【線形代数】3-3-1 連立一次方程式の解法｜問題集
1.次の連立一次方程式の解を求めなさい。
(1)\(\left\{\begin{array}{l}2x_1+3x_2+16x_3=26 \\ x_1+2x_3=7 \\ x_1+x_2+6x_3=12\end{array}\right.\)
拡大係数行列にすると、
\(\left(\begin{array}{ccc|c}2 & 3 & 16 & 26 \\ 1 & 0 & 2 & 7 \\ 1 & 1 & 6 & 12\end{array}\right)\)
\(\mapsto\left(\begin{array}{ccc|c}1 & 0 & 2 & 7 \\ 2 & 3 & 16 & 26 \\ 1 & 1 & 6 & 12\end{array}\right)\)
\(\mapsto\left(\begin{array}{ccc|c}1 & 0 & 2 & 7 \\ 0 & 3 & 12 & 12 \\ 0 & 1 & 4 & 5\end{array}\right)\)
\(\mapsto\left(\begin{array}{ccc|c}1 & 0 & 2 & 7 \\ 0 & 1 & 4 & 4 \\ 0 & 1 & 4 & 5\end{array}\right)\)
\(\mapsto\left(\begin{array}{ccc|c}1 & 0 & 2 & 7 \\ 0 & 1 & 4 & 4 \\ 0 & 0 & 0 & 1\end{array}\right)\)
一番右の列で階数が増えてるので、この連立一次方程式は解を持たない。
\(\left(\begin{array}{ccc|c}2 & 3 & 16 & 26 \\ 1 & 0 & 2 & 7 \\ 1 & 1 & 6 & 12\end{array}\right)\)
\(\mapsto\left(\begin{array}{ccc|c}1 & 0 & 2 & 7 \\ 2 & 3 & 16 & 26 \\ 1 & 1 & 6 & 12\end{array}\right)\)
\(\mapsto\left(\begin{array}{ccc|c}1 & 0 & 2 & 7 \\ 0 & 3 & 12 & 12 \\ 0 & 1 & 4 & 5\end{array}\right)\)
\(\mapsto\left(\begin{array}{ccc|c}1 & 0 & 2 & 7 \\ 0 & 1 & 4 & 4 \\ 0 & 1 & 4 & 5\end{array}\right)\)
\(\mapsto\left(\begin{array}{ccc|c}1 & 0 & 2 & 7 \\ 0 & 1 & 4 & 4 \\ 0 & 0 & 0 & 1\end{array}\right)\)
一番右の列で階数が増えてるので、この連立一次方程式は解を持たない。
(2)\(\left\{\begin{array}{l}x_1+x_2+x_3=3 \\ 2x_1+x_2+2x_3=5 \\ 4x_1+2x_2+6x_3=12\end{array}\right.\)
拡大係数行列にすると、
\(\left(\begin{array}{ccc|c}1 & 1 & 1 & 3 \\ 2 & 1 & 2 & 5 \\ 4 & 2 & 6 & 12\end{array}\right)\)
\(\mapsto\left(\begin{array}{ccc|c}1 & 1 & 1 & 3 \\ 0 & -1 & 0 & -1 \\ 0 & -1 & 1 & 0\end{array}\right)\)
\(\mapsto\left(\begin{array}{ccc|c}1 & 1 & 1 & 3 \\ 0 & 1 & 0 & 1 \\ 0 & -1 & 1 & 0\end{array}\right)\)
\(\mapsto\left(\begin{array}{ccc|c}1 & 0 & 1 & 2 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 1\end{array}\right)\)
\(\mapsto\left(\begin{array}{ccc|c}1 & 0 & 0 & 1 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 1\end{array}\right)\)
よって、
\((x_1,x_2,x_3)=(1,1,1)\)
\(\left(\begin{array}{ccc|c}1 & 1 & 1 & 3 \\ 2 & 1 & 2 & 5 \\ 4 & 2 & 6 & 12\end{array}\right)\)
\(\mapsto\left(\begin{array}{ccc|c}1 & 1 & 1 & 3 \\ 0 & -1 & 0 & -1 \\ 0 & -1 & 1 & 0\end{array}\right)\)
\(\mapsto\left(\begin{array}{ccc|c}1 & 1 & 1 & 3 \\ 0 & 1 & 0 & 1 \\ 0 & -1 & 1 & 0\end{array}\right)\)
\(\mapsto\left(\begin{array}{ccc|c}1 & 0 & 1 & 2 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 1\end{array}\right)\)
\(\mapsto\left(\begin{array}{ccc|c}1 & 0 & 0 & 1 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 1\end{array}\right)\)
よって、
\((x_1,x_2,x_3)=(1,1,1)\)
(3)\(\left\{\begin{array}{l}x_1+2x_3=7 \\ 2x_1+3x_2+16x_3=26 \\ x_1+x_2+6x_3=11\end{array}\right.\)
拡大係数行列にすると、
\(\left(\begin{array}{ccc|c}1 & 0 & 2 & 7 \\ 2 & 3 & 16 & 26 \\ 1 & 1 & 6 & 11\end{array}\right)\)
