114年成大工程科學碩士班-線性代數詳解

 國立成功大學114學年度碩士班招生考試

系所:工程科學系
科目:線性代數

解答:$$\text{if }\cases{a=b=e = f=g=i=0\\ c=-1\\ d=3\\ h=1} \Rightarrow \cases{\det \left(\begin{bmatrix} 0 & 0 & -1\\ 3& 0 & 0\\ 0 & 1& 0\end{bmatrix} \right) =-3 \\[1ex] \det\left( \begin{bmatrix} 0& 1& 0\\ 0& 0& 0\\ -1& 0 & 3\end{bmatrix}\right) =0\ne 3} \Rightarrow \bbox[red, 2pt]{(F)}$$

解答:$$\begin{vmatrix}x & 1 & x & 3 \\x & 1 & 1 & 2x \\1 & 2 & x & 3 \\2 & -3x & 2 & 2x \end{vmatrix} = \begin{vmatrix}0 & 0 & x-1 & 3-2x \\x & 1 & 1 & 2x \\1 & 2 & x & 3 \\2 & -3x & 2 & 2x \end{vmatrix}=(x-1) \begin{vmatrix} x& 1& 2x\\ 1& 2 &3 \\ 2 & -3x& 2x\end{vmatrix}-(3-2x) \begin{vmatrix} x& 1& 1\\ 1& 2& x\\ 2& -3x &2\end{vmatrix}\\ =(x-1)(7x^2-10x+6)-(3-2x)(3x^3+3x-6)=6x^4-2x^3-11x^2-5x+12\\ \Rightarrow \text{ the coefficient of term }x^4 \text{ is 6}\ne 24 \Rightarrow \bbox[red, 2pt]{(F)}  $$

解答:$$trace(4A-2B+6B^T-3A^T) =4trace(A)-2trace(B)+6trace(B)-3trace(A) \\=trace(A) +4trace(B)=-2+4\times3=10 \ne 12\Rightarrow \bbox[red, 2pt]{(F)}$$

解答:$$T((2,3))=T((2,4))=(2,0) \Rightarrow T \text{ is NOT 1-1} \Rightarrow \bbox[red, 2pt]{(F)}$$

解答:$$\mathbf a_1= \begin{pmatrix} a_{11} \\a_{12} \\a_{13} \\a_{14}\end{pmatrix} ,\mathbf a_2= \begin{pmatrix} a_{21} \\a_{22} \\a_{23} \\a_{24}\end{pmatrix} ,\mathbf a_3= \begin{pmatrix} a_{31} \\a_{32} \\a_{33} \\a_{34}\end{pmatrix} ,\mathbf a_4= \begin{pmatrix} a_{41} \\a_{42} \\a_{43} \\a_{44}\end{pmatrix} \\ \Rightarrow \mathbf b=(1)\mathbf a_1 +(-3)\mathbf a_2+(2) \mathbf a_3+(-2) \mathbf a_4 =\begin{pmatrix} a_{11} \\a_{12} \\a_{13} \\a_{14}\end{pmatrix} -3 \begin{pmatrix} a_{21} \\a_{22} \\a_{23} \\a_{24}\end{pmatrix} +2 \begin{pmatrix} a_{31} \\a_{32} \\a_{33} \\a_{34}\end{pmatrix} -2 \begin{pmatrix} a_{41} \\a_{42} \\a_{43} \\a_{44}\end{pmatrix} \\ \qquad =\begin{pmatrix} a_{11}- 3a_{21}+ 2a_{31}-2a_{41} \\a_{12} -3a_{22}+ 2a_{32}-2a_{42} \\a_{13}-3a_{23}+ 2a_{33}- 2a_{43} \\ a_{14}-3a_{24}+2 a_{34}-2a_{44}\end{pmatrix} =\begin{pmatrix}a_{11}& a_{21} & a_{31}& a_{41}  \\ a_{12}& a_{22} & a_{32}& a_{42}  \\a_{13}& a_{23} & a_{33}& a_{43}  \\a_{14}& a_{24} & a_{34}& a_{44}  \\\end{pmatrix} \begin{pmatrix} 1\\-3\\2\\-2\end{pmatrix} \\\Rightarrow \begin{pmatrix} 1\\-3\\2\\-2\end{pmatrix} \text{is the solution of }A\mathbf x=\mathbf b,\text{ but }\begin{pmatrix} 1\\-3 \\-2\\ 2\end{pmatrix} \text{ is NOT the solution} \Rightarrow \bbox[red, 2pt]{(F)}$$

