2.3 行列の階数（解答） 担当：市原

線形代数学

2.3 ^{行列の階数（解答） 担当：市原}

問題23 次の行列の階数を求めよ.

(1) 0 B@

1 −1 3 1 1

−3 4 −10 −2 −4

2 5 −3 11 −7

1 CA 0

B@

1 −1 3 1 1

−3 4 −10 −2 −4

2 5 −3 11 −7

1

CA °² + 1°×3

−−−−−−−−−→

0 B@

1 −1 3 1 1

0 1 −1 1 −1

2 5 −3 11 −7

1

CA °³ + 1°×(−2)

−−−−−−−−−−−−→

0 B@

1 −1 3 1 1

0 1 −1 1 −1

0 7 −9 9 −9

1 CA

°3 + 2°×(−7)

−−−−−−−−−−−−→

0 B@

1 −1 3 1 1

0 1 −1 1 −1

0 0 −2 2 −2

1

CA °^{3 ×(−1}₂⁾

−−−−−−−−→

0 B@

1 −1 3 1 1

0 1 −1 1 −1

0 0 1 −1 1

1 CA

よって、rank 0 B@

1 −1 3 1 1

−3 4 −10 −2 −4

2 5 −3 11 −7

1 CA= 3.

(2) 0 BB B@

1 2 3 4 5

6 7 8 9 0

1 2 3 4 5

3 4 5 6 3

1 CC CA

0 BB B@

1 2 3 4 5

6 7 8 9 0

1 2 3 4 5

3 4 5 6 3

1 CC CA

°2 + 1°×(−6)

−−−−−−−−−−−−→

0 BB B@

1 2 3 4 5

0 −5 −10 −15 −30

1 2 3 4 5

3 4 5 6 3

1 CC CA

°3 + 1°×(−1)

−−−−−−−−−−−−→

0 BB B@

1 2 3 4 5

0 −5 −10 −15 −30

0 0 0 0 0

3 4 5 6 3

1 CC CA

°4 + 1°×(−3)

−−−−−−−−−−−−→

0 BB B@

1 2 3 4 5

0 −5 −10 −15 −30

0 0 0 0 0

0 −2 −4 −6 −12

1 CC CA

°2 ^×(−1₅⁾

−−−−−−−−→

0 BB B@

1 2 3 4 5

0 1 2 3 6

0 0 0 0 0

0 −2 −4 −6 −12

1 CC CA

°3 ^↔°⁴

−−−−−−−−→ 0 BB B@

1 2 3 4 5

0 1 2 3 6

0 −2 −4 −6 −12

0 0 0 0 0

1 CC CA

°3 ^{+ 2}°^×2

−−−−−−−−−→

0 BB B@

1 2 3 4 5

0 1 2 3 6

0 0 0 0 0

0 0 0 0 0

1 CC CA

よって、rank 0 BB B@

1 2 3 4 5

6 7 8 9 0

1 2 3 4 5

3 4 5 6 3

1 CC CA= 2.

問題24 連立方程式 8>

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5x+ 2y+ 6z= 1 6x+ 8y+ 10z=−3 11x+ 6y+ 14z= 1 8x+ 4y+ 10z= 1

の係数行列をAとし,拡大係数行列をBとする.

(1)AとBの階数を求めなさい.

A= 0 BB B@

5 2 6

6 8 10 11 6 14 8 4 10 1 CC CA

°2 + 1°×(−1)

−−−−−−−−−−−−→

0 BB B@

5 2 6

1 6 4

11 6 14 8 4 10 1 CC CA

°1 ↔°2

−−−−−−−−→ 0 BB B@

1 6 4

5 2 6

11 6 14 8 4 10 1 CC CA

°2 ^{+ 1}°^×(−5)

−−−−−−−−−−−−→

0 BB B@

1 6 4

0 −28 −14

11 6 14

8 4 10

1 CC CA

°3 ^{+ 1}°^×(−11)

−−−−−−−−−−−−→

0 BB B@

1 6 4

0 −28 −14 0 −60 −30

8 4 10

1 CC CA

°4 ^{+ 1}°^×(−8)

−−−−−−−−−−−−→

0 BB B@

1 6 4

0 −28 −14 0 −60 −30 0 −44 −22 1 CC CA

°2 ×(−₁₄¹)

−−−−−−−−−→

0 BB B@

1 6 4

0 2 1

0 −60 −30 0 −44 −22 1 CC CA

°3 ×(−₃₀¹)

−−−−−−−−−→

0 BB B@

1 6 4

0 2 1

0 2 1

0 −44 −22 1 CC CA

°4 ×(−₂₂¹)

−−−−−−−−−→

0 BB B@

1 6 4 0 2 1 0 2 1 0 2 1 1 CC CA

°3 ^{+ 2}°^×(−1)

−−−−−−−−−−−−→

0 BB B@

1 6 4

0 2 1

0 0 0

0 2 1

1 CC CA

°4 ^{+ 2}°^×(−1)

−−−−−−−−−−−−→

0 BB B@

1 6 4

0 2 1

0 0 0

0 0 0

1 CC

CA よって、rankA= 2.

B= 0 BB B@

5 2 6 1

6 8 10 −3

11 6 14 1

8 4 10 1

1 CC CA

°2 ^{+ 1}°^×(−1)

−−−−−−−−−−−−→

0 BB B@

5 2 6 1

1 6 4 −4

11 6 14 1

8 4 10 1

1 CC CA

°1 ^↔°²

−−−−−−−−→ 0 BB B@

1 6 4 −4

5 2 6 1

11 6 14 1

8 4 10 1

1 CC CA

°2 + 1°×(−5)

−−−−−−−−−−−−→

0 BB B@

1 6 4 −4

0 −28 −14 21

11 6 14 1

8 4 10 1

1 CC CA

°3 + 1°×(−11)

−−−−−−−−−−−−→

0 BB B@

1 6 4 −4

0 −28 −14 21 0 −60 −30 45

8 4 10 1

1 CC CA

°4 + 1°×(−8)

−−−−−−−−−−−−→

0 BB B@

1 6 4 −4

0 −28 −14 21 0 −60 −30 45 0 −44 −22 33 1 CC CA

°2 ^×(−1₇⁾

−−−−−−−−→

0 BB B@

1 6 4 −4

0 4 2 −3

0 −60 −30 45 0 −44 −22 33 1 CC CA

°3 ^×(−₁₅¹⁾

−−−−−−−−−→

0 BB B@

1 6 4 −4

0 4 2 −3

0 4 2 −3

0 −44 −22 33 1 CC CA

°4 ^×(−₁₁¹⁾

−−−−−−−−−→

0 BB B@

1 6 4 −4

0 4 2 −3

0 4 2 −3

0 4 2 −3

1 CC CA

°3 + 2°×(−1)

−−−−−−−−−−−−→

0 BB B@

1 6 4 −4

0 4 2 −3

0 0 0 0

0 4 2 −3

1 CC CA

°4 + 2°×(−1)

−−−−−−−−−−−−→

0 BB B@

1 6 4 −4

0 4 2 −3

0 0 0 0

0 0 0 0

1 CC

CA よって、rankB= 2.

(2)「解なし」「解は唯１組」「解は無限個」のどれか答えなさい.

rankA= rankB= 2<3より,解は無限個となる.

実際, (1)のBに関する計算より,解は 0 B@ x y z 1 CA=

0 B@

−t+¹₂

−¹₂t−³₄ t

1

CA (ただしtは任意の実数)