\(\mapsto\left(\begin{array}{ccc|c}1 & 0 & 2 & 7 \\ 0 & 3 & 12 & 12 \\ 0 & 1 & 4 & 4\end{array}\right)\)
\(\mapsto\left(\begin{array}{ccc|c}1 & 0 & 2 & 7 \\ 0 & 1 & 4 & 4 \\ 0 & 1 & 4 & 4\end{array}\right)\)
\(\mapsto\left(\begin{array}{ccc|c}1 & 0 & 2 & 7 \\ 0 & 1 & 4 & 4 \\ 0 & 0 & 0 & 0\end{array}\right)\)
\(x_3=t\)とおくと、
\(\begin{pmatrix}x_1 \\ x_2 \\ x_3\end{pmatrix}=\begin{pmatrix}7 \\ 4 \\ 0\end{pmatrix}+t\begin{pmatrix}-2 \\ -4 \\ 1\end{pmatrix},\ \ (t\in\mathbb{R})\)
\(\left(\begin{array}{ccc|c}1 & 0 & 2 & 7 \\ 2 & 3 & 16 & 26 \\ 1 & 1 & 6 & 11\end{array}\right)\)
\(\mapsto\left(\begin{array}{ccc|c}1 & 0 & 2 & 7 \\ 0 & 3 & 12 & 12 \\ 0 & 1 & 4 & 4\end{array}\right)\)
\(\mapsto\left(\begin{array}{ccc|c}1 & 0 & 2 & 7 \\ 0 & 1 & 4 & 4 \\ 0 & 1 & 4 & 4\end{array}\right)\)
\(\mapsto\left(\begin{array}{ccc|c}1 & 0 & 2 & 7 \\ 0 & 1 & 4 & 4 \\ 0 & 0 & 0 & 0\end{array}\right)\)
\(x_3=t\)とおくと、
\(\begin{pmatrix}x_1 \\ x_2 \\ x_3\end{pmatrix}=\begin{pmatrix}7 \\ 4 \\ 0\end{pmatrix}+t\begin{pmatrix}-2 \\ -4 \\ 1\end{pmatrix},\ \ (t\in\mathbb{R})\)
(4)\(\left\{\begin{array}{l}x_1+2x_2-x_3+8x_4=2 \\ 2x_1+3x_2+2x_3+4x_4=-1 \\ x_1+x_2+x_3+2x_4=3\end{array}\right.\)
拡大係数行列にすると、
\(\left(\begin{array}{cccc|c}1 & 2 & -1 & 8 & 2 \\ 2 & 3 & 2 & 4 & -1 \\ 1 & 1 & 1 & 2 & 3\end{array}\right)\)
\(\mapsto\left(\begin{array}{cccc|c}1 & 2 & -1 & 8 & 2 \\ 0 & -1 & 4 & -12 & -5 \\ 0 & -1 & 2 & -6 & 1\end{array}\right)\)
\(\mapsto\left(\begin{array}{cccc|c}1 & 0 & 7 & -16 & -8 \\ 0 & -1 & 4 & -12 & -5 \\ 0 & 0 & -2 & 6 & 6\end{array}\right)\)
\(\mapsto\left(\begin{array}{cccc|c}1 & 0 & 7 & -16 & -8 \\ 0 & 1 & -4 & 12 & 5 \\ 0 & 0 & 1 & -3 & -3\end{array}\right)\)
\(\mapsto\left(\begin{array}{cccc|c}1 & 0 & 0 & 5 & 13 \\ 0 & 1 & 0 & 0 & -7 \\ 0 & 0 & 1 & -3 & -3\end{array}\right)\)
\(x_4=t\)とおくと、
\(\begin{pmatrix}x_1 \\ x_2 \\ x_3 \\ x_4\end{pmatrix}=\begin{pmatrix}13 \\ -7 \\ -3 \\ 0\end{pmatrix}+t\begin{pmatrix}-5 \\ 0 \\ 3 \\ 1\end{pmatrix},\ \ (t\in\mathbb{R})\)
\(\left(\begin{array}{cccc|c}1 & 2 & -1 & 8 & 2 \\ 2 & 3 & 2 & 4 & -1 \\ 1 & 1 & 1 & 2 & 3\end{array}\right)\)
\(\mapsto\left(\begin{array}{cccc|c}1 & 2 & -1 & 8 & 2 \\ 0 & -1 & 4 & -12 & -5 \\ 0 & -1 & 2 & -6 & 1\end{array}\right)\)
\(\mapsto\left(\begin{array}{cccc|c}1 & 0 & 7 & -16 & -8 \\ 0 & -1 & 4 & -12 & -5 \\ 0 & 0 & -2 & 6 & 6\end{array}\right)\)
\(\mapsto\left(\begin{array}{cccc|c}1 & 0 & 7 & -16 & -8 \\ 0 & 1 & -4 & 12 & 5 \\ 0 & 0 & 1 & -3 & -3\end{array}\right)\)
\(\mapsto\left(\begin{array}{cccc|c}1 & 0 & 0 & 5 & 13 \\ 0 & 1 & 0 & 0 & -7 \\ 0 & 0 & 1 & -3 & -3\end{array}\right)\)
\(x_4=t\)とおくと、
\(\begin{pmatrix}x_1 \\ x_2 \\ x_3 \\ x_4\end{pmatrix}=\begin{pmatrix}13 \\ -7 \\ -3 \\ 0\end{pmatrix}+t\begin{pmatrix}-5 \\ 0 \\ 3 \\ 1\end{pmatrix},\ \ (t\in\mathbb{R})\)
(5)\(\left\{\begin{array}{l}3x_1-4x_2-12x_3+11x_4=7 \\ x_1+2x_2+6x_3+7x_4=-1 \\ 2x_1-x_2-3x_3+9x_4=3\end{array}\right.\)
