114年中山大學通訊碩士班-線性代數詳解

 國立中山大學114學年碩士班考試入學招生考試

科目名稱：線性代數【通訊所碩士班甲組】

一、單選題(每題5分)

解答:$$(C) \text{If }n \gt m, \mathbf x \text{ will be a zero vector} \\ 故選\bbox[red, 2pt]{(C)}$$

解答:$$(A) \times: A^2=I_2 \Rightarrow A^4=A^2A^2=I_2I_2 =I_2 \\(B)\bigcirc: A=\begin{bmatrix}\cos 180^\circ & -\sin 180^\circ\\ \sin 180^\circ& \cos 180^\circ \end{bmatrix} =\begin{bmatrix}-1 & 0\\ 0& -1 \end{bmatrix} \Rightarrow A^2=I,\text{ but }A\ne I\\(C)\bigcirc: A=\begin{bmatrix}\cos 120^\circ & -\sin 120^\circ\\ \sin 120^\circ& \cos 120^\circ \end{bmatrix} =\begin{bmatrix}-1/2 & -\sqrt 3/2\\   \sqrt 3/2 & -1/2\end{bmatrix} \Rightarrow A^3=I,\text{ but }A\ne I \\(D) \bigcirc: A= \begin{bmatrix}0& 0\\ 0& 0 \end{bmatrix} \Rightarrow rank(A)=0 \\(E)\bigcirc: A=I_2 \Rightarrow rank(A)=2 \\故選\bbox[red, 2pt]{(A)}$$

解答:$$(A) \bigcirc: \mathbf v\in ker(\mathbf B) \Rightarrow \mathbf B(\mathbf v) =0 \Rightarrow \mathbf (\mathbf{AB})(\mathbf v) =\mathbf A(\mathbf B(\mathbf v)) =\mathbf A(0)=0\Rightarrow \mathbf v\in ker(\mathbf{AB}) \\(B) \bigcirc: \cases{\mathbf v \in im(\mathbf A) \Rightarrow \exists \mathbf x \ni \mathbf A(\mathbf x)=\mathbf v \\\mathbf w \in im(\mathbf A) \Rightarrow \exists \mathbf y \ni \mathbf A(\mathbf y)=\mathbf w } \Rightarrow 2\mathbf v-7\mathbf w=2\mathbf A(\mathbf x)-7\mathbf A(\mathbf y) = \mathbf A(2\mathbf x-7\mathbf y) \\\qquad \Rightarrow 2\mathbf v-7\mathbf w \in im(A)\\ (C)\times: \cases{Av=b\\ Aw=b} \Rightarrow Av+Aw=A(v+w) =2b \ne b \\(D) \bigcirc: Av=Aw \Rightarrow Av-Aw=A(v-w)=0 \Rightarrow v-w\in ker(A) \\(E)\bigcirc:\cases{A(0)=0 \Rightarrow 0 \in im(A) \\ \cases{b_1\in im(A)\\ b_2 \in im(A)} \Rightarrow \cases{A(v_1)=b_1\\ A(v_2)=b_2} \Rightarrow A(v_1)+A(v_2)=A(v_1+v_2)=b_1+b_2 \Rightarrow b_1+b_2\in im(A) \\ b\in im(A) \Rightarrow A(v)=b \Rightarrow cA(v)=A(cv)=cb \Rightarrow cb\in im(A)} \\\qquad \Rightarrow im(A) \text{ is a subspace of }\mathbb R^n\\ 故選\bbox[red, 2pt]{(C)}$$

解答:$$(E)\times: \cases{A = \begin{bmatrix}2&0\\0&0 \end{bmatrix} \\ B= \begin{bmatrix}0& 0\\0&2 \end{bmatrix}} \Rightarrow A= \begin{bmatrix}0&1\\ 1& 0 \end{bmatrix} B \begin{bmatrix}0&1\\ 1& 0 \end{bmatrix}^{-1} \Rightarrow A \text{ is similar to }B, \text{ but }A\ne B\\ 故選\bbox[red, 2pt]{(E)}$$

解答:$$(D)\times: A= \begin{bmatrix}1& 0\\ 0& 1 \end{bmatrix} \Rightarrow \cases{\det(-A) =1\\ -\det(A)=-1} \Rightarrow \det(-A) \ne -\det(A)\\ 故選\bbox[red, 2pt]{(D)}$$

