线性代数 02.03 逆矩阵

$\color{blue}{\S 第二章 第三节 逆矩阵}$

$\color{blue}{一、逆矩阵}$

$定义8.设A为n阶方阵,如果有一个n阶方阵B,使\\ AB = BA = E, \\ 则称矩阵A是可逆的,并把矩阵B称为A的逆矩阵.\\ A的逆矩阵记为A^{-1}$

$\color{blue}{二、逆矩阵是唯一的}$

$证明:设B和C都是A的逆矩阵,则\\ B = BE = B(AC)\\ =(AB) C = EC \\ = C$

$\color{blue}{三、逆矩阵的有关定理}$

$定理1.方阵A可逆的充分必要条件是|A| \neq 0,\\ 且 A^{-1} = \dfrac{1}{|A|}A^{*},其中\\ A^{*} = \begin{pmatrix} A_{11} & A_{21} & \cdots & A_{n1} \\ A_{12} & A_{22} & \cdots & A_{n2} \\ \cdots & \cdots & \cdots & \cdots \\ A_{1n} & A_{2n} & \cdots & A_{nn} \\ \end{pmatrix}称为A的伴随矩阵.\\ A^{*}中元素是A的所有元素的代数余子式.$
$证明: 必要性:因为A可逆,则有A^{-1},使AA^{-1} = E\\ |A|\cdot |A^{-1}| = |E| = 1 \\ 所以 |A| \neq 0. \\ 充分性:由于 AA^{*} = \begin{pmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \cdots & \cdots & \cdots & \cdots \\ a_{n1} & a_{n2} & \cdots & a_{nn} \\ \end{pmatrix} \begin{pmatrix} A_{11} & A_{21} & \cdots & A_{n1} \\ A_{12} & A_{22} & \cdots & A_{n2} \\ \cdots & \cdots & \cdots & \cdots \\ A_{1n} & A_{2n} & \cdots & A_{nn} \\ \end{pmatrix} \\ = \begin{pmatrix} |A| & 0 & \cdots & 0 \\ 0 & |A| & \cdots & 0 \\ \cdots & \cdots & \cdots & \cdots \\ 0 & 0 & \cdots & |A| \\ \end{pmatrix} \\ = |A| \begin{pmatrix} 1 & 0 & \cdots & 0 \\ 0 & 1 & \cdots & 0 \\ \cdots & \cdots & \cdots & \cdots \\ 0 & 0 & \cdots & 1 \\ \end{pmatrix} \\ = |A|E \\ 即AA^{ * } = |A|E \\ 同理 A^ { * }A = |A|E,所以\\ AA^ { * } = A^ { * }A = |A|E \\ 因为|A| \neq 0,所以\\ A \dfrac{A^ { * } }{|A|} = \dfrac{ A^ { * } }{|A|} A = E \\ [令B = \dfrac{A^ { * } }{|A|} ,即得 AB = BA = E] \\ 由定义,知A^{-1} = \dfrac{1}{|A|} A^{ * }.$

$推论：若AB = E(或BA = E), 则B = A^{-1}.$
$证明: |A| \cdot |B| = |E| = 1.故 |A| \neq 0, \\ 因而 A^{-1}存在,于是\\ B = EB = (A^{-1}A)B \\ = A^{-1}(AB) = A^{-1}E \\ = A^{-1}$

运算律
$1) 若A可逆,则A^{-1}亦可逆,且(A^{-1})^{-1} = A;$
$2) 若A可逆,数\lambda \neq 0, 则\lambda A 可逆,且 \\ (\lambda A)^{-1} = \dfrac{1}{\lambda} A^{-1};$
$3) 若A,B为同阶的可逆矩阵,则AB也可逆,且 \\ (AB)^{-1} = B^{-1}A^{-1}.$
$证明:(AB) (B^{-1}A^{-1}) = A(BB^{-1})A^{-1} \\ = AEA^{-1} = AA^{-1} = E$

$4)若A可逆,则A^T也可逆,且(A^T)^{-1} = (A^{-1})^T.$
$证明：\\ \because (A^T)(A^{-1})^T = (A^{-1}A)^T \\ = E^T = E \\ 又 \because (A^T)(A^T)^{-1} = E\\ \therefore (A^T)^{-1} = (A^{-1})^T$

$注\\ 1:当|A| \neq 0时, k为正整数,\lambda, \mu为整数,有\\ 1) A^0 = E \\ 2) A^{-k} = (A^{-1})^k \\ 3) A^{\lambda} A^{\mu} = A^{\lambda + \mu} \\ 4) (A^{\lambda})^{\mu} = A^{\lambda \mu} \\ 2:当|A| \neq 0时, A为可逆矩阵,也称为非奇异矩阵,\\ 当|A| = 0时,A为不可逆矩阵,也称为奇异矩阵.$

