114年國安情報三等-線性代數詳解

 114年國家安全局國家安全情報人員考試

考 試 別：國家安全情報人員考試
等 別：三等考試
類科組別：數理組（選試英文）
科 目：線性代數

解答:$$\cases{x_1+x_2-x_3+x_4+x_5=1\\ 2x_1-x_2+3x_3-x_4+x_5=2\\ 3x_1+2x_2-x_3+2x_4+3x_5=4} \Rightarrow \left[ \begin{array} {rrrrr|r}1& 1& -1& 1& 1& 1\\  2&-1& 3& -1& 1& 2\\ 3& 2& -1& 2& 3& 4\end{array}\right] \\ \underrightarrow{R_2-2R_1\to R_2, R_3-3R_1\to R_3} \left[ \begin{array} {rrrrr|r}1 & 1 & -1 & 1 & 1 & 1\\0 & -3 & 5 & -3 & -1 & 0\\0 & -1 & 2 & -1 & 0 & 1\end{array}\right] \underrightarrow{-R_2/3 \to R_2, -R_3\to R_3} \\ \left[ \begin{array} {rrrrr|r}1 & 1 & -1 & 1 & 1 & 1\\0 & 1 & - \frac{5}{3} & 1 & \frac{1}{3} & 0\\0 & 1 & -2 & 1 & 0 & -1\end{array}\right] \underrightarrow{R_1-R_2\to R_1, R_3-R_2\to R_3} \left[ \begin{array} {rrrrr|r}1 & 0 & \frac{2}{3} & 0 & \frac{2}{3} & 1\\0 & 1 & - \frac{5}{3} & 1 & \frac{1}{3} & 0\\0 & 0 & - \frac{1}{3} & 0 & - \frac{1}{3} & -1\end{array}\right] \\ \underrightarrow{-3R_3\to R_3}  \left[ \begin{array} {rrrrr|r}1 & 0 & \frac{2}{3} & 0 & \frac{2}{3} & 1\\0 & 1 & - \frac{5}{3} & 1 & \frac{1}{3} & 0\\0 & 0 & 1 & 0 & 1 & 3\end{array}\right] \underrightarrow{R_1-(2/3)R_3\to R_1, R_2+(5/3)R_3 \to R_2} \\ \left[ \begin{array} {rrrrr|r}1 & 0 & 0 & 0 & 0 & -1\\0 & 1 & 0 & 1 & 2 & 5\\0 & 0 & 1 & 0 & 1 & 3\end{array}\right] \Rightarrow \cases{x_1=-1\\ x_2+x_4+ 2x_5=5\\ x_3+x_5=3}\\ \Rightarrow 解集合: \bbox[red, 2pt]{\{(-1,s,3-t,5-s-2t,t)\mid s,t\in \mathbb R\}}$$

解答:$$\textbf{(一)}\cases{p(x)=1 \Rightarrow p'(x)=0 \Rightarrow T(p(x)) =0+p(x+2)=1\\ p(x)=x \Rightarrow p'(x)=1 \Rightarrow T(p(x)) =(x+1)+x+2=2x+3\\ p(x)=x^2 \Rightarrow p'(x)=2x \Rightarrow T(p(x)) =(x+1)(2x)+ (x+2)^2= 3x^2+6x+4 } \\ \Rightarrow \bbox[red, 2pt]{[T]_\beta =\begin{bmatrix}1& 3& 4\\ 0& 2& 6\\ 0& 0& 3 \end{bmatrix} }\\ \cases{p(x)=x-1 \Rightarrow T(p(x))= 2x+2=0(x-1)+2(x+1)+0(x^2-x)\\ p(x)=x+1 \Rightarrow T(p(x)) = 2x+4= -1(x-1)+3(x+1)+0(x^2-x)\\ p(x)=x^2-x \Rightarrow T(p(x)) = 3x^2+4x+1 =3(x-1)+4(x+1)+3(x^2-x)} \\ \Rightarrow \bbox[red, 2pt]{[T]_\gamma =\begin{bmatrix} 0& -1& 3\\ 2& 3& 4\\ 0& 0& 3 \end{bmatrix} } \\\textbf{(二) }\cases{1=-{1\over 2}(x-1)+{1\over 2}(x+1) \\ x={1\over 2}(x-1)+{1\over 2}(x+1) \\ x^2={1\over 2}(x-1)+{1\over 2}(x+1)+1(x^2-x)} \Rightarrow \bbox[red, 2pt]{C= \begin{bmatrix} -{1\over 2} &{1\over 2}& {1\over 2} \\ {1\over 2} & {1\over 2} & {1\over 2} \\0& 0& 1\end{bmatrix}}$$

