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\centerline{数学IV・練習問題}
\medskip
\rightline{1995年}
\rightline{河東泰之}

問題.次の5つの行列のJordan標準形を求めよ.

$$\left(\matrix
-4 & 4 & -2 \\
-3 & 4 & -1 \\
9 & -6 & 5 
\endmatrix\right)$$

$$\left(\matrix
-5/2 & 9/2 & -3 \\
-3/2 & 19/2 & -7 \\
0 & 9 & -7
\endmatrix\right)$$

$$\left(\matrix
-25 & -18 & 21\\
9 & 8 & -7 \\
-27 & -18 & 23
\endmatrix\right)$$

$$\left(\matrix
0 & 1 & 0\\
4 & 1 & -1\\
17 & 11 & -4
\endmatrix\right)$$

$$\left(\matrix
14 & -12 & 5\\
9 & -7 & 3\\
-12& 12&-5
\endmatrix\right)$$


解答.

$$
\left(\matrix
-5 & 4 & -2\\
12 & -9 & 5 \\
-3 & 2 & -1
\endmatrix\right)
\left(\matrix
-4 & 4 & -2 \\
-3 & 4 & -1 \\
9 & -6 & 5 
\endmatrix\right)
\left(\matrix
1 & 0 & -2\\
3 & 1 & -1\\
3 & 2 &3
\endmatrix\right)
=
\left(\matrix
2 & 0 & 0\\
0 & 2 &0\\
0 & 0 &1
\endmatrix\right)
$$

$$
\left(\matrix
3/2 & -5/2 & 2\\
3/2 & -3/2& 1\\
-1/2& 3/2& -1
\endmatrix\right)
\left(\matrix
-5/2 & 9/2 & -3 \\
-3/2 & 19/2 & -7 \\
0 & 9 & -7
\endmatrix\right)
\left(\matrix
0 & 1 & 1\\
2 & -1 & 3\\
3 & -2 &3
\endmatrix\right)
=
\left(\matrix
-1 & 1 & 0\\
0 & -1& 0\\
0 & 0 &2
\endmatrix\right)
$$

$$
\left(\matrix
-1 & -1 & 1\\
-9 & -6 & 7\\
4 & 3 & -3
\endmatrix\right)
\left(\matrix
-25 & -18 & 21\\
9 & 8 & -7 \\
-27 & -18 & 23
\endmatrix\right)
\left(\matrix
3 & 0 & 1\\
-1 & 1 & 2\\
3 & 1 & 3
\endmatrix\right)
=
\left(\matrix
2 & 1 & 0\\
0 & 2 & 0\\
0 & 0 & 2
\endmatrix\right)
$$

$$
\left(\matrix
-9 & -6 & 2\\
1 & 1 & 0\\
5 & 3 & -1
\endmatrix\right)
\left(\matrix
0 & 1 & 0\\
4 & 1 & -1\\
17 & 11 & -4
\endmatrix\right)
\left(\matrix
1 & 0 & 2\\
-1 & 1 & -2\\
2 & 3 & 3
\endmatrix\right)
=
\left(\matrix
-1 & 1 & 0\\
0 & -1 & 1 \\
0 & 0 &-1
\endmatrix\right)
$$

$$
\left(\matrix
2 & -2 & 1\\
-3 & 3 & -1\\
9 & -8 &3
\endmatrix\right)
\left(\matrix
14 & -12 & 5\\
9 & -7 & 3\\
-12& 12&-5
\endmatrix\right)
\left(\matrix
-1 & 2 & 1\\
0 & 3 & 1\\
3 & 2 & 0
\endmatrix\right)
=
\left(\matrix
-1 & 0 & 0\\
0 & 1 & 0\\
0 & 0 &2
\endmatrix\right)
$$

\bye