行列式按行(列)展开

节 3.3 行列式按行(列)展开

建设中！

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练习练习

基础题.

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1.

求\(\begin{vmatrix} 3&1&0&-1\\2&1&1&0\\-1&3&1&2\\0&6&5&0 \end{vmatrix}\)中第\((1,2)\)，第\((2,4)\)元素的余子式和代数余子式。

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解答.

\begin{equation*} M_{12}=\begin{vmatrix} 2&1&0\\-1&1&2\\0&5&0 \end{vmatrix}=5\times (-1)^{3+2}\begin{vmatrix} 2&0\\-1&2 \end{vmatrix}=-20, \end{equation*}

\begin{equation*} M_{24}=\begin{vmatrix} 3&1&0\\-1&3&1\\0&6&5 \end{vmatrix}=32, \end{equation*}

\begin{equation*} A_{12}=(-1)^{1+2}M_{12}=20,\ A_{24}=(-1)^{2+4}M_{24}=32. \end{equation*}

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2.

计算下列行列式：

\(\begin{vmatrix} 3&2&0&-1&9\\ 7&8&2&-7&6\\ -1&5&0&-1&4\\ 2&0&0&-2&0\\ -1&1&0&1&1 \end{vmatrix}\)；
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\(\begin{vmatrix} a_1&a_2&a_3&\cdots&a_{n-1}&a_n\\ 1&-1&0&\cdots&0&0\\ 0&2&-2&\cdots&0&0\\ \vdots&\vdots&\vdots&&\vdots&\vdots\\ 0&0&0&\cdots&n-1&1-n \end{vmatrix}\)；
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解答.

按第 \(3\)列展开，

\begin{equation*} \mbox{原式}=2\cdot (-1)^{2+3}\begin{vmatrix} 3&2&-1&9\\ -1&5&-1&4\\ 2&0&-2&0\\ -1&1&1&1 \end{vmatrix}, \end{equation*}

将第 \(1\)列加到第 \(3\)列，得

\begin{equation*} \mbox{原式}=(-2)\cdot\begin{vmatrix} 3&2&2&9\\ -1&5&-2&4\\ 2&0&0&0\\ -1&1&0&1 \end{vmatrix}, \end{equation*}

按第 \(3\)行展开，

\begin{equation*} \mbox{原式}=(-2)\cdot 2\cdot (-1)^{3+1}\begin{vmatrix} 2&2&9\\ 5&-2&4\\ 1&0&1 \end{vmatrix}, \end{equation*}

将第 \(1\)行加到第 \(2\)行，得

\begin{equation*} \mbox{原式}=(-4)\cdot\begin{vmatrix} 2&2&9\\ 7&0&13\\ 1&0&1 \end{vmatrix}, \end{equation*}

按第 \(2\)列展开，

\begin{equation*} \mbox{原式}=(-4)\cdot 2\cdot(-1)^{1+2}\begin{vmatrix} 7&13\\ 1&1 \end{vmatrix}=8\cdot (7-13)=-48. \end{equation*}

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从第\(n\)列起，自右而左，依次将每一列加到前一列，直到第\(2\)列为止，得

\begin{equation*} \mbox{原式} = \begin{vmatrix} \sum\limits_{i=1}^n a_i&\sum\limits_{i=2}^n a_i&\cdots&\sum\limits_{i=n-1}^n a_i&a_n\\ 0&-1&\cdots&0&0\\ \vdots&\vdots&&\vdots&\vdots\\ 0&0&\cdots&2-n&0\\ 0&0&\cdots&0&1-n \end{vmatrix}, \end{equation*}

按第一列展开，得

\begin{equation*} \begin{array}{ll} \mbox{原式} & =\left(\sum\limits_{i=1}^n a_i\right)\cdot\begin{vmatrix} -1&0&\cdots&0&0\\ 0&-2&\cdots&0&0\\ \vdots&\vdots&&\vdots&\vdots\\ 0&0&\cdots&2-n&0\\ 0&0&\cdots&0&1-n \end{vmatrix}\\ & =(-1)^{n-1}\cdot \left(\sum\limits_{i=1}^n a_i\right)\cdot (n-1)!.\end{array} \end{equation*}

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3.

计算下述行列式，并将结果因式分解：

\begin{equation*} \begin{vmatrix} \lambda-2&-2&2\\-2&\lambda-5&4\\2&4&\lambda-5 \end{vmatrix}. \end{equation*}

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解答.

将第\(2\)列加到第\(3\)列，再将新的行列式的第\(3\)行乘以\(-1\)加到第\(2\)行，

\begin{equation*} \mbox{原式}=\begin{vmatrix} \lambda-2&-2&0\\-2&\lambda-5&\lambda-1\\2&4&\lambda-1 \end{vmatrix}=\begin{vmatrix} \lambda-2&-2&0\\-4&\lambda-9&0\\2&4&\lambda-1 \end{vmatrix}, \end{equation*}

按第\(3\)列展开，得

\begin{equation*} \mbox{原式}=(\lambda-1) \begin{vmatrix} \lambda-2&-2\\-4&\lambda-9 \end{vmatrix} =(\lambda-1)^2(\lambda-10). \end{equation*}

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4.

计算\(n\)阶行列式

\begin{equation*} \begin{vmatrix} a_1^{n-1}&a_1^{n-2}b_1&a_1^{n-3}b_1^2&\cdots&a_1b_1^{n-2}&b_1^{n-1}\\ a_2^{n-1}&a_2^{n-2}b_2&a_2^{n-3}b_2^2&\cdots&a_2b_2^{n-2}&b_2^{n-1}\\ a_3^{n-1}&a_3^{n-2}b_3&a_3^{n-3}b_3^2&\cdots&a_3b_3^{n-2}&b_3^{n-1}\\ \vdots&\vdots&\vdots&&\vdots&\vdots\\ a_n^{n-1}&a_n^{n-2}b_n&a_n^{n-3}b_n^2&\cdots&a_nb_n^{n-2}&b_n^{n-1} \end{vmatrix}, \end{equation*}

其中\(a_1a_2\cdots a_{n}\neq 0\)。

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解答.