拡大係数行列にすると、
\(\left(\begin{array}{cccc|c}3 & -4 & -12 & 11 & 7 \\ 1 & 2 & 6 & 7 & -1 \\ 2 & -1 & -3 & 9 & 3\end{array}\right)\)
\(\mapsto\left(\begin{array}{cccc|c}0 & -10 & -30 & -10 & 10 \\ 1 & 2 & 6 & 7 & -1 \\ 0 & -5 & -15 & -5 & 5\end{array}\right)\)
\(\mapsto\left(\begin{array}{cccc|c}0 & 1 & 3 & 1 & -1 \\ 1 & 2 & 6 & 7 & -1 \\ 0 & -5 & -15 & -5 & 5\end{array}\right)\)
\(\mapsto\left(\begin{array}{cccc|c}0 & 1 & 3 & 1 & -1 \\ 1 & 0 & 0 & 5 & 1 \\ 0 & 0 & 0 & 0 & 0\end{array}\right)\)
\(\mapsto\left(\begin{array}{cccc|c}1 & 0 & 0 & 5 & 1 \\ 0 & 1 & 3 & 1 & -1 \\ 0 & 0 & 0 & 0 & 0\end{array}\right)\)
\(x_3=s,x_4=t\)とおくと、
\(\begin{pmatrix}x_1 \\ x_2 \\ x_3 \\ x_4\end{pmatrix}=\begin{pmatrix}1 \\ -1 \\ 0 \\ 0\end{pmatrix}+s\begin{pmatrix}0 \\ -3 \\ 1 \\ 0\end{pmatrix}+t\begin{pmatrix}-5 \\ -1 \\ 0 \\ 1\end{pmatrix},\ \ (s,t\in\mathbb{R})\)
\(\left(\begin{array}{cccc|c}3 & -4 & -12 & 11 & 7 \\ 1 & 2 & 6 & 7 & -1 \\ 2 & -1 & -3 & 9 & 3\end{array}\right)\)
\(\mapsto\left(\begin{array}{cccc|c}0 & -10 & -30 & -10 & 10 \\ 1 & 2 & 6 & 7 & -1 \\ 0 & -5 & -15 & -5 & 5\end{array}\right)\)
\(\mapsto\left(\begin{array}{cccc|c}0 & 1 & 3 & 1 & -1 \\ 1 & 2 & 6 & 7 & -1 \\ 0 & -5 & -15 & -5 & 5\end{array}\right)\)
\(\mapsto\left(\begin{array}{cccc|c}0 & 1 & 3 & 1 & -1 \\ 1 & 0 & 0 & 5 & 1 \\ 0 & 0 & 0 & 0 & 0\end{array}\right)\)
\(\mapsto\left(\begin{array}{cccc|c}1 & 0 & 0 & 5 & 1 \\ 0 & 1 & 3 & 1 & -1 \\ 0 & 0 & 0 & 0 & 0\end{array}\right)\)
\(x_3=s,x_4=t\)とおくと、
\(\begin{pmatrix}x_1 \\ x_2 \\ x_3 \\ x_4\end{pmatrix}=\begin{pmatrix}1 \\ -1 \\ 0 \\ 0\end{pmatrix}+s\begin{pmatrix}0 \\ -3 \\ 1 \\ 0\end{pmatrix}+t\begin{pmatrix}-5 \\ -1 \\ 0 \\ 1\end{pmatrix},\ \ (s,t\in\mathbb{R})\)
(6)\(\left\{\begin{array}{l}x_1+2x_2-x_3+7x_4=4 \\ 2x_1+4x_2+3x_3-x_4=3 \\ 3x_1+6x_2+2x_3+6x_4=7\end{array}\right.\)
拡大係数行列にすると、
\(\left(\begin{array}{cccc|c}1 & 2 & -1 & 7 & 4 \\ 2 & 4 & 3 & -1 & 3 \\ 3 & 6 & 2 & 6 & 7\end{array}\right)\)
\(\mapsto\left(\begin{array}{cccc|c}1 & 2 & -1 & 7 & 4 \\ 0 & 0 & 5 & -15 & -5 \\ 0 & 0 & 5 & -15 & -5\end{array}\right)\)
\(\mapsto\left(\begin{array}{cccc|c}1 & 2 & -1 & 7 & 4 \\ 0 & 0 & 1 & -3 & -1 \\ 0 & 0 & 5 & -15 & -5\end{array}\right)\)
\(\mapsto\left(\begin{array}{cccc|c}1 & 2 & 0 & 4 & 3 \\ 0 & 0 & 1 & -3 & -1 \\ 0 & 0 & 0 & 0 & 0\end{array}\right)\)
\(x_2=s,x_4=t\)とおくと、
\(\begin{pmatrix}x_1 \\ x_2 \\ x_3 \\ x_4\end{pmatrix}=\begin{pmatrix}3 \\ 0 \\ -1 \\ 0\end{pmatrix}+s\begin{pmatrix}-2 \\ 1 \\ 0 \\ 0\end{pmatrix}+t\begin{pmatrix}-4 \\ 0 \\ 3 \\ 1\end{pmatrix},\ \ (s,t\in\mathbb{R})\)
\(\left(\begin{array}{cccc|c}1 & 2 & -1 & 7 & 4 \\ 2 & 4 & 3 & -1 & 3 \\ 3 & 6 & 2 & 6 & 7\end{array}\right)\)
\(\mapsto\left(\begin{array}{cccc|c}1 & 2 & -1 & 7 & 4 \\ 0 & 0 & 5 & -15 & -5 \\ 0 & 0 & 5 & -15 & -5\end{array}\right)\)
