112年臺綜大轉學考-工程數學D37詳解

臺灣綜合大學系統112學年度學士班轉學生聯合招生

類組代碼: D37
科目名稱：工程數學

解答: $$w_1={u_1\over \Vert u_1\Vert} ={1\over \sqrt 3}\langle1,1,1 \rangle =\langle{1\over \sqrt 3},{1\over \sqrt 3},{1\over \sqrt 3} \rangle,　故選\bbox[red, 2pt]{(A)}$$

解答: $$w_2= u_2-cw_1,其中c={u_2\cdot w_1\over \Vert w_1 \Vert} =\langle 9,-1,1\rangle \cdot \langle {1\over \sqrt 3}, {1\over \sqrt 3}, {1\over \sqrt 3}\rangle ={9\over \sqrt 3}=3\sqrt 3\\ \Rightarrow w_2=\langle 9,-1,1\rangle- 3\sqrt 3\langle {1\over \sqrt 3}, {1\over \sqrt 3}, {1\over \sqrt 3}\rangle =\langle 6,-4,-2 \rangle 再正規化\\ \Rightarrow w_2={\langle 6,-4,-2 \rangle \over \Vert \langle 6,-4,-2 \rangle\Vert} ={1\over 2\sqrt{14}}\langle 6,-4,-2 \rangle =\langle {3\over \sqrt{14}},-{2\over \sqrt{14}},-{1\over \sqrt{14}} \rangle, 故選\bbox[red, 2pt]{(B)}$$

解答: $$w_1=(A), w_2=(B), 只需考慮選項(C),(D),(E),　也只有(C)與w_1,w_2的內積均為０, 故選\bbox[red, 2pt]{(C)}$$

解答: $$A=\begin{bmatrix}4 & 9 & 7 \\1 & 3 & 4\\ 2 & 5 & 3 \end{bmatrix} \xrightarrow{\bbox[cyan,2pt]{-1\over 4}r_1+r_2 \to r_2, \bbox[cyan,2pt]{-1\over 2}r_1+r_3\to r_2} \begin{bmatrix}4 & 9 & 7 \\0 & 3/4 & 9/4\\ 0 & 1/2 & -1/2 \end{bmatrix}\\ \xrightarrow{\bbox[cyan,2pt]{-2\over 3}r_2+r_3\to r_3}\begin{bmatrix}4 & 9 & 7 \\0 & 3/4 & 9/4\\ 0 & 0 & -2 \end{bmatrix} =U \Rightarrow L=\begin{bmatrix}1 & 0 & 0 \\\bbox[cyan,2pt]{1/4} & 1 & 0\\ \bbox[cyan, 2pt]{1 /2} & \bbox[cyan,2pt]{2/3} & 1 \end{bmatrix},　故選\bbox[red, 2pt]{(BE)}\\ \bbox[cyan,2pt]{註:}選項(B)與(E)相同$$

解答: $$過程如上題,　故選\bbox[red, 2pt]{(E)}$$

解答: $$\det(A-\lambda I)=0 \Rightarrow \lambda^2-9\lambda +20=0 \Rightarrow (\lambda-5)(\lambda-4)=0 \Rightarrow \lambda=5,4\\ \Rightarrow 對角矩陣=\begin{bmatrix}4 & 0 \\0 & 5 \end{bmatrix} 或\begin{bmatrix}5 & 0 \\0 & 4 \end{bmatrix},故選 \bbox[red, 2pt]{(BD)}$$

解答: $$\det(A^{-1})=12-9=3 \Rightarrow A={1\over 3}\begin{bmatrix}3 & -9 \\-1 & 4 \end{bmatrix} =\begin{bmatrix}1 & -3 \\-1/3 & 4/3 \end{bmatrix},故選 \bbox[red, 2pt]{(C)}$$

解答: $$\det(A-\lambda I)=0 \Rightarrow -\lambda^3+ 5\lambda^2+ \lambda-5=0 \Rightarrow -(\lambda-1)(\lambda+1)(\lambda-5)=0\\ \Rightarrow \lambda=\pm 1,5,　故選 \bbox[red, 2pt]{(A)}$$

解答: $$令\cases{P(x,y)=2x+3\\ Q(x,y)=6y^2} \Rightarrow \frac{\partial P}{\partial y} =\frac{\partial Q}{\partial y}=0 \Rightarrow 與積分路徑無關\\　取\cases{x(t)=2t+1\\ y(t)=2t+2} \Rightarrow \cases{dx=2dt\\ dy=2dt} \Rightarrow \int_{(1,2)}^{(3,4)} Pdx+Qdy = \int_0^1(4t+2+3)2dt + 6(2t+2)^22dt \\= \int_0^1 48t^2+104t+58\,dt =\left. \left[ 16t^3+52t^2+58t \right]\right|_0^1 =126, 故選\bbox[red, 2pt]{(D)}$$

