张祖锦2023年数学专业真题分类70天之第45天
1013、 4、 设 $\displaystyle A$ 是 $\displaystyle m\times n$ 实矩阵, $\displaystyle b$ 是 $\displaystyle m$ 维实列向量. 求证: 方程组 $\displaystyle Ax=b$ 有解的充分必要条件为方程组 $\displaystyle \left\{\begin{array}{llllllllllll}A^\mathrm{T} Y=0,\\\\ b^\mathrm{T} Y=1\end{array}\right.$ 无解. (哈尔滨工程大学2023年高等代数考研试题) [线性方程组 ]
纸质资料/答疑/pdf1/pdf2 / (1)、 $\displaystyle \Rightarrow$: 设 $\displaystyle Ax=b$ 有解, 往用反证法证明 $\displaystyle A^\mathrm{T} y=0, y^\mathrm{T} b=1$ 无解. 事实上,
$$\begin{aligned} &A^\mathrm{T} y=0, y^\mathrm{T} b=1\\\\ \Rightarrow&1=y^\mathrm{T} b=y^\mathrm{T} Ax=(A^\mathrm{T} y)^\mathrm{T} x=0^\mathrm{T} x=0,\\\\ &\mbox{这是一个矛盾, 故有结论}. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
(2)、 $\displaystyle \Leftarrow$: 也用反证法. 若 $\displaystyle Ax=b$ 无解, 则 $\displaystyle b$ 不能由 $\displaystyle A$ 的列向量组 $\displaystyle \alpha_1,\cdots,\alpha_n$ 线性表出, 即
$$\begin{aligned} \alpha_1,\cdots,\alpha_n,b \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
线性无关. 将 $\displaystyle \alpha_1,\cdots,\alpha_n,b$ 标准正交化为 $\displaystyle \beta_1,\cdots,\beta_n,c$, 则
$$\begin{aligned} L(\alpha_1,\cdots,\alpha_i)&=L(\beta_1,\cdots,\beta_i)\left(1\leq i\leq n\right),\\\\ L(\alpha_1,\cdots,\alpha_n,b)&=L(\beta_1,\cdots,\beta_n,c). \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
于是
$$\begin{aligned} &\beta_i^\mathrm{T} c=0\left(1\leq i\leq n\right)\\\\ \Rightarrow&\alpha_i^\mathrm{T} c=0\left(1\leq i\leq n\right)\\\\ \Rightarrow&A^\mathrm{T} c=0, \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
$$\begin{aligned} &L(\alpha_1,\cdots,\alpha_n,b)=L(\beta_1,\cdots,\beta_n,c)\\\\ \Rightarrow&b=k_1\beta_1+\cdots+k_n\beta_n+kc\\\\ &\left(k\neq 0,\mbox{否则 $\displaystyle b$ 可由 $\displaystyle \beta_1,\cdots,\beta_n$ 线性表出, 也可由 $\displaystyle \alpha_1,\cdots,\alpha_n$ 线性表出}\right)\\\\ \Rightarrow&c^\mathrm{T} b=k |c|^2\neq 0. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
取 $\displaystyle y=\frac{c}{c^\mathrm{T} b}$, 则
$$\begin{aligned} A^\mathrm{T} y=0, y^\mathrm{T} b=1. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
这与充分性假设矛盾. 故有结论.跟锦数学微信公众号. 在线资料/公众号/
1014、 6、 设线性方程组
$$\begin{aligned} \left\{\begin{array}{rrrrrrrrrrrrrrrr}x_1&+&x_2&+&x_3&=&0,\\\\ x_1&+&2x_2&+&ax_3&=&0,\\\\ x_1&+&4x_2&+&a^2x_3&=&0 \end{array}\right. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
与方程 $\displaystyle x_1+2x_2+x_3=a-1$ 有公共解, 求 $\displaystyle a$ 的值及所有公共解. (哈尔滨工程大学2023年高等代数考研试题) [线性方程组 ]
纸质资料/答疑/pdf1/pdf2 / 题中方程组分别编号为 $\displaystyle (I),(II)$, 则 $\displaystyle (I)$ 的系数矩阵
$$\begin{aligned} A=\left(\begin{array}{cccccccccccccccccccc}1&1&1\\\\ 1&2&a\\\\ 1&4&a^2\end{array}\right)\to\left(\begin{array}{cccccccccccccccccccc}1&1&1\\\\ 0&1&a-1\\\\ 0&3&a^2-1\end{array}\right)\to\left(\begin{array}{cccccccccccccccccccc}1&0&2-a\\\\ 0&1&a-1\\\\ 0&0&(a-2)(a-1)\end{array}\right). \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
(1)、 若 $\displaystyle a\neq 2$ 且 $\displaystyle a\neq 1$, 则 $\displaystyle (I)$ 只有零解. 由题设, $\displaystyle 0$ 是 $\displaystyle (II)$ 的解, $\displaystyle 0=a-1$. 这是已设矛盾. (2)、 故 $\displaystyle a=2, A\to\left(\begin{array}{cccccccccccccccccccc}1&0&0\\\\ 0&1&1\\\\ 0&0&0\end{array}\right)$. $\displaystyle (I)$ 的通解为 $\displaystyle k\left(\begin{array}{cccccccccccccccccccc}0\\\\-1\\\\1\end{array}\right), \forall\ k$. 而 $\displaystyle (II): x_1+2x_2+x_3=1$ 的通解为
$$\begin{aligned} l\left(\begin{array}{cccccccccccccccccccc}-2\\\\1\\\\0\end{array}\right)+m\left(\begin{array}{cccccccccccccccccccc}-1\\\\0\\\\1\end{array}\right)+\left(\begin{array}{cccccccccccccccccccc}1\\\\0\\\\0\end{array}\right),\quad \forall\ l,m. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
由
$$\begin{aligned} k\left(\begin{array}{cccccccccccccccccccc}0\\\\-1\\\\1\end{array}\right)=l\left(\begin{array}{cccccccccccccccccccc}-2\\\\1\\\\0\end{array}\right)+m\left(\begin{array}{cccccccccccccccccccc}-1\\\\0\\\\1\end{array}\right)+\left(\begin{array}{cccccccccccccccccccc}1\\\\0\\\\0\end{array}\right) \Rightarrow k=-1, l=1, m=-1 \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
知公共解为 $\displaystyle (0,1,-1)^\mathrm{T}$. (3)、 或 $\displaystyle a=1, A\to\left(\begin{array}{cccccccccccccccccccc}1&0&1\\\\ 0&1&0\\\\ 0&0&0\end{array}\right)$. $\displaystyle (I)$ 的通解为 $\displaystyle k\left(\begin{array}{cccccccccccccccccccc}-1\\\\0\\\\1\end{array}\right), \forall\ k$. 而 $\displaystyle (II): x_1+2x_2+x_3=0$ 的通解为
