Vector group (1)

已知向量组，证明 B组能由A组线性表示，但A组不能由B组表示。

A : a 1 = (\begin{matrix} 0 \\ 1 \\ 2 \\ 3 \end{matrix}), a 2 = (\begin{matrix} 3 \\ 0 \\ 1 \\ 2 \end{matrix}), a 3 = (\begin{matrix} 2 \\ 3 \\ 0 \\ 1 \end{matrix}); B : b 1 = (\begin{matrix} 2 \\ 1 \\ 1 \\ 2 \end{matrix}), b 2 = (\begin{matrix} 0 \\ - 2 \\ 1 \\ 1 \end{matrix}), b 3 = (\begin{matrix} 4 \\ 4 \\ 1 \\ 3 \end{matrix});

$A: a1=\begin{pmatrix} 0 \\ 1 \\ 2 \\ 3 \end{pmatrix}, a2=\begin{pmatrix} 3 \\ 0 \\ 1 \\ 2 \end{pmatrix}, a3=\begin{pmatrix} 2 \\ 3 \\ 0 \\ 1 \end{pmatrix}; B: b1=\begin{pmatrix} 2 \\ 1 \\ 1 \\ 2 \end{pmatrix}, b2=\begin{pmatrix} 0 \\ -2 \\ 1 \\ 1 \end{pmatrix}, b3=\begin{pmatrix} 4 \\ 4 \\ 1 \\ 3 \end{pmatrix};$

R (A, B) = (\begin{matrix} 1 & 0 & 3 & 1 & - 2 & 4 \\ 0 & 1 & - 6 & - 1 & 5 & - 7 \\ 0 & 2 & - 8 & - 1 & 7 & - 9 \\ 0 & 3 & 2 & 2 & 0 & 4 \end{matrix}) = (\begin{matrix} 1 & 0 & 3 & 1 & - 2 & 4 \\ 0 & 1 & - 6 & - 1 & 5 & - 7 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{matrix}) = 2

$R(A,B)=\begin{pmatrix} 1 & 0 & 3 & 1 & -2 & 4 \\ 0 & 1 & -6 & -1 & 5 & -7 \\ 0 & 2 & -8 & -1 & 7 & -9 \\ 0 & 3 & 2 & 2 & 0 & 4 \end{pmatrix} =\begin{pmatrix} 1 & 0 & 3 & 1 & -2 & 4 \\ 0 & 1 & -6 & -1 & 5 & -7 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{pmatrix}=2$

R (A) = (\begin{matrix} 1 & 0 & 3 \\ 0 & 1 & - 6 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}) = 2

$R(A)=\begin{pmatrix} 1 & 0 & 3 \\ 0 & 1 & -6 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}=2$

R(A, B) = R(A)
B 能由 A 线性表示

R (B, A) = (\begin{matrix} 1 & 1 & 1 & 2 & 1 & 0 \\ 0 & - 1 & 1 & - 1 & 0 & 1 \\ 0 & - 3 & 3 & - 1 & - 1 & 3 \\ 0 & - 2 & 2 & - 4 & 1 & 2 \end{matrix}) = (\begin{matrix} 1 & 1 & 1 & 2 & 1 & 4 \\ 0 & 1 & 1 & 3 & - 1 & 3 \\ 0 & 0 & 1 & - 2 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 \end{matrix}) = 4

$R(B,A)=\begin{pmatrix} 1 & 1 & 1 & 2 & 1 & 0 \\ 0 & -1 & 1 & -1 & 0 & 1 \\ 0 & -3 & 3 & -1 & -1 & 3 \\ 0 & -2 & 2 & -4 & 1 & 2 \end{pmatrix} =\begin{pmatrix} 1 & 1 & 1 & 2 & 1 & 4 \\ 0 & 1 & 1 & 3 & -1 & 3 \\ 0 & 0 & 1 & -2 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 \end{pmatrix}=4$

R (B) = (\begin{matrix} 1 & 1 & 1 \\ 0 & 1 & - 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}) = 2

$R(B)=\begin{pmatrix} 1 & 1 & 1 \\ 0 & 1 & -1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}=2$

A 不能由 B 线性表示

已知向量组A、B ，证明A、B等价

A : a 1 = (\begin{matrix} 0 \\ 1 \\ 1 \end{matrix}), a 2 = (\begin{matrix} 1 \\ 1 \\ 0 \end{matrix}); B : b 1 = (\begin{matrix} - 1 \\ 0 \\ 1 \end{matrix}), b 2 = (\begin{matrix} 1 \\ 2 \\ 1 \end{matrix}), b 2 = (\begin{matrix} 3 \\ 2 \\ - 1 \end{matrix});

$A: a1=\begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}, a2=\begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}; B: b1=\begin{pmatrix} -1 \\ 0 \\ 1 \end{pmatrix}, b2=\begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix}, b2=\begin{pmatrix} 3 \\ 2 \\ -1 \end{pmatrix};$

观察可知 R（A) = 2

R (B) = (\begin{matrix} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{matrix}) = 2

$R(B)=\begin{pmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{pmatrix}=2$

R(A) = R(B) = R(A, B) 即A、B等价

问a取什么值时下列向量组线性相关

A : a 1 = (\begin{matrix} a \\ 1 \\ 1 \end{matrix}), a 2 = (\begin{matrix} 1 \\ a \\ - 1 \end{matrix}), a 3 = (\begin{matrix} 1 \\ - 1 \\ a \end{matrix});

$A: a1=\begin{pmatrix} a \\ 1 \\ 1 \end{pmatrix}, a2=\begin{pmatrix} 1 \\ a \\ -1 \end{pmatrix}, a3=\begin{pmatrix} 1 \\ -1 \\ a \end{pmatrix};$

最简形：

A : (\begin{matrix} 1 & \frac{1}{a} & \frac{1}{a} \\ 0 & a - \frac{1}{a} & - 1 - \frac{1}{a} \\ 0 & - 1 - \frac{1}{a} & a - \frac{1}{a} \end{matrix})

$A: \begin{pmatrix} 1 & \frac{1}{a} & \frac{1}{a} \\ 0 & a-\frac{1}{a} & -1-\frac{1}{a} \\ 0 & -1-\frac{1}{a} & a-\frac{1}{a} \end{pmatrix}$

向量组线性相关即 R(A)<3，则有：

a - \frac{1}{a} = 1 + \frac{1}{a} (a + 1) (a - 2) = 0

$a-\frac{1}{a}=1+\frac{1}{a} (a+1)(a-2)=0$

a = -1 或 a = 2

设a1,a2线性无关，a1+b,a2+b线性相关，求b用a1,a2线性表示的表示式

设 b 可由 $xa_1 + ya_2$ 表示

因为a1+b,a2+b线性相关, 所以 $(x + 1)a_1 + ya_2 = z(xa_1 + (y + 1)a_2)$

(x + 1) a_{1} + y a_{2} = z (x a_{1} + (y + 1) a_{2})

$(x + 1)a_1 + ya_2 = z(xa_1 + (y + 1)a_2)$

有 $\frac{a+1}{zx} = \frac{y}{y+1}$

设 x = c, 有 y = -(1+c), 可得 $b = ca_1 - (1+c)a_2$

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