112年台大碩士班-線性代數C詳解

 國立臺灣大學112學年度碩士班招生考試

題號:295
科目:線性代數(C)

解答: $$\textbf{(a)}\; A=[u_1^T\mid u_2^t \mid u_3^T ] =\begin{bmatrix}1 & 2 & 3 \\-2 & 3& 8\\ 5& 1 & -3\\ -3& -4&-5 \end{bmatrix} \Rightarrow rref(A)= \left[\begin{matrix}1 & 0 & -1\\0 & 1 & 2\\0 & 0 & 0\\0 & 0 & 0\end{matrix}\right] \\\quad \Rightarrow \text{ basis of }W: \bbox[red, 2pt]{\{(1,-2,5,-3), (2,3,1,-4) \}} \text{ and dimension of }W=\bbox[red, 2pt]2 \\\textbf{(b)}\; B=[u_1^T\mid u_2^t \mid u_3^T \mid e_1^T\mid e_2^T \mid e_3^T \mid e_4^T] = \left[\begin{matrix}1 & 2 & 3 & 1 & 0 & 0 & 0\\-2 & 3 & 8 & 0 & 1 & 0 & 0\\5 & 1 & -3 & 0 & 0 & 1 & 0\\-3 & -4 & -5 & 0 & 0 & 0 & 1\end{matrix}\right] \\ \quad \Rightarrow rref(B)= \left[\begin{matrix}1 & 0 & -1 & 0 & 0 & \frac{4}{17} & \frac{1}{17}\\0 & 1 & 2 & 0 & 0 & - \frac{3}{17} & - \frac{5}{17}\\0 & 0 & 0 & 1 & 0 & \frac{2}{17} & \frac{9}{17}\\0 & 0 & 0 & 0 & 1 & 1 & 1\end{matrix}\right] \Rightarrow \text{extended basis: }\{u_1,u_2, e_1,e_2\} \\\quad =\bbox[red, 2pt]{\{(1,-2,5,-3), (2,3,1,-4), (1,0,0,0), (0,1, 0, 0)\}}$$

解答: $$A= \begin{bmatrix}\sin^2 \alpha & \sin^2 \beta & \sin^2 \gamma \\\cos^2 \alpha & \cos^2 \beta & \cos^2 \gamma \\ 1& 1& 1 \end{bmatrix}  \xrightarrow{R_1+ R_2 \to R_2} \begin{bmatrix}\sin^2 \alpha & \sin^2 \beta & \sin^2 \gamma \\ \sin^2\alpha +\cos^2 \alpha & \sin^2\beta+ \cos^2 \beta & \sin^2\gamma + \cos^2 \gamma \\ 1& 1& 1 \end{bmatrix} \\ =\begin{bmatrix}\sin^2 \alpha & \sin^2 \beta & \sin^2 \gamma \\1 & 1 & 1 \\ 1& 1& 1 \end{bmatrix} \Rightarrow \det(A)=0, \forall \alpha,\beta,\gamma  \Rightarrow A \text{ is not invertible}, \bbox[red, 2pt]{Q.E.D.}$$

解答: $$f(\lambda)=\begin{vmatrix}51-\lambda & -12 &-21\\60 & -40-\lambda & -28\\ 57& -68 & 1-\lambda \end{vmatrix} 為三次多項式,且f(\lambda)=0的三根為-48,24 及\alpha\\ f(\lambda)=-\lambda^3+b\lambda^2 +c\lambda+d =-(\lambda+48)(\lambda-24)(\lambda-\alpha)\\ 只需考慮\lambda^2的係數b＝－48＋24＋\alpha=\alpha-24 \\ \Rightarrow (51-\lambda)(-40-\lambda)(1-\lambda) =-(\lambda-51)(\lambda+40)(\lambda-1) 的\lambda^2係數＝-(-51+40-1)=12 \\ \Rightarrow \alpha-24=12 \Rightarrow \alpha= \bbox[red,2pt]{36}\\ 註:題目有誤, \bbox[cyan,2pt]{48應為－48}$$

解答: $$\left\Vert\begin{matrix}1 & -2 & 5 & -1 \\2 & 1 & -2 & 1 \\3 & 0 & 1 & -2 \\ 1 & -1 & 4 & -1 \end{matrix}\right \Vert =\bbox[red, 2pt]{17}$$