因为\(a_1a_2\cdots a_{n}\neq 0\)，所以将第\(i\)行提出\(a_i^{n-1}\)（\(1\leq i\leq n\)），得

\begin{equation*} \begin{array}{ccl} \mbox{原式}&=&\prod\limits_{i=1}^n a_i^{n-1}\cdot\begin{vmatrix} 1&\frac{b_1}{a_1}&\left(\frac{b_1}{a_1}\right)^2&\cdots&\left(\frac{b_1}{a_1}\right)^{n-1}\\ 1&\frac{b_2}{a_2}&\left(\frac{b_2}{a_2}\right)^2&\cdots&\left(\frac{b_2}{a_2}\right)^{n-1}\\ \vdots&\vdots&\vdots&&\vdots\\ 1&\frac{b_n}{a_n}&\left(\frac{b_n}{a_n}\right)^2&\cdots&\left(\frac{b_n}{a_n}\right)^{n-1} \end{vmatrix}\\ &=& \prod\limits_{i=1}^n a_i^{n-1}\cdot\prod\limits_{1\leq j<i\leq n}\left(\frac{b_i}{a_i}-\frac{b_j}{a_j}\right)\\ &=&\prod\limits_{1\leq j<i\leq n} (a_jb_i-a_ib_j). \end{array} \end{equation*}

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5.

设\(\det A=\begin{vmatrix} -1 & -1 & 1 & -1\\ 4 & 2 & 0 & 0\\ 3 & 0 & 1 & 0\\ 5 & 0 & 0 & 4 \\ \end{vmatrix}\)，求

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\(A_{11}+A_{12}+A_{13}+A_{14}\)；
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\(M_{11}+2M_{21}+3M_{31}+4M_{41}\)。
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解答.

\begin{equation*} A_{11}+A_{12}+A_{13}+A_{14}=\begin{vmatrix} 1 & 1 & 1 & 1\\ 4 & 2 & 0 & 0\\ 3 & 0 & 1 & 0\\ 5 & 0 & 0 & 4 \end{vmatrix}. \end{equation*}

将第\(2\)、\(3\)、\(4\)列分别乘以\(-2\)、\(-3\)、\(-\frac{5}{4}\)加第\(1\)列，得

\begin{equation*} \text{原式}=\begin{vmatrix} -\frac{21}{4} & 1 & 1 & 1\\ 0 & 2 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 4 \end{vmatrix}=-42. \end{equation*}

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由于\(M_{ij}=(-1)^{i+j}A_{ij}\)，所以

\begin{equation*} \begin{array}{ll} M_{11}+2M_{21}+3M_{31}+4M_{41} & =A_{11}-2A_{21}+3A_{31}-4A_{41}\\ & =\begin{vmatrix} 1 & -1 & 1 & -1\\ -2 & 2 & 0 & 0\\ 3 & 0 & 1 & 0\\ -4 & 0 & 0 & 4 \\ \end{vmatrix}\\ & =\begin{vmatrix} -4 & -1 & 1 & -1\\ 0 & 2 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 4 \\ \end{vmatrix}\\ & =-32. \end{array} \end{equation*}

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6.

计算行列式\(\left|\begin{array}{ccccc} 2&3&0&0&0\\ -1&4&0&0&0\\ 37&25&1&2&0\\ 11&49&0&3&4\\ 19&52&1&0&2 \end{array}\right|\)。

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解答.

\begin{equation*} \left|\begin{array}{cc:ccc} 2&3&0&0&0\\ -1&4&0&0&0\\\hdashline 37&25&1&2&0\\ 11&49&0&3&4\\ 19&52&1&0&2 \end{array}\right|=\begin{vmatrix} 2 & 3 \\-1 & 4 \end{vmatrix}\cdot \begin{vmatrix} 1 & 2 & 0\\ 0& 3 & 4\\1 & 0 & 2 \end{vmatrix}=11\times 14=154. \end{equation*}

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提高题.

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7.

设\(A\)为数域\(\F\)上\(n\)阶方阵，\(\alpha\)为数域\(\F\)上\(n\)维列向量。证明：若\(\begin{vmatrix}A&\alpha\\ \alpha^T&a\end{vmatrix}=0\)，则

\begin{equation*} \begin{vmatrix}A&\alpha\\ \alpha^T&b\end{vmatrix}=(b-a)\cdot |A|. \end{equation*}

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解答.

将最后一行拆项得

\begin{equation*} \begin{vmatrix}A&\alpha\\ \alpha^T&b\end{vmatrix}=\begin{vmatrix}A&\alpha\\ \alpha^T&a\end{vmatrix}+\begin{vmatrix}A&\alpha\\ 0&b-a\end{vmatrix}. \end{equation*}

由于\(\begin{vmatrix}A&\alpha\\ \alpha^T&a\end{vmatrix}=0\)，\(\begin{vmatrix}A&\alpha\\ 0&b-a\end{vmatrix}=(b-a)\cdot |A|\)，因此

\begin{equation*} \begin{vmatrix}A&\alpha\\ \alpha^T&b\end{vmatrix}=(b-a)\cdot |A|. \end{equation*}

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8.

设 \(n\)阶行列式

\begin{equation*} D_n=\begin{vmatrix} \cos\alpha&1&&&&&\\ 1&2\cos\alpha&1&&&&\\ &1&2\cos\alpha&\ddots&&&\\ &&\ddots&\ddots&\ddots&&\\ &&&\ddots&2\cos\alpha&1&\\ &&&&1&2\cos\alpha&1\\ &&&&&1&2\cos\alpha \end{vmatrix}, \end{equation*}

证明：\(D_n=\cos (n \alpha)\)。

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解答.

