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)AA−1=I⇒det(AA−1)=det(I)⇒det(A)⋅det(A−1)=1⇒det(A−1)=1det(A)Q.E.D.(b)det[2I3A0CB]=det(2I3)⋅det(CB)=8det(B)det(C)=−8⋅det(A)2=−8解答:
A=[012101]⇒B=ATA=[101011125]det(B−λI)=0⇒eigenvalues: λ=6,1,0⇒normailized eigenvectors: v1=[1/√302/√305/√30],v2=[−2/√51/√50],v3=[−1/√6−2/√61/√6]⇒V=[v1∣v2∣v3]=[1/√30−2/√5−1/√62/√301/√5−2/√65/√3001/√6]⇒square root of the nonzero eigenvalues: σ1=√6,σ2=1⇒∑=[σ1000σ20]=[√600010]{u1=1σ1Av1=[2/√51/√5]u2=1σ1Av2=[1/√5−2/√5]⇒U=[2/√51/√51/√5−2/√5]⇒A=U∑VT, where U=[2/√51/√51/√5−2/√5],∑=[√600010],V=[1/√30−2/√5−1/√62/√301/√5−2/√65/√3001/√6]

解答:
(a)A=[011101110]⇒det(A−λI)=−λ3+3λ+2=−(λ+1)2(λ−2)⇒eigenvalues: 2,−1⇒indefinite(b)λ1=−1⇒(A−λ1I)v=0⇒[111111111][x1x2x3]=0⇒x1+x2+x3=0λ2=2⇒(A−λ2I)v=0⇒[−2111−2111−2][x1x2x3]=0⇒{x1−x3=0x2−x3=0⇒there is only one eigenvalue −1 on the subspace M⇒possitive definite(c)f(x1,x2,x3)=[x1,x2,x3][1β−1β12−125][x1x2x3]=xTAxA is positive defininte⇒{|1ββ1|=1−β2>0|1β−1β12−125|=a(−5a−4)>0⇒{−1<β<1−4/5<β<0⇒−45<β<0

解答:
T(x)=0⇒{a1−2a2=03a3=0⇒x=(2a2a20)⇒Null(T)={k(210)∣k∈R}⇒Nullity(A)=1,Rank(A)=3−Nullity(A)=2⇒Rank(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 ==================== END ======================
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