114年台綜大轉學考-線性代數詳解

臺灣綜合大學系統114學年度學士班轉學生聯合招生考試

科目名稱:線性代數
類組代碼:A07/C11 

Single-choice questions

解答:$$\begin{vmatrix}1 &2  &0&0  \\0 &1  &3 &0  \\0 & 0 & 1&4 \\ 5& 0& 0& 1\end{vmatrix} =1\begin{vmatrix}  1  &3 &0  \\ 0 & 1&4 \\  0& 0& 1 \end{vmatrix} -2 \begin{vmatrix}0    &3 &0  \\0   & 1&4 \\ 5 & 0& 1\end{vmatrix} =1-2\cdot 60=-119，故選\bbox[red, 2pt]{(A)}$$

解答:$$A= \begin{bmatrix} 0 & -2 & a & 0 \\b & 0 & c & 8 \\2 & -7 & 0 & d \\0 & e & 6 & 0\end{bmatrix} \Rightarrow \cases{A^T= \begin{bmatrix}0 & b & 2 & 0 \\-2 & 0 & -7 & e \\a & c & 0 & 6 \\ 0 & 8 & d & 0 \end{bmatrix} \\-A= \begin{bmatrix}0 & 2 & -a & 0 \\-b & 0 & -c & -8 \\-2 & 7 & 0 & -d \\0 & -e & -6 & 0 \end{bmatrix} } \\ A^T=-A \Rightarrow \cases{a=-2\\ b=2\\c=7\\ d=-6\\ e=-8}，故選\bbox[red, 2pt]{(B)}$$

解答:$$(A)\times: a_{11}=0\\ (B)\times: \begin{vmatrix}1 & 2 \\2 & 4\end{vmatrix} =0\\ (C)\times: \begin{vmatrix}2 & 2 \\2 & 2\end{vmatrix} =0\\，故選\bbox[red, 2pt]{(D)}$$

解答:$$\left[ \begin{matrix}-1 & 0 & 1 \\1 & 1 & 0 \\3 & 1 & -1 \end{matrix} \right] X= \left[ \begin{matrix} 1 & 2 \\-3 & 1 \\1 & -1\end{matrix} \right] \Rightarrow X=\left[ \begin{matrix}-1 & 0 & 1 \\1 & 1 & 0 \\3 & 1 & -1 \end{matrix} \right]^{-1}\left[ \begin{matrix} 1 & 2 \\-3 & 1 \\1 & -1\end{matrix} \right] \\ \Rightarrow \begin{bmatrix} x_{11}& x_{12} \\ x_{21}& x_{22} \\ x_{31}& x_{32}\end{bmatrix} = \begin{bmatrix} 1 & -1 & 1 \\-1 & 2 & -1 \\2 & -1 & 1\end{bmatrix} \left[ \begin{matrix} 1 & 2 \\-3 & 1 \\1 & -1\end{matrix} \right] = \begin{bmatrix} 5 & 0 \\-8 & 1 \\6 & 2\end{bmatrix} \Rightarrow \cases{(A)\times: x_{12}=0 \ne 2\\ (B) \times: x_{21}=-8 \ne 5 \\(C) \times:x_{11}=5\ne -1\\(D)\bigcirc: x_{32}=2}\\，故選\bbox[red, 2pt]{(D)}$$

解答:$$A = \begin{bmatrix} -1 & 3& 2 \\0 & -2& 1 \\1&0 & -2\end{bmatrix}  \Rightarrow \text{ cofactor matrix of }A= \begin{bmatrix} 4 & 1& 2 \\6 & 0& 3 \\7 &1 & 2\end{bmatrix} \Rightarrow adj(A)= \begin{bmatrix} 4 & 6& 7 \\1 & 0& 1 \\2 &3 & 2\end{bmatrix}\\，故選\bbox[red, 2pt]{(D)}$$

解答:$$rref\left( \begin{bmatrix} 1 & 2 & 3 & 4\\0 & 1 & 4 & 1\\0 & 0 & 1 & 2\\0 & 0 & 0 & 2\end{bmatrix} \right)= \begin{bmatrix} 1 & 0 & 0 & - \frac{1}{9}\\0 & 1 & 0 & \frac{4}{9}\\0 & 0 & 1 & \frac{4}{9}\\0 & 0 & 0 & 0\end{bmatrix} \Rightarrow rank\left( \begin{bmatrix} 1 & 2 & 3 & 4\\0 & 1 & 4 & 1\\0 & 0 & 1 & 2\\0 & 0 & 0 & 2\end{bmatrix} \right)=3\\ 其它三個矩陣皆為上三角或下三角矩陣，其rank均為4，因此rank(A)=3，故選\bbox[red, 2pt]{(C)}$$

