线代强化NO6|矩阵|例题|小结

$$A^2+A-4E=0$$ $$(A-E)(A+2E)-2E=0$$ $$(A-E)(A+2E)=2E$$ $$(A-E{)}^{-1}=\frac{A+2E}{2}=\frac{A}{2}+E$$
【小结】已知某矩阵方程求某个矩阵的逆矩阵，用因式分解法。

$$A^{\ast }=\left(\begin{array}{cccc}{A}_{11}& A_{21}& \cdots & A_{n1}\\ A_{12}& A_{22}& \cdots & A_{n2}\\ ⋮& ⋮& & ⋮\\ A_{1n}& A_{2n}& \cdots & A_{nn}\end{array}\right)=\left(\begin{array}{cccc}-a_{11}& -a_{21}& \cdots & -a_{n1}\\ -a_{12}& -a_{22}& \cdots & -a_{n2}\\ ⋮& ⋮& & ⋮\\ -a_{1n}& -a_{n2}& \cdots & -a_{nn}\end{array}\right)=-A^T$$ $$|-A^T|=(-1{)}^3|A^T|=-|A|,|A^{\ast }|=|A{|}^2,|A{|}^2=-|A|$$ $$|A|(|A|+1)=0$$ $$A\ne 0,不妨设第一个行不为0.按第一行展开$$ $$|A|=a_{11}{A}_{11}+a_{12}{A}_{12}+a_{13}{A}_{13}=-a_{11}^2-a_{12}^2-a_{13}^2\ne 0$$ $$故|A|=-1$$

【总结公式】
$$若A_{ij}=a_{ij},A\ne 0,n>2,则有.$$ $$①A^{\ast }=A^T②|A|>0③|A|=|A{|}^{n-1}④|A|=1$$ $$⑤A为正交矩阵.$$
$$证:A\ne 0,不妨设A的第一行不等于0,$$ $$|A|=a_{11}{A}_{11}+a_{12}{A}_{12}+\cdots +a_{1n}{A}_{1n}$$ $$=a_{11}^2+a_{12}^2+\cdots +a_{1n}^2>0$$ $$|A^{\ast }|=|A{|}^{n-1},|A^{\ast }|=|A^T|=|A|.故|A|=|A{|}^{n-1}$$ $$因为|A|>0,|A{|}^{n-2}=1,|A|=1$$
$$A{A}^{\ast }=|A|E=E,A{A}^T=E,故A为正交矩阵.$$

$$【小结】1.已知a_{ij}与A_{ij}的关系求|A|，首先考虑用伴随矩阵的性质。$$ $$2.本题的结论可以推广：若a_{ij}=A_{ij}（A^{\ast }=A^T），则有$$ $$①|A|=\sum _{j=1}^n{a}_{ij}^2=\sum _{i=1}^n{a}_{ij}^2；$$ $$②|A|=|A{|}^{n-1}，如果A\ne 0，则有|A|>0，且当n>2时，|A|=1（此时A为正交矩阵）。$$

$$不推荐:$$ $$思路1:A^{-1}\to A\to A^{\ast }\to (A^{\ast }{)}^{-1},思路2:(A^{\ast }{)}^{-1}=(A^{-1}{)}^{\ast }$$ $$A{A}^{\ast }=|A|E⟹(A^{\ast }{)}^{-1}=\frac{A}{|A|}$$ $$推荐解:$$ $$\left(\begin{array}{ccccccc}1& 1& 1& |& 1& 0& 0\\ 1& 2& 1& |& 0& 1& 0\\ 1& 1& 3& |& 0& 0& 1\end{array}\right)\to \left(\begin{array}{ccccccc}1& 1& 1& |& 1& 0& 0\\ 0& 1& 0& |& -1& 1& 0\\ 0& 0& 2& |& -1& 0& 1\end{array}\right)\to \left(\begin{array}{ccccccc}1& 0& 0& |& \frac{5}{2}& -1& -\frac{1}{2}\\ 0& 1& 0& |& -1& 1& 0\\ 0& 0& 1& |& -\frac{1}{2}& 0& \frac{1}{2}\end{array}\right)$$ $$|A^{-1}|=\left|\begin{array}{ccc}1& 1& 1\\ 1& 2& 1\\ 1& 1& 3\end{array}\right|=\left|\begin{array}{ccc}1& 1& 1\\ 0& 1& 0\\ 0& 0& 2\end{array}\right|=2,|A|=\frac{1}{2}$$ $$(A^{\ast }{)}^{-1}=\left(\begin{array}{ccc}5& -2& -1\\ -2& 2& 0\\ -1& 0& 1\end{array}\right)$$

