115年高師大電機碩士班-工程數學詳解

國立高雄師範大學 115 學年度碩士班招生考試

系所別： 電機工程學系
科 目： 工程數學（線性代數與微分方程）

PART I. Multiple Choice Questions (10%) (單選題， 只需填寫答案不用計算過程)

解答:$$(A)\times:  \mathbf{a} \cdot \mathbf{b} = (1)(2) + (2)(-1) + (3)(1) = 2 - 2 + 3 = 3 \ne 0 \\(B)\times:   \mathbf{a} \times \mathbf{b} \text{ results in a vector not a scalar} \\  (D)\times:   \vert{}\mathbf{a} \times \mathbf{b}\vert{} = \vert{}\mathbf{a} \vert{}\vert{}\mathbf{b} \vert{}\sin\theta \ne \vert{}\mathbf{a} \vert{}\vert{}\mathbf{b} \vert{}\cos\theta \\ \Rightarrow \text{Correct statement: }\bbox[red, 2pt]{(C)}$$

解答:$$(A)\bigcirc: T \text{ is linear }\Rightarrow    T(\mathbf{x}_1 + \mathbf{x}_2) = T(\mathbf{x}_1) + T(\mathbf{x}_2) \\(B)\times: \text{If $A$ is a scaling matrix like $A = \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix}$, it doubles the length of every non-zero vector.} \\(C)\times: T \text{ is 1-1 iff the columns of $A$ are linearly independent. }m=n \text{ is not required.} \\(D)\times : A = \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \end{bmatrix} \Rightarrow T \text{ is onto , but the columns are NOT linearly independent} \\ \Rightarrow \bbox[red, 2pt]{(A)}$$

解答:$$(B)\times: A= \begin{bmatrix}i&0\\0& i \end{bmatrix} \Rightarrow A^* = \begin{bmatrix}-i& 0\\0& -i \end{bmatrix} \Rightarrow A^*A= I \Rightarrow A \text{ is Unitary, but }A\ne A^* \\(C)\times: A= \begin{bmatrix}0& 1\\-1& 0 \end{bmatrix} \Rightarrow A^*=A^T= \begin{bmatrix}0& -1\\1& 0 \end{bmatrix} \text{ and }A^*A= AA^* =I \Rightarrow A \text{ is Normal, but }A\ne A^* \\(D)\times:A= \begin{bmatrix}2& 0\\0&2 \end{bmatrix} \Rightarrow A=A^* \Rightarrow A\text{ is Hermitian, but }A^*A= 4I \ne I \\ \Rightarrow \text{Correct statement: }\bbox[red, 2pt]{(A)}$$

解答:$$ A = \begin{bmatrix} 1 & 0 \\ 0 & 2 \\ 0 & 0 \end{bmatrix} \Rightarrow B=A^TA = \begin{bmatrix}1& 0\\0& 4 \end{bmatrix} \Rightarrow B \text{ has eigenvalues: }\lambda_1=4, \lambda_2=1 \\ \Rightarrow \text{singular values:}\cases{\sigma_1= \sqrt{\lambda_1}=2 \\\sigma_2= \sqrt{\lambda_2} =1} \Rightarrow \bbox[red, 2pt]{(A)}$$

解答:$$(A)\times: \text{Both vectors in }\{(1,0), (2,0)\} \text{ are nonzero, but they are linealy dependent. } \\(C)\times: \{(1,0), (0,1), (1,1)\} \text{ can span }\mathbb R^2, \text{ but they are linealy dependent.} \\(D)\times: \{(1,0), (0,1), (1,1)\} \text{ has 3 vectors,  but they are linealy dependent.} \\ \text{True statement: }\bbox[red, 2pt]{(B)}$$

PART II. Calculation Problems (90%) (計算題， 此大題均需寫出計算過程)