解答: $$A=\begin{bmatrix}1 & 1 \\-2 & 4 \end{bmatrix} =PDP^{-1} =\begin{bmatrix}1/2 & 1 \\1 & 1 \end{bmatrix} \begin{bmatrix}3 & 0 \\0 & 2 \end{bmatrix} \begin{bmatrix}-2 & 2 \\2 & -1 \end{bmatrix} \\ \Rightarrow e^A=\begin{bmatrix}1/2 & 1 \\1 & 1 \end{bmatrix} \begin{bmatrix}e^3 & 0 \\0 & e^2 \end{bmatrix} \begin{bmatrix}-2 & 2 \\2 & -1 \end{bmatrix} = \bbox[red, 2pt] { \begin{bmatrix} e^2(2-e) & e^2(e-1) \\2e^2 (1-e) & e^2(2e-1) \end{bmatrix}}$$

解答: $$(A-I)(A-2I)=0 \Rightarrow \bbox[red, 2pt]{m_A(t)=(t-2)(t-1)} \\ (B-I)(B-2I) \ne 0 \Rightarrow \bbox[red, 2pt]{m_B(t)=(t-1)(t-1)^2}$$

解答: $$\begin{bmatrix}A\mid I \end{bmatrix} = \left[\begin{matrix}1 & -3 & 2 & 1 & 0 & 0\\-3 & 7 & -5 & 0 & 1 & 0\\2 & -5 & 8 & 0 & 0 & 1\end{matrix}\right] \xrightarrow{3R_1+R_2 \to R_2,3C_1+C_2\to C_2}\\ \left[\begin{matrix}1 & 0 & 2 & 1 & 0 & 0\\0 & -2 & 1 & 3 & 1 & 0\\2 & 1 & 8 & 0 & 0 & 1\end{matrix}\right] \xrightarrow{-2R_1+R_3\to R_3, -2C_1+C_3 \to C_3} \left[\begin{matrix}1 & 0 & 0 & 1 & 0 & 0\\0 & -2 & 1 & 3 & 1 & 0\\0 & 1 & 4 & -2 & 0 & 1\end{matrix}\right] \\ \xrightarrow{R_3+0.5R_2 \to R_3, C_3+ 0.5C_2 \to C_3}\left[\begin{matrix}1 & 0 & 0 & 1 & 0 & 0\\0 & -2 & 0 & 3 & 1 & 0\\0 & 0 & \frac{9}{2} & - \frac{1}{2} & \frac{1}{2} & 1\end{matrix}\right] \\ \Rightarrow \left[\begin{matrix}1 & 0 & 0\\3 & 1 & 0\\- \frac{1}{2} & \frac{1}{2} & 1\end{matrix}\right] \left[\begin{matrix}1 & -3 & 2\\-3 & 7 & -5\\2 & -5 & 8\end{matrix}\right]  \left[\begin{matrix}1 & 0 & 0\\3 & 1 & 0\\- \frac{1}{2} & \frac{1}{2} & 1\end{matrix}\right]^T =\left[\begin{matrix}1 & 0 & 0\\0 & -2 & 0\\0 & 0 & \frac{9}{2}\end{matrix}\right] \text{ is  diagonal}\\ \Rightarrow \bbox[red, 2pt]{P=\left[\begin{matrix}1 & 3 & - \frac{1}{2}\\0 & 1 & \frac{1}{2}\\0 & 0 & 1\end{matrix}\right]} \Rightarrow \text{signature of }A =\bbox[red, 2pt]{(2,1)}$$

解答: $$A=\begin{bmatrix}\langle u_1, u_1\rangle & \langle u_1, u_2\rangle & \langle u_1, u_3\rangle \\\langle u_2, u_1\rangle & \langle u_2, u_2\rangle & \langle u_2, u_3\rangle \\ \langle u_3, u_1\rangle & \langle u_3, u_2\rangle & \langle u_3, u_3\rangle\end{bmatrix} = \bbox[red,2pt]{ \begin{bmatrix}2& 3& 4 \\3 & 14 & 22 \\ 4& 22& 35 \end{bmatrix}}$$

解答: $$P=\begin{bmatrix}1/3 & 2/3  & 2/3\\2/3 & a & c\\ 2/3 &c & b\end{bmatrix} \Rightarrow \cases{(1/3,2/3,2/3) \cdot (2/3, a,c)=0\\ (1/3,2/3, 2/3) \cdot (2/3,c,b)=0\\ (2/3,a,c)\cdot (2/3,c,b)=0} \Rightarrow \cases{a=b\\ a+c=-1/3\\ ac=-2/9} \\ \text{choose} \cases{a=-2/3\\ b=-2/3\\ c=1/3} \Rightarrow P= \bbox[red, 2pt]{\begin{bmatrix}1/3 & 2/3  & 2/3\\2/3 & -2/3 & 1/3\\ 2/3 &1/3 & -2/3 \end{bmatrix}}$$

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