A| A^{-1} \Rightarrow\left(A^{*}\right)^{\top}=|A| \cdot\left(A^{-1}\right)^{\top}=|A^{T}| \cdot\left(A^{\top}\right)^{-1}=\left(\mid A^{\top}\mid\right)^{*} A=AA1(A)=A(A1)=AT(A)1=(A)

  5. A , B A, B A,B 为 3 阶矩阵,且 ∣ A ∣ = 3 , ∣ B ∣ = 2 , ∣ A − 1 + B ∣ = 2. |A|=3,|B|=2,\left|A^{-1}+B\right|=2 . A=3,B=2,A1+B=2. 计算 det ⁡ ( A + B − 1 ) \operatorname{det}\left(A+B^{-1}\right) det(A+B1).

    ∣ A + B − 1 ∣ = ∣ E A + B − 1 E ∣ = ∣ ( B − 1 B ) A + B − 1 ( A − 1 A ) ∣ = ∣ B − 1 ( B + A − 1 ) A ∣ \begin{aligned}\left|\boldsymbol{A}+\boldsymbol{B}^{-1}\right| &=\left|\boldsymbol{E} \boldsymbol{A}+\boldsymbol{B}^{-1} \boldsymbol{E}\right| \\ &\left.=\mid \boldsymbol{( B}^{-1} \boldsymbol{B}\right) \boldsymbol{A}+\boldsymbol{B}^{-1}\left(\boldsymbol{A}^{-1} \boldsymbol{A}\right)|=| \boldsymbol{B}^{-1}\left(\boldsymbol{B}+\boldsymbol{A}^{-1}\right) \boldsymbol{A} \mid \end{aligned} A+B1=EA+B1E=(B1B)A+B1(A1A)=B1(B+A1)A.

    = ∣ B − 1 ∣ ⋅ ∣ B + A − 1 ∣ ⋅ ∣ A ∣ = 1 2 ⋅ 2 ⋅ 3 = 3 =\left|\boldsymbol{B}^{-1}\right| \cdot\left|\boldsymbol{B}+\boldsymbol{A}^{-1}\right| \cdot|\boldsymbol{A}|=\frac{1}{2} \cdot 2 \cdot 3=3 =B1B+A1A=2123=3.

  6. 设 A , B 为 n 阶 方 阵 , 且 I − A B 可 逆 . 证 明 : I − B A 也 可 逆 设A,B为n阶方阵,且I-AB可逆.证明:I-BA也可逆 A,Bn,IAB.:IBA.

    I − A B = E − A B I-A B=E-A B IAB=EAB.
    B × ( E − A B ) = B − B A B = ( E − B A ) B \quad B \times(E-A B)=B-B A B=(E-B A) B B×(EAB)=BBAB=(EBA)B
    ⇒ B = ( E − B A ) B ( E − A B ) − 1 \Rightarrow \quad B=(E-B A) B(E-A B)^{-1} B=(EBA)B(EAB)1

    E − B A ( 代 入 B ) E-B A \quad(代入B) EBA(B).
    = E − [ ( E − B A ) B ( E − A B ) − 1 ] A \left.=E-[(E-B A) B(E-A B)^{-1}\right] A =E[(EBA)B(EAB)1]A.

    移项: E − B A + [ ( E − B A ) B ( E − A B ) − 1 ] A = E \left.E-B A+[(E-B A) B(E-A B)^{-1}\right] A=E EBA+[(EBA)B(EAB)1]A=E

    提取 E − B A E-BA EBA,得到: ( E − B A ) ( E + B ( E − A B ) − 1 A ) = E ( E - B A ) ( E + B ( E - A B ) ^ { - 1 } A ) = E (EBA)(E+B(EAB)1A)=E.

    证得 E − B A 可 逆 E-BA可逆 EBA.

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