解答:$$\begin{vmatrix} 1 & 2& -3 \\3 & -1& 5 \\4& 1 & k^2-14\end{vmatrix} =0 \Rightarrow k^2=16 \Rightarrow k=\pm 4 \\ k=4 \Rightarrow \cases{x_1+2x_2-3x_3=4\\ 3x_1-x_2+5x_3=2\\ 4x_1+x_2+2x_3=6} \Rightarrow rref \left(\left[\begin{array}{ccc|c} 1 & 2& -3 &4\\3 & -1& 5 &2\\4& 1 & 2& 6\end{array} \right] \right) = \left[\begin{array}{ccc|c} 1 & 0 & 1 & \frac{8}{7}\\0 & 1 & -2 & \frac{10}{7}\\0 & 0 & 0 & 0\end{array} \right] \\\qquad \Rightarrow \text{infinity many solution} \\ k=-4\Rightarrow \cases{x_1+2x_2-3x_3=4\\ 3x_1-x_2+5x_3=2\\ 4x_1+x_2+2x_3=-2} \Rightarrow rref \left(\left[\begin{array}{ccc|c} 1 & 2& -3 &4\\3 & -1& 5 &2\\4& 1 & 2& -2\end{array} \right] \right) = \left[\begin{array}{ccc|c}1 & 0 & 1 & 0\\0 & 1 & -2 & 0\\0 & 0 & 0 & 1\end{array} \right] \\\qquad \Rightarrow \text{no solution}\\，故選\bbox[red, 2pt]{(B)}$$

解答:$$\begin{bmatrix}1 \\-1 \\2 \end{bmatrix} \times \begin{bmatrix}-2 \\3 \\1 \end{bmatrix} =\begin{bmatrix}-7 \\-5 \\1 \end{bmatrix}，故選\bbox[red, 2pt]{(D)}$$

解答:$$\cases{T(u)=[-2,1,0] \\ T(v)=[5,-7,1]} \Rightarrow \cases{\begin{bmatrix}a_{11} &a_{12}\\a_{21}& a_{22} \\a_{31} &a_{32} \end{bmatrix}  \begin{bmatrix}-1 \\2  \end{bmatrix} =\begin{bmatrix}-2 \\1 \\0 \end{bmatrix} \\ \begin{bmatrix}a_{11} &a_{12}\\a_{21}& a_{22} \\a_{31} &a_{32} \end{bmatrix}  \begin{bmatrix}3 \\-5  \end{bmatrix} =\begin{bmatrix} 5 \\-7 \\1 \end{bmatrix} } \Rightarrow \cases{\cases{-a_{11}+2a_{12}=-2 \\3a_{11}-5a_{12}=5} \\ \cases{-a_{21}+2a_{22}=1\\ 3a_{21}-5a_{22}=-7}} \\ \Rightarrow \cases{a_{11}=0\\ a_{12}=-1 \\a_{21}=-9\\ a_{22}=-4}，故選\bbox[red, 2pt]{(A)}$$

解答:$$\begin{vmatrix}e^x & e^{2x} &e^{3x}  \\ e^x & 2e^{2x} & 3e^{3x} \\ e^x & 4e^{2x} & 9e^{3x} \end{vmatrix} =e^{6x}(18+4+3-2-9-12) =2e^{6x} \ne 0，故選\bbox[red, 2pt]{(D)}$$

解答:$$rank(T) =dim(R(T)) =dim\left( \begin{bmatrix} 2a_{11}-a_{12}& a_{13}+a_{12} \\0 & 0\end{bmatrix}\right) =2，故選\bbox[red, 2pt]{(B)}$$