当\(n=1\)时，\(D_n=\cos\alpha\)，结论成立。
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当\(n=2\)时，

\begin{equation*} D_n=\begin{vmatrix} \cos\alpha&1\\1&2\cos\alpha \end{vmatrix}=2\cos^2 \alpha-1=\cos (2 \alpha), \end{equation*}

结论成立。

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假设当\(k\leq n-1\)时，\(D_{k}=\cos (k\alpha)\)。对于\(n\)阶行列式\(D_n\)，按最后一行展开，有

\begin{equation*} D_n=2\cos\alpha\cdot D_{n-1}-\begin{vmatrix} \cos\alpha&1&&&&\\ 1&2\cos\alpha&&&&\\ &\ddots&\ddots&\ddots&&\\ &&\ddots&2\cos\alpha&1&\\ &&&1&2\cos\alpha&\\ &&&&1&1 \end{vmatrix}, \end{equation*}

将后面一个\(n-1\)阶行列式按最后一列展开，得

\begin{equation*} D_n=2\cos\alpha\cdot D_{n-1}-D_{n-2}, \end{equation*}

由归纳假设

\begin{equation*} D_{n-1}=\cos [(n-1)\alpha],\ D_{n-2}=\cos [(n-2) \alpha], \end{equation*}

代入得

\begin{equation*} D_n=2\cos\alpha\cdot\cos [(n-1)\alpha]-\cos [(n-2) \alpha]=\cos (n \alpha). \end{equation*}

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9.

计算下列\(n\)阶行列式：

\(\begin{vmatrix} x+1&x+2&x+3&\cdots&x+n-1&x+n\\ x+2&x+3&x+4&\cdots&x+n&x+1\\ x+3&x+4&x+5&\cdots&x+1&x+2\\ \vdots&\vdots&\vdots&&\vdots&\vdots\\ x+n-1&x+n&x+1&\cdots&x+n-3&x+n-2\\ x+n&x+1&x+2&\cdots&x+n-2&x+n-1 \end{vmatrix}\)；
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\(\begin{vmatrix} 1&1&1&\cdots&1\\ 1&C_2^1&C_3^1&\cdots&C_n^1\\ 1&C_3^2&C_4^2&\cdots&C_{n+1}^2\\ \vdots&\vdots&\vdots&&\vdots\\ 1&C_{n}^{n-1}&C_{n+1}^{n-1}&\cdots&C_{2n-2}^{n-1} \end{vmatrix}\)。
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解答.

从第\(n-1\)行起，自下而上，依次将每一行乘以\(-1\)加到下面一行，直至第一行，得

\begin{equation*} \mbox{原式}= \begin{vmatrix} x+1&x+2&x+3&\cdots&x+n-1&x+n\\ 1&1&1&\cdots&1&1-n\\ 1&1&1&\cdots&1-n&1\\ \vdots&\vdots&\vdots&&\vdots&\vdots\\ 1&1&1-n&\cdots&1&1\\ 1&1-n&1&\cdots&1&1 \end{vmatrix}, \end{equation*}

将每一列都加到第一列，得

\begin{equation*} \mbox{原式}= \begin{vmatrix} nx+\frac{n(n+1)}{2}&x+2&x+3&\cdots&x+n-1&x+n\\ 0&1&1&\cdots&1&1-n\\ 0&1&1&\cdots&1-n&1\\ \vdots&\vdots&\vdots&&\vdots&\vdots\\ 0&1&1-n&\cdots&1&1\\ 0&1-n&1&\cdots&1&1 \end{vmatrix}, \end{equation*}

按第一列展开，可降为\(n-1\)阶行列式

\begin{equation*} \text{原式} =\left[nx+\frac{n(n+1)}{2}\right] \begin{vmatrix} 1&1&\cdots&1&1-n\\ 1&1&\cdots&1-n&1\\ \vdots&\vdots&&\vdots&\vdots\\ 1&1-n&\cdots&1&1\\ 1-n&1&\cdots&1&1 \end{vmatrix}, \end{equation*}

先将上述行列式第\(2\)至第\(n-1\)行都加到第\(1\)行，得

\begin{equation*} \text{原式}=\left[nx+\frac{n(n+1)}{2}\right]\begin{vmatrix} -1&-1&\cdots&-1&-1\\ 1&1&\cdots&1-n&1\\ \vdots&\vdots&&\vdots&\vdots\\ 1&1-n&\cdots&1&1\\ 1-n&1&\cdots&1&1 \end{vmatrix}, \end{equation*}

再将第\(1\)行加到第\(i\)行(\(2\leq i\leq n-1\))，得

\begin{align*} \text{原式} & =\left[nx+\frac{n(n+1)}{2}\right]\begin{vmatrix} -1&-1&\cdots&-1&-1\\ 0&0&\cdots&-n&0\\ \vdots&\vdots&&\vdots&\vdots\\ 0&-n&\cdots&0&0\\ -n&0&\cdots&0&0 \end{vmatrix} \\ & =\left[nx+\frac{n(n+1)}{2}\right]\cdot (-1)^{\frac{(n-1)(n-2)}{2}}\cdot (-1)\cdot (-n)^{n-2}\\ & = (-1)^{\frac{n(n-1)}{2}}\left(x+\frac{n+1}{2}\right)n^{n-1}. \end{align*}

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因为

\begin{equation*} C_{k+1}^{i+1}-C_k^i=C_k^{i+1}, \end{equation*}

所以从倒数第二行起，自下而上，依次将每一行乘以 \(-1\)加到下一行，直至第一行，得

\begin{equation*} \mbox{原式}=\begin{vmatrix} 1&1&1&\cdots&1\\ 0&C_1^1&C_2^1&\cdots&C_{n-1}^1\\ 0&C_2^2&C_3^2&\cdots&C_{n}^2\\ \vdots&\vdots&\vdots&&\vdots\\ 0&C_{n-1}^{n-1}&C_{n}^{n-1}&\cdots&C_{2n-3}^{n-1} \end{vmatrix}, \end{equation*}

按第一列展开，可降为 \(n-1\)阶行列式

\begin{equation*} \mbox{原式}=\begin{vmatrix} C_1^1&C_2^1&C_3^1&\cdots&C_{n-1}^1\\ C_2^2&C_3^2&C_4^2&\cdots&C_{n}^2\\ C_3^3&C_4^3&C_5^3&\cdots&C_{n+1}^3\\ \vdots&\vdots&\vdots&&\vdots\\ C_{n-1}^{n-1}&C_{n}^{n-1}&C_{n+1}^{n-1}&\cdots&C_{2n-3}^{n-1} \end{vmatrix}. \end{equation*}

