2023-2024春季线代期中仅供参考

一

D D C C A

二

2-1

A^{- 1} = [\begin{array}{ccc} 1 \\ - a & 1 \\ 3 a - b / 2 & - 3 / 2 & 1 / 2 \end{array}]

analysis:

[\begin{array}{ll} A & I \end{array}] = [\begin{array}{cccccc} 1 & 0 & 0 & 1 & 0 & 0 \\ a & 1 & 0 & 0 & 1 & 0 \\ b & 3 & 2 & 0 & 0 & 1 \end{array}]

$\to$

[\begin{array}{cccccc} 1 & 0 & 0 & 1 & 0 & 0 \\ 0 & 1 & 0 & - a & 1 & 0 \\ 0 & 3 & 2 & - b & 0 & 1 \end{array}]

$\to$

[\begin{array}{cccccc} 1 & 0 & 0 & 1 & 0 & 0 \\ 0 & 1 & 0 & - a & 1 & 0 \\ 0 & 0 & 2 & 3 a - b & - 3 & 1 \end{array}]

$\to$

[\begin{array}{cccccc} 1 & 0 & 0 & 1 & 0 & 0 \\ 0 & 1 & 0 & - a & 1 & 0 \\ 0 & 0 & 1 & (3 a - b) / 2 & - 3 / 2 & 1 / 2 \end{array}] = [\begin{array}{ll} I & A^{- 1} \end{array}]

2-2

解析:

法1:

A = [\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}]

$A B$ =

[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}] [\begin{array}{ccc} 1 & 0 & 2 \\ 0 & 2 & 0 \\ - 1 & 0 & 3 \end{array}] =

[\begin{array}{ccc} 1 & 0 & 2 \\ - 1 & 0 & 3 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}]

因此 $R (A B) = 2$

法2: $R (A) = 2, R (B) = 3$

\begin{aligned} R (A B) \geq R (A) + R (B) - n = 2 + 3 - 3 = 2 \Rightarrow R (A B) = 2 \\ R (A B) \leq min [R (A), R (B) = 2 \end{aligned}

R (A B) \leq min [R (A), R (B)] = 2

(3)

4^{2023} A = 4^{2023} [\begin{array}{ccc} 1 & - 1 & 1 \\ - 1 & 1 & - 1 \\ 2 & - 2 & 2 \end{array}]

解析: 算! $A^{2}, A^{3}$ , 即得出规律

(4)

$[\begin{matrix} \frac{7}{3} \\ \frac{10}{3} \end{matrix}]$

解析:

A = [\begin{array}{ll} 1 & 0 \\ 0 & 1 \\ 1 & 1 \end{array}] b = [\begin{array}{l} 2 \\ 3 \\ 6 \end{array}]

三

\begin{aligned} A = [\begin{array}{lll} 1 & - 2 & 3 \\ 2 & - 5 & 1 \\ 1 & - 4 & - 7 \end{array}] \\ E_{21} (- 2) A = [\begin{array}{lll} 1 & - 2 & 3 \\ 0 & - 1 & - 5 \\ 1 & - 4 & - 7 \end{array}] \\ E_{31} (- 1) E_{21} (- 2) A = [\begin{array}{ccc} 1 & - 2 & 3 \\ 0 & - 1 & - 5 \\ 0 & - 2 & - 10 \end{array}] \\ E_{32} (- 2) E_{31} (- 1) E_{21} (- 2) A = [\begin{array}{ccc} 1 & - 2 & 3 \\ 0 & - 1 & - 5 \\ 0 & 0 & 0 \end{array}] \\ A = E_{21}^{- 1} (- 2) E_{31} (- 1)^{- 1} E_{32}^{- 1} (- 2) [\begin{array}{ccc} 1 & - 2 & 3 \\ 0 & - 1 & - 5 \\ 0 & 0 & 0 \end{array}] \\ = [\begin{array}{lll} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 1 & 2 & 1 \end{array}] [\begin{array}{ccc} 1 & - 2 & 3 \\ 0 & - 1 & - 5 \\ 0 & 0 & 0 \end{array}] = L U \end{aligned}