\(\mapsto\left(\begin{array}{cccc|c}1 & 2 & -1 & 7 & 4 \\ 0 & 0 & 1 & -3 & -1 \\ 0 & 0 & 5 & -15 & -5\end{array}\right)\)
\(\mapsto\left(\begin{array}{cccc|c}1 & 2 & 0 & 4 & 3 \\ 0 & 0 & 1 & -3 & -1 \\ 0 & 0 & 0 & 0 & 0\end{array}\right)\)
\(x_2=s,x_4=t\)とおくと、
\(\begin{pmatrix}x_1 \\ x_2 \\ x_3 \\ x_4\end{pmatrix}=\begin{pmatrix}3 \\ 0 \\ -1 \\ 0\end{pmatrix}+s\begin{pmatrix}-2 \\ 1 \\ 0 \\ 0\end{pmatrix}+t\begin{pmatrix}-4 \\ 0 \\ 3 \\ 1\end{pmatrix},\ \ (s,t\in\mathbb{R})\)
(7)\(\left\{\begin{array}{l}x_1-5x_2=-4 \\ 2x_1-3x_2-x_3=-2 \\ 3x_1-x_2-2x_3=0\end{array}\right.\)
拡大係数行列にすると、
\(\left(\begin{array}{ccc|c}1 & -5 & 0 & -4 \\ 2 & -3 & -1 & -2 \\ 3 & -1 & -2 & 0\end{array}\right)\)
\(\mapsto\left(\begin{array}{ccc|c}1 & -5 & 0 & -4 \\ 0 & 7 & -1 & 6 \\ 0 & 14 & -2 & 12\end{array}\right)\)
\(\mapsto\left(\begin{array}{ccc|c}1 & -5 & 0 & -4 \\ 0 & 1 & -\frac{1}{7} & \frac{6}{7} \\ 0 & 14 & -2 & 12\end{array}\right)\)
\(\mapsto\left(\begin{array}{ccc|c}1 & 0 & -\frac{5}{7} & \frac{2}{7} \\ 0 & 1 & -\frac{1}{7} & \frac{6}{7} \\ 0 & 0 & 0 & 0\end{array}\right)\)
\(x_3=t\)とおくと、
\(\begin{pmatrix}x_1 \\ x_2 \\ x_3 \end{pmatrix}=\begin{pmatrix}\frac{2}{7} \\ \frac{6}{7} \\ 0 \end{pmatrix}+t\begin{pmatrix}\frac{5}{7} \\ \frac{1}{7} \\ 1 \end{pmatrix},\ \ (t\in\mathbb{R})\)
\(\left(\begin{array}{ccc|c}1 & -5 & 0 & -4 \\ 2 & -3 & -1 & -2 \\ 3 & -1 & -2 & 0\end{array}\right)\)
\(\mapsto\left(\begin{array}{ccc|c}1 & -5 & 0 & -4 \\ 0 & 7 & -1 & 6 \\ 0 & 14 & -2 & 12\end{array}\right)\)
\(\mapsto\left(\begin{array}{ccc|c}1 & -5 & 0 & -4 \\ 0 & 1 & -\frac{1}{7} & \frac{6}{7} \\ 0 & 14 & -2 & 12\end{array}\right)\)
\(\mapsto\left(\begin{array}{ccc|c}1 & 0 & -\frac{5}{7} & \frac{2}{7} \\ 0 & 1 & -\frac{1}{7} & \frac{6}{7} \\ 0 & 0 & 0 & 0\end{array}\right)\)
\(x_3=t\)とおくと、
\(\begin{pmatrix}x_1 \\ x_2 \\ x_3 \end{pmatrix}=\begin{pmatrix}\frac{2}{7} \\ \frac{6}{7} \\ 0 \end{pmatrix}+t\begin{pmatrix}\frac{5}{7} \\ \frac{1}{7} \\ 1 \end{pmatrix},\ \ (t\in\mathbb{R})\)
(8)\(\left\{\begin{array}{l}x_1+x_2-3x_3+x_4=2 \\ 2x_1+x_2-2x_3=3 \\ 3x_1+x_2-6x_3+x_4=4 \\ 4x_1+2x_2-9x_3+2x_4=6\end{array}\right.\)
拡大係数行列にすると、
\(\left(\begin{array}{cccc|c}1 & 1 & -3 & 1 & 2 \\ 2 & 1 & -2 & 0 & 3 \\ 3 & 1 & -6 & 1 & 4 \\ 4 & 2 & -9 & 2 & 6\end{array}\right)\)
\(\mapsto\left(\begin{array}{cccc|c}1 & 1 & -3 & 1 & 2 \\ 0 & -1 & 4 & -2 & -1 \\ 0 & -2 & 3 & -2 & -2 \\ 0 & -2 & 3 & -2 & -2\end{array}\right)\)
\(\mapsto\left(\begin{array}{cccc|c}1 & 0 & 1 & -1 & 1 \\ 0 & -1 & 4 & -2 & -1 \\ 0 & 0 & -5 & 2 & 0 \\ 0 & 0 & -5 & 2 & 0\end{array}\right)\)
\(\mapsto\left(\begin{array}{cccc|c}1 & 0 & 1 & -1 & 1 \\ 0 & 1 & -4 & 2 & 1 \\ 0 & 0 & 1 & -\frac{2}{5} & 0 \\ 0 & 0 & -5 & 2 & 0\end{array}\right)\)
\(\mapsto\left(\begin{array}{cccc|c}1 & 0 & 0 & -\frac{3}{5} & 1 \\ 0 & 1 & 0 & \frac{2}{5} & 1 \\ 0 & 0 & 1 & -\frac{2}{5} & 0 \\ 0 & 0 & 0 & 0 & 0\end{array}\right)\)