$$\begin{aligned} l\left(\begin{array}{cccccccccccccccccccc}-2\\\\1\\\\0\end{array}\right)+m\left(\begin{array}{cccccccccccccccccccc}-1\\\\0\\\\1\end{array}\right),\quad \forall\ l,m. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
由
$$\begin{aligned} k\left(\begin{array}{cccccccccccccccccccc}-1\\\\0\\\\1\end{array}\right)=l\left(\begin{array}{cccccccccccccccccccc}-2\\\\1\\\\0\end{array}\right)+m\left(\begin{array}{cccccccccccccccccccc}-1\\\\0\\\\1\end{array}\right) \Rightarrow k=m\mbox{任意}, l=0 \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
知公共解为 $\displaystyle k(-1,0,1)^\mathrm{T}, \forall\ k$.跟锦数学微信公众号. 在线资料/公众号/
1015、 5、 对以下方程组
$$\begin{aligned} \left\{\begin{array}{rrrrrrrrrrrrrrrr} \lambda x_1&+&2x_2&+&3x_3&=&1,\\\\ x_1&+&2\lambda x_2&+&3x_3&=&\lambda,\\\\ x_1&+&2x_2&+&3\lambda x_3&=&\lambda. \end{array}\right. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
求 $\displaystyle \lambda$ 为何值时, 方程组无解, 有唯一解, 有无穷解? 并写出一般解的形式. (哈尔滨工业大学2023年高等代数考研试题) [线性方程组 ]
纸质资料/答疑/pdf1/pdf2 / 增广矩阵
$$\begin{aligned} &(A,\beta)=\left(\begin{array}{cccccccccccccccccccc}\lambda&2&3&1\\\\ 1&2\lambda&3&\lambda\\\\ 1&2&3\lambda&\lambda\end{array}\right)\to\left(\begin{array}{cccccccccccccccccccc}0&2-2\lambda&3-3\lambda^2&1-\lambda^2\\\\ 0&2\lambda-2&3-3\lambda&0\\\\ 1&2&3\lambda&\lambda\end{array}\right). \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
(1)、 当 $\displaystyle \lambda=1$ 时, $\displaystyle (A,\beta)\to\left(\begin{array}{cccccccccccccccccccc}1&2&3&1\\\\ 0&0&0&0\\\\ 0&0&0&0\end{array}\right)$, 而 $\displaystyle Ax=\beta$ 有无穷多解, 且通解为
$$\begin{aligned} k\left(\begin{array}{cccccccccccccccccccc}-2\\\\1\\\\0\end{array}\right)+l\left(\begin{array}{cccccccccccccccccccc}-3\\\\0\\\\1\end{array}\right)+\left(\begin{array}{cccccccccccccccccccc}1\\\\0\\\\0\end{array}\right),\quad \forall\ k,l. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
(2)、 当 $\displaystyle \lambda\neq 1$ 时,
$$\begin{aligned} (A,\beta)\to\left(\begin{array}{cccccccccccccccccccc}0&2&3(\lambda+1)&\lambda+1\\\\ 0&2&-3&0\\\\ 1&2&3\lambda&\lambda\end{array}\right)\to\left(\begin{array}{cccccccccccccccccccc}0&0&3(\lambda+2)&\lambda+1\\\\ 0&2&-3&0\\\\ 1&2&3\lambda&\lambda\end{array}\right). \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
(2-1)、 当 $\displaystyle \lambda\neq -2$ 时, $\displaystyle |A|\neq 0$, $\displaystyle Ax=\beta$ 有唯一解, 经过计算为
$$\begin{aligned} \left(-\frac{1}{\lambda+2},\frac{\lambda+1}{2(\lambda+2)},\frac{\lambda+1}{3(\lambda+2)}\right)^\mathrm{T}. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
(2-2)、 当 $\displaystyle \lambda=-2$ 时, $\displaystyle \mathrm{rank} A=2 < 3=\mathrm{rank}(A,\beta)$, $\displaystyle Ax=\beta$ 无解.跟锦数学微信公众号. 在线资料/公众号/
1016、 4、 当 $\displaystyle a,b$ 为何值时, 方程组
$$\begin{aligned} \left\{\begin{array}{rrrrrrrrrrrrrrrr} x_1&+&2x_2&-&x_3&+&x_4&=&0,\\\\ 2x_1&-&x_2&+&x_3&+&x_4&=&1,\\\\ x_1&+&x_2&+&ax_3&+&x_4&=&1,\\\\ 3x_1&+&x_2&+&2x_2&+&x_4&=&b \end{array}\right. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
有唯一解, 无解, 有无穷多解, 并写出解的结构通式. (黑龙江大学2023年高等代数考研试题) [线性方程组 ]
纸质资料/答疑/pdf1/pdf2 / 增广矩阵
$$\begin{aligned} &(A,\beta)=\left(\begin{array}{cccccccccccccccccccc}1&2&-1&1&0\\\\ 2&-1&1&1&1\\\\ 1&1&a&1&1\\\\ 3&1&2&1&b\end{array}\right)\to\left(\begin{array}{cccccccccccccccccccc}1&2&-1&1&0\\\\ 0&-5&3&-1&1\\\\ 0&-1&a+1&0&1\\\\ 0&-5&5&-2&b\end{array}\right)\\\\ \to&\left(\begin{array}{cccccccccccccccccccc}1&2&-1&1&0\\\\ 0&0&-5a-2&-1&-4\\\\ 0&-1&a+1&0&1\\\\ 0&0&-5a&-2&b-5\end{array}\right)\to\left(\begin{array}{cccccccccccccccccccc}1&2&-1&1&0\\\\ 0&0&-5a-2&-1&-4\\\\ 0&-1&a+1&0&-1\\\\ 0&0&2&-1&b-1\end{array}\right)\\\\ \to&\left(\begin{array}{cccccccccccccccccccc}1&0&3+2a&0&b+1\\\\ 0&1&-a-1&0&1\\\\ 0&0&1&-\frac{1}{2}&\frac{b-1}{2}\\\\ 0&0&0&\frac{-5a-4}{2}&\frac{5ab-5a+2b-10}{2}\end{array}\right). \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
(1)、 若 $\displaystyle a\neq -\frac{4}{5}$, 则 $\displaystyle |A|\neq 0$, 而 $\displaystyle Ax=\beta$ 有唯一解, 经过计算后为
$$\begin{aligned} \left(\frac{-5-a+b+3ab}{4+5a}, \frac{-1-2a+b+ab}{4+5a}, \frac{3+b}{4+5a}, \frac{10+5a-2b-5ab}{4+5a}\right)^\mathrm{T}. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
(2)、 若 $\displaystyle a=-\frac{4}{5}$, 则
$$\begin{aligned} (A,\beta)\to\left(\begin{array}{cccccccccccccccccccc}1&0&\frac{7}{5}&0&b+1\\\\ 0&1&-\frac{1}{5}&0&1\\\\ 0&0&1&-\frac{1}{2}&\frac{b-1}{2}\\\\ 0&0&0&0&-b-3\end{array}\right). \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