解答:$$\textbf{(a) }   \cases{ x = x'\cos\theta - y'\sin\theta \\ y = x'\sin\theta + y'\cos\theta} \Rightarrow  \bbox[red, 2pt]{\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix} \begin{bmatrix} x' \\ y' \end{bmatrix} } \\ \textbf{(b) } \left[\begin{array}{cc\|c} \cos\theta & -\sin\theta & x \\ \sin\theta & \cos\theta & y \end{array}\right] \xrightarrow{\cos \theta R_1+ \sin \theta R_2 \to R_1}  \left[\begin{array}{cc\|c} 1 & 0 & x\cos\theta + y\sin\theta \\ \sin\theta & \cos\theta & y \end{array}\right] \xrightarrow{R_2-\sin \theta R_1\to R_2} \\ \left[\begin{array}{cc\|c} 1 & 0 & x\cos\theta + y\sin\theta \\ 0 & \cos\theta & y - (x\cos\theta + y\sin\theta)\sin\theta \end{array}\right] = \left[ \begin{array}{cc\|c} 1 & 0 & x\cos\theta + y\sin\theta \\ 0 & \cos\theta & y\cos^2 \theta - x\cos\theta  \sin\theta  \end{array}\right]  \\ \xrightarrow{R_2/\cos \theta \to R_2} \left[\begin{array}{cc\|c} 1 & 0 & x\cos\theta + y\sin\theta \\ 0 & 1 & y\cos \theta - x   \sin\theta  \end{array}\right]  \Rightarrow \bbox[red, 2pt]{\cases{ x' = x\cos\theta + y\sin\theta \\ y' = -x\sin\theta + y\cos\theta }} \\ \textbf{(c) }A= \begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix} \Rightarrow  \Delta = \det (A) =1 \Rightarrow  \Delta_{x'} = \det \begin{bmatrix} x & -\sin\theta \\ y & \cos\theta \end{bmatrix}   = x\cos\theta + y\sin\theta  \\  \Delta_{y'} = \det \begin{bmatrix} \cos\theta & x \\ \sin\theta & y \end{bmatrix}  = -x\sin\theta + y\cos\theta  \Rightarrow \cases{x'= \Delta_{x'}/\Delta \\ y'= \Delta_{y'}/ \Delta}  \Rightarrow \bbox[red, 2pt]{\cases{ x' = x\cos\theta + y\sin\theta \\ y' = -x\sin\theta + y\cos\theta }}$$

