## 张祖锦2023年数学专业真题分类70天之第45天
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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 ]
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https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / (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](
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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 ]
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https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 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](
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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 ]
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https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 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](
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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](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
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https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 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](
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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](
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https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 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\}
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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](
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\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](
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---
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](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 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](
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---
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](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 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](
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---
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](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 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](
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---
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](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
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https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) /
\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](
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---
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](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / (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](
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---
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](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 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\}
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---
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](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 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](
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---
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](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 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](
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---
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](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 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](
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---
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](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) /
\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\}
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---
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](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 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\}
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---
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](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / [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](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/
---
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](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / $\displaystyle -ie\_1+e\_i, 2\leq i\leq n$.跟锦数学微信公众号. [在线资料](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
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---
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](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 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](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
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---
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](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 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\}
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---
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](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 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](
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---
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](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / (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](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/
发表于 2023-3-5 09:17:39
