111年北科大電機碩士班丁組-線性代數詳解

 國立臺北科技大學 111學年度碩士班 招 生考試

系所組別 :2142電機工程系碩士班丁組
第一節 線性代數 試題 (選考)

解答： $$\textbf{(1)}\;A\textbf x=0 \Rightarrow \begin{bmatrix}1 & -3& 2& -4\\ 0 & 0 & 5 & -7 \\0 & 0 & 0 &5\\ 0 & 0 & 0 & 0 \end{bmatrix} \begin{bmatrix}x_1 \\x_2\\ x_3\\x_4 \end{bmatrix} =0 \Rightarrow \cases{x_1-3x_2+ 3x_3-4x_4=0 \\5x_3=7x_4\\x_4=0}\\ \Rightarrow \textbf x= x_2\begin{pmatrix}3 \\1\\0\\0 \end{pmatrix} \Rightarrow \text{a basis of }NULL(A) = \bbox[red, 2pt]{ \left\{ \begin{pmatrix}3 \\1\\0\\0 \end{pmatrix}\right\} } \\\textbf{(2)}\; Rank(A)=\bbox[red, 2pt] 3\\ \textbf{(3)}\; \det(A)= \bbox[red, 2pt] 0 \\\textbf{(4)}\; \text{Suppose }B=A^{-1}, \text{then  }AB=I \Rightarrow \det(AB)=\det(I)=1. \\\quad \text{ But }\det(AB)=\det(A)\det(B) =0\cdot \det(B)=0 \ne 1. \text{That is }A \text{ is } \bbox[red, 2pt]{\text{Not invertible}}$$

解答： $$\textbf{(1)}\; \Vert ACQx\Vert ^2= (ACQx)^T ACQx =x^TQ^T C^TA^T ACQx =x^TQ^T C^TIC Qx =x^TQ^TIQx = x^Tx \\\quad = \Vert x\Vert^2 \Rightarrow \Vert ACQx \Vert=\Vert x\Vert ,\bbox[red, 2pt]{QED}\\ \textbf{(2)}\; \text{If one diagonal element of }B \text{ is zero, then} \det(H) =\det(ABC)= \det(A)\det(B) \det(C) =0 \\ \quad \Rightarrow \det(H)=0 \Rightarrow H\text{ is invertible}, \bbox[red, 2pt]{QED} \\ \textbf{(3)}\; A^{-1}=A^T =A^H \Rightarrow A^{-1}=A^H, \bbox[red, 2pt]{QED}$$

解答： $$\textbf{(1)}\; A^TA=\left( \begin{matrix} 1 & -1 & 0 & 2 \\-2 & 2 & 3 & 5 \end{matrix} \right) \left(\begin{matrix} 1 & -2 \\-1 & 2 \\ 0 & 3 \\2 & 5\end{matrix} \right)= \left( \begin{matrix} 6 & 6 \\6 & 42\end{matrix} \right) \\ A^Tb= \left( \begin{matrix} 1 & -1 & 0 & 2 \\-2 & 2 & 3 & 5 \end{matrix} \right) \left(\begin{matrix}3 \\1 \\-4 \\2\end{matrix} \right) =\left(\begin{matrix}6 \\-6\end{matrix} \right) \\ \Rightarrow \left( \begin{matrix} 6 & 6 \\6 & 42\end{matrix} \right) \left( \begin{matrix} x_1 \\x_2\end{matrix} \right) =\left(\begin{matrix}6 \\-6 \end{matrix} \right) \Rightarrow \cases{x_1+x_2=1 \\x_1+7x_2=-1} \Rightarrow \cases{x_1=4/3\\ x_2=-1/3} \Rightarrow \textbf{x}= \bbox[red, 2pt]{ \left( \begin{matrix} 4/3 \\-1/3\end{matrix} \right)} \\\textbf{(b)}\; \text{To minimize the distance from }A \hat x \text{ to } b, i.e., \Vert b-A\hat x \Vert$$

解答： $$"\Rightarrow": Ax=0 \Rightarrow A^H(Ax)=A^H(0)=0 \Rightarrow A^HAx=0\\ "\Leftarrow":A^HAx=0 \Rightarrow x^HA^HAx=0 \Rightarrow \Vert Ax\Vert^2=0 \Rightarrow Ax=0\\ \bbox[red, 2pt]{Q.E.D.}$$

解答： $$\textbf{(1)}\; 假設A的N個\text{ row vectors: }\vec a_1, \vec a_2,\dots \vec a_N, 即A=\begin{bmatrix}\vec a_1\\ \vec a_2\\ \vdots\\ \vec a_N \end{bmatrix} 及x= \begin{bmatrix}x_1\\ x_2\\ \vdots\\ x_n \end{bmatrix} \\ \quad Ax=0 \Rightarrow x_1\vec a_1+x_2 \vec a_2+\cdots +x_n\vec a_N =0\\ 由於\vec a_1, \vec a_2,\dots \vec a_N為線性獨立,因此x_1=x_2=\cdots =x_N=0, 即x=0為唯一的明顯解. \bbox[red, 2pt]{Q.E.D.}$$

解答： $$T(a_0+a_1t +a_2t^2)= 2a_0+(5a_0-2a_1) t+(4a_1+a_2)t^2\\ \Rightarrow A\begin{bmatrix}a_0   \\a_1\\ a_2 \end{bmatrix} = \begin{bmatrix}2a_0   \\5a_0-2a_1\\ 4a_1+a_2 \end{bmatrix} \Rightarrow A= \bbox[red, 2pt] {\begin{bmatrix}2 & 0& 0 \\5& -2& 0 \\ 0& 4& 1 \end{bmatrix}}$$

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解題僅供參考, 其他歷年試題及詳解