同样，再从倒数第二行起，自下而上，依次将每一行乘以 \(-1\)加到下一行，直至第一行，得

\begin{equation*} \mbox{原式}=\begin{vmatrix} C_1^1&C_2^1&C_3^1&\cdots&C_{n-1}^1\\ 0&C_2^2&C_3^2&\cdots&C_{n-1}^2\\ 0&C_3^3&C_4^3&\cdots&C_{n}^3\\ \vdots&\vdots&\vdots&&\vdots\\ 0&C_{n-1}^{n-1}&C_{n}^{n-1}&\cdots&C_{2n-4}^{n-1} \end{vmatrix}, \end{equation*}

按第一列展开，降为 \(n-2\)阶行列式

\begin{equation*} \mbox{原式}=\begin{vmatrix} C_2^2&C_3^2&\cdots&C_{n-1}^2\\ C_3^3&C_4^3&\cdots&C_{n}^3\\ \vdots&\vdots&&\vdots\\ C_{n-1}^{n-1}&C_{n}^{n-1}&\cdots&C_{2n-4}^{n-1} \end{vmatrix}. \end{equation*}

依此类推，不断进行下去，得

\begin{align*} \mbox{原式}&=\begin{vmatrix} C_1^1&C_2^1&\cdots&C_{n-1}^1\\ C_2^2&C_3^2&\cdots&C_{n}^2\\ C_3^3&C_4^3&\cdots&C_{n+1}^3\\ \vdots&\vdots&&\vdots\\ C_{n-1}^{n-1}&C_{n}^{n-1}&\cdots&C_{2n-3}^{n-1} \end{vmatrix} \\ &= \begin{vmatrix} C_2^2&C_3^2&\cdots&C_{n-1}^2\\ C_3^3&C_4^3&\cdots&C_{n}^3\\ \vdots&\vdots&&\vdots\\ C_{n-1}^{n-1}&C_{n}^{n-1}&\cdots&C_{2n-4}^{n-1} \end{vmatrix}\\ &=\cdots \\ &=C_{n-1}^{n-1} =1. \end{align*}

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10.

设\(n\geq 2\)，计算\(n\)阶行列式

\begin{equation*} D_n=\begin{vmatrix} a_1&x&\cdots&x\\ x&a_2&\cdots&x\\ \vdots&\vdots&&\vdots\\ x&x&\cdots&a_n \end{vmatrix}, \end{equation*}

其中\(x\neq a_i,\ i=1,\ldots ,n\)。

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解答.

将最后一列分拆，得

\begin{equation*} D_n=\begin{vmatrix} a_1&x&\cdots&x&x\\ x&a_2&\cdots&x&x\\ \vdots&\vdots&&\vdots&\vdots\\ x&x&\cdots&a_{n-1}&x\\ x&x&\cdots&x&x \end{vmatrix}+\begin{vmatrix} a_1&x&\cdots&x&0\\ x&a_2&\cdots&x&0\\ \vdots&\vdots&&\vdots&\vdots\\ x&x&\cdots&a_{n-1}&0\\ x&x&\cdots&x&a_n-x \end{vmatrix}, \end{equation*}

则

\begin{equation*} D_n=\begin{vmatrix} a_1-x&0&\cdots&0&x\\ 0&a_2-x&\cdots&0&x\\ \vdots&\vdots&&\vdots&\vdots\\ 0&0&\cdots&a_{n-1}-x&x\\ 0&0&\cdots&0&x \end{vmatrix}+(a_n-x)D_{n-1}, \end{equation*}

即

\begin{equation*} D_n=x\cdot\prod\limits_{i=1}^{n-1}(a_i-x)+(a_n-x)D_{n-1}. \end{equation*}

依此类推，

\begin{equation*} \begin{array}{ccl} D_n&=&x\cdot\prod\limits_{i=1}^{n-1}(a_i-x)+(a_n-x)\left[ x\cdot\prod\limits_{i=1}^{n-2}(a_i-x)+(a_{n-1}-x)D_{n-2}\right]\\ &=&x\cdot\frac{\prod\limits_{i=1}^n (a_i-x)}{a_n-x}+x\cdot\frac{\prod\limits_{i=1}^n (a_i-x)}{a_{n-1}-x}+(a_n-x)(a_{n-1}-x)D_{n-2}\\ &=&\cdots\cdots\cdots\\ &=&x\cdot\frac{\prod\limits_{i=1}^n (a_i-x)}{a_n-x}+x\cdot\frac{\prod\limits_{i=1}^n (a_i-x)}{a_{n-1}-x}+\cdots+x\cdot\frac{\prod\limits_{i=1}^n (a_i-x)}{a_{2}-x}+\prod\limits_{i=2}^n(a_i-x)D_{1}. \end{array} \end{equation*}

由\(D_1=a_1\)可知

\begin{equation*} D_n=x\cdot\sum\limits_{j=2}^n\frac{\prod\limits_{i=1}^n (a_i-x)}{a_j-x}+a_1\cdot\prod\limits_{i=2}^n(a_i-x). \end{equation*}

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11.

计算\(n\)阶行列式

\begin{equation*} D_n= \begin{vmatrix} x&y&y&\cdots&y&y\\ z&x&y&\cdots&y&y\\ z&z&x&\cdots&y&y\\ \vdots&\vdots&\vdots&&\vdots&\vdots\\ z&z&z&\cdots&x&y\\ z&z&z&\cdots&z&x \end{vmatrix}, \end{equation*}

其中\(y\neq z\)。

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解答.

