國立台北科技大學110學年度碩士班招生考試
系所組別: 2121電機工程系碩士班戊組
第一節 線性代數(選考)
解答:(1)A=[24−21−2−57337−86]R1/2→R1→[12−11/2−2−57337−86]2R1+R2→R2,−3R1+R3→R3→[12−1120−15401−592]2R2+R1→R1,R2+R3→R3→[1091720−154000172]−R2→R2,R3−R1→R1→[109001−5−4000172]2R3/17→R3→[109001−5−40001]4R3+R2→R2→[109001−500001]⇒B=rref(A)=[109001−500001]⇒Ax=0⇒Bx=0⇒[109001−500001][x1x2x3x4]=0⇒{x1+9x3=0x2=5x3x4=0⇒x=x3(−9510)={k(−9510)∣k∈R}(2)Au=[24−21−2−57337−86][3−2−10]=[3−2−10]=[0−33]≠0⇒u∉Null(A)
解答:\textbf{(1)}\; \bbox[red, 2pt]{Prove}A\text{ is diagonalizable and invertible }\Rightarrow A=PDP^{-1} \text{ where D is a diagonal matrix}\\\qquad i.e., d_{i,j}=0, \forall i\ne j \Rightarrow D^{-1}=E, e_{i,j}=1/d_{i,j} \Rightarrow D^{-1} =E\text{ is also diagonal} \\\qquad A=PDP^{-1} \Rightarrow A^{-1} =(PDP^{-1})^{-1} =PD^{-1}P^{-1} =PEP^{-1}\\\qquad \Rightarrow A^{-1} \text{ is also diagonalizable and invertible} \\\textbf{(2)}\; \bbox[red, 2pt]{Prove}:\text{If }a=0, \det(A-a I)=\det(A)=0 \Rightarrow A \text{ is NOT invertible.}\\\qquad \text{ That is } a\ne 0 \Rightarrow Av=av \Rightarrow A^{-1}Av =A^{-1}av \Rightarrow v=aA^{-1}v\\\qquad \Rightarrow A^{-1}v ={1\over a}v \Rightarrow a^{-1} \text{ is an eigenvalue of }A^{-1} \\\textbf{(3)}\; \bbox[red, 2pt]{Prove}: A=QR \Rightarrow Q^{-1}A=R \Rightarrow Q^{-1}AQ=RQ \Rightarrow A \text{ is similar to }RQ \\ \textbf{(4)}\; \bbox[red, 2pt]{Prove}: \text{Let }\lambda \text{ be a eigenvalue of }A, \text{ and v be the corresponding eigenvector.} \\ \qquad Av=\lambda v \Rightarrow AAv =A\lambda v \Rightarrow A^2v =\lambda Av \Rightarrow 0=\lambda \lambda v=\lambda^2v \Rightarrow \lambda =0\\ \textbf{(5)} \; \bbox[red, 2pt]{Prove}: 1\cdot 0+0\cdot v_1+ 0\cdot v_2+ 0\cdot v_3=0 \Rightarrow 0,v_1,v_2,v_3\text{ are linear dependent} \\\textbf{(6)}\; \bbox[red, 2pt]{Disprove}: \text{Let }\cases{v_1=(1,2)\\ v_2=(3,4)} \Rightarrow av_1+bv_2=0 \Rightarrow \cases{a+3b=0\\ 2a+4b=0} \Rightarrow a=b=0\\ \qquad \Rightarrow v_1\text{ and }v_2 \text{ are independent} \Rightarrow v_1\cdot v_2=3+8=11\ne 0 \Rightarrow \text{ Not orthogoanl} \\\textbf{(7)}\; \bbox[red, 2pt]{Disprove}: v_1=0 \Rightarrow v_1 \text{ and }v_1 \text{ are orthogonal but dependent}
解答:略
解答:A=\left[ \begin{matrix}4 & 0\\0 & 2\\1 & 1\end{matrix} \right] \Rightarrow \cases{ A^T A=\left[\begin{matrix}17 & 1\\1 & 5\end{matrix} \right] \\[1ex] A^Tb= \left[ \begin{matrix} 19 \\ 11 \end{matrix} \right]} \Rightarrow (A^TA)x=A^Tb \Rightarrow x=(A^TA)^{-1}(A^Tb) = \bbox[red, 2pt]{\left[ \begin{matrix} 1 \\ 2 \end{matrix} \right]}
解答:\textbf{(1)} \;H=\left[\begin{matrix}3 & -1 & 2 & -5\\0 & 5 & -3 & -6\\-6 & 7 & -7 & 4\\-5 & -8 & 0 & 9\end{matrix}\right] \\\Rightarrow \det(H)=3\left| \begin{matrix}5 & -3 & -6\\7 & -7 & 4\\-8 & 0 & 9\end{matrix}\right| +\left|\begin{matrix}0 & -3 & -6\\-6 & -7 & 4\\-5 & 0 & 9\end{matrix} \right| +2\left| \begin{matrix}0 & 5 & -6\\-6 & 7 & 4\\-5 & -8 & 9\end{matrix}\right| +5 \left| \begin{matrix}0 & 5 & -3\\-6 & 7 & -7\\-5 & -8 & 0\end{matrix}\right| \\=3\times 306+108+2\times (-328)+5\times (-74)= \bbox[red, 2pt]0 \\ \textbf{(2)} \; \det(H)=0 \Rightarrow H\text{ is }\bbox[red, 2pt] {\text{not invertible}}\\ \bbox[cyan,2pt]註:A應該是H
==================== END ======================
解題僅供參考,其他歷年試題及詳解
沒有留言:
張貼留言