解答:$$\alpha A+\beta B=\begin{bmatrix} \alpha+3\beta& -2\alpha+2\beta\\ 2\alpha& 4\alpha+5\beta\end{bmatrix} =0 \Rightarrow \alpha=\beta=0 \Rightarrow \text{linearly independent} \Rightarrow \bbox[red, 2pt]{(T)}$$

解答:$$\cases{A \text{ is }4\times 3 \Rightarrow A^T \text{ is }3\times 4\\ \text{rank}(A)=1 \Rightarrow \text{ rank}(A^T)=1} \Rightarrow \text{rank}(A^T)+ N(A^T) = \text{ num. of columns of }A^T \\ \Rightarrow 1+N(A^T) =4 \Rightarrow N(A^T)=4-1=3 \Rightarrow \bbox[red, 2pt]T$$

解答:$$\text{RREF} \left( \left[\begin{matrix}1 & 4 & 5 & 2\\2 & 1 & 3 & 0\\-1 & 3 & 2 & 2\end{matrix} \right] \right) =\left[\begin{matrix}1 & 0 & 1 & - \frac{2}{7}\\0 & 1 & 1 & \frac{4}{7}\\0 & 0 & 0 & 0\end{matrix}\right]\\ \textbf{(1)} \bbox[red, 2pt]{\text{ Rol }(A)= \text{span}\left\{ \left(1,0,1,-{2\over 7} \right), \left(0,1,1,{4\over 7} \right) \right\} }\\\textbf{(2) } \bbox[red, 2pt]{\text{Col}(A) = \text{span}\left\{  \begin{bmatrix}1\\2\\-1 \end{bmatrix},  \begin{bmatrix} 4\\ 1\\3 \end{bmatrix} \right\}} \\\textbf{(3) } \left[ \begin{matrix}1 & 0 & 1 & - \frac{2}{7}\\0 & 1 & 1 & \frac{4}{7}\\0 & 0 & 0 & 0\end{matrix}\right] \begin{bmatrix}x_1\\ x_2\\ x_3 \\x_4\end{bmatrix} =0 \Rightarrow \cases{x_1+x_3-{2\over 7}x_4=0\\ x_2+x_3+{4\over 7}x_4 =0} \Rightarrow \begin{bmatrix}x_1\\ x_2\\ x_3 \\x_4\end{bmatrix} = \begin{bmatrix}-x_3+{2\over 7}x_4\\ -x_3-{4\over 7}x_4\\ x_3\\ x_4 \end{bmatrix} \\\quad = x_3 \begin{bmatrix}-1 \\ -1\\ 1\\ 0 \end{bmatrix} +x_4 \begin{bmatrix}2/7\\ -4/7\\ 0\\ 1 \end{bmatrix} \Rightarrow \bbox[red, 2pt]{N(A) = \text{span} \left\{\begin{bmatrix}-1\\ -1\\ 1\\ 0 \end{bmatrix},  \begin{bmatrix}2/7\\ -4/7\\ 0\\ 1 \end{bmatrix}\right\}} \\\textbf{(4) } \text{RREF}(A) = \left[\begin{matrix}1 & 0 & 1 & - \frac{2}{7}\\0 & 1 & 1 & \frac{4}{7}\\0 & 0 & 0 & 0\end{matrix}\right] \Rightarrow \text{rank}(A)=\bbox[red, 2pt]2 \\ \textbf{(5) } N(A) = \text{span} \left\{\begin{bmatrix}-1\\ -1\\ 1\\ 0 \end{bmatrix},  \begin{bmatrix}2/7\\ -4/7\\ 0\\ 1 \end{bmatrix}\right\} \Rightarrow \text{nullity}(A)= \bbox[red, 2pt]2$$