(2-1)、 若 $\displaystyle b\neq -3$, 则 $\displaystyle \mathrm{rank} A=3 < 4=\mathrm{rank}(A,\beta)$, 而 $\displaystyle Ax=\beta$ 无解. (2-2)、 若 $\displaystyle b=-3$, 则 $\displaystyle (A,\beta)\to\left(\begin{array}{cccccccccccccccccccc}1&0&0&\frac{7}{10}&\frac{4}{5}\\\\ 0&1&0&-\frac{1}{10}&\frac{7}{5}\\\\ 0&0&1&-\frac{1}{2}&-2\\\\ 0&0&0&0&0\end{array}\right)$, 而 $\displaystyle Ax=\beta$ 有无穷多解, 且通解为
$$\begin{aligned} \left(\frac{4}{5},-\frac{7}{5},-2,0\right)^\mathrm{T}+k\left(\begin{array}{cccccccccccccccccccc}-7,1,5,10\end{array}\right)^\mathrm{T},\quad \forall\ k. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
跟锦数学微信公众号. 在线资料/公众号/
1017、 2、 当 $\displaystyle \lambda$ 为何值时, 下列方程组有唯一解, 无穷解, 无解? 并在方程组有解时求一般解:
$$\begin{aligned} \left\{\begin{array}{rrrrrrrrrrrrrrrr} \lambda x_1&+&x_2&+&x_3&+&x_4&=&1,\\\\ x_1&+&\lambda x_2&+&x_3&+&x_4&=&1,\\\\ x_1&+&x_2&+&\lambda x_3&+&x_4&=&1,\\\\ x_1&+&x_2&+&x_3&+&\lambda x_4&=&1. \end{array}\right. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
(湖南大学2023年高等代数考研试题) [线性方程组 ]
纸质资料/答疑/pdf1/pdf2 / 增广矩阵
$$\begin{aligned} (A,\beta)=&\left(\begin{array}{cccccccccccccccccccc}\lambda&1&1&1&1\\\\ 1&\lambda&1&1&1\\\\ 1&1&\lambda&1&1\\\\ 1&1&1&\lambda&1\end{array}\right) \to\left(\begin{array}{cccccccccccccccccccc}0&1-\lambda&1-\lambda&1-\lambda^2&1-\lambda\\\\ 0&\lambda-1&0&1-\lambda&0\\\\ 0&0&\lambda-1&1-\lambda&0\\\\ 1&1&1&1&1\end{array}\right)\\\\ \to&\left(\begin{array}{cccccccccccccccccccc}0&0&1-\lambda&-(\lambda-1)(\lambda+2)&1-\lambda\\\\ 0&\lambda-1&0&1-\lambda&0\\\\ 0&0&\lambda-1&1-\lambda&0\\\\ 1&1&1&\lambda&1\end{array}\right)\\\\ \to&\left(\begin{array}{cccccccccccccccccccc}0&0&0&-(\lambda-1)(\lambda+3)&1-\lambda\\\\ 0&\lambda-1&0&1-\lambda&0\\\\ 0&0&\lambda-1&1-\lambda&0\\\\ 1&1&1&\lambda&1\end{array}\right). \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
(1)、 若 $\displaystyle \lambda=1$, 则 $\displaystyle (A,\beta)\to\left(\begin{array}{cccccccccccccccccccc}1&1&1&1&1\\\\ 0&0&0&0&0\\\\ 0&0&0&0&0\\\\ 0&0&0&0&0\end{array}\right)$, 而 $\displaystyle Ax=\beta$ 有无穷多解, 且通解为
$$\begin{aligned} k_1\left(\begin{array}{cccccccccccccccccccc}-1\\\\1\\\\0\\\\0\end{array}\right)+k_2\left(\begin{array}{cccccccccccccccccccc}-1\\\\0\\\\1\\\\0\end{array}\right)+k_3\left(\begin{array}{cccccccccccccccccccc}-1\\\\0\\\\0\\\\1\end{array}\right)+\left(\begin{array}{cccccccccccccccccccc}1\\\\0\\\\0\\\\0\end{array}\right),\quad \forall\ k_1,k_2,k_3. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
(2)、 若 $\displaystyle \lambda=-3$, 则 $\displaystyle \mathrm{rank} A=3 < 4=\mathrm{rank}(A,\beta)$, 而 $\displaystyle Ax=\beta$ 无解. (3)、 若 $\displaystyle \lambda\neq 1\mbox{且} \lambda\neq -3$, 则 $\displaystyle Ax=\beta$ 有唯一解, 经过计算知为
$$\begin{aligned} \frac{1}{\lambda+3}(1,1,1,1)^\mathrm{T}. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
跟锦数学微信公众号. 在线资料/公众号/
1018、 (5)、 齐次线性方程组 $\displaystyle AX=0$ 的解空间维数是 $\displaystyle \underline{\ \ \ \ \ \ \ \ \ \ }$, 其中
$$\begin{aligned} A=\left(\begin{array}{cccccccccccccccccccc}1&3&1&5\\\\ 0&2&1&3\\\\ 2&0&-1&1\\\\ -1&3&2&4\end{array}\right). \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
(华东师范大学2023年高等代数考研试题) [线性方程组 ]
纸质资料/答疑/pdf1/pdf2 / 由
$$\begin{aligned} A\to \left(\begin{array}{cccccccccccccccccccc}1&0&-\frac{1}{2}&\frac{1}{2}\\\\ 0&1&\frac{1}{2}&\frac{3}{2}\\\\ 0&0&0&0\\\\ 0&0&0&0\end{array}\right)\Rightarrow \mathrm{rank} A=2 \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
知 $\displaystyle Ax=0$ 的解空间维数 $\displaystyle =4-2=2$.跟锦数学微信公众号. 在线资料/公众号/
1019、 3、 当 $\displaystyle \lambda,\mu$ 为何值时, 线性方程组
$$\begin{aligned} \left\{\begin{array}{rrrrrrrrrrrrrrrr} x_1&+&x_2&+&x_3&+&x_4&=&0,\\\\ x_1&+&3x_2&+&5x_3&+&5x_4&=&2,\\\\ &-&x_2&+&(\lambda-3)x_3&-&2x_4&=&\mu,\\\\ x_1&+&2x_2&+&\lambda x_3&+&x_4&=&0 \end{array}\right. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
无解? 有唯一解? 有无穷多解? 并求出有无穷多解时的特解. (华南理工大学2023年高等代数考研试题) [线性方程组 ]
纸质资料/答疑/pdf1/pdf2 / 增广矩阵
$$\begin{aligned} (A,\beta)=&\left(\begin{array}{cccccccccccccccccccc}1&1&1&1&0\\\\ 1&3&5&5&2\\\\ 0&-1&\lambda-3&-2&\mu\\\\ 1&2&\lambda&1&0\end{array}\right)\to\left(\begin{array}{cccccccccccccccccccc}1&1&1&1&0\\\\ 0&2&4&4&2\\\\ 0&-1&\lambda-3&-2&\mu\\\\ 0&1&\lambda-1&0&0\end{array}\right)\\\\ \to&\left(\begin{array}{cccccccccccccccccccc}1&0&-1&-1&-1\\\\ 0&1&2&2&1\\\\ 0&0&\lambda-1&0&\mu+1\\\\ 0&0&\lambda-3&-2&-1\end{array}\right)\to\left(\begin{array}{cccccccccccccccccccc}1&0&-1&-1&-1\\\\ 0&1&2&2&1\\\\ 0&0&\lambda-1&0&\mu+1\\\\ 0&0&-2&-2&-\mu-2\end{array}\right)\\\\ \to&\left(\begin{array}{cccccccccccccccccccc}1&0&0&0&\frac{\mu}{2}\\\\ 0&1&0&0&-\mu-1\\\\ 0&0&1&1&\frac{\mu+2}{2}\\\\ 0&0&0&1-\lambda&\frac{4-2\lambda+3\mu-\lambda\mu}{2}\end{array}\right). \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