解答:$$A=\begin{bmatrix} 0 & 1 & 1 \\1 & 3 & 1 \\ 1 & 1 & 0\end{bmatrix}  \Rightarrow \det(A-\lambda I) = -(\lambda+1)(\lambda^2- 4\lambda+1) =0 \Rightarrow \lambda= -1,2\pm \sqrt 3\\ \cases{\lambda_1=-1 \Rightarrow \text{ eigenvector }v_1=[-1,0,1]^T \\ \lambda_2=2-\sqrt 3 \Rightarrow \text{ eigenvector }v_2=[1,1-\sqrt 3,1]^T \\ \lambda_3=2+\sqrt 3 \Rightarrow \text{ eigenvector }v_3=[1,1+\sqrt 3,1]^T } \\ \text{By Gram-Schmidt process, we have }\cases{e_1=[-{\sqrt 2\over 2},0,{\sqrt 2\over 2}]^T \\ e_2=[{\sqrt 2\over 2\sqrt{3-\sqrt 3}},{\sqrt 2-\sqrt 6\over 2\sqrt{3-\sqrt 3}}, {\sqrt 2\over 2\sqrt{3-\sqrt 3}}]^T \\ e_3=[{\sqrt 2\over 2\sqrt{3+\sqrt 3}}, {\sqrt 2+\sqrt 6\over 2\sqrt{3+\sqrt 3}},{\sqrt 2\over 2\sqrt{3+\sqrt 3}}]^T} \Rightarrow P=[e_1\; e_2\; e_3]\\ \Rightarrow P= \bbox[red, 2pt]{\begin{bmatrix}-{\sqrt 2\over 2}& {\sqrt 2\over 2\sqrt{3-\sqrt 3} } & {\sqrt 2\over 2\sqrt{3+\sqrt 3}} \\ 0&{\sqrt 2-\sqrt 6\over 2\sqrt{3-\sqrt 3}} & {\sqrt 2+\sqrt 6\over 2\sqrt{3+\sqrt 3}} \\ {\sqrt 2\over 2}& {\sqrt 2\over 2\sqrt{3-\sqrt 3}} & {\sqrt 2\over 2\sqrt{3+\sqrt 3}}\end{bmatrix}}$$

解答:$$u\in W \Rightarrow u=\begin{pmatrix} x_1\\ x_2\\ x_3\\ -x_1-2x_2-3x_3\end{pmatrix} =x_1\begin{pmatrix} 1\\ 0\\ 0\\ -1\end{pmatrix} +x_2 \begin{pmatrix} 0\\ 1\\ 0\\ -2\end{pmatrix} +x_3 \begin{pmatrix} 0\\ 0\\ 1\\ -3\end{pmatrix} =x_1\vec v_1+ x_2 \vec v_2+ x_3 \vec v_3\\ \Rightarrow \left\{ \vec v_1, \vec v_2, \vec v_3\right\} \text{ is a basis of W} \Rightarrow \text{By Gram-Schmit process, we have} \\ \vec u_1= \vec v_1 \Rightarrow \vec e_1={\vec u_1\over |\vec u_1|} =\begin{pmatrix} \sqrt 2/2\\ 0\\ 0\\ -\sqrt 2/2\end{pmatrix} \\ \vec u_2=\vec v_2-(\vec v_2\cdot \vec e_1)\vec e_1 =\begin{pmatrix} -1\\ 1\\ 0\\ -1\end{pmatrix} \Rightarrow \vec e_2={\vec u_2\over |\vec u_2|} =\begin{pmatrix} -\sqrt 3/3\\ \sqrt 3/3\\ 0\\ -\sqrt 3/3\end{pmatrix}  \\ \vec u_3=\vec v_3-(\vec v_3\cdot \vec e_1)\vec e_1-(\vec v_3\cdot \vec e_2)\vec e_2 =\begin{pmatrix} -1/2\\ -1\\ 1\\ -1/2\end{pmatrix} \Rightarrow \vec e_3={\vec u_3\over |\vec u_3|} =\begin{pmatrix} -\sqrt{10}/10\\ -\sqrt{10}/5\\ \sqrt{10}/5\\ -\sqrt {10}/10\end{pmatrix} \\ \Rightarrow \text{ an orthogonal basis: }\{\vec e_1, \vec e_2, \vec e_3\} = \bbox[red, 2pt]{\left\{\begin{pmatrix} \sqrt 2/2\\ 0\\ 0\\ -\sqrt 2/2\end{pmatrix}, \begin{pmatrix} -\sqrt 3/3\\ \sqrt 3/3\\ 0\\ -\sqrt 3/3\end{pmatrix}, \begin{pmatrix} -\sqrt{10}/10\\ -\sqrt{10}/5\\ \sqrt{10}/5\\ -\sqrt {10}/10\end{pmatrix}\right\}}$$

解答:$$u\in \text{span}(\{v_1-2v_2,2v_1+3v_2\}) \Rightarrow u=a(v_1-2v_2)+b(2v_1+3v_2), a,b\in \mathbb R \\=(a+2b)v_1 +(-2a+3b)v_2 =(-a+{8\over 3}b)(v_1+v_2)+(a-{b\over 3})(2v_1-v_2)\in \text{span}(\{v_1+ v_2,2v_1 -v_2\}) \\ \Rightarrow   \text{span}(\{v_1- 2v_2,2v_1+3v_2\})  \subseteq \text{span}(\{v_1+v_2,2v_1 -v_2\}) \cdots(1)\\ u \in\text{span}(\{v_1+v_2,2v_1-v_2\}) \Rightarrow u= a(v_1+v_2)+ b(2v_1-v_2), a,b\in \mathbb R \\ \Rightarrow u=(a+2b)v_1+ (a-b)v_2 ={a+ 8b\over 7}(v_1-2v_2)+{3a+ 3b\over 7}(2v_1+3v_2) \in \text{span}(\{v_1- 2v_2,2v_1 +3v_2\}) \\ \Rightarrow \text{span}(\{v_1+v_2,2v_1-v_2\}) \subseteq\text{span}(\{v_1 -2v_2,2v_1 +3v_2\}) \cdots(2) \\ 由(1)及(2)可知:\text{span}(\{v_1+v_2,2v_1-v_2\})  = \text{span}(\{v_1 -2v_2,2v_1+3v_2\}) \quad \bbox[red, 2pt]{故得證}$$

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