解答:$$\mathbf u=\begin{bmatrix}-1 \\8\\9 \end{bmatrix} =3\mathbf v-2\mathbf w \Rightarrow \mathbf A^5 \mathbf u= \mathbf A^4  (\mathbf A\mathbf u)= \mathbf A^4  (\mathbf A(3\mathbf v-2\mathbf w)) = \mathbf A^4  (3\mathbf A(\mathbf v)-2 \mathbf A(\mathbf w)) \\=\mathbf A^4  ( -3\mathbf v-4  \mathbf w) =\mathbf A^3  (-3\mathbf A(\mathbf v)-4 \mathbf A(\mathbf w)) =\mathbf A^3  (3 \mathbf v-8  \mathbf w)  =\mathbf A^2  (3 \mathbf A (\mathbf v)-8 \mathbf A (\mathbf w)) \\ =\mathbf A^2  (-3  \mathbf v-16  \mathbf w) = \mathbf A(-3\mathbf A(\mathbf v)-16 \mathbf A(\mathbf w)) = \mathbf A( 3\mathbf v-32 \mathbf   w) = 3\mathbf A(\mathbf v)-32\mathbf A(  \mathbf w)=-3 \mathbf v-64\mathbf w \\ =-3\begin{bmatrix}1 \\2 \\-1 \end{bmatrix}-64 \begin{bmatrix}2 \\-1 \\3 \end{bmatrix} =\begin{bmatrix}-131 \\58 \\-189 \end{bmatrix}, 故選\bbox[red, 2pt]{(A)}$$ 

解答:$$A(A+B)^{-1}B = A(A+B)^{-1}((A+B)-A) =  A-A(A+B)^{-1}A \\=A -((A+B)-B)(A+B)^{-1}A =A -(A-B(A+B)^{-1}A) =B(A+B)^{-1}A \\ \Rightarrow A(A+B)^{-1}B =B(A+B)^{-1}A,故選 \bbox[red, 2pt]{(E)}$$

解答:$$J=[1]_{n\times n} \Rightarrow rank(J)=1 \Rightarrow 0 \text{ is an eigenvalue with multiplicity }n-1\\ \text{tr}(J)=n = \text{ sum of eigenvalues }\Rightarrow n \text{ is an eigenvalue} \\ \Rightarrow J \text{ has eigenvalues }\lambda_{J_1}=n, \lambda_{J_2}=0 \\ A=(a-b)I +bJ \Rightarrow A \text{ has eigenvalue } \lambda_A =(a-b)\cdot 1+ b\cdot \lambda_J =\cases{a-b+nb\\ a-b+0} \\ \Rightarrow \text{eigenvalues of A: }a+(n-1)b, a-b, 故選\bbox[red, 2pt]{(C)}$$