(1)、 当 $\displaystyle \lambda=1, \mu\neq-1$ 时, $\displaystyle \mathrm{rank} A=3 < 4=\mathrm{rank}(A,\beta)$, 而 $\displaystyle Ax=\beta$ 无解. (2)、 当 $\displaystyle \lambda=1, \mu=-1$ 时, $\displaystyle \mathrm{rank} A=3=\mathrm{rank}(A,\beta)$, 而 $\displaystyle Ax=\beta$ 有无穷多解, 且有一个特解为 $\displaystyle \left(-\frac{1}{2},0,\frac{1}{2},0\right)^\mathrm{T}$. (3)、 当 $\displaystyle \lambda\neq 1$ 时, $\displaystyle |A|\neq 0$, 而 $\displaystyle Ax=\beta$ 有唯一解.跟锦数学微信公众号. 在线资料/公众号/
1020、 (3)、 设四元非齐次线性方程组的系数矩阵的秩是 $\displaystyle 3$, 而 $\displaystyle \eta_1,\eta_2,\eta_3$ 是它的三个解向量, 且满足
$$\begin{aligned} \eta_1+\eta_2=(2,2,0,4)^\mathrm{T}, \eta_1+\eta_3=(1,0,1,3)^\mathrm{T}, \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
则方程组的一般解为 $\displaystyle \underline{\ \ \ \ \ \ \ \ \ \ }$. (华南师范大学2023年高等代数考研试题) [线性方程组 ]
纸质资料/答疑/pdf1/pdf2 / 设方程组为 $\displaystyle Ax=\beta$, 则 $\displaystyle \mathrm{rank} A=3, x\in\mathbb{F}^{4\times 1}$, 而它有 $\displaystyle 4-3+1=2$ 个线性无关的解向量. 设
$$\begin{aligned} \gamma_1=\eta_1+\eta_2=(2,2,0,4)^\mathrm{T}, \gamma_2= \eta_1+\eta_3=(1,0,1,3)^\mathrm{T}, \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
则 $\displaystyle A\gamma_1=2\beta, A\gamma_2=2\beta$. 于是 $\displaystyle \gamma_1-\gamma_2=(1,2,-1,1)^\mathrm{T}$ 是 $\displaystyle Ax=0$ 的基础解系, $\displaystyle \frac{\gamma_1}{2}=(1,1,0,2)^\mathrm{T}$ 是 $\displaystyle Ax=\beta$ 的一个特解. 最终, 原方程组的通解为
$$\begin{aligned} (1,1,0,2)^\mathrm{T}+k(1,2,-1,1)^\mathrm{T}, \forall\ k. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
跟锦数学微信公众号. 在线资料/公众号/
1021、 (2)、 设 $\displaystyle \lambda_1,\cdots,\lambda_n$ 是 $\displaystyle n$ 个复数, 且满足
$$\begin{aligned} \lambda_1+\cdots+\lambda_n=\lambda_1^2+\cdots+\lambda_n^2=\cdots =\lambda_1^n+\cdots+\lambda_n^n=0. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
证明: $\displaystyle \lambda_1=\lambda_2=\cdots=\lambda_n=0$. (华中师范大学2023年高等代数考研试题) [线性方程组 ]
纸质资料/答疑/pdf1/pdf2 / 用反证法. 若存在某个 $\displaystyle \lambda_i\neq 0$, 则设 $\displaystyle \lambda_1,\cdots,\lambda_n$ 中非零且互异的为 $\displaystyle \mu_1,\cdots,\mu_s$, $\displaystyle \mu_i$ 出现了 $\displaystyle n_i\geq 1$ 次, 则 $\displaystyle \sum_{i=1}^n n_i=n$, 且
$$\begin{aligned} \left(\begin{array}{cccccccccccccccccccc}\mu_1&\cdots&\mu_s\\\\ \vdots&&\vdots\\\\ \mu_1^s&\cdots&\mu_s^s\end{array}\right)\left(\begin{array}{cccccccccccccccccccc}n_1\\\\\vdots\\\\nabla _s\end{array}\right)=\left(\begin{array}{cccccccccccccccccccc}0\\\\\vdots\\\\0\end{array}\right). \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
由 $\displaystyle n_i\geq 1$ 知上述关于 $\displaystyle n_1,\cdots,n_s$ 的方程组有非零解, 而系数矩阵行列式
$$\begin{aligned} \mu_1\cdots\mu_s\prod_{1\leq i < j\leq s}(\mu_j-\mu_i)=0. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
这与 $\displaystyle \mu_i$ 非零且互异矛盾. 故有结论.跟锦数学微信公众号. 在线资料/公众号/
1022、 2、 设 $\displaystyle \beta_1,\beta_2,\beta_3,\beta_4$ 是线性方程组 $\displaystyle AX=0$ 的一个基础解系, 记
$$\begin{aligned} \gamma_1=\beta_1+a\beta_2, \gamma_2=\beta_2+a\beta_3, \gamma_3=\beta_3+a\beta_4, \gamma_4=\beta_4+a\beta_1. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
问 $\displaystyle a$ 在什么条件下, $\displaystyle \gamma_1,\gamma_2,\gamma_3,\gamma_4$ 也是 $\displaystyle AX=0$ 的一个基础解系. (暨南大学2023年高等代数考研试题) [线性方程组 ]
纸质资料/答疑/pdf1/pdf2 /
$$\begin{aligned} (\gamma_1,\cdots,\gamma_4)=(\beta_1,\cdots,\beta_4)A, A=\left(\begin{array}{cccccccccccccccccccc}1&&&a\\\\ a&1&&\\\\ &a&1&\\\\ &&a&1\end{array}\right), \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
按第 $\displaystyle 1$ 行展开知 $\displaystyle |A|=1-a^4$. 故当且仅当 $\displaystyle a\not\in \left\{1,-1,\mathrm{ i},-\mathrm{ i}\right\}$ 时, $\displaystyle \gamma_1,\gamma_2,\gamma_3,\gamma_4$ 也是 $\displaystyle AX=0$ 的一个基础解系.跟锦数学微信公众号. 在线资料/公众号/
1023、 3、 设矩阵 $\displaystyle A=\left(\begin{array}{cccccccccccccccccccc}a_{11}&a_{12}&\cdots&a_{1n}\\\\ a_{21}&a_{22}&\cdots&a_{2n}\\\\ \vdots&\vdots&&\vdots\\\\ a_{n-1,1}&a_{n-1,2}&\cdots&a_{n-1,n}\end{array}\right)$ 的行向量组是线性方程
$$\begin{aligned} x_1+\cdots+x_n=0 \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