解答:$$\textbf{(a) } \cases{x_1+ 2x_2- x_3+ x_5+3x_6=0\\ 2x_1+4x_2-2x_3+ x_4+3x_5+ 7x_6=0 \\ x_1+ 2x_2-x_3+ x_4+ 2x_5+ 4x_6=0\\ 3x_1+6x_2-3x_3 +x_4+4x_5+8x_6=0 } \Rightarrow  \begin{bmatrix}  1 & 2 & -1 & 0 & 1 & 3 \\  2 & 4 & -2 & 1 & 3 & 7 \\  1 & 2 & -1 & 1 & 2 & 4 \\  3 & 6 & -3 & 1 & 4 & 8  \end{bmatrix} \begin{bmatrix}x_1\\x_2\\ x_3\\ x_4\\ x_5\\ x_6  \end{bmatrix}=0 \\  A = \begin{bmatrix}  1 & 2 & -1 & 0 & 1 & 3 \\  2 & 4 & -2 & 1 & 3 & 7 \\  1 & 2 & -1 & 1 & 2 & 4 \\  3 & 6 & -3 & 1 & 4 & 8  \end{bmatrix} \Rightarrow \text{RREF}(A)= \begin{bmatrix}  1 & 2 & -1 & 0 & 1 & 0 \\  0 & 0 & 0 & 1 & 1 & 0 \\  0 & 0 & 0 & 0 & 0 & 1 \\  0 & 0 & 0 & 0 & 0 & 0  \end{bmatrix} \Rightarrow \text{rank}(A) = 3 \\ \Rightarrow  \text{nullity}(A) = n - \text{rank}(A) = 6 - 3 = 3 \Rightarrow \bbox[red, 2pt] {\cases{\text{rank}(A) = 3 \\\text{nullity}(A)=3}} \\\textbf{(b) }   \text{From RREF}(A), \text{ we have } \cases{x_1 + 2x_2 - x_3 + x_5 = 0\\ x_4 + x_5 = 0\\ x_6 = 0} \Rightarrow \cases{ x_1 = -2s + t - u\\ x_4 = -u\\ x_6 = 0 },s,t,u \in \mathbb R \\ \Rightarrow  \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \\ x_5 \\ x_6 \end{bmatrix} = \begin{bmatrix} -2s + t - u \\ s \\ t \\ -u \\ u \\ 0 \end{bmatrix} \Rightarrow  \bbox[red, 2pt]{\begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \\ x_5 \\ x_6 \end{bmatrix} =  s \begin{bmatrix} -2 \\ 1 \\ 0 \\ 0 \\ 0 \\ 0 \end{bmatrix} + t \begin{bmatrix} 1 \\ 0 \\ 1 \\ 0 \\ 0 \\ 0 \end{bmatrix} + u \begin{bmatrix} -1 \\ 0 \\ 0 \\ -1 \\ 1 \\ 0 \end{bmatrix},s,t,u \in \mathbb R } \\ \textbf{(c) }\text{From (b), we have a basis for Null(A): }\bbox[red, 2pt]{\left\{ \begin{bmatrix} -2 \\ 1 \\ 0 \\ 0 \\ 0 \\ 0 \end{bmatrix}, \begin{bmatrix} 1 \\ 0 \\ 1 \\ 0 \\ 0 \\ 0 \end{bmatrix}, \begin{bmatrix} -1 \\ 0 \\ 0 \\ -1 \\ 1 \\ 0 \end{bmatrix} \right\}} $$

解答:$$ \mathbf{v}_1=(1,1,0) 、 \mathbf{v}_2=(1,0,1)、 \mathbf{v}_3=(0,1,1)  \Rightarrow  \mathbf{u}_1 = \mathbf{v}_1 = (1, 1, 0) \\ \Rightarrow  \mathbf{u}_2 = \mathbf{v}_2 - \frac{\mathbf{v}_2 \cdot \mathbf{u}_1}{\mathbf{u}_1 \cdot \mathbf{u}_1}\mathbf{u}_1  = (1, 0, 1) - \frac{1}{2}(1, 1, 0)  = \left(\frac{1}{2}, -\frac{1}{2}, 1\right)  \\ \Rightarrow  \mathbf{u}_3 = \mathbf{v}_3 - \frac{\mathbf{v}_3 \cdot \mathbf{u}_1}{\mathbf{u}_1 \cdot \mathbf{u}_1}\mathbf{u}_1 - \frac{\mathbf{v}_3 \cdot \mathbf{u}_2}{\mathbf{u}_2 \cdot \mathbf{u}_2}\mathbf{u}_2   = (0, 1, 1) - \frac{1}{2}(1, 1, 0) - \frac{1/2}{3/2}\left(\frac{1}{2}, -\frac{1}{2}, 1\right)  =   \left(-\frac{2}{3}, \frac{2}{3}, \frac{2}{3}\right)  \\ \Rightarrow \cases{ \mathbf e_1= \mathbf u_1/\Vert u_1\Vert = \left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, 0\right) \\ \mathbf e_2= \mathbf u_2/\Vert u_2\Vert =   \left(\frac{1}{\sqrt{6}}, -\frac{1}{\sqrt{6}}, \frac{2}{\sqrt{6}}\right) \\ \mathbf e_3= \mathbf u_3/\Vert u_3\Vert =  \left(-\frac{\sqrt{3}}{3}, \frac{\sqrt{3}}{3}, \frac{\sqrt{3}}{3}\right) } \Rightarrow \text{an orthonormal basis for span {$\mathbf v_1,\mathbf v_2, \mathbf v_3$}: }\\ \bbox[red, 2pt]{\left\{ \left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, 0\right), \left(\frac{1}{\sqrt{6}}, -\frac{1}{\sqrt{6}}, \frac{2}{\sqrt{6}}\right), \left(-\frac{\sqrt{3}}{3}, \frac{\sqrt{3}}{3}, \frac{\sqrt{3}}{3}\right) \right\} }$$