解答:$$A = \begin{bmatrix} -2& 1 \\1 & -3\end{bmatrix}  \Rightarrow \cases{A^T=A\\ \text{eigenvalues of A}:(-5\pm \sqrt 5)/2 \lt 0} \Rightarrow A \text{ is Negative Definite} \\ B=\begin{bmatrix} 1& 0 \\0 & 0\end{bmatrix}  \Rightarrow \cases{B^T=B\\ \text{eigenvalues of B}:0,1\ge 0} \Rightarrow B \text{ is Positive Semi-Definite} \\C=\begin{bmatrix} -2& 9 \\9 & 9\end{bmatrix}  \Rightarrow \cases{C^T=C\\ \text{eigenvalues of C}:-2,0 \le 0} \Rightarrow C \text{ is Negative Semi-Definite} \\ D=\begin{bmatrix} -2& -1&0 \\-1 & 2&-1\\ 0&-1&2\end{bmatrix}  \Rightarrow \cases{D^T=D\\ \text{eigenvalues of D}:2,2\pm \sqrt 2\gt 0} \Rightarrow D \text{ is Positive Definite}\\，故選\bbox[red, 2pt]{(D)}$$

解答:$$A= \begin{bmatrix}2& 1& 0\\ -1& 0 & 1\\ 1& 3& 1\end{bmatrix} \Rightarrow \det(A-\lambda I) =-(\lambda+1) (\lambda-2)^2 =0 \Rightarrow \cases{\text{algebraic multiplicity of }\lambda=-1 \text{ is 1} \\\text{algebraic multiplicity of }\lambda=2 \text{ is 2}} \\ \lambda=2 \Rightarrow (A-\lambda I)v=0 \Rightarrow \begin{bmatrix}0 & 1 & 0 \\-1 & -2 & 1 \\1 & 3 & -1\end{bmatrix} \begin{bmatrix} x_1   \\x_2  \\x_3 \end{bmatrix} =0 \Rightarrow \cases{x_1=x_3 \\x_2=0 } \Rightarrow v =x_3 \begin{bmatrix} 1   \\ 0  \\1 \end{bmatrix} \\ \Rightarrow \text{geometric multiplicity of }\lambda=2 \text{ is }1，故選\bbox[red, 2pt]{(D)}$$

解答:$$W_1\cap W_2 =\left\{ \begin{bmatrix}0& a\\ -a& 0 \end{bmatrix}, a\in \mathbb R\right\} =span\left( \left\{\begin{bmatrix}0& 1\\ -1& 0 \end{bmatrix}, a\in \mathbb R\right\} \right) \Rightarrow dim(W_1\cap W_2)=1\\，故選\bbox[red, 2pt]{(A)}$$

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

解答:$$(A)\times: T(0,0) \ne (0,0) \\(B)\times \cases{T(0,1)+T(0,1)=(0,1)+(0,1) =(0,2)\\T(0,2) =(0,2^2)=(0,4)} \Rightarrow T(0,2) \ne 2T(0,1) \\(C)\times T(0,0)=(1,0) \ne (0,0)\\，故選\bbox[red, 2pt]{(D)}$$

解答:$$(D)\times: \text{Let }T(x,y)=(0,0), \forall v \in \mathbb R^2. \text{ Here }m=n=2, \text{but }T \text{ is NOT invertible}，故選\bbox[red, 2pt]{(D)}$$

Multi-choice questions

解答:$$(A)\bigcirc: A={1\over \sqrt 2} \begin{bmatrix} 1& 1\\ i& -i\end{bmatrix} \Rightarrow A^H ={1\over \sqrt 2} \begin{bmatrix} 1& -i\\ 1& i\end{bmatrix} \Rightarrow AA^H ={1\over 2} \begin{bmatrix} 2& 0\\ 0& 2 \end{bmatrix} =I \\(B)\times: B= \begin{bmatrix} 1& i\\ 0& i\end{bmatrix} \Rightarrow B^H= \begin{bmatrix} 1& 0\\ -i& -i\end{bmatrix} \Rightarrow BB^H =\begin{bmatrix} 2& 1\\ 1& 1\end{bmatrix} \ne I \\(C) \bigcirc:C= {1\over 3}\begin{bmatrix}2& -2+i\\ 2+i& 2\end{bmatrix} \Rightarrow C^H= {1\over 3} \begin{bmatrix}2&  2-i\\ -2-i& 2\end{bmatrix} \Rightarrow CC^H={1\over 9} \begin{bmatrix}9 &  0\\ 0& 9\end{bmatrix} =I \\(D) \times: D= \begin{bmatrix} 0& 1\\ 1& 1\end{bmatrix} \Rightarrow D^H=D \Rightarrow DD^H= \begin{bmatrix} 1& 1\\ 1& 2\end{bmatrix} \ne I\\，故選\bbox[red, 2pt]{(AC)}$$