四

A

4-A-1

$C (A)$ , 对 $A$ 行变换

A \to [\begin{array}{ccccc} 0 & 2 & 0 & 3 & 0 \\ 0 & 0 & 2 & - 1 & 0 \\ 0 & 0 & 0 & 0 & 10 \\ 0 & 0 & 0 & 0 & 0 \end{array}] \Rightarrow C (A) = span {[\begin{matrix} 2 \\ 1 \\ 4 \\ - 1 \end{matrix}], [\begin{matrix} 4 \\ 1 \\ 10 \\ - 5 \end{matrix}], [\begin{array}{l} 6 \\ 3 \\ 2 \\ 7 \end{array}]}

4-A-2

$C (A^{⊤})$ 对 $A^{⊤}$ 行变换

PS: 注意本行很长

A^{⊤} = [\begin{array}{cccc} 0 & 0 & 0 & 0 \\ 2 & 1 & 4 & - 1 \\ 4 & 1 & 10 & - 5 \\ 1 & 1 & 1 & 1 \\ 6 & 3 & 2 & 7 \end{array}] \to [\begin{array}{cccc} 1 & 1 & 1 & 1 \\ 0 & 1 & - 2 & 3 \\ 0 & 3 & - 6 & 9 \\ 0 & 3 & 4 & - 1 \\ 0 & 0 & 0 & 0 \end{array}] \to [\begin{array}{rrrr} 1 & 1 & 1 & 1 \\ 1 & - 2 & 3 \\ 10 & - 10 \\ 0 \\ 0 \end{array}] \Rightarrow C (A^{⊤}) = span {[\begin{array}{l} 0 \\ 2 \\ 4 \\ 1 \\ 6 \end{array}], [\begin{array}{l} 0 \\ 1 \\ 1 \\ 1 \\ 3 \end{array}], [\begin{array}{l} 0 \\ 4 \\ 10 \\ 1 \\ 2 \end{array}]}

4-A-3

$N (A)$

A x = 0 \Rightarrow [\begin{array}{ccccc} 0 & 2 & 0 & 3 & 0 \\ 0 & 0 & 2 & - 1 & 0 \\ 0 & 0 & 0 & 0 & 10 \\ 0 & 0 & 0 & 0 & 0 \end{array}] [\begin{matrix} x_{1} \\ ⋮ \\ x_{5} \end{matrix}] = 0 \Rightarrow {\begin{cases} x_{1}, x_{4} \in R \\ x_{2} = - \frac{3}{2} x_{4} \\ x_{3} = \frac{1}{2} x_{4} \\ x_{5} = 0 \end{cases}

\Rightarrow x = k_{1} [\begin{matrix} 1 \\ 0 \\ 0 \\ 0 \\ 0 \end{matrix}] + k_{4} [\begin{matrix} 0 \\ - 3 / 12 \\ 1 / 2 \\ 1 \\ 0 \end{matrix}] (k_{1}, k_{2} \in R)

4-A-4

$N (A^{⊤})$

A^{⊤} y = 0 \Rightarrow [\begin{array}{cccc} 1 & 1 & 1 & 1 \\ 0 & 1 & - 2 & 3 \\ 0 & 0 & 10 & - 10 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}] [\begin{array}{l} y_{1} \\ y_{2} \\ y_{3} \\ y_{4} \end{array}] = 0

\Rightarrow {\begin{cases} y_{1} = - y_{4} \\ y_{2} = - y_{4} \\ y_{3} = y_{4} \\ y_{4} \in R \end{cases}

\Rightarrow y = k_{0} [\begin{matrix} - 1 \\ - 1 \\ 1 \\ 1 \end{matrix}], k_{0} \in R

(b)

$A x = b$ 时利用高斯消元化简增广矩阵(略)