解答:$$\textbf{(1) }v_1= \begin{bmatrix}2\\ 1\\2 \end{bmatrix}, v_2= \begin{bmatrix}0\\1\\1 \end{bmatrix} \\ \quad \Rightarrow u_1=v_1 \Rightarrow e_1= {u_1\over |u_1|} ={1\over 3} \begin{bmatrix}2\\ 1\\2 \end{bmatrix} =  \begin{bmatrix}{2\over 3}\\ {1\over 3}\\{2\over 3} \end{bmatrix} \\ \quad \Rightarrow u_2=v_2-\text{proj}_{u_1}v_2 =\begin{bmatrix}0\\1\\1 \end{bmatrix} -{3\over 9}\begin{bmatrix}2\\ 1\\2 \end{bmatrix} = \begin{bmatrix}{2\over 3}\\ {1\over 3}\\{2\over 3} \end{bmatrix}= \begin{bmatrix}-{2\over 3}\\ {2\over 3}\\{1\over 3} \end{bmatrix}\Rightarrow e_2={u_2\over |u_2|} = \begin{bmatrix} -{2\over 3}\\ {2\over 3}\\{1\over 3}\end{bmatrix} \\\quad \Rightarrow \text{orthonormal basis: }\{e_1, e_2\} = \bbox[red, 2pt]{\left\{ \begin{bmatrix}{2\over 3}\\ {1\over 3}\\{2\over 3} \end{bmatrix}, \begin{bmatrix} -{2\over 3}\\ {2\over 3}\\{1\over 3}\end{bmatrix}\right\}} \\\textbf{(2) }Q=[e_1\; e_2] = \begin{bmatrix}{2\over 3} &-{2\over 3} \\ {1\over 3} &{2\over 3} \\ {2\over 3}& {1\over 3} \end{bmatrix} \Rightarrow R=Q^T A = \begin{bmatrix}{2\over 3}& {1\over 3}& {2\over 3}\\ -{2\over 3} &{2\over 3}& {1\over 3} \end{bmatrix} \begin{bmatrix}2& 0\\ 1& 1\\ 2& 1 \end{bmatrix} =\begin{bmatrix} 3& 1\\ 0& 1 \end{bmatrix} \\\quad \Rightarrow \bbox[red, 2pt]{Q=\begin{bmatrix}{2\over 3} &-{2\over 3} \\ {1\over 3} &{2\over 3} \\ {2\over 3}& {1\over 3} \end{bmatrix}, R=\begin{bmatrix} 3& 1\\ 0& 1 \end{bmatrix}} \\\textbf{(3) } \text{RREF}([A^TA \mid A^T b]) =\text{RREF} \left( \begin{bmatrix}9& 3 & 8\\ 3& 2& 3 \end{bmatrix} \right) = \left[\begin{matrix}1 & 0 & \frac{7}{9}\\0 & 1 & \frac{1}{3}\end{matrix}\right] \\\quad \Rightarrow \text{the least square solution }\hat x= \bbox[red, 2pt]{\begin{bmatrix}{7\over 9}\\{1\over 3} \end{bmatrix}}$$