$$解:法1:特殊值法.令n=2,(A^{\ast }{)}^{\ast }=A,排除ABD,选C.$$ $$法2:(A^{\ast }{)}^{\ast }=|A^{\ast }|(A^{\ast }{)}^{-1}=|A{|}^{n-1}\cdot \frac{A}{|A|}=|A{|}^{n-2}A,选C.$$

$$【小结】求伴随矩阵有两种情况：$$ $$第一、A可逆，则计算或是讨论A^{\ast }的基本思路是用公式A^{\ast }=|A|A^{-1},$$ $$第二、如果矩阵不可逆或是否可逆未知，处理或计算其伴随矩阵的基本思路是用定义或是用公式A{A}^{\ast }=A^{\ast }A=|A|E。$$

$$解:法1:特殊值.A=\left(\begin{array}{ccc}2& 0& 0\\ 1& 1& 0\\ \frac{1}{2}& 0& \frac{3}{2}\end{array}\right)$$ $$A_{11}=(-1{)}^{1+1}\left|\begin{array}{cc}1& 0\\ 0& \frac{3}{2}\end{array}\right|=\frac{3}{2}$$ $$A_{21}=0,A_{31}=0$$
$$法2:A的属于特征值2的特征向量为(1,1,1{)}^T$$ $$A^{\ast }的属于特征值\frac{3}{2}的特征向量为(1,1,1{)}^T$$ $$A^{\ast }的每行元素之和为\frac{3}{2}.$$ $$A_{11}+A_{21}+A_{31}=\frac{3}{2}.$$

$$解:(A-2E)B=A$$ $$法1:|A-2E|=\left|\begin{array}{ccc}2& 2& 3\\ 1& -1& 0\\ -1& 2& 1\end{array}\right|=\left|\begin{array}{ccc}4& 2& 3\\ 0& -1& 0\\ 1& 2& 1\end{array}\right|=-1\ne 0,A-2E可逆,B=(A-2E{)}^{-1}A$$ $$法2:(A-2E\mid A)\stackrel{初等行变换}{\to }(E\mid B)$$
小结：
$$小结:三种矩阵方程A,B可逆$$ $$①AX=B法1:X=A^{-1}B,法2:(A\mid B)\to (E\mid X)$$ $$②XA=B法1:X=B{A}^{-1},法2:(A^T\mid B^T)\to (E\mid X^T)$$$$(XA{)}^T=B^T,A^T{X}^T=B^T$$$$③AXB=C,X=A^{-1}C{B}^{-1}$$ $$\begin{gathered} 因式分解6个字. \\ 从左看,从右看. \end{gathered}$$

$$解:AXA+BXB-AXB-BXA=E$$ $$AX(A-B)-BX(A-B)=E$$ $$(A-B)X(A-B)=E①$$ $$X=(A-B{)}^{-1}\cdot (A-B{)}^{-1}=[(A-B{)}^{-1}{]}^2=[(A-B{)}^2{]}^{-1}②$$ $$A-B=\left(\begin{array}{ccc}1& -1& -1\\ 0& 1& -1\\ 0& 0& 1\end{array}\right)$$ $$法1:推荐$$ $$\left(\begin{array}{ccccccc}1& -1& -1& |& 1& 0& 0\\ 0& 1& -1& |& 0& 1& 0\\ 0& 0& 1& |& 0& 0& 1\end{array}\right)\to \left(\begin{array}{ccccccc}1& -1& 0& |& 1& 1& 2\\ 0& 1& 0& |& 0& 1& 1\\ 0& 0& 1& |& 0& 0& 1\end{array}\right)$$ $$X=\left(\begin{array}{ccc}1& 1& 2\\ 0& 1& 1\\ 0& 0& 1\end{array}\right)\left(\begin{array}{ccc}1& 1& 2\\ 0& 1& 1\\ 0& 0& 1\end{array}\right)=\left(\begin{array}{ccc}1& 2& 5\\ 0& 1& 2\\ 0& 0& 1\end{array}\right)$$ $$法2:$$ $$(A-B{)}^2=\left(\begin{array}{ccc}1& -1& -1\\ 0& 1& -1\\ 0& 0& 1\end{array}\right)\left(\begin{array}{ccc}1& -1& -1\\ 0& 1& -1\\ 0& 0& 1\end{array}\right)=\left(\begin{array}{ccc}1& -2& -1\\ 0& 1& -2\\ 0& 0& 1\end{array}\right)$$ $$\left(\begin{array}{ccccccc}1& -2& -1& |& 1& 0& 0\\ 0& 1& -2& |& 0& 1& 0\\ 0& 0& 1& |& 0& 0& 1\end{array}\right)\to \left(\begin{array}{ccccccc}1& -2& 0& |& 1& 2& 5\\ 0& 1& 0& |& 0& 1& 2\\ 0& 0& 1& |& 0& 0& 1\end{array}\right)$$