将最后一列分拆，得

\begin{equation*} \begin{array}{ccl} D_n& =& \begin{vmatrix} x&y&y&\cdots&y&y\\ z&x&y&\cdots&y&y\\ z&z&x&\cdots&y&y\\ \vdots&\vdots&\vdots&&\vdots&\vdots\\ z&z&z&\cdots&x&y\\ z&z&z&\cdots&z&y \end{vmatrix}+\begin{vmatrix} x&y&y&\cdots&y&0\\ z&x&y&\cdots&y&0\\ z&z&x&\cdots&y&0\\ \vdots&\vdots&\vdots&&\vdots&\vdots\\ z&z&z&\cdots&x&0\\ z&z&z&\cdots&z&x-y \end{vmatrix}\\ &=&y\begin{vmatrix} x-z&y-z&y-z&\cdots&y-z&1\\ 0&x-z&y-z&\cdots&y-z&1\\ 0&0&x-z&\cdots&y-z&1\\ \vdots&\vdots&\vdots&&\vdots&\vdots\\ 0&0&0&\cdots&x-z&1\\ 0&0&0&\cdots&0&1 \end{vmatrix}+(x-y)D_{n-1}\end{array} \end{equation*}

即

\begin{equation} D_n= y(x-z)^{n-1}+(x-y)D_{n-1}.\tag{3.3.1} \end{equation}

同理，

\begin{equation} D_n=z(x-y)^{n-1}+(x-z)D_{n-1}.\tag{3.3.2} \end{equation}

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(3.3.1)\(\times (x-z)-\)(3.3.2) \(\times (x-y)\)得

\begin{equation*} D_n=\frac{y(x-z)^{n}-z(x-y)^n}{y-z}. \end{equation*}

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12.

设\(A=\left(a_{ij}\right)_{n\times n}\)是\(n\)阶方阵，\(A_{ij}\)是\(a_{ij}\)的代数余子式，证明：

\begin{equation*} \begin{vmatrix} a_{11}+x_1&a_{12}+x_2&\cdots&a_{1n}+x_n\\ a_{21}+x_1&a_{22}+x_2&\cdots&a_{2n}+x_n\\ \vdots&\vdots&&\vdots\\ a_{n1}+x_1&a_{n2}+x_2&\cdots&a_{nn}+x_n \end{vmatrix}=|A|+\sum\limits_{j=1}^n\sum\limits_{i=1}^nx_jA_{ij}. \end{equation*}

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解答.

将行列式按列依次拆项，可拆成\(2^n\)个行列式的和。注意到其中有些行列式两列成比例，值为\(0\)，因此

\begin{equation*} \mbox{原式}=\begin{vmatrix} a_{11}&a_{12}&\cdots&a_{1n}\\ a_{21}&a_{22}&\cdots&a_{2n}\\ \vdots&\vdots&&\vdots\\ a_{n1}&a_{n2}&\cdots&a_{nn} \end{vmatrix}+\sum\limits_{j=1}^n\begin{vmatrix} a_{11}&\cdots&x_{j}&\cdots&a_{1n}\\ a_{21}&\cdots&x_{j}&\cdots&a_{2n}\\ \vdots&&\vdots&&\vdots\\ a_{n1}&\cdots&x_{j}&\cdots&a_{nn} \end{vmatrix}. \end{equation*}

由于\(\begin{vmatrix} a_{11}&\cdots&x_{j}&\cdots&a_{1n}\\ a_{21}&\cdots&x_{j}&\cdots&a_{2n}\\ \vdots&&\vdots&&\vdots\\ a_{n1}&\cdots&x_{j}&\cdots&a_{nn} \end{vmatrix}\)第\(j\)列元素的代数余子式与\(|A|\)第\(j\)列对应元素的代数余子式一致，因此按第\(j\)列展开，有

\begin{equation*} \begin{vmatrix} a_{11}&\cdots&x_{j}&\cdots&a_{1n}\\ a_{21}&\cdots&x_{j}&\cdots&a_{2n}\\ \vdots&&\vdots&&\vdots\\ a_{n1}&\cdots&x_{j}&\cdots&a_{nn} \end{vmatrix}=\sum\limits_{i=1}^n x_jA_{ij}, \end{equation*}

从而

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13.

设 \(A=\left(a_{ij}\right)_{n\times n}\) 是 \(n\)阶方阵， \(x\)是未定元，

\begin{equation*} f(x)=\det\left(xE_n-A\right), \end{equation*}

证明：

\begin{equation*} f(x)=x^n+a_1x^{n-1}+\cdots +a_{n-1}x+a_n, \end{equation*}

其中

\begin{equation*} a_k=\left(-1\right)^k\sum\limits_{1\leq i_1< i_2<\cdots<i_k\leq n}A\begin{bmatrix} i_1& i_2&\cdots&i_k\\ i_1& i_2&\cdots&i_k \end{bmatrix}. \end{equation*}

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解答.

将 \(A\)按列分块： \(A=\left(\alpha_1,\alpha_2,\ldots ,\alpha_n\right)\)，则

\begin{equation*} f(x)=\left|x\varepsilon_1-\alpha_1,x\varepsilon_2-\alpha_2,\ldots ,x\varepsilon_n-\alpha_n\right|. \end{equation*}

