3.33 演習問題 〜 解空間

3.135 (解空間)   次の解空間の基底と次元を求めよ. ただし,(12), (13), (19)では $ \dim(W)\ge1$ となるよう $ a$ の値を定めよ.

(1) $ \displaystyle{W=\left\{\left.\,{\vec{x}\in\mathbb{R}^{n}}\,\,\right\vert\,\,{A\vec{x}=\vec{0}\,\text{ただし $\mathrm{rank}\,(A)=r$}}\,\right\}}$         (2) $ \displaystyle{W=
\left\{\vec{x}\in\mathbb{R}^2
\left\vert 2x_{1}-x_{2}=0 \right.\right\}}$

(3) \begin{displaymath}\displaystyle{W=
\left\{\vec{x}\in\mathbb{R}^2
\left\vert
\be...
...}
2x_{1}-x_{2}=0 \\
x_{1}+x_{2}=0
\end{array}\right.\right\}}\end{displaymath}         (4) \begin{displaymath}\displaystyle{W=
\left\{\vec{x}\in\mathbb{R}^2
\left\vert
\be...
...
2x_{1}-x_{2}=0 \\
-2x_{1}+x_{2}=0
\end{array}\right.\right\}}\end{displaymath}

(5) $ \displaystyle{W=
\left\{\vec{x}\in\mathbb{R}^2
\quad\left\vert\quad
\begin{bma...
...だし $a_{11}a_{22}-a_{12}a_{21}=0$, $a_{11}\neq0,a_{22}\neq0$}
\right.\right\}}$

(6) $ \displaystyle{W=
\left\{\vec{x}\in\mathbb{R}^2
\quad\left\vert\quad
\begin{bma...
... $a_{11}a_{22}-a_{12}a_{21}\neq0$, $a_{11}\neq0,a_{22}\neq0$}
\right.\right\}}$

(7) $ \displaystyle{W=
\left\{\left.\,{\vec{x}\in\mathbb{R}^3}\,\,\right\vert\,\,{x_1+x_2+x_3=0}\,\right\}}$          (8) $ \displaystyle{W=
\left\{\vec{x}\in\mathbb{R}^3
\,\,\left\vert\,\,
\begin{bmatrix}
1 & 2 & -1 \\
3 &-3 & 2
\end{bmatrix}\vec{x}=\vec{0}
\right.\right\}}$

(9) $ \displaystyle{W=
\left\{\vec{x}\in\mathbb{R}^3
\,\,\left\vert\,\,
\begin{bmatrix}
1 & 0 & -1 \\
2 &-1 & 3
\end{bmatrix}\vec{x}=\vec{0}
\right.\right\}}$          (10) $ \displaystyle{W=
\left\{\vec{x}\in\mathbb{R}^3
\,\,\left\vert\,\,
\begin{bmatrix}
1 & 1 & -1 \\
2 & 2 & -2
\end{bmatrix}\vec{x}=\vec{0}
\right.\right\}}$

(11) $ \displaystyle{W=
\left\{\vec{x}\in\mathbb{R}^3
\,\,\left\vert\,\,
\begin{bmatr...
...& 0 \\
0 & 3 & -1 \\
2 & -2 & 0
\end{bmatrix}\vec{x}=\vec{0}
\right.\right\}}$          (12) $ \displaystyle{W=
\left\{\vec{x}\in\mathbb{R}^3
\,\,\left\vert\,\,
\begin{bmatr...
...-a \\
2 & -3 & 1 \\
3 & -1 & -2
\end{bmatrix}\vec{x}=\vec{0}
\right.\right\}}$

(13) $ \displaystyle{W=
\left\{\vec{x}\in\mathbb{R}^3
\,\,\left\vert\,\,
\begin{bmatr...
... -1 \\
3 & a & -2 \\
1 & 5a & 0
\end{bmatrix}\vec{x}=\vec{0}
\right.\right\}}$          (14) $ \displaystyle{W=
\left\{\vec{x}\in\mathbb{R}^3
\,\,\left\vert\,\,
\begin{bmatr...
...& 0 \\
0 & 3 & -1 \\
1 & 2 & -1
\end{bmatrix}\vec{x}=\vec{0}
\right.\right\}}$

(15) $ \displaystyle{W=
\left\{\vec{x}\in\mathbb{R}^4
\,\,\left\vert\,\,
\begin{bmatr...
...1 & -2 & -3 & 4 \\
0 & 2 & 2 & 0
\end{bmatrix}\vec{x}=\vec{0}
\right.\right\}}$      (16) $ \displaystyle{W=
\left\{\vec{x}\in\mathbb{R}^4
\,\,\left\vert\,\,
\begin{bmatr...
...1 & 1 & -1 & 1 \\
3 & 1 & 2 & -1
\end{bmatrix}\vec{x}=\vec{0}
\right.\right\}}$

(17) $ \displaystyle{W=
\left\{\vec{x}\in\mathbb{R}^4
\,\,\left\vert\,\,
\begin{bmatr...
... & 2 & -2 & -1 \\
3 & 2 & 6 & 11
\end{bmatrix}\vec{x}=\vec{0}
\right.\right\}}$      (18) $ \displaystyle{W=
\left\{\vec{x}\in\mathbb{R}^4
\,\,\left\vert\,\,
\begin{bmatr...
...2 & -2 & 0 & 2 \\
1 & -2 & 0 & 2
\end{bmatrix}\vec{x}=\vec{0}
\right.\right\}}$

(19) $ \displaystyle{W=
\left\{\vec{x}\in\mathbb{R}^4
\left\vert
\begin{bmatrix}
1 & ...
... & -6 & 2 & 13 \\
2 & -4 & 1 & 9
\end{bmatrix}\vec{x}=\vec{0}
\right.\right\}}$      (20) $ \displaystyle{W=
\left\{\vec{x}\in\mathbb{R}^5
\left\vert
\begin{bmatrix}
1 & ...
... 1 & 0 & 2 \\
2 & 1 & 2 & -1 & 5
\end{bmatrix}\vec{x}=\vec{0}
\right.\right\}}$

(21) $ \displaystyle{W=
\left\{\vec{x}\in\mathbb{R}^5
\,\,\left\vert\,\,
\begin{bmatr...
... & 1 & -5 \\
3 & 1 & 4 & -7 & 10
\end{bmatrix}\vec{x}=\vec{0}
\right.\right\}}$

(22) $ W=\left\{\left.\,{f(x)\in\mathbb{R}[x]_3}\,\,\right\vert\,\,{f(1)=0,\,\,f'(1)=0}\,\right\}$      (23) $ W=\left\{\left.\,{f(x)\in\mathbb{R}[x]_3}\,\,\right\vert\,\,{f(1)=0,\,\,f(-1)=0}\,\right\}$




平成20年2月2日