\(x_4=5t\)とおくと、
\(\begin{pmatrix}x_1 \\ x_2 \\ x_3 \\ x_4\end{pmatrix}=\begin{pmatrix}1 \\ 1 \\ 0 \\ 0\end{pmatrix}+t\begin{pmatrix}3 \\ -2 \\ 2 \\ 5\end{pmatrix},\ \ (t\in\mathbb{R})\)
\(\left(\begin{array}{cccc|c}1 & 1 & -3 & 1 & 2 \\ 2 & 1 & -2 & 0 & 3 \\ 3 & 1 & -6 & 1 & 4 \\ 4 & 2 & -9 & 2 & 6\end{array}\right)\)
\(\mapsto\left(\begin{array}{cccc|c}1 & 1 & -3 & 1 & 2 \\ 0 & -1 & 4 & -2 & -1 \\ 0 & -2 & 3 & -2 & -2 \\ 0 & -2 & 3 & -2 & -2\end{array}\right)\)
\(\mapsto\left(\begin{array}{cccc|c}1 & 0 & 1 & -1 & 1 \\ 0 & -1 & 4 & -2 & -1 \\ 0 & 0 & -5 & 2 & 0 \\ 0 & 0 & -5 & 2 & 0\end{array}\right)\)
\(\mapsto\left(\begin{array}{cccc|c}1 & 0 & 1 & -1 & 1 \\ 0 & 1 & -4 & 2 & 1 \\ 0 & 0 & 1 & -\frac{2}{5} & 0 \\ 0 & 0 & -5 & 2 & 0\end{array}\right)\)
\(\mapsto\left(\begin{array}{cccc|c}1 & 0 & 0 & -\frac{3}{5} & 1 \\ 0 & 1 & 0 & \frac{2}{5} & 1 \\ 0 & 0 & 1 & -\frac{2}{5} & 0 \\ 0 & 0 & 0 & 0 & 0\end{array}\right)\)
\(x_4=5t\)とおくと、
\(\begin{pmatrix}x_1 \\ x_2 \\ x_3 \\ x_4\end{pmatrix}=\begin{pmatrix}1 \\ 1 \\ 0 \\ 0\end{pmatrix}+t\begin{pmatrix}3 \\ -2 \\ 2 \\ 5\end{pmatrix},\ \ (t\in\mathbb{R})\)
(9)\(\left\{\begin{array}{l}2x_1+4x_2+x_3+x_4-3x_5=5 \\ x_1+2x_2+x_3+x_5=3 \\ -x_1-2x_2+2x_3+x_4+2x_5=0 \\ x_1+2x_2+x_3+x_4-2x_5=3\end{array}\right.\)
拡大係数行列にすると、
\(\left(\begin{array}{ccccc|c}2 & 4 & 1 & 1 & -3 & 5 \\ 1 & 2 & 1 & 0 & 1 & 3 \\ -1 & -2 & 2 & 1 & 2 & 0 \\ 1 & 2 & 1 & 1 & -2 & 3\end{array}\right)\)
\(\mapsto\left(\begin{array}{ccccc|c}0 & 0 & -1 & 1 & -5 & -1 \\ 1 & 2 & 1 & 0 & 1 & 3 \\ 0 & 0 & 3 & 1 & 3 & 3 \\ 0 & 0 & 0 & 1 & -3 & 0\end{array}\right)\)
\(\mapsto\left(\begin{array}{ccccc|c}1 & 2 & 1 & 0 & 1 & 3 \\ 0 & 0 & -1 & 1 & -5 & -1 \\ 0 & 0 & 3 & 1 & 3 & 3 \\ 0 & 0 & 0 & 1 & -3 & 0\end{array}\right)\)
\(\mapsto\left(\begin{array}{ccccc|c}1 & 2 & 0 & 1 & -4 & 2 \\ 0 & 0 & -1 & 1 & -5 & -1 \\ 0 & 0 & 0 & 4 & -12 & 0 \\ 0 & 0 & 0 & 1 & -3 & 0\end{array}\right)\)
\(\mapsto\left(\begin{array}{ccccc|c}1 & 2 & 0 & 1 & -4 & 2 \\ 0 & 0 & 1 & -1 & 5 & 1 \\ 0 & 0 & 0 & 1 & -3 & 0 \\ 0 & 0 & 0 & 1 & -3 & 0\end{array}\right)\)
\(\mapsto\left(\begin{array}{ccccc|c}1 & 2 & 0 & 0 & -1 & 2 \\ 0 & 0 & 1 & 0 & 2 & 1 \\ 0 & 0 & 0 & 1 & -3 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0\end{array}\right)\)
\(x_2=s,x_5=t\)とおくと、
\(\begin{pmatrix}x_1 \\ x_2 \\ x_3 \\ x_4 \\ x_5\end{pmatrix}=\begin{pmatrix}2 \\ 0 \\ 1 \\ 0 \\ 0\end{pmatrix}+s\begin{pmatrix}-2 \\ 1 \\ 0 \\ 0 \\ 0\end{pmatrix}+t\begin{pmatrix}1 \\ 0 \\ -2 \\ 3 \\ 1\end{pmatrix},\ \ (s,t\in\mathbb{R})\)
\(\left(\begin{array}{ccccc|c}2 & 4 & 1 & 1 & -3 & 5 \\ 1 & 2 & 1 & 0 & 1 & 3 \\ -1 & -2 & 2 & 1 & 2 & 0 \\ 1 & 2 & 1 & 1 & -2 & 3\end{array}\right)\)