根据行列式性质，将列依次分拆，得

\begin{equation*} \begin{array}{ccl} f(x)&=&\left|x\varepsilon_1,x\varepsilon_2,\ldots ,x\varepsilon_n\right|+\sum\limits_{i=1}^n\left|x\varepsilon_1,\ldots ,x\varepsilon_{i-1},-\alpha_i,x\varepsilon_{i+1},\ldots ,x\varepsilon_n\right|\\ &&+\sum\limits_{1\leq i_1< i_2\leq n}\left|x\varepsilon_1,\ldots ,x\varepsilon_{i_1-1},-\alpha_{i_1},x\varepsilon_{i_1+1},\ldots ,x\varepsilon_{i_2-1},-\alpha_{i_2},x\varepsilon_{i_2+1},\ldots ,x\varepsilon_n\right|\\ &&+\sum\limits_{1\leq i_1< i_2<i_3\leq n}\left|x\varepsilon_1,\ldots ,x\varepsilon_{i_1-1},-\alpha_{i_1},x\varepsilon_{i_1+1},\ldots ,x\varepsilon_{i_2-1},-\alpha_{i_2},x\varepsilon_{i_2+1},\ldots ,x\varepsilon_{i_3-1},-\alpha_{i_3},x\varepsilon_{i_3+1},\ldots ,x\varepsilon_n\right|\\ &&+\cdots+\left|-\alpha_1,-\alpha_2,\ldots ,-\alpha_n\right|\\ &=&x^n-\sum\limits_{i=1}^nx^{n-1}\left|\varepsilon_1,\ldots ,\varepsilon_{i-1},\alpha_i,\varepsilon_{i+1},\ldots ,\varepsilon_n\right|\\ &&+\sum\limits_{1\leq i_1< i_2\leq n}x^{n-2}\left|\varepsilon_1,\ldots ,\varepsilon_{i_1-1},\alpha_{i_1},\varepsilon_{i_1+1},\ldots ,\varepsilon_{i_2-1},\alpha_{i_2},\varepsilon_{i_2+1},\ldots ,\varepsilon_n\right|\\ &&-\sum\limits_{1\leq i_1< i_2<i_3\leq n}x^{n-3}\left|\varepsilon_1,\ldots ,\varepsilon_{i_1-1},\alpha_{i_1},\varepsilon_{i_1+1},\ldots ,\varepsilon_{i_2-1},\alpha_{i_2},\varepsilon_{i_2+1},\ldots ,\varepsilon_{i_3-1},\alpha_{i_3},\varepsilon_{i_3+1},\ldots ,\varepsilon_n\right|\\ &&+\cdots+(-1)^n|A|. \end{array} \end{equation*}

因此 \(f(x)\)是一个首项系数为 \(1\)的 \(n\)次多项式，其中\(n-k\)次项系数为

\begin{equation*} a_k=(-1)^k\sum\limits_{1\leq i_1< \cdots<i_k\leq n}\left|\varepsilon_1,\ldots ,\varepsilon_{i_1-1},\alpha_{i_1},\varepsilon_{i_1+1},\ldots ,\varepsilon_{i_2-1},\alpha_{i_2},\varepsilon_{i_2+1},\ldots ,\varepsilon_{i_{k}-1},\alpha_{i_k},\varepsilon_{i_k+1},\ldots ,\varepsilon_n\right|. \end{equation*}

由Laplace定理，按第\(i_1,i_2,\ldots ,i_k\)列展开，得

\begin{equation*} a_k=(-1)^k\sum\limits_{1\leq i_1< \cdots<i_k\leq n}A\begin{bmatrix} i_1& i_2&\cdots&i_k\\ i_1& i_2&\cdots&i_k \end{bmatrix}. \end{equation*}

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挑战题.

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14.

计算下列 \(n\)阶行列式 \((1\leq j\leq n-1)\)：

\begin{equation*} \det A=\begin{vmatrix} 1&a_1&\cdots&a_1^{j-1}&a_1^{j+1}&\cdots&a_1^{n}\\ 1&a_2&\cdots&a_2^{j-1}&a_2^{j+1}&\cdots&a_2^{n}\\ \vdots&\vdots&&\vdots&\vdots&&\vdots\\ 1&a_n&\cdots&a_n^{j-1}&a_n^{j+1}&\cdots&a_n^{n} \end{vmatrix}. \end{equation*}

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解答.

注意到\(\det A\)与Vandermonde行列式的区别在于它缺少 \(j\)次幂的列，所以再适当增加一行一列即可成为 \(n+1\)阶Vandermonde行列式。令

\begin{equation*} \det B=\begin{vmatrix} 1&a_1&\cdots&a_1^{j-1}&{\color{blue}{a_1^j}}&a_1^{j+1}&\cdots&a_1^{n}\\ 1&a_2&\cdots&a_2^{j-1}&{\color{blue}{a_2^j}}&a_2^{j+1}&\cdots&a_2^{n}\\ \vdots&\vdots&&\vdots&{\color{blue}\vdots}&\vdots&&\vdots\\ 1&a_n&\cdots&a_n^{j-1}&{\color{blue}{a_n^j}}&a_n^{j+1}&\cdots&a_n^{n}\\ {\color{blue}1}&{\color{blue}x}&{\color{blue}\cdots}&{\color{blue}x^{j-1}}&{\color{blue}{x^j}}&{\color{blue}x^{j+1}}&{\color{blue}\cdots}&{\color{blue}x^{n}}\\ \end{vmatrix}, \end{equation*}

按最后一行展开可知， \(\det B\)是关于 \(x\)的多项式，其中 \(x^j\)的系数为

\begin{equation*} (-1)^{n+1+j+1}\det A=(-1)^{n-j}\det A. \end{equation*}

由于

\begin{equation*} \det B=\left(\prod\limits_{i=1}^n(x-a_i)\right)\cdot \left(\prod\limits_{1\leq k<l\leq n}(a_l-a_k)\right) \end{equation*}

中 \(x^j\)的系数为

\begin{equation*} \left(\prod\limits_{1\leq k<l\leq n}(a_l-a_k)\right)\cdot\left(\sum\limits_{1\leq i_1< i_2<\cdots<i_{n-j}\leq n}(-1)^{n-j}a_{i_1}a_{i_2}\cdots a_{i_{n-j}}\right), \end{equation*}

因此

\begin{equation*} \det A=\left(\prod\limits_{1\leq k<l\leq n}(a_l-a_k)\right)\cdot\left(\sum\limits_{1\leq i_1< i_2<\cdots<i_{n-j}\leq n}a_{i_1}a_{i_2}\cdots a_{i_{n-j}}\right). \end{equation*}

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15.

设\(b_0,\ldots ,b_{n-1}\in\F\)互不相同。对任意\(0\leq i\leq n-1\)，令

\begin{equation*} f_i(x)=\prod\limits_{\begin{array}{c}0\leq j\leq n-1\\j\neq i \end{array}}\left(x-b_j\right)=a_{i0}+a_{i1}x+\cdots +a_{i,n-2}x^{n-2}+x^{n-1}, \end{equation*}

计算\(n\)阶行列式

\begin{equation*} \begin{vmatrix} a_{00}&a_{01}&\cdots&a_{0,n-2}&1\\ a_{10}&a_{11}&\cdots&a_{1,n-2}&1\\ \vdots&\vdots&&\vdots&\vdots\\ a_{n-1,0}&a_{n-1,1}&\cdots&a_{n-1,n-2}&1 \end{vmatrix}. \end{equation*}

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解答.