的解. 令 $\displaystyle M_i$ 表示自 $\displaystyle A$ 中划去第 $\displaystyle i$ 列所成 $\displaystyle n-1$ 阶子式. (0-26)、 证明: $\displaystyle \sum_{i=1}^n (-1)^iM_i=0$ 的充要条件是 $\displaystyle A$ 的行向量组不是
$$\begin{aligned} x_1+\cdots+x_n=0 \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
的基础解系. (0-27)、 若 $\displaystyle \sum_{i=1}^n (-1)^iM_i=0$, 求 $\displaystyle M_i, i=1,2,\cdots,n$. (南昌大学2023年高等代数考研试题) [线性方程组 ]
纸质资料/答疑/pdf1/pdf2 / (0-28)、 设 $\displaystyle B=\left(\begin{array}{cccccccccccccccccccc}1&1&\cdots&1\\\\ a_{11}&a_{12}&\cdots&a_{1n}\\\\ a_{21}&a_{22}&\cdots&a_{2n}\\\\ \vdots&\vdots&&\vdots\\\\ a_{n-1,1}&a_{n-1,2}&\cdots&a_{n-1,n}\end{array}\right)=\left(\begin{array}{cccccccccccccccccccc}e^\mathrm{T}\\\\ \alpha_1^\mathrm{T}\\\\ \vdots\\\\\alpha_{n-1}^\mathrm{T}\end{array}\right)$, 则按第一行展开知
$$\begin{aligned} |B|=M_1-M_2+\cdots+(-1)^{1+n}M_n. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
设 $\displaystyle A$ 的行向量组 $\displaystyle \alpha_1^\mathrm{T},\cdots,\alpha_{n-1}^\mathrm{T}$ 的一个极大无关组为 $\displaystyle \alpha_{i_1}^\mathrm{T}, \cdots,\alpha_{i_r}^\mathrm{T}$, 则由题设知 $\displaystyle Ae=0\Rightarrow \alpha_{i_j}^\mathrm{T} e=0$, 而
$$\begin{aligned} &\sum_j x_j\alpha_{i_j}+xe=0\stackrel{\alpha_{i_k}^\mathrm{T}\cdot}{\Rightarrow}0=\alpha_{i_k}^\mathrm{T}\cdot \sum_j x_j\alpha_{i_j}\\\\ \Rightarrow&0=\sum_k x_k\alpha_{i_k}\cdot \sum_j x_j\alpha_{i_j} =\left|\sum_j x_j\alpha_{i_j}\right|^2\\\\ \Rightarrow&s\sum_j x_j\alpha_{i_j}=0\Rightarrow x_j=0\Rightarrow x=0. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
故 $\displaystyle \alpha_{i_1},\cdots,\alpha_{i_r}, e$ 线性无关, 而
$$\begin{aligned} \mathrm{rank} B=r+1=\mathrm{rank} A+1. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
进而
$$\begin{aligned} &\sum_{i=1}^n (-1)^iM_i=0 \Leftrightarrow |B|=0\Leftrightarrow \mathrm{rank} B\leq n-1\Leftrightarrow \mathrm{rank} A\leq n-2\\\\ \Leftrightarrow&\mbox{$A$ 的行向量组不是方程的基础解系}. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
最后一步是因为 $\displaystyle x_1+\cdots+x_n=0$ 的基础解系有 $\displaystyle n-1$ 个线性无关的向量. (0-29)、 由第 1 步知
$$\begin{aligned} -1=\sum_{i=1}^n (-1)^{1+i}M_i=|B|=\left|\begin{array}{cccccccccc}1&1&\cdots&1\\\\ a_{11}&a_{12}&\cdots&a_{1n}\\\\ a_{21}&a_{22}&\cdots&a_{2n}\\\\ \vdots&\vdots&&\vdots\\\\ a_{n-1,1}&a_{n-1,2}&\cdots&a_{n-1,n}\end{array}\right|. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
将除了第 $\displaystyle i$ 列外的所有列都加到第 $\displaystyle i$ 列得
$$\begin{aligned} -1=&\left|\begin{array}{cccccccccc}1&\cdots&1&n&1&\cdots&1\\\\ a_{11}&\cdots&a_{1,i-1}&0&a_{1,i+1}&\cdots&a_{1n}\\\\ \vdots&&\vdots&\vdots&\vdots&&\vdots\\\\ a_{n-1,1}&\cdots&a_{n-1,i-1}&0&a_{n-1,i+1}&\cdots&a_{n-1,n}\end{array}\right|\\\\ =&n\cdot (-1)^{1+i}M_i. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
故 $\displaystyle M_i=\frac{(-1)^i}{n}$.跟锦数学微信公众号. 在线资料/公众号/
1024、 1、 线性方程组
$$\begin{aligned} \left\{\begin{array}{rrrrrrrrrrrrrrrr} x_1&+&4x_2&+&2x_3&+&3x_4&=&b,\\\\ 3x_1&+&12x_2&+&6x_3&+&8x_4&=&9,\\\\ 2x_1&+&8x_2&+&4x_3&+&8x_4&=&12. \end{array}\right. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
(1)、 $\displaystyle b$ 为何值时, 方程组有解; (2)、 有解时求一般解. (南方科技大学2023年高等代数考研试题) [线性方程组 ]
纸质资料/答疑/pdf1/pdf2 / 增广矩阵
$$\begin{aligned} (A,\beta)\to&\left(\begin{array}{cccccccccccccccccccc}1&4&2&3&b\\\\ 3&12&6&8&9\\\\ 2&8&4&8&12\end{array}\right) \to\left(\begin{array}{cccccccccccccccccccc}1&4&2&3&b\\\\ 0&0&0&-1&9-3b\\\\ 0&0&0&2&12-2b\end{array}\right)\\\\ \to&\left(\begin{array}{cccccccccccccccccccc}1&4&2&0&27-8b\\\\ 0&0&0&1&3b-9\\\\ 0&0&0&0&30-8b\end{array}\right). \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
(1)、 当 $\displaystyle b\neq \frac{15}{4}$ 时, $\displaystyle \mathrm{rank} A=2 < 3=\mathrm{rank}(A,\beta)$, 而 $\displaystyle Ax=\beta$ 无解. (2)、 当 $\displaystyle b=\frac{15}{4}$ 时,
$$\begin{aligned} (A,\beta)\to\left(\begin{array}{cccccccccccccccccccc}1&4&2&0&-3\\\\ 0&0&0&1&\frac{9}{4}\\\\ 0&0&0&0&0\end{array}\right). \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
取 $\displaystyle x_2,x_3$ 为自由变量知 $\displaystyle Ax=\beta$ 的通解为
$$\begin{aligned} \left(-3,0,0,\frac{9}{4}\right)^\mathrm{T}+k\left(-4,1,0,0\right)^\mathrm{T}+l\left(-2,0,1,0\right)^\mathrm{T},\quad \forall\ k,l. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
跟锦数学微信公众号. 在线资料/公众号/
1025、 (5)、 非齐次线性方程组
$$\begin{aligned} \left\{\begin{array}{rrrrrrrrrrrrrrrr} \lambda x_1&+&x_2&+&x_3&=&0,\\\\ x_1&+&\lambda x_2&+&x_3&=&\lambda,\\\\ x_1&+&x_2&+&\lambda x_3&=&-\lambda \end{array}\right. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