\(\mapsto\left(\begin{array}{ccccc|c}0 & 0 & -1 & 1 & -5 & -1 \\ 1 & 2 & 1 & 0 & 1 & 3 \\ 0 & 0 & 3 & 1 & 3 & 3 \\ 0 & 0 & 0 & 1 & -3 & 0\end{array}\right)\)
\(\mapsto\left(\begin{array}{ccccc|c}1 & 2 & 1 & 0 & 1 & 3 \\ 0 & 0 & -1 & 1 & -5 & -1 \\ 0 & 0 & 3 & 1 & 3 & 3 \\ 0 & 0 & 0 & 1 & -3 & 0\end{array}\right)\)
\(\mapsto\left(\begin{array}{ccccc|c}1 & 2 & 0 & 1 & -4 & 2 \\ 0 & 0 & -1 & 1 & -5 & -1 \\ 0 & 0 & 0 & 4 & -12 & 0 \\ 0 & 0 & 0 & 1 & -3 & 0\end{array}\right)\)
\(\mapsto\left(\begin{array}{ccccc|c}1 & 2 & 0 & 1 & -4 & 2 \\ 0 & 0 & 1 & -1 & 5 & 1 \\ 0 & 0 & 0 & 1 & -3 & 0 \\ 0 & 0 & 0 & 1 & -3 & 0\end{array}\right)\)
\(\mapsto\left(\begin{array}{ccccc|c}1 & 2 & 0 & 0 & -1 & 2 \\ 0 & 0 & 1 & 0 & 2 & 1 \\ 0 & 0 & 0 & 1 & -3 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0\end{array}\right)\)
\(x_2=s,x_5=t\)とおくと、
\(\begin{pmatrix}x_1 \\ x_2 \\ x_3 \\ x_4 \\ x_5\end{pmatrix}=\begin{pmatrix}2 \\ 0 \\ 1 \\ 0 \\ 0\end{pmatrix}+s\begin{pmatrix}-2 \\ 1 \\ 0 \\ 0 \\ 0\end{pmatrix}+t\begin{pmatrix}1 \\ 0 \\ -2 \\ 3 \\ 1\end{pmatrix},\ \ (s,t\in\mathbb{R})\)
2.\(a\)を実数とするとき、次の連立一次方程式の解を求めなさい。
(1)\(\left\{\begin{array}{l}x+y+(a+1)z=2 \\ x+az=3 \\ x+2y+(a+2)z=a\end{array}\right.\)
拡大係数行列にすると、
\(\left(\begin{array}{ccc|c}1 & 1 & a+1 & 2 \\ 1 & 0 & a & 3 \\ 1 & 2 & a+2 & a\end{array}\right)\)
\(\mapsto\left(\begin{array}{ccc|c}1 & 0 & a & 3 \\ 1 & 1 & a+1 & 2 \\ 1 & 2 & a+2 & a\end{array}\right)\)
\(\mapsto\left(\begin{array}{ccc|c}1 & 0 & a & 3 \\ 0 & 1 & 1 & -1 \\ 0 & 2 & 2 & a-3\end{array}\right)\)
\(\mapsto\left(\begin{array}{ccc|c}1 & 0 & a & 3 \\ 0 & 1 & 1 & -1 \\ 0 & 0 & 0 & a-1\end{array}\right)\)
・\(a\neq1\)のとき、
一番右の列で階数が増えてるので、この連立一次方程式は解を持たない。
・\(a=1\)のとき、
\(z=t\)とおくと、
\(\begin{pmatrix}x \\ y \\ z\end{pmatrix}=\begin{pmatrix}3 \\ -1 \\ 0\end{pmatrix}+t\begin{pmatrix}-1 \\ -1 \\ 1\end{pmatrix},\ \ (t\in\mathbb{R})\)
\(\left(\begin{array}{ccc|c}1 & 1 & a+1 & 2 \\ 1 & 0 & a & 3 \\ 1 & 2 & a+2 & a\end{array}\right)\)
\(\mapsto\left(\begin{array}{ccc|c}1 & 0 & a & 3 \\ 1 & 1 & a+1 & 2 \\ 1 & 2 & a+2 & a\end{array}\right)\)
\(\mapsto\left(\begin{array}{ccc|c}1 & 0 & a & 3 \\ 0 & 1 & 1 & -1 \\ 0 & 2 & 2 & a-3\end{array}\right)\)
\(\mapsto\left(\begin{array}{ccc|c}1 & 0 & a & 3 \\ 0 & 1 & 1 & -1 \\ 0 & 0 & 0 & a-1\end{array}\right)\)
・\(a\neq1\)のとき、
一番右の列で階数が増えてるので、この連立一次方程式は解を持たない。
・\(a=1\)のとき、
\(z=t\)とおくと、
\(\begin{pmatrix}x \\ y \\ z\end{pmatrix}=\begin{pmatrix}3 \\ -1 \\ 0\end{pmatrix}+t\begin{pmatrix}-1 \\ -1 \\ 1\end{pmatrix},\ \ (t\in\mathbb{R})\)
(2)\(\left\{\begin{array}{l}x-ay-2z=2 \\ ax+2y+z=1 \\ 4x-ay-3z=5\end{array}\right.\)
拡大係数行列にすると、
\(\left(\begin{array}{ccc|c}1 & -a & -2 & 2 \\ a & 2 & 1 & 1 \\ 4 & -a & -3 & 5\end{array}\right)\)
\(\mapsto\left(\begin{array}{ccc|c}1 & -a & -2 & 2 \\ 0 & a^2+2 & 2a+1 & -2a+1 \\ 0 & 3a & 5 & -3\end{array}\right)\)