令

\begin{equation*} A=\begin{pmatrix} a_{00}&a_{01}&\cdots&a_{0,n-2}&1\\ a_{10}&a_{11}&\cdots&a_{1,n-2}&1\\ \vdots&\vdots&&\vdots&\vdots\\ a_{n-1,0}&a_{n-1,1}&\cdots&a_{n-1,n-2}&1 \end{pmatrix}, \end{equation*}

\begin{equation*} B=\begin{pmatrix} 1&1&\cdots&1\\ b_0&b_1&\cdots&b_{n-1}\\ \vdots&\vdots&\cdots&\vdots\\ b_{0}^{n-2}&b_1^{n-2}&\cdots&b_{n-1}^{n-2}\\ b_{0}^{n-1}&b_1^{n-1}&\cdots&b_{n-1}^{n-1} \end{pmatrix}, \end{equation*}

则

\begin{equation*} AB=\begin{pmatrix} f_{0}(b_0)&f_{0}(b_1)&\cdots&f_{0}(b_{n-1})\\ f_{1}(b_0)&f_{1}(b_1)&\cdots&f_{1}(b_{n-1})\\ \vdots&\vdots&&\vdots\\ f_{n-1}(b_0)&f_{n-1}(b_1)&\cdots&f_{n-1}(b_{n-1}) \end{pmatrix}. \end{equation*}

当\(i\neq j\)时，\(f_i(b_j)=0\)，所以

\begin{equation*} AB=\begin{pmatrix} f_{0}(b_0)&&&\\ &f_{1}(b_1)&&\\ &&\ddots&\\ &&&f_{n-1}(b_{n-1}) \end{pmatrix}, \end{equation*}

两边同取行列式得

\begin{equation*} \det A\cdot\prod\limits_{0\leq i< j\leq n-1}\left(b_j-b_i\right)=\prod\limits_{i=0}^{n-1}f_i(b_i). \end{equation*}

由此推出

\begin{equation*} \det A=\frac{\prod\limits_{0\leq i\neq j\leq n-1}\left(b_j-b_i\right)}{\prod\limits_{0\leq i< j\leq n-1}\left(b_j-b_i\right)}=\prod\limits_{0\leq j< i\leq n-1}\left(b_j-b_i\right). \end{equation*}

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16.

设 \(f_0(x),f_1(x),\ldots ,f_{n-1}(x)\in\Z[x]\)，其中

\begin{equation*} f_i(x)=a_{i0}+a_{i1}x+\cdots +a_{i,n-1}x^{n-1}(i=0,\ldots ,n-1). \end{equation*}

记 \(A\)为 \(n\)阶方阵

\begin{equation*} \begin{pmatrix} a_{00}&a_{01}&\cdots&a_{0,n-1}\\ a_{10}&a_{11}&\cdots&a_{1,n-1}\\ \vdots&\vdots& &\vdots\\ a_{n-1,0}&a_{n-1,1}&\cdots&a_{n-1,n-1} \end{pmatrix}, \end{equation*}

证明：存在 \(n\)阶整数矩阵 \(C\)使得 \(AC=E_{n}\) 的充分必要条件是存在 \(n\) 个互不相同的整数 \(b_0,b_1,\ldots ,b_{n-1}\)使得

\begin{equation*} \begin{vmatrix} f_0(b_0)&f_0(b_1)&\cdots&f_0(b_{n-1})\\ f_1(b_0)&f_1(b_1)&\cdots&f_1(b_{n-1})\\ \vdots&\vdots& &\vdots\\ f_{n-1}(b_0)&f_{n-1}(b_1)&\cdots&f_{n-1}(b_{n-1})\\ \end{vmatrix}=\pm\prod\limits_{0\leq i < j\leq n-1}(b_j-b_i). \end{equation*}

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解答.

充分性：令\(B=\begin{pmatrix} 1&1&\cdots&1\\ b_0&b_1&\cdots&b_{n-1}\\ \vdots&\vdots&\cdots&\vdots\\ b_{0}^{n-2}&b_1^{n-2}&\cdots&b_{n-1}^{n-2}\\ b_{0}^{n-1}&b_1^{n-1}&\cdots&b_{n-1}^{n-1} \end{pmatrix}\)，则

两边同取行列式得

\begin{equation*} \det A\cdot \det B=\pm\prod\limits_{0\leq i < j\leq n-1}(b_j-b_i), \end{equation*}

即

\begin{equation*} \det A\cdot\prod\limits_{0\leq i < j\leq n-1}(b_j-b_i)=\pm\prod\limits_{0\leq i < j\leq n-1}(b_j-b_i), \end{equation*}

由此推出\(\det A=\pm 1\)，因此\(A\)可逆且

\begin{equation*} A^{-1}=\frac{1}{\det A}A^*=\pm A^* \end{equation*}

是整数矩阵。从而存在 \(n\)阶整数矩阵 \(C\)，使得\(AC=E_{n}\)。

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必要性：因\(A,C\)是整数矩阵，所以\(\det A, \det C\in\Z\)。由\(AC=E_n\)可知

\begin{equation*} \det A\cdot \det C=1, \end{equation*}

因此

\begin{equation*} \det A=\det C=\pm 1. \end{equation*}

任意取定\(n\)个互不相同的整数 \(b_0,b_1,\ldots ,b_{n-1}\)，令

则

\begin{equation*} \begin{array}{ll} \begin{vmatrix} f_0(b_0)&f_0(b_1)&\cdots&f_0(b_{n-1})\\ f_1(b_0)&f_1(b_1)&\cdots&f_1(b_{n-1})\\ \vdots&\vdots& &\vdots\\ f_{n-1}(b_0)&f_{n-1}(b_1)&\cdots&f_{n-1}(b_{n-1})\\ \end{vmatrix} & = \det (AB)\\ & = \det A\cdot \det B\\ & = \pm\prod\limits_{0\leq i < j\leq n-1}(b_j-b_i). \end{array} \end{equation*}

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17.