有无穷多解, 则 $\displaystyle \lambda=\underline{\ \ \ \ \ \ \ \ \ \ }$. (南京理工大学2023年高等代数考研试题) [线性方程组 ]
纸质资料/答疑/pdf1/pdf2 / 由增广矩阵
$$\begin{aligned} &(A,\beta)=\left(\begin{array}{cccccccccccccccccccc}\lambda&1&1&0\\\\ 1&\lambda&1&\lambda\\\\ 1&1&\lambda&-\lambda\end{array}\right)\to\left(\begin{array}{cccccccccccccccccccc}0&1-\lambda&1-\lambda^2&\lambda^2\\\\ 0&\lambda-1&1-\lambda&2\lambda\\\\ 1&1&\lambda&-\lambda\end{array}\right)\\\\ \to&\left(\begin{array}{cccccccccccccccccccc}0&0&-(\lambda-1)(\lambda+2)&\lambda(\lambda+2)\\\\ 0&\lambda-1&1-\lambda&2\lambda\\\\ 1&1&\lambda&-\lambda\end{array}\right)\\\\ \to&\left(\begin{array}{cccccccccccccccccccc}1&1&\lambda&-\lambda\\\\ 0&\lambda-1&1-\lambda&2\lambda\\\\ 0&0&-(\lambda-1)(\lambda+2)&\lambda(\lambda+2)\end{array}\right) \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
知方程组有无穷多解 $\displaystyle \Leftrightarrow \lambda=-2$.跟锦数学微信公众号. 在线资料/公众号/
1026、 4、 设有齐次线性方程组
$$\begin{aligned} \left(\begin{array}{cccccccccccccccccccc}1+a&1&\cdots&1\\\\ 2&2+a&\cdots&2\\\\ \vdots&\vdots&&\vdots\\\\ n&n&\cdots&n+a\end{array}\right)\left(\begin{array}{cccccccccccccccccccc}x_1\\\\x_2\\\\\vdots\\\\x_n\end{array}\right)=\left(\begin{array}{cccccccccccccccccccc}0\\\\0\\\\\vdots\\\\0\end{array}\right). \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
试问 $\displaystyle a$ 取何值时, 该方程组有非零解, 并求出通解. (南京师范大学2023年高等代数考研试题) [线性方程组 ]
纸质资料/答疑/pdf1/pdf2 / 齐次线性方程组的系数矩阵
$$\begin{aligned} A=\left(\begin{array}{cccccccccccccccccccc}1+a&1&\cdots&1\\\\ 2&2+a&\cdots&2\\\\ \vdots&\vdots&&\vdots\\\\ n&n&\cdots&n+a\end{array}\right)\to\left(\begin{array}{cccccccccccccccccccc}1+a&1&\cdots&1\\\\ -2a&a&\cdots&0\\\\ \vdots&\vdots&&\vdots\\\\ -na&0&\cdots&a\end{array}\right). \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
(1)、 若 $\displaystyle a=0$, 则
$$\begin{aligned} A\to\left(\begin{array}{cccccccccccccccccccc}1&1&\cdots&1\\\\ 0&0&\cdots&0\\\\ \vdots&\vdots&&\vdots\\\\ 0&0&\cdots&0\end{array}\right). \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
故 $\displaystyle Ax=0$ 有无穷多解, 且有基础解系
$$\begin{aligned} -e_1+e_2, -e_1+e_3,\cdots, -e_1+e_n, \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
通解为
$$\begin{aligned} &k_2(-e_1+e_2)+k_3(-e_1+e_3)+\cdots+k_n(-e_1+e_n)\\\\ =&(-k_2-\cdots-k_n,k_2,\cdots,k_n)^\mathrm{T}, \quad \forall\ k_2,\cdots,k_n. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
(2)、 若 $\displaystyle a\neq 0$, 则
$$\begin{aligned} A\to\left(\begin{array}{cccccccccccccccccccc}1+a&1&\cdots&1\\\\ -2&1&\cdots&0\\\\ \vdots&\vdots&&\vdots\\\\ -n&0&\cdots&1\end{array}\right)\to\left(\begin{array}{cccccccccccccccccccc}a+\frac{n(n+1)}{2}&0&\cdots&0\\\\ -2&1&\cdots&0\\\\ \vdots&\vdots&&\vdots\\\\ -n&0&\cdots&1\end{array}\right). \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
(2-1)、 若 $\displaystyle a=-\frac{n(n+1)}{2}$, 则 $\displaystyle Ax=0$ 有无穷多解, 且有基础解系
$$\begin{aligned} (1,2,\cdots,n)^\mathrm{T}, \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
通解为
$$\begin{aligned} k(1,2,\cdots,n)^\mathrm{T},\quad \forall\ k. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
(2-2)、 若 $\displaystyle a\neq -\frac{n(n+1)}{2}$, 则 $\displaystyle |A|\neq 0$, 而 $\displaystyle Ax=0$ 只有零解.跟锦数学微信公众号. 在线资料/公众号/
1027、 3、 (20 分) 设向量组 $\displaystyle \alpha_1,\cdots,\alpha_s\in\mathbb{P}^n$ 是某齐次线性方程组的一个基础解系, 向量 $\displaystyle \beta\in\mathbb{P}^n$ 不是该方程组的解. 证明: 向量组
$$\begin{aligned} \beta,\beta+\alpha_1,\cdots,\beta+\alpha_s \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
线性无关. (南开大学2023年高等代数考研试题) [线性方程组 ]
纸质资料/答疑/pdf1/pdf2 / 由题设, $\displaystyle A\beta\neq 0\Rightarrow \beta\neq 0$. 设
$$\begin{aligned} k\beta+\sum_{i=1}^s k_i(\beta+\alpha_i)=0,\qquad(I) \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
则用 $\displaystyle A$ 作用得
$$\begin{aligned} \left(k+\sum_{i=1}^s k_i\right)A\beta=0. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
由 $\displaystyle A\beta\neq 0$ 知 $\displaystyle k+\sum_{i=1}^s k_i=0$. 代入 $\displaystyle (I)$ 得
$$\begin{aligned} \sum_{i=1}^s k_i\alpha_i=0\Rightarrow k_i=0, 1\leq i\leq s. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
再代入 $\displaystyle (I)$ 得 $\displaystyle k=0$. 故
$$\begin{aligned} \beta,\beta+\alpha_1,\cdots,\beta+\alpha_s \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
线性无关.跟锦数学微信公众号. 在线资料/公众号/