\(\mapsto\left(\begin{array}{ccc|c}1 & -a & -2 & 2 \\ 0 & 1 & \frac{2a+1}{a^2+2} & \frac{-2a+1}{a^2+2} \\ 0 & 3a & 5 & -3\end{array}\right)\)
\(\mapsto\left(\begin{array}{ccc|c}1 & 0 & \frac{-a-4}{a^2+2} & \frac{-a+4}{a^2+2} \\ 0 & 1 & \frac{2a+1}{a^2+2} & \frac{-2a+1}{a^2+2} \\ 0 & 0 & \frac{-(a-2)(a+5)}{a^2+2} & \frac{3(a-2)(a+1)}{a^2+2}\end{array}\right)\)
・\(a=-5\)のとき、
一番右の列で階数が増えてるので、この連立一次方程式は解を持たない。
・\(a=2\)のとき、
\(z=6t\)とおくと、
\(\begin{pmatrix}x \\ y \\ z\end{pmatrix}=\begin{pmatrix}\frac{1}{3} \\ -\frac{1}{2} \\ 0\end{pmatrix}+t\begin{pmatrix}6 \\ -5 \\ 6\end{pmatrix},\ \ (t\in\mathbb{R})\)
・\(a\neq-5,2\)のとき、
\(\displaystyle (x,y,z)=\left(\frac{4}{a+5},\frac{4}{a+5},\frac{-3a-3}{a+5}\right)\)
\(\left(\begin{array}{ccc|c}1 & -a & -2 & 2 \\ a & 2 & 1 & 1 \\ 4 & -a & -3 & 5\end{array}\right)\)
\(\mapsto\left(\begin{array}{ccc|c}1 & -a & -2 & 2 \\ 0 & a^2+2 & 2a+1 & -2a+1 \\ 0 & 3a & 5 & -3\end{array}\right)\)
\(\mapsto\left(\begin{array}{ccc|c}1 & -a & -2 & 2 \\ 0 & 1 & \frac{2a+1}{a^2+2} & \frac{-2a+1}{a^2+2} \\ 0 & 3a & 5 & -3\end{array}\right)\)
\(\mapsto\left(\begin{array}{ccc|c}1 & 0 & \frac{-a-4}{a^2+2} & \frac{-a+4}{a^2+2} \\ 0 & 1 & \frac{2a+1}{a^2+2} & \frac{-2a+1}{a^2+2} \\ 0 & 0 & \frac{-(a-2)(a+5)}{a^2+2} & \frac{3(a-2)(a+1)}{a^2+2}\end{array}\right)\)
・\(a=-5\)のとき、
一番右の列で階数が増えてるので、この連立一次方程式は解を持たない。
・\(a=2\)のとき、
\(z=6t\)とおくと、
\(\begin{pmatrix}x \\ y \\ z\end{pmatrix}=\begin{pmatrix}\frac{1}{3} \\ -\frac{1}{2} \\ 0\end{pmatrix}+t\begin{pmatrix}6 \\ -5 \\ 6\end{pmatrix},\ \ (t\in\mathbb{R})\)
・\(a\neq-5,2\)のとき、
\(\displaystyle (x,y,z)=\left(\frac{4}{a+5},\frac{4}{a+5},\frac{-3a-3}{a+5}\right)\)
3.次の連立一次方程式が解をもつような\(a\)の値を求めなさい。また、そのときの解を求めなさい。
(1)\(\left\{\begin{array}{l}x+y-2z+u=2 \\ -x-2y+3z-u=3 \\ 2x+y-3z+2u=a\end{array}\right.\)
拡大係数行列にすると、
\(\left(\begin{array}{cccc|c}1 & 1 & -2 & 1 & 2 \\ -1 & -2 & 3 & -1 & 3 \\ 2 & 1 & -3 & 2 & a\end{array}\right)\)
\(\mapsto\left(\begin{array}{cccc|c}1 & 1 & -2 & 1 & 2 \\ 0 & -1 & 1 & 0 & 5 \\ 0 & -1 & 1 & 0 & a-4\end{array}\right)\)
\(\mapsto\left(\begin{array}{cccc|c}1 & 0 & -1 & 1 & 7 \\ 0 & -1 & 1 & 0 & 5 \\ 0 & 0 & 0 & 0 & a-9\end{array}\right)\)
\(\mapsto\left(\begin{array}{cccc|c}1 & 0 & -1 & 1 & 7 \\ 0 & 1 & -1 & 0 & -5 \\ 0 & 0 & 0 & 0 & a-9\end{array}\right)\)
よって、\(a=9\)のとき、連立一次方程式が解をもち、
\(z=s,u=t\)とおくと、
\(\begin{pmatrix}x \\ y \\ z \\ u\end{pmatrix}=\begin{pmatrix}7 \\ -5 \\ 0 \\ 0\end{pmatrix}+s\begin{pmatrix}1 \\ 1 \\ 1 \\ 0\end{pmatrix}+t\begin{pmatrix}-1 \\ 0 \\ 0 \\ 1\end{pmatrix},\ \ (s,t\in\mathbb{R})\)
\(\left(\begin{array}{cccc|c}1 & 1 & -2 & 1 & 2 \\ -1 & -2 & 3 & -1 & 3 \\ 2 & 1 & -3 & 2 & a\end{array}\right)\)
\(\mapsto\left(\begin{array}{cccc|c}1 & 1 & -2 & 1 & 2 \\ 0 & -1 & 1 & 0 & 5 \\ 0 & -1 & 1 & 0 & a-4\end{array}\right)\)
\(\mapsto\left(\begin{array}{cccc|c}1 & 0 & -1 & 1 & 7 \\ 0 & -1 & 1 & 0 & 5 \\ 0 & 0 & 0 & 0 & a-9\end{array}\right)\)