解答:$$A = \begin{bmatrix} \frac{1}{7} & \frac{3}{7} & \frac{3}{7} \\\frac{3}{7} & \frac{1}{7} & \frac{3}{7} \\ \frac{3}{7} & \frac{3}{7} & \frac{1}{7}\end{bmatrix} = \begin{bmatrix}  1 & -1 & -1 \\1 & 1 & 0 \\1 & 0 & 1\end{bmatrix} \begin{bmatrix}1 & 0 & 0 \\0 & \frac{-2}{7} & 0 \\0 & 0 & \frac{-2}{7} \end{bmatrix} \begin{bmatrix} \frac{1}{3} & \frac{1}{3} & \frac{1}{3} \\\frac{-1}{3} & \frac{2}{3} & \frac{-1}{3} \\\frac{-1}{3} & \frac{-1}{3} & \frac{2}{3} \end{bmatrix} \\ \Rightarrow A^n =\begin{bmatrix} \frac{1}{7} & \frac{3}{7} & \frac{3}{7} \\\frac{3}{7} & \frac{1}{7} & \frac{3}{7} \\ \frac{3}{7} & \frac{3}{7} & \frac{1}{7}\end{bmatrix} = \begin{bmatrix}  1 & -1 & -1 \\ 1 & 1 & 0 \\1 & 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 \\0 & (\frac{-2}{7})^n & 0 \\0 & 0 & (\frac{-2}{7})^n \end{bmatrix} \begin{bmatrix}\frac{1}{3} & \frac{1}{3} & \frac{1}{3} \\\frac{-1}{3} & \frac{2}{3} & \frac{-1}{3} \\\frac{-1}{3} & \frac{-1}{3} & \frac{2}{3} \end{bmatrix} \\ \Rightarrow \lim_{n\to \infty} A^n = \begin{bmatrix} \frac{1}{7} & \frac{3}{7} & \frac{3}{7} \\\frac{3}{7} & \frac{1}{7} & \frac{3}{7} \\ \frac{3}{7} & \frac{3}{7} & \frac{1}{7}\end{bmatrix} = \begin{bmatrix}  1 & -1 & -1 \\ 1 & 1 & 0 \\1 & 0 & 1\end{bmatrix} \begin{bmatrix}1 & 0 & 0 \\0 & 0 & 0 \\0 & 0 & 0 \end{bmatrix} \begin{bmatrix}\frac{1}{3} & \frac{1}{3} & \frac{1}{3} \\\frac{-1}{3} & \frac{2}{3} & \frac{-1}{3} \\\frac{-1}{3} & \frac{-1}{3} & \frac{2}{3} \end{bmatrix} \\\qquad = \begin{bmatrix}1 & 0 & 0 \\1 & 0 & 0 \\1 & 0 & 0 \end{bmatrix} \begin{bmatrix} \frac{1}{3} & \frac{1}{3} & \frac{1}{3} \\\frac{-1}{3} & \frac{2}{3} & \frac{-1}{3} \\\frac{-1}{3} & \frac{-1}{3} & \frac{2}{3} \end{bmatrix} =  \begin{bmatrix} \frac{1}{3} & \frac{1}{3} & \frac{1}{3} \\\frac{1}{3} & \frac{1}{3} & \frac{1}{3} \\\frac{1}{3} & \frac{1}{3} & \frac{1}{3}\end{bmatrix}，故選\bbox[red, 2pt]{(B)}$$

解答:$$\begin{vmatrix} 3a& -7& 8&9& -6\\ 0& 2& -5& 7& 7\\ 0& 0& 1& 5& 0\\  0& 0& 2& 4& -1\\ 0& 0& 0& 2&0\end{vmatrix} =3a \begin{vmatrix}   2& -5& 7& 7\\   0& 1& 5& 0\\  0& 2& 4& -1\\   0& 0& 2&0\end{vmatrix} =3a \cdot 2 \begin{vmatrix}  1& 5 & 0\\  2& 4& -1\\ 0& 2&0\end{vmatrix} =6a\cdot 2=12a，故選\bbox[red, 2pt]{(D)}$$

解答:$$\text{RREF} \left( \left[\begin{matrix}1 & 0 & -1 & 2\\0 & 1 & -1 & 1\\1 & -1 & 0 & 1\end{matrix}\right] \right) = \left[\begin{matrix}1 & 0 & -1 & 2\\0 & 1 & -1 & 1\\0 & 0 & 0 & 0\end{matrix}\right] \Rightarrow (0,0,0,0),(1,1,1,0 )\in Null(A)，故選\bbox[red, 2pt]{(D)}$$

解答:$$\text{The dimension of the row space  = rank}(A) =3, 故選\bbox[red, 2pt]{(C)}$$

解答:$$Ax= \lambda x \Rightarrow A^2(x) =A(\lambda x) =\lambda A(x)= \lambda^2 x \Rightarrow Ax= \lambda^2 x \Rightarrow \lambda=\lambda^2 \Rightarrow \lambda=0,1\\ \Rightarrow \lambda \ne -1，故選\bbox[red, 2pt]{(A)}$$

