9月19日线性代数（B1）作业

$$A=\begin{pmatrix} -3 & -1 & -2 \\ 1 & 3 & 4 \\\end{pmatrix} , B=\begin{pmatrix} 2 & 2 & -2 \\ 4 & -1 & -4 \\ 4 & 3 & -3 \\\end{pmatrix} , C=\begin{pmatrix} 1 & 1 \\ -4 & 1 \\ -1 & -2 \\\end{pmatrix} \\ AB=\begin{pmatrix}-18 & -11 & 16\\30 & 11 & -26\end{pmatrix} , BC=\begin{pmatrix}-4 & 8\\12 & 11\\-5 & 13\end{pmatrix} , ABC=\begin{pmatrix}10 & -61\\12 & 93\end{pmatrix} \\ B^2=\begin{pmatrix}4 & -4 & -6\\-12 & -3 & 8\\8 & -4 & -11\end{pmatrix} , AC=\begin{pmatrix}3 & 0\\-15 & -4\end{pmatrix} , CA=\begin{pmatrix}-2 & 2 & 2\\13 & 7 & 12\\1 & -5 & -6\end{pmatrix}$$

P113 第5题

$$\begin{aligned} &\ \ \ \ \begin{pmatrix} x_1 & x_2 & \cdots & x_m \end{pmatrix} \begin{pmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{m1} & a_{m2} & \cdots & a_{mn} \\\end{pmatrix} \begin{pmatrix} y_1 \\ y_2 \\ \vdots \\ y_n \end{pmatrix} \\&= \begin{pmatrix} x_1 & x_2 & \cdots & x_m \end{pmatrix} \begin{pmatrix} a_{11}y_1+a_{12}y_2+ \cdots + a_{1n}y_n \\ a_{21}y_1 + a_{22}y_2 + \cdots + a_{2n}y_n \\ \vdots \\ a_{m1}y_1 + a_{m2}y_2 + \cdots + a_{mn}y_n \\\end{pmatrix} \\&=\sum_{i=1}^m\sum_{j=1}^nx_{i}a_{ij}y_j \end{aligned}$$

P113 第6题

( 1 )

$$设A=\begin{pmatrix} a & b \\ c & d \\\end{pmatrix}=\begin{pmatrix} 0 & 1 \\ 1 & 0 \\\end{pmatrix} \\ 则有 \begin{cases} a^2+bc=0 &(1) \\ ab+bd=1 &(2)\\ ac+cd=1 &(3)\\ bc+d^2=0 &(4)\end{cases} \\ 由(1),(4)知a=\pm d，又由(2),(3)知a=d, b=c=\frac{1}{2a}=\frac{1}{2d} 代入至(1)得a^2+\frac{1}{4a^2}=0，矛盾 \\ 故不存在这样的A。$$

( 2 )

$$设A=\begin{pmatrix} a & b \\ c & d \\\end{pmatrix}，A^2=\begin{pmatrix} 0 & 1 \\ -1 & 0 \\\end{pmatrix} \\ 则有 \begin{cases} a^2+bc=0 &(1) \\ ab+bd=1 &(2)\\ ac+cd=-1 &(3)\\ bc+d^2=0 &(4)\end{cases} \\ 由(1),(4)知a=\pm d，又由(2),(3)知a=d, b=\frac{1}{2a}, c=-\frac{1}{2a} \\代入至(1)得a^2=\frac{1}{4a^2} \\ 得a=\pm \frac{\sqrt2}{2} 即A=\begin{pmatrix} \frac{\sqrt2}{2} & \frac{\sqrt2}{2} \\ -\frac{\sqrt2}{2} & \frac{\sqrt2}{2} \\\end{pmatrix}或\begin{pmatrix} -\frac{\sqrt2}{2} & -\frac{\sqrt2}{2} \\ \frac{\sqrt2}{2} & -\frac{\sqrt2}{2} \\\end{pmatrix}$$

( 3 )

$$当A=\begin{pmatrix} -\frac{1}{2} & \frac{\sqrt{3}}{2} \\ -\frac{\sqrt{3}}{2} & -\frac{1}{2} \\\end{pmatrix}时， \begin{pmatrix} -\frac{1}{2} & \frac{\sqrt{3}}{2} \\ -\frac{\sqrt{3}}{2} & -\frac{1}{2} \\\end{pmatrix}^3=I$$