解答:$$\textbf{(1) }D= \begin{bmatrix}\lambda_1 &0&0\\ 0& \lambda_2& 0\\ 0& 0& \lambda_3 \end{bmatrix} = \begin{bmatrix} 0 &0&0\\ 0& -1& 0\\ 0& 0& 1 \end{bmatrix}, P= [v_1 \;v_2\; v_3] = \begin{bmatrix}1& 1& 0\\ -1&0& 0\\ 0& 2& -1 \end{bmatrix} \\ \Rightarrow A= PDP^{-1} =\begin{bmatrix}1& 1& 0\\ -1&0& 0\\ 0& 2& -1 \end{bmatrix} \begin{bmatrix} 0 &0&0\\ 0& -1& 0\\ 0& 0& 1 \end{bmatrix} \begin{bmatrix} 0 & -1 & 0 \\1 & 1 & 0 \\2 & 2 & -1 \end{bmatrix} = \bbox[red, 2pt]{\begin{bmatrix} -1 & -1 & 0 \\0 & 0 & 0 \\ -4 & -4 & 1\end{bmatrix}} \\ \textbf{(2) }A^{10} =PD^{10}P^{-1} =\begin{bmatrix}1& 1& 0\\ -1&0& 0\\ 0& 2& -1 \end{bmatrix} \begin{bmatrix} 0 &0&0\\ 0& 1& 0\\ 0& 0& 1 \end{bmatrix} \begin{bmatrix}0 & -1 & 0 \\1 & 1 & 0 \\2 & 2 & -1 \end{bmatrix} = \bbox[red, 2pt]{\begin{bmatrix}1 & 1 & 0 \\0 & 0 & 0 \\0 & 0 & 1 \end{bmatrix}} \\\textbf{(3) }Y'(t)=AY(t) \Rightarrow Y(t)= e^{At}Y(0) = \begin{bmatrix}1& 1& 0\\ -1&0& 0\\ 0& 2& -1 \end{bmatrix} \begin{bmatrix} 1 &0&0\\ 0& e^{-t}& 0\\ 0& 0& e^t \end{bmatrix} \begin{bmatrix} 0 & -1 & 0 \\1 & 1 & 0 \\2 & 2 & -1 \end{bmatrix} \begin{bmatrix} 1\\ 0\\ 0 \end{bmatrix} \\\qquad \Rightarrow \bbox[red, 2pt]{Y(t)= \begin{bmatrix}e^{-t}\\0 \\ -2e^t+2e^{-t} \end{bmatrix}} \\\textbf{(4) } \bbox[red, 2pt]{Y(\infty) = \begin{bmatrix}0\\0 \\ -\infty \end{bmatrix}}$$

解答:$$\text{For a vector space }V, \text{if }x\in v, \text{ there exists }y \text{ such that }x+y=0 \\ \Rightarrow y=-x \le 0 \Rightarrow y \text{ does not exist if }x\gt 0. \text{ That is Failure of Additive Inverses} \\ \Rightarrow \bbox[red, 2pt]{No}, (V, \oplus, \otimes) \text{ is NOT a vector space}$$

解答:$$p = \text{proj}_y x ={x\cdot y\over y\cdot y} y ={1 \over 1} y= \begin{bmatrix}0\\ 0 \\ 1 \end{bmatrix} \Rightarrow z= x-p = \begin{bmatrix}2\\ 0\\ 1 \end{bmatrix}-\begin{bmatrix}0\\ 0 \\ 1 \end{bmatrix} =\begin{bmatrix}2\\ 0 \\ 0 \end{bmatrix} \\ \Rightarrow \bbox[red, 2pt]{p= \begin{bmatrix}0\\ 0 \\ 1 \end{bmatrix} , z= \begin{bmatrix}2\\ 0 \\ 0 \end{bmatrix}}$$

解答:$$\textbf{(1) } \cases{L(x^2) =x^2+2x \\ L(x)=x^2+1\\ L(1)=x^2} \Rightarrow \bbox[red, 2pt]{A = \begin{bmatrix}1& 1& 1\\ 2& 0& 0\\ 0& 1& 0 \end{bmatrix}} \\\textbf{(2) } \cases{L(x^2+1)= 2x^2+2x =2(x^2+1)+(2x+1)-3\\ L(2x+1)=3x^2+2 =3(x^2+1)+0\cdot(2x+1)-1\\ L(1) =x^2 =1(x^2+1) +0\cdot(2x+1)-1} \Rightarrow \bbox[red, 2pt]{B= \begin{bmatrix}2& 3& 1\\ 1& 0& 0\\ -3& -1& -1 \end{bmatrix}} \\\textbf{(3) } \cases{x^2+1 =1\cdot x^2+0\cdot x+ 1\\ 2x+1 =0\cdot x^2+2\cdot x+1\\ 1= 0\cdot x^2+0\cdot x+1} \Rightarrow \bbox[red, 2pt]{S = \begin{bmatrix}1& 0& 0\\ 0& 2& 0\\ 1& 1& 1 \end{bmatrix}}$$

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