解答:$$\textbf{ (a) }P = x_1^2 + 5x_2^2 - 4x_3^2 + 6x_1x_2 - 8x_1x_3 + 10x_2x_3 =   \bbox[red, 2pt]{\begin{bmatrix} x_1 & x_2 & x_3 \end{bmatrix} \begin{bmatrix} 1 & 3 & -4 \\ 3 & 5 & 5 \\ -4 & 5 & -4 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} } \\ \textbf{(b)} A = \begin{bmatrix}  1 & 3 & -4 \\  3 & 5 & 5 \\  -4 & 5 & -4  \end{bmatrix} \Rightarrow \det(A-\lambda I)=0 \Rightarrow  \lambda^3 - 2\lambda^2 - 33\lambda + 90 = 0\\ \Rightarrow \text{ eigenvalues: }\lambda =5,2,-9 \Rightarrow \text{eigenvectors: } \begin{bmatrix} 1/3 \\ 2/3 \\ 2/3 \end{bmatrix}, \begin{bmatrix} 2/3 \\ 1/3 \\ -2/3 \end{bmatrix},\begin{bmatrix} -2/3 \\ 2/3 \\ -1/3 \end{bmatrix}  \\ \Rightarrow  Q = \begin{bmatrix}  1/3 & 2/3 & -2/3 \\  2/3 & 1/3 & 2/3 \\  2/3 & -2/3 & -1/3  \end{bmatrix} \Rightarrow \begin{bmatrix}x_1\\ x_2\\ x_3 \end{bmatrix} =\begin{bmatrix}  1/3 & 2/3 & -2/3 \\  2/3 & 1/3 & 2/3 \\  2/3 & -2/3 & -1/3  \end{bmatrix} \begin{bmatrix}y_1\\ y_2\\ y_3 \end{bmatrix} \\ \Rightarrow  \bbox[red, 2pt]{ \cases{x_1= (y_1+2y_2-2y_3)/3\\ x_2= (2y_1+y_2 +2y_3)/3\\ x_3=(2y_1-2y_2-y_3)/3 } \Rightarrow  P = 5y_1^2 + 2y_2^2 - 9y_3^2 }$$

解答:$$y'''+y''=0 \Rightarrow r^3+r^2=0 \Rightarrow r^2(r+1)=0 \Rightarrow r=0,-1 \Rightarrow y_h(x) = c_1 + c_2 x + c_3 e^{-x} \\  y_p= e^x(A\cos x+B\sin x) \Rightarrow y_p'=  e^x [(A + B) \cos x + (B - A) \sin x] \\ \Rightarrow y_p'' = e^x (2B\cos x-2A\sin x)  \Rightarrow y_p'''= e^x \left[ (2B-2A)\cos x+(-2A-2B)\sin x \right] \\ \Rightarrow y_p'''+y_p'' = e^x [(4B - 2A) \cos x + (-4A - 2B) \sin x] = e^x \cos x \Rightarrow \cases{A=-1/10\\ B=1/5} \\ \Rightarrow  y_p(x) = e^x \left(-\frac{1}{10} \cos x + \frac{1}{5} \sin x\right)  \Rightarrow y=y_h+ y_p \\ \Rightarrow \bbox[red, 2pt]{ y(x) = c_1 + c_2 x + c_3 e^{-x} + e^x \left(\frac{1}{5} \sin x - \frac{1}{10} \cos x\right) }$$