解答:$$[A\mid I] = \left[\begin{array}{rrr|rrr} a& b& c& 1& 0& 0\\ 0& d& e& 0& 1& 0\\ 0& 0& f& 0& 0& 1\end{array} \right] \xrightarrow{R_1/a\to R_1, R_2/d\to R_2, R_3/f\to R_3} \left[\begin{array}{rrr|rrr} 1 & b/a& c/a& 1/a& 0& 0\\ 0& 1& e/d& 0& 1/d& 0\\ 0& 0& 1& 0& 0& 1/f\end{array} \right] \\ \xrightarrow{R_1-(b/a)R_2 \to R_1} \left[\begin{array}{rrr|rrr} 1 & 0& (cd-be)/ad& 1/a& -b/ad& 0\\ 0& 1& e/d& 0& 1/d& 0\\ 0& 0& 1& 0& 0& 1/f\end{array} \right] \xrightarrow{R_1-(cd-be)R_3/ad \to R_1, R_2-(e/d)R_3 \to R_2} \\  \left[\begin{array}{rrr|rrr} 1 & 0& 0& 1/a& -b/ad& (be-cd)/adf\\ 0& 1& 0& 0& 1/d& -e/df\\ 0& 0& 1& 0& 0& 1/f\end{array} \right]  \Rightarrow A^{-1} =\begin{bmatrix}1/a & -b/ad& (be-cd)/adf\\ 0& 1/d& -e/df\\ 0& 0& 1/f \end{bmatrix} \\={1\over adf} \begin{bmatrix}df& -bf&  be-cd\\ 0 & af& -ae\\ 0& 0& ad\end{bmatrix}, 故選\bbox[red, 2pt]{(E)}$$

解答:$$(B)\times: A = \begin{bmatrix}3& 0& 0\\ 0& 2& 0\\ 0& 0& 1 \end{bmatrix} \Rightarrow A \text{ is symmetric postive definite} \\ \Rightarrow \text{eigenvalues of }A:\cases{\lambda_1=3\\ \lambda_2=2 \\\lambda_3 =1} \text{ and the corresponding eigenvectors: }\cases{u_1=(1,0,0)^T\\ u_2=(0,1,0)^T\\ u_3=(0,0,1)^T} \\ \Rightarrow \cases{A^HA =A^2 = \begin{bmatrix}9&0& 0\\ 0& 4& 0\\ 0& 0& 1 \end{bmatrix} \\ \lambda_1u_1u_1^H + \lambda_2 u_2u_2^H + \lambda_3 u_3u_3^H = \begin{bmatrix}3&0& 0\\ 0& 2& 0\\ 0& 0& 1 \end{bmatrix}} \Rightarrow A^HA \ne \lambda_1u_1u_1^H + \lambda_2 u_2u_2^H + \lambda_3 u_3u_3^H \\ 故選\bbox[red, 2pt]{(B)}$$


解答:$$(A) \bigcirc: A^{-1} =\left[\begin{matrix}0 & 1 & 0 & 0\\0 & 0 & 1 & 0\\0 & 0 & 0 & 1\\1 & 0 & 0 & 0\end{matrix}\right] =A^T \\ (B) \bigcirc: A^2 = \left[\begin{matrix}0 & 0 & 1 & 0\\0 & 0 & 0 & 1\\1 & 0 & 0 & 0\\0 & 1 & 0 & 0\end{matrix}\right] \Rightarrow (A^2)^{-1} =\left[\begin{matrix}0 & 0 & 1 & 0\\0 & 0 & 0 & 1\\1 & 0 & 0 & 0\\0 & 1 & 0 & 0\end{matrix}\right] =A^2\\ (C)\times: \det(A-\lambda I) =\lambda^4-1=0 \Rightarrow \lambda =\pm 1, \pm i \\(D)\bigcirc: A+I= \left[\begin{matrix}1 & 0 & 0 & 1\\1 & 1 & 0 & 0\\0 & 1 & 1 & 0\\0 & 0 & 1 & 1\end{matrix}\right] \Rightarrow \det(A+I)=0 \\(E) \bigcirc: A+2I = \left[\begin{matrix}2 & 0 & 0 & 1\\1 & 2 & 0 & 0\\0 & 1 & 2 & 0\\0 & 0 & 1 & 2\end{matrix}\right] \Rightarrow \det(A+2I)=15\\ 故選\bbox[red, 2pt]{(C)}$$