解答:$$\det(A-\lambda I) =-\lambda^3+ 14\lambda^2-55\lambda+42 =-(\lambda-1)(\lambda-6) (\lambda-7) =0 \\ \Rightarrow \lambda=1,6,7，故選\bbox[red, 2pt]{(BCD)}$$

解答:$$(A)\times: x_1+x_2=5 \text{ has infinity solutions} \\(B) \times: \cases{x_1+x_2=1\\ 2x_1+2x_2=2}  \text{ has infinity solutions} \\，故選\bbox[red, 2pt]{(CDE)}$$

解答:$$(A)\bigcirc: A^2 \text{ is invertible} \Rightarrow \det(A^2) =\det(A)\cdot \det(A) \ne0 \Rightarrow \det(A) \ne 0 \\\qquad \Rightarrow \cases{\det(A^T)= \det(A) \ne 0\\ \det(A^3) =(\det(A))^3 \ne 0} \Rightarrow \cases{A^T \text{ is invertible} \\A^3 \text{ is invertible}} \\(B) \times:\cases{A= \begin{bmatrix} 1& 0\\0& 1\end{bmatrix} \\B= \begin{bmatrix} -1& 0\\0& -1\end{bmatrix}} \Rightarrow \cases{A^{-1} =A\\ B^{-1} =B} \Rightarrow A+B=0 \Rightarrow A+B\text{ is NOT invertible} \\(C)\bigcirc: \operatorname{tr}(AB) =\sum_{i=1}^m(AB)_{ii}= \sum_{i=1}^m\sum_{k=1}^n A_{ik}B_{ki} =\sum_{k=1}^n\sum_{i =1}^mB_{ki}A_{ik} =\sum_{k =1}^n(BA)_{kk} =\operatorname{tr}(BA). \\(D) \bigcirc:\\ (E)\times: \cases{A= \begin{bmatrix} 0& 1\\0& 1\end{bmatrix} \\B= \begin{bmatrix} 1& 0\\ 1& 0\end{bmatrix}} \Rightarrow  \cases{AB= \begin{bmatrix} 1& 0\\1& 0\end{bmatrix} \\BA =\begin{bmatrix} 0& 1\\ 0& 1\end{bmatrix}} \Rightarrow AB\ne BA \\，故選\bbox[red, 2pt]{(BE)}$$

解答:$$(A)\times: A=\begin{bmatrix}1&0& 0\\ 0& 1& 0\\0&   0& 1 \end{bmatrix} \Rightarrow A=IAI^{-1} \Rightarrow A \text{ is diagonalizable and has only one eigenvalue.} \\(B)\times: Av=\lambda v \Rightarrow A(2v)=\lambda (2v) \Rightarrow v \text{ and }2v \text{ are eigenvectors corresponding to the same}\\\quad \text{eigenvalue }\lambda, \text{ but they are linearly dependent.} \\(C)\times: E_\lambda \text{ is a subspace and contains zero vector, but zero is NOT a eigenvector.}\\ (D)\bigcirc \\ (E) \bigcirc: A=PDP^{-1} \Rightarrow A^T =(PDP^{-1})^{T} =(P^{T})^{-1}DP^T \Rightarrow A^T \text{ is diagonalizable}\\，故選\bbox[red, 2pt]{(ABC)}$$

解答:$$(C)\times: \vec 0+\vec 0=\vec 0\\ (D)\times: \text{A vector space can have an infinite basis (eg. Hamel basis) and still be finitely generated.}\\，故選\bbox[red, 2pt]{(CD)}$$

解答:$$(B)\times: \text{Let }\cases{W_1=\{(x,0)\mid x\in \mathbb R\} \\W_2=\{(0,y)\mid y\in \mathbb R\}} \text{ and }\cases{u=(1,0) \in W_1\\ v=(0,1) \in W_2} \Rightarrow u+v=(1,1) \not \in W_1\cup W_2\\，故選\bbox[red, 2pt]{(ACDE)}$$

解答:$$(D)\times: T(v)=-v \Rightarrow T=-I \Rightarrow (-I)^*=-I \Rightarrow T \text{ is self-adjoint} \Rightarrow \text{eigenvalue of }T=-1 \not \gt 0\\(E)\times: T(v)=v \Rightarrow T=I \Rightarrow T  \text{ is self-adjoint} \Rightarrow \text{eigenvalue of }T=1 \not \lt 0\\，故選\bbox[red, 2pt]{(ABC)}$$

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解題僅供參考，轉學考歷年試題及詳解