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2024年3月5日 星期二

111年成大測量-線性代數詳解

國立成功大學111學年度碩士班招生考試

系所:測量及空間資訊學系
科目:線性代數




解答:(a)AA1=Idet(AA1)=det(I)det(A)det(A1)=1det(A1)=1det(A)Q.E.D.(b)det[2I3A0CB]=det(2I3)det(CB)=8det(B)det(C)=8det(A)2=8
解答:A=[012101]B=ATA=[101011125]det(BλI)=0eigenvalues: λ=6,1,0normailized eigenvectors: v1=[1/302/305/30],v2=[2/51/50],v3=[1/62/61/6]V=[v1v2v3]=[1/302/51/62/301/52/65/3001/6]square root of the nonzero eigenvalues: σ1=6,σ2=1=[σ1000σ20]=[600010]{u1=1σ1Av1=[2/51/5]u2=1σ1Av2=[1/52/5]U=[2/51/51/52/5]A=UVT, where U=[2/51/51/52/5],=[600010],V=[1/302/51/62/301/52/65/3001/6]

解答:(a)A[011101110]det(AλI)=λ3+3λ+2=(λ+1)2(λ2)eigenvalues: 2,1indefinite(b)λ1=1(Aλ1I)v=0[111111111][x1x2x3]=0x1+x2+x3=0λ2=2(Aλ2I)v=0[211121112][x1x2x3]=0{x1x3=0x2x3=0there is only one eigenvalue 1 on the subspace Mpossitive definite(c)f(x1,x2,x3)=[x1,x2,x3][1β1β12125][x1x2x3]=xTAxA is positive defininte{|1ββ1|=1β2>0|1β1β12125|=a(5a4)>0{1<β<14/5<β<045<β<0


解答:T(x)=0{a12a2=03a3=0x=(2a2a20)Null(T)={k(210)kR}Nullity(A)=1,Rank(A)=3Nullity(A)=2Rank(A)=2

解答:\textbf{(a)}\; T(\mathbf u)=\begin{bmatrix}1 & -3 \\3 & 5\\ -1 & 7 \end{bmatrix} \begin{bmatrix}2 \\-1 \end{bmatrix} = \bbox[red, 2pt]{\begin{bmatrix}5 \\ 1 \\ -9 \end{bmatrix}} \\\textbf{(b)}\; T(\mathbf x)=b \Rightarrow B=[A\mid b]= \left[\begin{matrix}1 & -3 & 3\\3 & 5 & 2\\-1 & 7 & -5\end{matrix} \right] \Rightarrow rref(B)= \left[\begin{matrix}1 & 0 & \frac{3}{2}\\0 & 1 & - \frac{1}{2}\\0 & 0 & 0\end{matrix}\right]\\\qquad  \Rightarrow \bbox[red, 2pt]{\mathbf  x=\begin{bmatrix} 3/2 \\-1/2 \end{bmatrix}}\\ \textbf{(c)}\; rref(A) = \begin{bmatrix}1` & 0 \\0 & 1\\ 0 & 0 \end{bmatrix} \Rightarrow rank(A)=2 \Rightarrow  \bbox[red, 2pt]{NO} \\ \textbf{(d)}\; T(\mathbf x)=\mathbf c \Rightarrow B=[A\mid c] = \left[ \begin{matrix}1 & -3 & 3\\3 & 5 & 2\\-1 & 7 & 5\end{matrix}\right] \Rightarrow rref(B)=\left[\begin{matrix}1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1\end{matrix}\right] \Rightarrow\bbox[red, 2pt]{ \mathbf c \not \in T(\mathbf x)}

解答:\textbf{(a)}\; Az=b \equiv \begin{bmatrix}1 & 0 \\1 & 1 \\1 & 2 \\ 1& 3\end{bmatrix} \begin{bmatrix}\alpha \\ \beta \end{bmatrix}=\begin{bmatrix}0 \\2 \\ 4\\ 6 \end{bmatrix} \\ \textbf{(b)}\; \cases{A^TA= \left[\begin{matrix}1 & 1 & 1 & 1\\0 & 1 & 2 & 3\end{matrix}\right] \left[\begin{matrix}1 & 0\\1 & 1\\1 & 2\\1 & 3\end{matrix} \right] = \left[\begin{matrix}4 & 6\\6 & 14\end{matrix}\right] \\ A^Tb = \left[\begin{matrix}1 & 1 & 1 & 1\\0 & 1 & 2 & 3\end{matrix}\right] \left[\begin{matrix}0\\2\\4\\6\end{matrix} \right] =\left[ \begin{matrix} 12\\28\end{matrix} \right]} \\\Rightarrow \mathbf z= (A^TA)^{-1}(A^Tb)   =\left[\begin{matrix} \frac{7}{10} & - \frac{3}{10}\\- \frac{3}{10} & \frac{1}{5} \end{matrix} \right] \left[ \begin{matrix} 12\\28 \end{matrix}\right] = \begin{bmatrix}0 \\2 \end{bmatrix} \Rightarrow \cases{\alpha=0\\ \beta=2}\\ f=2x \Rightarrow \cases{f(0)=0\\ f(1)=2\\ f(2)=4\\ f(3)=6} \Rightarrow \text{square error }=0+0+0+0=\bbox[red, 2pt]0
 
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解題僅供參考,其他歷年試題及詳解

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