$\color{blue}{四、逆矩阵的应用}$

$例1.解矩阵方程,求矩阵X.\\ \begin{pmatrix} 1 & 2 & 3 \\ 0 & 1 & 2 \\ 0 & 0 & 1 \\ \end{pmatrix} X \begin{pmatrix} 2 & 1 \\ 5 & 3 \\ \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \\ 1 & 0 \\ \end{pmatrix}$
$解:设\\ A = \begin{pmatrix} 1 & 2 & 3 \\ 0 & 1 & 2 \\ 0 & 0 & 1 \\ \end{pmatrix}, B = \begin{pmatrix} 2 & 1 \\ 5 & 3 \\ \end{pmatrix}, C = \begin{pmatrix} 1 & 0 \\ 0 & 1 \\ 1 & 0 \\ \end{pmatrix} \\ 则上式变成:AXB = C \\ 因为|A| = 1 \neq 0, |B| = 1 \neq 0.\\ 所以A^{-1},B^{-1}均存在,且\\ A^{-1} = \begin{pmatrix} 1 & -2 & 1 \\ 0 & 1 & -2 \\ 0 & 0 & 1 \\ \end{pmatrix}, B^{-1} = \begin{pmatrix} 3 & -1 \\ -5 & 2 \\ \end{pmatrix} \\ 于是 X = A^{-1}CB^{-1} = \begin{pmatrix} 1 & -2 & 1 \\ 0 & 1 & -2 \\ 0 & 0 & 1 \\ \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & 1 \\ 1 & 0 \\ \end{pmatrix} \begin{pmatrix} 3 & -1 \\ -5 & 2 \\ \end{pmatrix} \\ = \begin{pmatrix} 2 & -2 \\ -2 & 1 \\ 1 & 0 \\ \end{pmatrix} \begin{pmatrix} 3 & -1 \\ -5 & 2 \\ \end{pmatrix} = \begin{pmatrix} 16 & -6 \\ -11 & 4 \\ 3 & -1 \\ \end{pmatrix}$

$例2.设 A = \begin{pmatrix} 1 & 0 & 0 & 0 \\ -2 & 3 & 0 & 0 \\ 0 & -4 & 5 & 0 \\ 0 & 0 & -6 & 7 \\ \end{pmatrix},且B = (E + A)^{-1}(E - A)\\ 求(E + B)^{-1}$
$解:由 B = (E + A)^{-1}(E - A)得\\ (E + A)B = E - A \\ 从而, B + AB = E - A \\ E + B + A + AB = 2E \\ (E + B) + A(E + B) = 2E \\ (E + A)(E + B) = 2E \\ \dfrac{(E + A)}{2}(E + B) = E \\ 故(E + B)^{-1} = \dfrac{1}{2}(A + E) = \begin{pmatrix} 1 & 0 & 0 & 0 \\ -1 & 2 & 0 & 0 \\ 0 & -2 & 3 & 0 \\ 0 & 0 &-3 & 4 \\ \end{pmatrix}$

$例3.设A,B均为n阶方矩阵,若E-AB可逆,\\ 则E-BA也可逆,并求:(E - BA)^{-1}$
$证明: A - ABA = A- ABA \\ (E - AB)A = A(E - BA) \\ 两端乘以(E - AB)^{-1} \\ A = (E - AB)^{-1} A (E - BA) \quad ①\\ 又因为 \\ E = E -BA + BA \quad(把①代入)\\ =(E - BA) + B[(E - AB)^{-1} A (E - BA)] \\ = [E + B(E -AB)^{-1}A](E - BA) \\ 所以E-BA可逆,且 (E-BA)^{-1} = E + B(E -AB)^{-1}A$

$\color{blue}{五、几个常用公式}$

$1) AA ^ { * } = A ^ { * } A = |A|E$
$2) A^{*} = |A|A^{-1}$
$3) |A^{-1}| = |A|^{-1}$
$4) |\lambda A| = \lambda^n |A|$
$5) (\lambda A)^{-1} = \lambda^{-1} A^{-1}$

$例4.若|A| \neq 0,试证$
$(1) |A^{*}| = |A|^{n - 1};$
$(2) (A^{*})^{-1} = (A^{-1})^{*};$
$(3) (A^{*})^T = (A^T)^{*};$
$(4) (A^{*})^{*} = |A|^{n - 2}A;$
$(5) (kA)^{*} = k^{n - 1}A^{*}.$
证:
$(1) |A^{*}| = ||A|A^{-1}| = |A|^n |A^{-1}| = |A|^{n - 1};$
$(2) (A^{*})^{-1} = (|A|A^{-1})^{-1} = |A^{-1}|(A^{-1})^{-1} = (A^{-1})^{*};$
$(3) (A^{*})^T = (|A|A^{-1})^T = |A^T|(A^{-1})^T = |A^T|(A^T)^{-1} = (A^T)^{*};$
$(4) (A^{*})^{*} = |A^{*}|(A^{*})^{-1} = |A|^{n - 1}(|A|A^{-1})^{-1} = |A|^{n -2} A;$
$(5) (kA)^{*} = |kA|(kA)^{-1} = k^n|A|k^{-1}A^{-1} = k^{n - 1}|A|A^{-1} = k^{n - 1}A^{*}.$

$例5.设矩阵A、B满足A^{*}BA = 2BA - 8E,其中\\ A = \begin{pmatrix} 1 & 0 & 0 \\ 0 & -2 & 0 \\ 0 & 0 & 1 \\ \end{pmatrix},求B$
$解:由于|A| \neq 0,所以A可逆,在\\ A^{*}BA = 2BA - 8E\\ 两端分别左乘A,右乘A^{-1}得\\ |A|B = 2AB - 8E \\ 把|A| = -2代入 即2AB + 2B = 8E \\ AB + B = 4E\\ (A + E)B = 4E \\ B = 4(A + E)^{-1} \\ = 4 \left[\begin{pmatrix} 1 & 0 & 0 \\ 0 & -2 & 0 \\ 0 & 0 & 1 \\ \end{pmatrix} + \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{pmatrix} \right]^{-1} \\ = 4 \begin{pmatrix} 1 & 0 & 0 \\ 0 & -2 & 0 \\ 0 & 0 & 1 \\ \end{pmatrix} ^{-1} \\ = 4 \begin{pmatrix} \frac{1}{2} & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & \frac{1}{2} \\ \end{pmatrix} = \begin{pmatrix} 2 & 0 & 0 \\ 0 & -4 & 0 \\ 0 & 0 & 2 \\ \end{pmatrix}$