解答:$$\cases{x+y+2z=k \\ 3x+y+7z=3k+4\\ 4x+2y+(k+7)z= 5k} \Rightarrow \begin{bmatrix}1&1& 2\\ 3& 1&7\\ 4& 2& k+7 \end{bmatrix} \begin{bmatrix}x\\ y\\ z \end{bmatrix} = \begin{bmatrix}k\\ 3k+4\\ 5k \end{bmatrix} \\ \Rightarrow  \det \left( \begin{bmatrix}1&1& 2\\ 3& 1&7\\ 4& 2& k+7 \end{bmatrix}  \right) =-2k+4=0 \Rightarrow k=2 \Rightarrow \begin{bmatrix}1&1& 2\\ 3& 1&7\\ 4& 2& 9 \end{bmatrix} \begin{bmatrix}x\\ y\\ z \end{bmatrix} = \begin{bmatrix}2\\ 10\\ 10 \end{bmatrix} \\ \Rightarrow \text{RREF} \left(  \left[ \begin{array}{rrr|r} 1 & 1 & 2 & 2\\3 & 1 & 7 & 10\\4 & 2 & 9 & 10\end{array} \right] \right) =\left[ \begin{array}{rrr|r} 1 & 0 & \frac{5}{2} & 0\\0 & 1 & - \frac{1}{2} & 0\\0 & 0 & 0 & 1\end{array} \right] \Rightarrow \text{No solution},故選\bbox[red, 2pt]{(D)}$$

解答:$$(A)\bigcirc: u_1={ v_1 \over |v_1|} ={1\over 2} [1,1,1,1]^T\\ (B)\bigcirc: u_2= {v_2-(v_2 \cdot u_1)u_1 \over |v_2-(v_2 \cdot u_1)u_1|} ={[-5/2, 5/2, 5/2,-5/2]^T \over |[-5/2, 5/2, 5/2,-5/2]^T|} = {1\over 2}[-1, 1, 1,-1]^T\\ (C)\bigcirc: u_3={v_3-(v_3\cdot u_1)u_1-(v_3\cdot u_2)u_2 \over |v_3-(v_3\cdot u_1)u_1-(v_3\cdot u_2)u_2|} = {[2,-2,2,-2]^T \over |[2,-2,2,-2]^T|} ={1\over 2}[1,-1,1, -1]^T\\ (D)\bigcirc: v_1,v_2, v_3 \text{ are linearly independent} \Rightarrow Dim(S)=3\\ 故選\bbox[red, 2pt]{(E)}$$

解答:$$A=\begin{bmatrix}1 & -1 & 4 \\1 & 4 & -2 \\1 & 4 & 2 \\1 & -1 & 0 \end{bmatrix} \Rightarrow v_1=\begin{bmatrix} 1\\ 1\\ 1\\ 1 \end{bmatrix}, v_2= \begin{bmatrix} -1\\4\\4\\-1\end{bmatrix},  v_3=\begin{bmatrix} 4\\ -2\\2 \\0 \end{bmatrix} \\ \text{Applying Gram-Schmidt processes, we have }e_1= \begin{bmatrix} 1/2\\ 1/2\\ 1/2\\ 1/2 \end{bmatrix}, e_2= \begin{bmatrix} -1/2\\1/2 \\ 1/2 \\-1/2 \end{bmatrix},  e_3=\begin{bmatrix} 1/2 \\ -1/2\\ 1/2\\ -1/2 \end{bmatrix} \\ \Rightarrow Q=[ e_1 e_2 e_3] = \begin{bmatrix} 1/2 &-1/2& 1/2\\ 1/2 & 1/2 &-1/2\\1/2 & 1/2& 1/2\\ 1/2& -1/2& -1/2\end{bmatrix} \Rightarrow Q^T= \begin{bmatrix} 1/2 & 1/2& 1/2 & 1/2\\ -1/2 & 1/2 & 1/2 & -1/2\\1/2 & -1/2& 1/2 &-1/2 \end{bmatrix} \\ \Rightarrow R =  Q^TA = \begin{bmatrix} 2& 3& 2\\ 0& 5& -2\\ 0& 0& 4\end{bmatrix}, 故選\bbox[red, 2pt]{(B)}$$

解答:$$\cases{A=  \begin{bmatrix}1 & -1 & 4 \\1 & 4 & -2 \\1 & 4 & 2 \\1 & -1 & 0 \end{bmatrix} \\ \mathbf x= \begin{bmatrix}1 & -2 & 2 & 1  \end{bmatrix}^T  }\Rightarrow A(A^TA)^{-1}A \mathbf x = \begin{bmatrix}2 & -1 & 1 & 0  \end{bmatrix}^T ，故選\bbox[red, 2pt]{(A)}$$

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解題僅供參考，碩士班歷年試題及詳解