解答:$$y=x^m \Rightarrow y'=mx^{m-1} \Rightarrow y''=m(m-1)x^{m-2} \Rightarrow y'''=m(m-1)(m-2)x^{m-3} \\ \Rightarrow  x^3 [m(m-1)(m-2)x^{m-3}] - 6x^m = 0 \Rightarrow [m(m-1)(m-2) - 6]x^m = 0 \\ \Rightarrow m(m-1)(m-2)-6=(m^2+2)(m-3)=0 \Rightarrow m=3, \pm \sqrt 2i \\ \Rightarrow \bbox[red, 2pt]{ y(x) = c_1 x^3 + c_2 \cos(\sqrt{2} \ln x) + c_3 \sin(\sqrt{2} \ln x) }$$

解答:$$y''-y=0 \Rightarrow r^2-1=0 \Rightarrow r=\pm 1 \Rightarrow y_h= c_1e^x+c_2e^{-x}\\ \text{Let }\cases{y_1=e^x\\ y_2=e^{-x}} \Rightarrow W= \begin{vmatrix} y_1& y_2\\ y_1'& y_2' \end{vmatrix} = \begin{vmatrix} e^x& e^{-x}  \\e^x& -e^{-x}\end{vmatrix} =-2; \text{ Using Variation of Parameters,} \\y_p= -e^{x} \int{ e^{-x} \cdot {2e^x\over e^x+e^{-x}} \over -2}\,dx+ e^{-x} \int {e^x\cdot {2e^x\over e^x+e^{-x}} \over -2} \,dx =e^x \int{1\over e^x+e^{-x}}\,dx-e^{-x} \int {e^{2x} \over e^x+e^{-x}}\,dx \\=e^x \tan^{-1}(e^x)+e^{-x}(\tan^{-1}x-e^x)= (e^x+e^{-x}) \tan^{-1}(e^x)-1 \Rightarrow y=y_h+y_p \\ \Rightarrow \bbox[red, 2pt]{y(x) = c_1 e^x + c_2 e^{-x} + (e^x + e^{-x})\arctan(e^x) - 1}$$

 解答:$$ L \frac{d^2q}{dt^2} + R \frac{dq}{dt} + \frac{1}{C} q = E(t)  \Rightarrow  0.5 q'' + 10 q' + \frac{1}{0.01} q = E(t) =10-10u(t-5) \\ \Rightarrow q'' + 20 q' + 200 q = 20-20u(t-5) \Rightarrow  \mathcal{L}\{q'' + 20 q' + 200 q\} = \mathcal{L}\{20 - 20 u(t-5)\}  \\ \Rightarrow (s^2 + 20s + 200) Q(s) = \frac{20}{s} (1 - e^{-5s}) \Rightarrow  Q(s) = \frac{20}{s(s^2 + 20s + 200)} (1 - e^{-5s})  \\\text{Let } H(s) = \frac{20}{s(s^2 + 20s + 200)} = \frac{0.1}{s} - \frac{0.1s + 2}{s^2 + 20s + 200}  \\  = 0.1 \left[ \frac{1}{s} - \frac{s + 10}{(s + 10)^2 + 10^2} - \frac{10}{(s + 10)^2 + 10^2} \right]\\ \Rightarrow h(t)=L^{-1}\{H(s)\}  = 0.1 [1 - e^{-10t} \cos(10t) - e^{-10t} \sin(10t)] \Rightarrow Q(s)=H(s)-H(s)e^{-5s} \\ \Rightarrow q(t)= L^{-1}\{ Q(s)\}= h(t)-h(t-5)u(t-5) \\ \Rightarrow \bbox[red, 2pt]{q(t) = 0.1 \left[ 1 - e^{-10t}(\cos(10t) + \sin(10t)) \right] - 0.1 \left[ 1 - e^{-10(t-5)}(\cos(10(t-5)) + \sin(10(t-5))) \right] u(t-5)}$$

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解題僅供參考，碩士班歷年試題及詳解