線形代数

行列式の計算

次の行列式を求めよ。(3), (4) については因数分解した形で答えを示せ。

(1)
\[
\begin{array}{|ccc|}
2016 & 2017 & 2018 \\
2010 & 2010 & 2010 \\
2021 & 2020 & 2019 \\
\end{array}
\]

(2)
\[
\begin{array}{|cccc|}
1 & 1 & 2 & 1 \\
3 & 1 & 4 & 3 \\
7 & 6 & 1 & 2 \\
1 & 1 & 3 & 4 \\
\end{array}
\]

(3)
\[
\begin{array}{|cccc|}
1 & 1 & 1 & 1 \\
r & 1 & 1 & 1 \\
r & r & 1 & 1 \\
r & r & r & 1 \\
\end{array}
\]

(4)
\[
\begin{array}{|ccc|}
2 a & 2 b & b – c \\
2 b & 2 a & a + c \\
a + b & a + b & b \\
\end{array}
\]

(1)
\begin{eqnarray*}
\begin{array}{|ccc|}
2016 & 2017 & 2018 \\
2010 & 2010 & 2010 \\
2021 & 2020 & 2019 \\
\end{array}
&=&
2010 \times
\begin{array}{|ccc|}
2016 & 2017 & 2018 \\
1 & 1 & 1 \\
2021 & 2020 & 2019 \\
\end{array} \\
&=&
2010 \times
\begin{array}{|ccc|}
0 & 1 & 2 \\
1 & 1 & 1 \\
2 & 1 & 0 \\
\end{array} \\
&=& 2010 \times(2 + 2 – 4) \\
&=& 0
\end{eqnarray*}

(2)
\begin{eqnarray*}
\begin{array}{|cccc|}
1 & 1 & 2 & 1 \\
3 & 1 & 4 & 3 \\
7 & 6 & 1 & 2 \\
1 & 1 & 3 & 4 \\
\end{array}
&=&
\begin{array}{|cccc|}
1 & 0 & 0 & 0 \\
3 & -2 & -2 & 0 \\
7 & -1 & -13 & -5 \\
1 & 0 & 1 & 3 \\
\end{array} \\
&=&
\begin{array}{|ccc|}
-2 & -2 & 0 \\
-1 & -13 & -5 \\
0 & 1 & 3 \\
\end{array} \\
&=&
\begin{array}{|ccc|}
-2 & 0 & 0 \\
-1 & – 12 & -5 \\
0 & 1 & 3 \\
\end{array} \\
&=&
-2 \times
\begin{array}{|cc|}
-12 & -5 \\
1 & 3 \\
\end{array} \\
&=& 62
\end{eqnarray*}

(3)
\begin{eqnarray*}
\begin{array}{|cccc|}
1 & 1 & 1 & 1 \\
r & 1 & 1 & 1 \\
r & r & 1 & 1 \\
r & r & r & 1 \\
\end{array}
&=&
\begin{array}{|cccc|}
0 & 0 & 0 & 1 \\
r – 1 & 0 & 0 & 1 \\
r – 1 & r – 1 & 0 & 1 \\
r – 1 & r – 1 & r – 1 & 1 \\
\end{array} \\
&=&
-(r – 1) \times
\begin{array}{|cc|}
r – 1 & 0 \\
r – 1 & r – 1 \\
\end{array} \\
&=&
– (r – 1)^3
\end{eqnarray*}

(4)
\begin{eqnarray*}
\begin{array}{|ccc|}
2 a & 2 b & b – c \\
2 b & 2 a & a + c \\
a + b & a + b & b \\
\end{array}
&=&
\begin{array}{|ccc|}
2 a & 2 (b – a) & b – c \\
2 b & 2 (a – b) & a + c \\
a + b & 0 & b \\
\end{array} \\
&=&
\begin{array}{|ccc|}
2(a + b) & 0 & a + b \\
2 b & 2 (a – b) & a + c \\
a + b & 0 & b \\
\end{array} \\
&=&
\begin{array}{|cc|}
2 (a + b) & a + b \\
a + b & b \\
\end{array} \\
&=&
(a + b) \times
\begin{array}{|cc|}
2 & 1 \\
a + b & b \\
\end{array} \\
&=&
– (a – b)^2
\end{eqnarray*}