1028、 2、 已知 $\displaystyle A=\left(\begin{array}{cccccccccccccccccccc}1&a&0&0\\\\ 0&1&a&0\\\\ 0&0&1&a\\\\ a&0&0&1\end{array}\right), \beta=\left(\begin{array}{cccccccccccccccccccc}1\\\\-1\\\\0\\\\0\end{array}\right)$, $\displaystyle Ax=\beta$ 有无穷多解. (1)、 求 $\displaystyle a$; (2)、 求 $\displaystyle Ax=\beta$ 的通解. (厦门大学2023年高等代数考研试题) [线性方程组 ]
纸质资料/答疑/pdf1/pdf2 /
$$\begin{aligned} (A,\beta)=&\left(\begin{array}{cccccccccccccccccccc}1&a&0&0&1\\\\ 0&1&a&0&-1\\\\ 0&0&1&a&0\\\\ a&0&0&1&0\end{array}\right)\to\left(\begin{array}{cccccccccccccccccccc}1&a&0&0&1\\\\ 0&1&a&0&-1\\\\ 0&0&1&a&0\\\\ 0&-a^2&0&1&-a\end{array}\right)\\\\ \to&\left(\begin{array}{cccccccccccccccccccc}1&a&0&0&1\\\\ 0&1&a&0&-1\\\\ 0&0&1&a&0\\\\ 0&0&a^3&1&-a-a^2\end{array}\right)\to \left(\begin{array}{cccccccccccccccccccc}1&a&0&0&1\\\\ 0&1&a&0&-1\\\\ 0&0&1&a&0\\\\ 0&0&0&1-a^4&-a(a+1)\end{array}\right). \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
由题设知 $\displaystyle a^4=1, a(a+1)=0\Rightarrow a=-1$. 此时,
$$\begin{aligned} (A,\beta)\to\left(\begin{array}{cccccccccccccccccccc}1&0&0&-1&0\\\\ 0&1&0&-1&-1\\\\ 0&0&1&-1&0\\\\ 0&0&0&0&0\end{array}\right), \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
而 $\displaystyle Ax=\beta$ 的通解为
$$\begin{aligned} k(1,1,1,1)^\mathrm{T}+(0,-1,0,0)^\mathrm{T}, \forall\ k. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
跟锦数学微信公众号. 在线资料/公众号/
1029、 2、 求下列方程组的一个基础解系:
$$\begin{aligned} \left\{\begin{array}{rrrrrrrrrrrrrrrr} 3x_1&+&2x_2&+&x_3&+&3x_4&+&5x_5&=&0,\\\\ 6x_1&+&4x_2&+&3x_3&+&5x_4&+&7x_5&=&0,\\\\ 9x_1&+&6x_2&+&5x_3&+&7x_4&+&9x_5&=&0,\\\\ 3x_1&+&2x_2&&&+&4x_4&+&8x_5&=&0. \end{array}\right. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
(山东大学2023年高等代数与常微分方程考研试题) [线性方程组 ]
纸质资料/答疑/pdf1/pdf2 / 系数矩阵
$$\begin{aligned} A=&\left(\begin{array}{cccccccccccccccccccc}3&2&1&3&5\\\\ 6&4&3&5&7\\\\ 9&6&5&7&9\\\\ 3&2&0&4&8\end{array}\right)\to\left(\begin{array}{cccccccccccccccccccc}1&\frac{2}{3}&0&\frac{4}{3}&\frac{8}{3}\\\\ 0&0&1&-1&-3\\\\ 0&0&0&0&0\\\\ 0&0&0&0&0\end{array}\right). \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
取 $\displaystyle x_2,x_4,x_5$ 为自由变量知 $\displaystyle Ax=0$ 的基础解系为
$$\begin{aligned} (-2,3,0,0,0)^\mathrm{T}, (-4,0,3,3,0)^\mathrm{T}, (-8,0,9,0,3)^\mathrm{T}. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
跟锦数学微信公众号. 在线资料/公众号/
1030、 7、 考虑线性方程组
$$\begin{aligned} \left\{\begin{array}{rrrrrrrrrrrrrrrr} 2x_1&+&4x_2&-&4x_3&=&2,\\\\ 4x_1&+&2x_2&-&\lambda x_3&=&-1,\\\\ 6x_1&+&7x_2&+&x_3&=&3. \end{array}\right. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
讨论 $\displaystyle \lambda$ 取何值时, 方程组无解, 有唯一解, 无穷多解, 并在有解时求解. [张祖锦注: 回忆的题目可能有问题, 就不做了. 以后如果有, 再补上.] (上海财经大学2023年高等代数考研试题) [线性方程组 ]
纸质资料/答疑/pdf1/pdf2 / [张祖锦注: 回忆的题目可能有问题, 就不做了. 以后如果有, 再补上.]跟锦数学微信公众号. 在线资料/公众号/
1031、 (2)、 设 $\displaystyle \alpha=(1,2,\cdots,n), A=\alpha^\mathrm{T} \alpha$, 则线性方程组 $\displaystyle AX=0$ 解空间的一组基为 $\displaystyle \underline{\ \ \ \ \ \ \ \ \ \ }$. (上海大学2023年高等代数考研试题) [线性方程组 ]
纸质资料/答疑/pdf1/pdf2 / $\displaystyle -ie_1+e_i, 2\leq i\leq n$.跟锦数学微信公众号. 在线资料/公众号/
1032、 3、 简答题. (1)、 (20 分) 求解线性方程组
$$\begin{aligned} \left\{\begin{array}{rrrrrrrrrrrrrrrr} 3x_1&+&7x_2&+&5x_3&+&7x_4&=&2,\\\\ 2x_1&+&5x_2&+&2x_3&+&3x_4&=&1,\\\\ 2x_1&+&4x_2&+&ax_3&+&(a^2-6a+8)x_4&=&2, \end{array}\right. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
其中 $\displaystyle a$ 为常数. (上海大学2023年高等代数考研试题) [线性方程组 ]
纸质资料/答疑/pdf1/pdf2 / 增广矩阵
$$\begin{aligned} &(A,\beta)=\left(\begin{array}{cccccccccccccccccccc}3&7&5&7&2\\\\ 2&5&2&3&1\\\\ 2&4&a&a^2-6a+8&2\end{array}\right)\to\left(\begin{array}{cccccccccccccccccccc}1&2&3&4&1\\\\ 2&5&2&3&1\\\\ 2&4&a&a^2-6a+8&2\end{array}\right)\\\\ \to&\left(\begin{array}{cccccccccccccccccccc}1&2&3&4&1\\\\ 0&1&-4&-5&-1\\\\ 0&0&a-6&a^2-6a&0\end{array}\right) \to\left(\begin{array}{cccccccccccccccccccc}1&0&11&14&3\\\\ 0&1&-4&-5&-1\\\\ 0&0&a-6&a^2-6a&0\end{array}\right). \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
(1-1)、 若 $\displaystyle a=6$, 则 $\displaystyle Ax=\beta$ 的通解为
$$\begin{aligned} k\left(\begin{array}{cccccccccccccccccccc}-11\\\\4\\\\1\\\\0\end{array}\right)+l\left(\begin{array}{cccccccccccccccccccc}-14\\\\5\\\\0\\\\1\end{array}\right)+\left(\begin{array}{cccccccccccccccccccc}3\\\\-1\\\\0\\\\0\end{array}\right), \forall\ k,l. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
(1-2)、 若 $\displaystyle a=0$, 则 $\displaystyle (A,\beta)\to\left(\begin{array}{cccccccccccccccccccc}1&0&11&14&3\\\\ 0&1&-4&-5&-1\\\\ 0&0&1&0&0\end{array}\right)$, 而 $\displaystyle Ax=\beta$ 的通解为