\(\mapsto\left(\begin{array}{cccc|c}1 & 0 & -1 & 1 & 7 \\ 0 & 1 & -1 & 0 & -5 \\ 0 & 0 & 0 & 0 & a-9\end{array}\right)\)
よって、\(a=9\)のとき、連立一次方程式が解をもち、
\(z=s,u=t\)とおくと、
\(\begin{pmatrix}x \\ y \\ z \\ u\end{pmatrix}=\begin{pmatrix}7 \\ -5 \\ 0 \\ 0\end{pmatrix}+s\begin{pmatrix}1 \\ 1 \\ 1 \\ 0\end{pmatrix}+t\begin{pmatrix}-1 \\ 0 \\ 0 \\ 1\end{pmatrix},\ \ (s,t\in\mathbb{R})\)
4.次の連立一次方程式が無限個の解をもつような\(a,b\)の値を求めなさい。また、そのときの解を求めなさい。
(1)\(\left\{\begin{array}{l}x-y+z+u=2 \\ x+2y-2z+4u=-1 \\ 2x+y+az-u=1 \\ 2x-2y+3z+u=b\end{array}\right.\)
拡大係数行列にすると、
\(\left(\begin{array}{cccc|c}1 & -1 & 1 & 1 & 2 \\ 1 & 2 & -2 & 4 & -1 \\ 2 & 1 & a & -1 & 1 \\ 2 & -2 & 3 & 1 & b\end{array}\right)\)
\(\mapsto\left(\begin{array}{cccc|c}1 & -1 & 1 & 1 & 2 \\ 0 & 3 & -3 & 3 & -3 \\ 0 & 3 & a-2 & -3 & -3 \\ 0 & 0 & 1 & -1 & b-4\end{array}\right)\)
\(\mapsto\left(\begin{array}{cccc|c}1 & -1 & 1 & 1 & 2 \\ 0 & 1 & -1 & 1 & -1 \\ 0 & 3 & a-2 & -3 & -3 \\ 0 & 0 & 1 & -1 & b-4\end{array}\right)\)
\(\mapsto\left(\begin{array}{cccc|c}1 & 0 & 0 & 2 & 1 \\ 0 & 1 & -1 & 1 & -1 \\ 0 & 0 & a+1 & -6 & 0 \\ 0 & 0 & 1 & -1 & b-4\end{array}\right)\)
\(\mapsto\left(\begin{array}{cccc|c}1 & 0 & 0 & 2 & 1 \\ 0 & 1 & 0 & 0 & b-5 \\ 0 & 0 & 0 & a-5 & -(a+1)(b-4) \\ 0 & 0 & 1 & -1 & b-4\end{array}\right)\)
\(\mapsto\left(\begin{array}{cccc|c}1 & 0 & 0 & 2 & 1 \\ 0 & 1 & 0 & 0 & b-5 \\ 0 & 0 & 1 & -1 & b-4 \\ 0 & 0 & 0 & a-5 & -(a+1)(b-4)\end{array}\right)\)
よって、\(a=5,b=4\)のとき、連立一次方程式が無限個の解をもち、
\(u=t\)とおくと、
\(\begin{pmatrix}x \\ y \\ z \\ u\end{pmatrix}=\begin{pmatrix}1 \\ -1 \\ 0 \\ 0\end{pmatrix}+t\begin{pmatrix}-2 \\ 0 \\ 1 \\ 1\end{pmatrix},\ \ (t\in\mathbb{R})\)
\(\left(\begin{array}{cccc|c}1 & -1 & 1 & 1 & 2 \\ 1 & 2 & -2 & 4 & -1 \\ 2 & 1 & a & -1 & 1 \\ 2 & -2 & 3 & 1 & b\end{array}\right)\)
\(\mapsto\left(\begin{array}{cccc|c}1 & -1 & 1 & 1 & 2 \\ 0 & 3 & -3 & 3 & -3 \\ 0 & 3 & a-2 & -3 & -3 \\ 0 & 0 & 1 & -1 & b-4\end{array}\right)\)
\(\mapsto\left(\begin{array}{cccc|c}1 & -1 & 1 & 1 & 2 \\ 0 & 1 & -1 & 1 & -1 \\ 0 & 3 & a-2 & -3 & -3 \\ 0 & 0 & 1 & -1 & b-4\end{array}\right)\)
\(\mapsto\left(\begin{array}{cccc|c}1 & 0 & 0 & 2 & 1 \\ 0 & 1 & -1 & 1 & -1 \\ 0 & 0 & a+1 & -6 & 0 \\ 0 & 0 & 1 & -1 & b-4\end{array}\right)\)
\(\mapsto\left(\begin{array}{cccc|c}1 & 0 & 0 & 2 & 1 \\ 0 & 1 & 0 & 0 & b-5 \\ 0 & 0 & 0 & a-5 & -(a+1)(b-4) \\ 0 & 0 & 1 & -1 & b-4\end{array}\right)\)
\(\mapsto\left(\begin{array}{cccc|c}1 & 0 & 0 & 2 & 1 \\ 0 & 1 & 0 & 0 & b-5 \\ 0 & 0 & 1 & -1 & b-4 \\ 0 & 0 & 0 & a-5 & -(a+1)(b-4)\end{array}\right)\)
よって、\(a=5,b=4\)のとき、連立一次方程式が無限個の解をもち、
\(u=t\)とおくと、
\(\begin{pmatrix}x \\ y \\ z \\ u\end{pmatrix}=\begin{pmatrix}1 \\ -1 \\ 0 \\ 0\end{pmatrix}+t\begin{pmatrix}-2 \\ 0 \\ 1 \\ 1\end{pmatrix},\ \ (t\in\mathbb{R})\)
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