114年台大碩士班-線性代數(C)詳解

 國立臺灣大學114學年度碩士班招生考試

科目:線性代數(C)

解答: $$(a)\bbox[red, 2pt]{\text{false}}:\text{Suppose }\vec u \text{ and }\vec v \text{ are linearly independent, then }S=\{\vec v, 2\vec v, \vec u\} \text{ is a linearly dependent set.} \\ \qquad \text{ But }\vec u \text{ cannot be  a combination of }\vec v \text{ and }2\vec v \\(b) \bbox[red, 2pt]{\text{true}}: A=PBP^{-1} \Rightarrow  A\vec v = PBP^{-1  }\vec v=\lambda v \Rightarrow B(P^{-1} \vec v) =\lambda (P^{-1}\vec v)  \\\qquad \Rightarrow A \text{ and }B\text{ have the same eigenvalue }\lambda\\ (c)\bbox[red, 2pt]{\text{false}}:這題在考英文，不是考數學. 無論\text{system of linear equations}有幾個解，\\\qquad \text{ homogeneous system} 一定有一個解\vec x= \vec 0 \\(d)\bbox[red, 2pt]{\text{false}}: \text{If }ker(A-\lambda I)\gt 1(有重根), \text{ then you can   find at least two independent vectors.} \\(e)\bbox[red, 2pt] {\text{false}}: \text{Only the first (not both) component}$$

解答: $$2M-A= \begin{bmatrix}8 & 4 & 2&6 \\10 & 2 &2& 8  \\6 & 2 &4& 12 \end{bmatrix} -\begin{bmatrix}5 & 3 & 1& 2 \\6 & 2 &1& 5  \\1 & 0 &3& 3 \end{bmatrix} =\begin{bmatrix}3 & 1 & 1& 4 \\4 & 0 &1 & 3  \\5 & 2 &1 & 9 \end{bmatrix} \\ \Rightarrow  \sum_{ij} (2M-A)_{ij} = 3+1+1+4+ 4+1+3+5+2+1+9 =\bbox[red, 2pt]{34}$$

解答: $$A=\begin{bmatrix}1 & 1 & 0 \\ 0& -1& 0  \\1 & 2 & 0 \end{bmatrix} \Rightarrow \cases{ A\begin{bmatrix} 0 \\ 0 \\  0 \end{bmatrix} =  A\begin{bmatrix} 0 \\ 0 \\  1 \end{bmatrix} =\begin{bmatrix} 0 \\ 0 \\  0 \end{bmatrix} \\[1ex] A\begin{bmatrix} 1 \\ 0 \\  0 \end{bmatrix} =\begin{bmatrix} 1 \\ 0 \\  1 \end{bmatrix} \\[1ex] A\begin{bmatrix} 0 \\ 1 \\  0 \end{bmatrix} =\begin{bmatrix} 1 \\ -1 \\  2 \end{bmatrix}} \\\text{Let }\cases{\vec u=(1,0,1) \\ \vec v=(1,-1,2)} \Rightarrow \text{Area of the image }C={1\over 2} \sqrt{|\vec u|^2|\vec v|^2 -(\vec u\cdot \vec v)^2} = \bbox[red, 2pt]{{1\over 2}\sqrt 3}$$

解答: $$A= \left(\begin{matrix}1 & 4 \\2 & 3 \end{matrix} \right)= \left(\begin{matrix}-2 & 1 \\ 1 & 1 \end{matrix} \right) \left(\begin{matrix}-1 & 0 \\0 & 5\end{matrix}\right) \left(\begin{matrix} \frac{-1}{3} & \frac{1}{3} \\\frac{1}{3} & \frac{2}{3} \end{matrix}\right) \\ \Rightarrow A^n = \left(\begin{matrix}1 & 4 \\2 & 3 \end{matrix} \right)= \left(\begin{matrix}-2 & 1 \\ 1 & 1 \end{matrix} \right) \left(\begin{matrix}(-1)^n & 0 \\0 & 5^n\end{matrix}\right) \left(\begin{matrix} \frac{-1}{3} & \frac{1}{3} \\\frac{1}{3} & \frac{2}{3} \end{matrix}\right) \\= \bbox[red, 2pt]{ \left(\begin{matrix} \displaystyle {1\over 3}(2(-1)^n+5^n) & \displaystyle \frac{2}{3}((-1)^{n +1}+ 5^n) \\\displaystyle \frac{1}{3} ((-1)^{n+1})+5^n& \displaystyle \frac{1}{3}((-1)^n+2\cdot 5^n) \end{matrix}\right) }$$

解答: $$(2,3) =-(1,0)+ 3(1,1) \Rightarrow T(2,3)= -T(1,0)+3T(1,1) =(-1,-4)+(6,15) \Rightarrow \bbox[red, 2pt]{T(2,3)=(5,11)} \\ \bbox[red, 2pt]{\text{Yes, T is one-to-one}}, \text{because }(1,0) \text{ and }(1,1) \text{ are linearly independent}$$

解答: $$A= \begin{bmatrix} 1& 0& a& 1 &d\\ -1& -1& b& -2&e\\ 3& 1& c& 0 &f\end{bmatrix} \xrightarrow{R_1+R_2\to R_2, R_3-3R_1\to R_3} \begin{bmatrix} 1& 0& a& 1 &d\\ 0& -1& a+b& -1 &d+e\\ 0& 1& c-3a& -3 &f-3d\end{bmatrix} \xrightarrow{R_2+R_3 \to R_3} \\\begin{bmatrix} 1& 0& a& 1 &d\\ 0& -1& a+b& -1 &d+e\\ 0& 0& -2a+ b+ c & -4 &-2d+e+f\end{bmatrix} \xrightarrow{R_1 +R_3/4 \to R_1, -R_2 \to R_2}\\ \begin{bmatrix} 1& 0& (2a+ b+ c)/4& 0 &(2d+ e+f)/4\\ 0& 1& -a-b& 1 &-d -e\\ 0& 0& -2a+ b+ c & -4 &-2d+e+f\end{bmatrix} \xrightarrow{R_2 +R_3/4 \to R_2}\\ \begin{bmatrix} 1& 0& (2a+ b+ c)/4& 0 &(2d+ e+f)/4\\ 0& 1& (-6a-3b+c)/4& 0 & (-6d-3e+f)/4\\ 0& 0& -2a+ b+ c & -4 &-2d+e+f\end{bmatrix} \xrightarrow{R_3/(-4) \to R_3} \\\begin{bmatrix} 1& 0& (2a+ b+ c)/4& 0 &(2d+ e+f)/4\\ 0& 1& (-6a-3b+c)/4& 0 & (-6d-3e+f)/4\\ 0& 0& (-2a+ b+ c)/(-4) & 1 &(-2d+e+f)/-4\end{bmatrix} = \begin{bmatrix} 1 & 0 & 2& 0& -2\\ 0& 1& -5& 0& -3\\ 0& 0& 0&   1& 6\end{bmatrix} \\ \Rightarrow \cases{2a+b+c= 8\\ 2d+e +f=-8\\ -6a-3b+c=-20\\ -6d-3e+f=-12\\ -2a+b+c=0\\ -2d+e+f=-24} \Rightarrow \cases{a=2\\ b=3\\ c=1 \\d=4\\ e=-7\\ f=-9} \Rightarrow A=\bbox[red, 2pt]{ \begin{bmatrix} 1 & 0 & 2 & 1 & 4\\-1 & -1 & 3 & -2 & -7\\3 & 1 & 1 & 0 & -9\end{bmatrix}}$$

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