$$\begin{aligned} k\left(\begin{array}{cccccccccccccccccccc}-14\\\\5\\\\0\\\\1\end{array}\right)+\left(\begin{array}{cccccccccccccccccccc}3\\\\-1\\\\0\\\\0\end{array}\right),\forall\ k. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
(1-3)、 若 $\displaystyle a\neq 6, a\neq 0$, 则 $\displaystyle (A,\beta)\to\left(\begin{array}{cccccccccccccccccccc}1&0&0&14-11a&3\\\\ 0&1&0&4a-5&-1\\\\ 0&0&1&a&0\end{array}\right)$, 而 $\displaystyle Ax=\beta$ 的通解为
$$\begin{aligned} k\left(\begin{array}{cccccccccccccccccccc}11a-14\\\\ 5-4a\\\\-a\\\\1\end{array}\right)+\left(\begin{array}{cccccccccccccccccccc}3\\\\-1\\\\0\\\\0\end{array}\right), \forall\ k. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
跟锦数学微信公众号. 在线资料/公众号/
1033、 4、 设 $\displaystyle \mathbb{P}$ 为一个数域, $\displaystyle A\in \mathbb{P}^{n\times m}$, $\displaystyle \mathrm{rank} A=n$, $\displaystyle B\in \mathbb{P}^{m\times(m-n)}$, $\displaystyle \mathrm{rank} B=m-n$, 且 $\displaystyle AB=0$, 同时 $\displaystyle \eta$ 为 $\displaystyle Ax=0$ 的解, 证明方程组 $\displaystyle By=\eta$ 存在唯一解. (首都师范大学2023年高等代数考研试题) [线性方程组 ]
纸质资料/答疑/pdf1/pdf2 / 设 $\displaystyle B=(\eta_1,\cdots\eta_{m-n})$, 则由 $\displaystyle \mathrm{rank} B=m-n$ 知 $\displaystyle \eta_1,\cdots,\eta_{n-m}$ 线性无关. 又由
$$\begin{aligned} 0=AB=(A\eta_1,\cdots, A\eta_{m-n})\Rightarrow A\eta_i=0, 1\leq i\leq m-n \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
知 $\displaystyle \eta_1,\cdots,\eta_{n-m}$ 是 $\displaystyle Ax=0$ 的基础解系. 故对 $\displaystyle Ax=0$ 的解 $\displaystyle \eta$, 存在唯一的 $\displaystyle y=(y_1,\cdots,y_{n-m})^\mathrm{T}$, 使得
$$\begin{aligned} \eta=\sum_{i=1}^{m-n} y_i\eta_i=(\eta_1,\cdots,\eta_{m-n})\left(\begin{array}{cccccccccccccccccccc}y_1\\\\\vdots\\\\y_{n-m}\end{array}\right) =By. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
跟锦数学微信公众号. 在线资料/公众号/
1034、 2、 解答如下问题. (1)、 设 $\displaystyle a_1,\cdots,a_n$ 是数域 $\displaystyle \mathbb{F}$ 上的任意数, 在 $\displaystyle \mathbb{F}$ 上解线性方程组 (有无穷解用基础解系表示)
$$\begin{aligned} a_1x_1+\cdots+a_nx_n=0. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
(四川大学2023年高等代数考研试题) [线性方程组 ]
纸质资料/答疑/pdf1/pdf2 / 若 $\displaystyle \forall\ i, a_i=0$, 则通解为 $\displaystyle \mathbb{F}^n$. 若 $\displaystyle \exists\ k,\mathrm{ s.t.} a_k\neq 0$, 则取 $\displaystyle x_1,\cdots,x_{k-1},x_{k+1},\cdots,x_n$ 为自由变量知通解为
$$\begin{aligned} &x_1(-a_ke_1+a_1e_k)+\cdots+x_{k-1}(-a_ke_{k-1}+a_{k-1}e_k)\\\\ &+x_{k+1}(-a_ke_{k+1}+a_{k+1}e_k) +\cdots+x_n(-a_ke_n+a_ne_k), \forall\ x_i\in\mathbb{F}. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
跟锦数学微信公众号. 在线资料/公众号/
1035、 (2)、 设 $\displaystyle A$ 是 $\displaystyle m\times n$ 实矩阵, $\displaystyle \beta$ 是 $\displaystyle m$ 维实列向量. (2-1)、 证明: $\displaystyle \mathrm{rank}(A^\mathrm{T} A)=\mathrm{rank} A$, 其中 $\displaystyle A^\mathrm{T}$ 是 $\displaystyle A$ 的转置; (2-2)、 证明: 线性方程组 $\displaystyle A^\mathrm{T} AX=A^\mathrm{T}\beta$ 有解且它的任意解 $\displaystyle X_1,X_2$ 满足 $\displaystyle AX_1=AX_2$. (四川大学2023年高等代数考研试题) [线性方程组 ]
纸质资料/答疑/pdf1/pdf2 / (2-1)、 设
$$\begin{aligned} V_1&=\left\{\alpha\in\mathbb{R}^n; A^\mathrm{T} A\alpha=0\right\},\\\\ V_2&=\left\{\alpha\in\mathbb{R}^n; A\alpha=0\right\}. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
则显然有 $\displaystyle V_2\subset V_1$. 又由
$$\begin{aligned} A^\mathrm{T} A\alpha =0&\Rightarrow (A\alpha)^\mathrm{T} A\alpha=0\Rightarrow A\alpha=0 \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
知 $\displaystyle V_1\subset V_2$. 故
$$\begin{aligned} V_1=V_2\Rightarrow&\dim V_1=\dim V_2\\\\ \Rightarrow&\mathrm{rank}(A^\mathrm{T} A)=\mathrm{rank} A . \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
于是
$$\begin{aligned} \mathrm{rank}(A^\mathrm{T} A)=\mathrm{rank} A=\mathrm{rank} A^\mathrm{T} =\mathrm{rank}\left((A^\mathrm{T})^\mathrm{T} A^\mathrm{T}\right) =\mathrm{rank}(AA^\mathrm{T}). \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
(2-2)、 由第 1 步即知 $\displaystyle A^\mathrm{T} AX=A^\mathrm{T}\beta$ 有解. 再者, 若 $\displaystyle A^\mathrm{T} AX_i=A^\mathrm{T} \beta, i=1,2$, 则
$$\begin{aligned} &A^\mathrm{T} A(X_1-X_2)\Rightarrow (X_1-X_2)^\mathrm{T} A^\mathrm{T} A(X_1-X_2)\\\\ \stackrel{Y=A(X_1-X_2)}{\Rightarrow}&Y^\mathrm{T} Y=0\Rightarrow 0=Y=A(X_1-X_2)\Rightarrow AX_1=AX_2. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
跟锦数学微信公众号. 在线资料/公众号/
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