$$解:(1)A^2=(E-\xi {\xi }^T)(E-\xi {\xi }^T)=E-2\xi {\xi }^T+\xi \underline{{\xi }^T\xi }{\xi }^T$$ $$=E+({\xi }^T\xi -2)\xi {\xi }^T=A的充要条件为{\xi }^T\xi =1$$ $$(2)法1:假设A可逆.A^2=A,A^{-1}A\cdot A=A^{-1}A,A=E⟹E-\xi {\xi }^T=E$$ $$⟹-\xi {\xi }^T=0,而\xi 为n维非零列向量,矛盾.故A不可逆.$$ $$法2:用特征值.\xi {\xi }^T的特征值0(n-1重),tr(\xi {\xi }^T)={\xi }^T\xi =1$$ $$A的特征值为1(n-1重),0,|A|=0,A不可逆.$$
小结
$$【小结】1.假设n阶矩阵A的秩为1，则有$$ $$(1)存在n维列向量\alpha ,\beta ，使得A=\alpha {\beta }^T；$$ $$(2)对于上述\alpha ,\beta ，有{\alpha }^T\beta ={\beta }^T\alpha =tr(A)；$$ $$(3)A^n={\left[tr(A)\right]}^{n-1}A；$$ $$(4)A的特征值为0(n-1重)，tr(A)。$$$$\alpha {\beta }^T的特征值为0(n-1重)，{\beta }^T\alpha 。$$$$(5)0的特征向量只看A的第一行tr(A)的特征向量为A的第一列$$
$$2.矩阵可逆的充要条件，假设A为n阶矩阵，则有$$ $$A可逆\iff |A|\ne 0$$ $$\iff r(A)=n$$ $$\iff 齐次线性方程组Ax=0仅有零解$$ $$\iff 非齐次线性方程组Ax=b有唯一解$$ $$\iff A的特征值中不含零$$
拓：
$$已知r(A)=1,证明:tr(A)=1时,有(A-E{)}^3=A-E$$ $$解:(A-E{)}^3=A^3-3A^2+3A-E$$ $$={\left[tr(A)\right]}^{2}A-3tr(A)\cdot A+3A-E$$ $$=A-3A+3A-E=A-E.$$