P113 第7题

(1)

$$设A=\begin{pmatrix} \cos\theta & \sin\theta \\ -\sin \theta & \cos\theta \end{pmatrix} \\ 下证A^k=\begin{pmatrix} \cos{k}\theta & \sin{k}\theta \\ -\sin{k} \theta & \cos{k}\theta \end{pmatrix} \\ A^1=\begin{pmatrix} \cos1\theta & \sin{1}\theta \\ -\sin{1} \theta & \cos{1}\theta \end{pmatrix} 故上式对于k=1时成立。 \\ 设A^k=\begin{pmatrix} \cos{k}\theta & \sin{k}\theta \\ -\sin{k} \theta & \cos{k}\theta \end{pmatrix} \\ 则A^{k+1}=A^k\cdot A=\begin{pmatrix} \cos{k}\theta & \sin{k}\theta \\ -\sin{k} \theta & \cos{k}\theta \end{pmatrix}\begin{pmatrix} \cos\theta & \sin\theta \\ -\sin \theta & \cos\theta \end{pmatrix}=\begin{pmatrix} \cos{(k+1)}\theta & \sin{(k+1)}\theta \\ -\sin{(k+1)} \theta & \cos{(k+1)}\theta \end{pmatrix} \\ 得证 \\ 故A^k=\begin{pmatrix} \cos{k}\theta & \sin{k}\theta \\ -\sin{k} \theta & \cos{k}\theta \end{pmatrix}$$

(2)

$$设A=\begin{pmatrix} a & b \\ -b & a \\\end{pmatrix}=\sqrt{a^2+b^2}\begin{pmatrix} \frac{a}{\sqrt{a^2+b^2}} & \frac{b}{\sqrt{a^2+b^2}} \\ -\frac{b}{\sqrt{a^2+b^2}} & \frac{a}{\sqrt{a^2+b^2}} \\\end{pmatrix}=\sqrt{a^2+b^2}\begin{pmatrix} \cos\theta & \sin\theta \\ -\sin \theta & \cos\theta \end{pmatrix}，其中\theta=\arccos{\frac{a}{\sqrt{a^2+b^2}}} \\ 由(1)结论可知，A^k={(a^2+b^2)}^{\frac{k}{2}}\begin{pmatrix} \cos{k}\theta & \sin{k}\theta \\ -\sin{k} \theta & \cos{k}\theta \end{pmatrix} \\ 故A^k={(a^2+b^2)}^{\frac{k}{2}}\begin{pmatrix} \cos{k}\theta & \sin{k}\theta \\ -\sin{k} \theta & \cos{k}\theta \end{pmatrix} \\ 即A^k=\begin{pmatrix} a & b \\ -b & a \\\end{pmatrix}^k={(a^2+b^2)}^{\frac{k}{2}}\begin{pmatrix} \cos{(k\arccos{\frac{a}{\sqrt{a^2+b^2}}})} & \sin{(k\arccos{\frac{a}{\sqrt{a^2+b^2}}})} \\ -\sin {(k\arccos{\frac{a}{\sqrt{a^2+b^2}}})} & \cos{(k\arccos{\frac{a}{\sqrt{a^2+b^2}}})} \end{pmatrix}$$

(5)

$$设A= \begin{pmatrix} a_1 \\ a_2 \\ \vdots \\ a_n \end{pmatrix} ，B= \begin{pmatrix} b_1 & b_2 & \cdots & b_n \end{pmatrix} \\ 则有C=\begin{pmatrix} a_{1}b_1 & a_{1}b_2 & \cdots & a_{1}b_n \\ a_{2}b_1 & a_{2}b_2 & \cdots & a_{2}b_n \\ \vdots & \vdots & \ddots & \vdots \\ a_{n}b_1 & a_{n}b_2 & \cdots & a_{n}b_n \\\end{pmatrix}=AB \\ C^k=(AB)^k=A(BA)^{k-1}B \\ 又BA=\begin{pmatrix}b_1 & b_2 & \cdots & b_n\end{pmatrix}\begin{pmatrix}a_1 \\ a_2 \\ \vdots \\ a_n\end{pmatrix}=\sum_{i=1}^n a_ib_i \\ \ \\ 故C^k={(\sum_{i=1}^n a_ib_i)}^{k-1} \cdot AB \\ C^k={(\sum_{i=1}^n a_ib_i)}^{k-1}\begin{pmatrix} a_{1}b_1 & a_{1}b_2 & \cdots & a_{1}b_n \\ a_{2}b_1 & a_{2}b_2 & \cdots & a_{2}b_n \\ \vdots & \vdots & \ddots & \vdots \\ a_{n}b_1 & a_{n}b_2 & \cdots & a_{n}b_n \\\end{pmatrix}$$