解答: $$\cases{x_1-2x_2+x_3=2 \\ 3x_1-x_2+2x_3=5\\ 2x_1+x_2+x_3=1} \Rightarrow Ax=b,其中A=\begin{bmatrix}1 & -2& 1 \\3 & -1& 2\\ 2& 1& 1 \end{bmatrix}, x=\begin{bmatrix}x_1 \\x_2 \\x_3 \end{bmatrix}, b=\begin{bmatrix}2 \\5 \\1 \end{bmatrix}\\ \Rightarrow rref(A)= \begin{bmatrix}5 & 0& 3 \\0 & 5& -1\\ 0& 0 & 0 \end{bmatrix} \Rightarrow rank(A)=2 \Rightarrow 無限多解, 故選\bbox[red, 2pt]{(D)}$$

解答: $$y''-8y'+16y=e^{4x} \Rightarrow \lambda^2-8\lambda+16=0 \Rightarrow (\lambda-4)^2=0\\ \Rightarrow 齊次解y_h=c_1e^{4x}+c_2xe^{4x}\\ 又r(x)=e^{4x} \Rightarrow y_p=c_3x^2e^{4x} \Rightarrow y=y_h+y_p = c_1e^{4x}+ c_2xe^{4x} +c_3x^2e^{4x} \\ \Rightarrow y'= (4c_1+c_2)e^{4x}  +(4c_2+ 2c_3)xe^{4x}+ + 4c_3x^2e^{4x} \\ 由\cases{y(0)=0\\ y'(0)=0} \Rightarrow \cases{c_1=0\\ 4c_1+c_2=0 } \Rightarrow c_1=c_2=0 \Rightarrow y=c_3x^2e^{4x} \\ \Rightarrow \cases{y'=2c_3xe^{4x}+ 4c_3x^2e^{4x} \\ y''=2c_3e^{4x}+16c_3xe^{4x} +16c_3x^2e^{4x}} \\ \Rightarrow y''-8y'+16y=2c_3e^{4x} =e^{4x} \Rightarrow c_3={1\over 2} \Rightarrow y={1\over 2}x^2e^{4x}, 故選\bbox[red, 2pt]{(D)}$$

解答: $$r(t)= 6\cos t\mathbf i+6\sin t\mathbf j-2t\mathbf k \Rightarrow r'(t)= -6\sin t\mathbf i+6\cos t\mathbf j-2\mathbf k \\ \Rightarrow T={r'(t)\over \Vert r'(t) \Vert} ={1\over \sqrt{40}}\langle -6\sin t,6\cos t,-2\rangle ={1\over \sqrt{10}} \langle -3\sin t,3\cos t,-1\rangle, 故選\bbox[red, 2pt]{(D)}$$

解答: $$T'(t)={1\over \sqrt{10}} \langle -3\cos t,-3\sin t,0\rangle \Rightarrow N={T'(t)\over \Vert T'(t) \Vert} =\langle \cos t,\sin t,0 \rangle, \bbox[cyan, 2pt]{無此選項}$$

解答: $$\mathbf F(x,y,z)=x^3y \mathbf i +(y-z)\mathbf j+ yz^2\mathbf k =(x^3y, y-z,yz^2) =(F_1,F_2,F_3) \\ \Rightarrow \text{curl } \mathbf F =\begin{vmatrix}\mathbf i & \mathbf j &\mathbf k \\ \frac{\partial }{\partial x} &\frac{\partial }{\partial y} &\frac{\partial }{\partial z} \\ x^3y & y-z& yz^2\end{vmatrix} =(z^2+1, 0,-x^3), 故選\bbox[red, 2pt]{(D)}$$

解答: $$\mathbf F(x,y,z)=x^3y \mathbf i +(y-z)\mathbf j+ yz^2\mathbf k =(x^3y, y-z,yz^2) =(F_1,F_2,F_3) \\ \Rightarrow \text{div } \mathbf F =\frac{\partial }{\partial x}F_1 +\frac{\partial }{\partial y}F_2+ \frac{\partial }{\partial z}F_3 =3x^2y+1+2yz, 故選\bbox[red, 2pt]{(D)}$$

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