选D
$$【小结】初等变换会改变行列式的值，但是不会改变行列式的零性。$$

$$求A.$$ $$法1:\left(\begin{array}{ccc}1& 0& 0\\ 0& 0& 1\\ 0& 1& 0\end{array}\right)A\left(\begin{array}{ccc}1& 0& 0\\ -1& 1& 0\\ 0& 0& 1\end{array}\right)=\left(\begin{array}{ccc}-2& 1& -1\\ 1& -1& 0\\ -1& 0& 0\end{array}\right),$$ $$A={\left(\begin{array}{ccc}1& 0& 0\\ 0& 0& 1\\ 0& 1& 0\end{array}\right)}^{-1}\left(\begin{array}{ccc}-2& 1& -1\\ 1& -1& 0\\ -1& 0& 0\end{array}\right){\left(\begin{array}{ccc}1& 0& 0\\ -1& 1& 0\\ 0& 0& 1\end{array}\right)}^{-1}$$ $$法2:A=\left(\begin{array}{ccc}1& 0& 0\\ 0& 0& 1\\ 0& 1& 0\end{array}\right)\left(\begin{array}{ccc}-2& 1& -1\\ 1& -1& 0\\ -1& 0& 0\end{array}\right)\left(\begin{array}{ccc}1& 0& 0\\ 1& 1& 0\\ 0& 0& 1\end{array}\right)=\left(\begin{array}{ccc}-1& 1& -1\\ -1& 0& 0\\ 0& -1& 0\end{array}\right)$$
$$已知A=\left(\begin{array}{ccc}-1& 1& -1\\ -1& 0& 0\\ 0& -1& 0\end{array}\right),求tr(A^{-1})$$ $$法1:$$ $$\left(\begin{array}{ccccccc}-1& 1& -1& |& 1& 0& 0\\ -1& 0& 0& |& 0& 1& 0\\ 0& -1& 0& |& 0& 0& 1\end{array}\right)\to \left(\begin{array}{ccccccc}1& 0& 0& |& 0& -1& 0\\ 0& 1& 0& |& 0& 0& -1\\ -1& 1& -1& |& 1& 0& 0\end{array}\right)\to \left(\begin{array}{ccccccc}1& 0& 0& |& 0& -1& 0\\ 0& 1& 0& |& 0& 0& -1\\ 0& 0& 1& |& -1& 1& -1\end{array}\right)$$ $$tr(A^{-1})=0+0+(-1)=-1$$ $$法2:用伴随矩阵.设B=A^{-1},|A|=\left|\begin{array}{ccc}-1& 1& -1\\ -1& 0& 0\\ 0& -1& 0\end{array}\right|=-1,$$ $$A^{-1}=\frac{A^{\ast }}{|A|}=-A^{\ast }$$ $$tr(B)=tr(-A^{\ast })=-A_{11}-A_{22}-A_{33}=0+0-1=-1$$ $$法3:用特征值.$$ $$|A-\lambda E|=\left|\begin{array}{ccc}-1-\lambda & 1& -1\\ -1& -\lambda & 0\\ 0& -1& -\lambda \end{array}\right|=\left|\begin{array}{ccc}-1-\lambda & 1& -1\\ -1& -\lambda & 0\\ -1-\lambda & 0& -1+\lambda \end{array}\right|$$ $$=\left|\begin{array}{ccc}-\lambda & 1& -1\\ -1& -\lambda & 0\\ 0& 0& -1+\lambda \end{array}\right|=(-1-\lambda )({\lambda }^2+1)=0,特征值-1,i,-i$$ $$A^{-1}的特征值-1,\frac{1}{i},\frac{1}{-i},tr(A^{-1})=-1+\frac{1}{i}+\frac{1}{-i}=-1$$
改：
$$改:A=\left(\begin{array}{ccc}29& 1& 311\\ 0& 31& 0\\ 3& 7& 25\end{array}\right),求tr(A^{-1})$$ $$解:|A|=31\cdot (29\times 25-933)=\alpha$$ $$A^{-1}=\frac{1}{\alpha }{A}^{\ast }$$ $$tr(A^{-1})=tr\left(\frac{1}{\alpha }{A}^{\ast }\right)=\frac{1}{\alpha }(A_{11}+A_{22}+A_{33})$$ $$=\frac{1}{\alpha }\left|\begin{array}{cc}31& 0\\ 7& 25\end{array}\right|+\frac{1}{\alpha }\left|\begin{array}{cc}29& 311\\ 3& 25\end{array}\right|+\frac{1}{\alpha }\left|\begin{array}{cc}29& 1\\ 0& 31\end{array}\right|$$ $$改:A=\left(\begin{array}{ccc}29& 1& 311\\ 0& 31& 0\\ 315& 7& 25\end{array}\right),求tr(A^{-1})$$ $$解:用特征值:$$ $$\left(\begin{array}{cc}29& 311\\ 315& 25\end{array}\right)\left(\begin{array}{c}1\\ 1\end{array}\right)=340\left(\begin{array}{c}1\\ 1\end{array}\right)$$ $$用快速计算特征值方法得出A的特征值为31,340,-286$$ $$A^{-1}的特征值为\frac{1}{31},\frac{1}{340},\frac{1}{-286}$$ $$tr(A^{-1})=\frac{1}{31}+\frac{1}{340}-\frac{1}{286}=?$$

$$法1:Q=P\left(\begin{array}{ccc}1& 0& 0\\ 0& 0& 1\\ 0& 0& 1\end{array}\right),Q^{-1}AQ={\left(\begin{array}{ccc}1& 0& 0\\ 1& 1& 0\\ 0& 0& 1\end{array}\right)}^{-1}{P}^{-1}AP\left(\begin{array}{ccc}1& 0& 0\\ 1& 1& 0\\ 0& 0& 1\end{array}\right)$$ $$=\left(\begin{array}{ccc}1& 0& 0\\ -1& 1& 0\\ 0& 0& 1\end{array}\right)\left(\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 2\end{array}\right)\left(\begin{array}{ccc}1& 0& 0\\ 1& 1& 1\\ 0& 0& 1\end{array}\right)=\left(\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 2\end{array}\right)$$ $$法2:AP=P\left(\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 2\end{array}\right),A({\alpha }_1,{\alpha }_2,{\alpha }_3)=({\alpha }_1,{\alpha }_2,{\alpha }_3)\left(\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 2\end{array}\right)=({\alpha }_1,{\alpha }_2,2{\alpha }_3)$$ $$\{\begin{array}{l}A{\alpha }_1={\alpha }_1\\ A{\alpha }_2={\alpha }_2\\ A{\alpha }_3=2{\alpha }_3\end{array}$$ $${\alpha }_1+{\alpha }_2,{\alpha }_2为A的特征值1的两个线性无关的特征向量.$$ $${\alpha }_3为A的特征值2的特征向量.$$ $$Q^{-1}AQ=\left(\begin{array}{ccc}1& 0& 0\\ 1& 1& 0\\ 0& 0& 2\end{array}\right)$$ $$小结:若Q的列向量仍为A的特征向量,能用发2,若不是,只能用法1$$