令出一个特解

x_{p} = [\begin{matrix} 1 \\ - 5 \\ 1 \\ 3 \\ 1 \end{matrix}]

\begin{array}{r} x = x_{p} + x_{n} = [\begin{array}{c} 1 \\ - 5 \\ 1 \\ 3 \\ 1 \end{array}] + k_{1} [\begin{array}{c} 1 \\ 0 \\ 0 \\ 0 \\ 0 \end{array}] + k_{4} [\begin{array}{c} 1 \\ - 3 / 2 \\ 1 / 2 \\ 1 \\ 0 \end{array}], k_{1}, k_{4} \in R \end{array}

五

令 $X = [\begin{array}{ll} a & b \\ c & d \end{array}]$ ,

\begin{aligned} 则 X A = [\begin{array}{ll} a & b \\ c & d \end{array}] [\begin{array}{ll} 1 & 1 \\ 0 & 2 \end{array}] = [\begin{array}{ll} a & a + 2 b \\ c & c + 2 d \end{array}] \\ A X = [\begin{array}{ll} 1 & 1 \\ 0 & 2 \end{array}] [\begin{array}{ll} a & b \\ c & d \end{array}] = [\begin{array}{ll} a + c & b + d \\ 2 c & 2 d \end{array}] \\ T (X) = X A + A X = [\begin{array}{cc} 2 a + c & a + 3 b + d \\ 3 c & c + 4 d \end{array}] = (2 a + c) v_{1} + (a + 3 b + d) v_{2} + (3 c) v_{3} + (c + 4 d) v_{4} \\ T ([\begin{array}{ll} a & b \\ c & d \end{array}]) = (v_{1} v_{2} v_{3} v_{4}) (\begin{array}{c} 2 a + c \\ a + 3 b + d \\ 3 c \\ c + 4 d \end{array}) \end{aligned}

六

证明: $(A + B)^{2}$

\begin{aligned} = (A + B) (A + B) = A^{2} + A B + B A + B^{2} = A + B \\ ∵ \\ A^{2} = A, B^{2} = B \\ ∴ \\ A B + B A = 0 (1) \\ B (A B + B A) = B A B + B^{2} A = (B A) B + B^{2} A = - A B^{2} + B^{2} A = 0 \end{aligned}

a n d ∵ B^{2} = B

∴ - A B^{2} + B^{2} A = - A B + B A = 0 (2)

联立(1) (2)

{\begin{cases} A B + B A = 0 \\ - A B + B A = 0 \end{cases} \Rightarrow A B = 0

得证

七

(A)

(A B)^{2} = (A B) (A B) = [\begin{array}{ccc} 8 & 0 & - 4 \\ - \frac{3}{2} & 9 & - 6 \\ - 2 & 0 & 1 \end{array}] [\begin{array}{ccc} 8 & 0 & - 4 \\ - \frac{3}{2} & 9 & - 6 \\ - 2 & 0 & 1 \end{array}]

= [\begin{array}{ccc} 72 & 0 & - 36 \\ - \frac{27}{2} & 81 & - 54 \\ - 18 & 0 & 9 \end{array}]

(B) $(A B)^{2} = 9 (A B)$

$R (A) \geq R (A B) = 2 \Rightarrow R (A) = 2$ , 同理 $R (B) = 2$

$A$ 行满秩, 令 $X A = I_{2}$ $X$ 为 $A$ 左逆

$B$ 列满秩, 令 $B Y = I_{2}$ $Y$ 为 $B$ 右逆

∴ B A = (X A) (B A) (B Y) = X (A B)^{2} Y = 9 X A B Y = 9 I

2023-2024春季线代期中仅供参考 ​

一 ​

二 ​

2-1 ​

2-2 ​

(3) ​

(4) ​

三 ​

四 ​

A ​

4-A-1 ​

4-A-2 ​

4-A-3 ​

4-A-4 ​

(b) ​

五 ​

六 ​

七 ​

2023-2024春季线代期中仅供参考

一

二

2-1

2-2

(3)

(4)

三

四

A

4-A-1

4-A-2

4-A-3

4-A-4

(b)

五

六

七