设\(A,B\)是\(n\)阶方阵，\(\alpha_1,\ldots,\alpha_n\)、\(\beta_1,\ldots,\beta_n\)分别是\(A,B\)的列向量组，证明：

\begin{equation*} |A|\cdot |B|=\sum\limits_{i=1}^n|\alpha_1,\ldots,\alpha_{i-1},\beta_1,\alpha_{i+1},\ldots,\alpha_n|\cdot|\alpha_i,\beta_2,\ldots,\beta_n|. \end{equation*}

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解答.

令\(C=(\beta_1,0,\ldots ,0)\)，由于

\begin{equation*} \begin{pmatrix} E_n&0\\ -E_n&E_n \end{pmatrix}\begin{pmatrix} A&C\\ 0&B \end{pmatrix}=\begin{pmatrix}A&C\\ -A&B-C\end{pmatrix}, \end{equation*}

两边同取行列式得

\begin{equation*} \begin{vmatrix}A&C\\ 0&B\end{vmatrix}=\begin{vmatrix}A&C\\ -A&B-C\end{vmatrix}=\left|\begin{array}{ccccccc} \alpha_1&\cdots&\alpha_n&\beta_1&0&\cdots&0\\ -\alpha_1&\cdots&-\alpha_n&0&\beta_2&\cdots&\beta_n \end{array}\right|. \end{equation*}

对于右边的行列式，取定前\(n\)行，利用Laplace定理，由

\begin{equation*} M\begin{bmatrix}1&\cdots&n\\1&\cdots&n\end{bmatrix}=\begin{vmatrix}0&\beta_2&\cdots&\beta_n\end{vmatrix}=0 \end{equation*}

可知

\begin{align*} \begin{vmatrix}A&C\\ 0&B\end{vmatrix}&=\sum\limits_{i=1}^n(-1)^{i}|\alpha_1,\ldots,\alpha_{i-1},\alpha_{i+1},\ldots,\alpha_n,\beta_1|\cdot|-\alpha_i,\beta_2,\ldots,\beta_n| \\ &=\sum\limits_{i=1}^n|\alpha_1,\ldots,\alpha_{i-1},\beta_1,\alpha_{i+1},\ldots,\alpha_n|\cdot|\alpha_i,\beta_2,\ldots,\beta_n|. \end{align*}

又

\begin{equation*} \begin{vmatrix}A&C\\ 0&B\end{vmatrix}=|A|\cdot |B|, \end{equation*}

因此

\begin{equation*} |A|\cdot |B|=\sum\limits_{i=1}^n|\alpha_1,\ldots,\alpha_{i-1},\beta_1,\alpha_{i+1},\ldots,\alpha_n|\cdot|\alpha_i,\beta_2,\ldots,\beta_n|. \end{equation*}

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18.

（Hermite定理）设 \(f_1(x), f_2(x), \ldots , f_n(x)\)是 \(\mathbb{F} [x]\) 中 \(n\)个非零多项式 \((n\geq 2)\)，证明：存在 \(n(n-1)\)个多项式

\begin{equation*} g_{ij}(x)\in\mathbb{F}[x](2\leq i\leq n, 1\leq j\leq n), \end{equation*}

使得

\begin{equation*} \begin{vmatrix} f_1(x)&f_2(x)&\cdots&f_n(x)\\ g_{21}(x)&g_{22}(x)&\cdots&g_{2n}(x)\\ \vdots&\vdots& &\vdots\\ g_{n1}(x)&g_{n2}(x)&\cdots&g_{nn}(x)\\ \end{vmatrix}=\left(f_1(x),f_2(x),\dots ,f_n(x)\right). \end{equation*}

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解答.

对 \(n\)用数学归纳法。

当 \(n=2\)时，根据定理 1.3.3，存在 \(u(x),v(x)\in\F[x]\)，使得

\begin{equation*} u(x)f_1(x)+v(x)f_2(x)=\left(f_1(x),f_2(x)\right). \end{equation*}

取 \(g_{21}(x)=-v(x),g_{22}(x)=u(x)\)，则

\begin{equation*} \begin{vmatrix} f_1(x)&f_2(x)\\ g_{21}(x)&g_{22}(x) \end{vmatrix}=\left(f_1(x),f_2(x)\right). \end{equation*}

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假设当 \(n=k\)时结论成立，考虑\(k+1\)个多项式

\begin{equation*} f_1(x),\ldots ,f_k(x),f_{k+1}(x). \end{equation*}

记 \(d_k(x)=\left(f_1(x),\ldots ,f_k(x)\right)\)，则

\begin{equation*} \left(f_1(x),\ldots ,f_k(x),f_{k+1}(x)\right)=\left(d_k(x),f_{k+1}(x)\right), \end{equation*}

于是存在 \(u(x),v(x)\in\F[x]\)，使得

\begin{equation*} u(x)d_k(x)+v(x)f_{k+1}(x)=\left(f_1(x),\ldots ,f_k(x),f_{k+1}(x)\right). \end{equation*}

由归纳假设存在

\begin{equation*} g_{ij}(x)\in\mathbb{F}[x](2\leq i\leq k, 1\leq j\leq k), \end{equation*}

使得

\begin{equation} \begin{vmatrix} f_1(x)&f_2(x)&\cdots&f_k(x)\\ g_{21}(x)&g_{22}(x)&\cdots&g_{2k}(x)\\ \vdots&\vdots& &\vdots\\ g_{k1}(x)&g_{k2}(x)&\cdots&g_{kk}(x)\\ \end{vmatrix}=d_k(x).\tag{3.3.3} \end{equation}

记 (3.3.3) 左式 \((i,j)\)元的余子式为 \(M_{ij}\)，则按第一行展开

\begin{equation*} d_k(x)=f_1(x)M_{11}-f_2(x)M_{12}+\cdots +(-1)^{k+1}f_k(x)M_{1k}. \end{equation*}

因此存在 \(\frac{f_1(x)v(x)}{d_k(x)},\ldots ,\frac{f_k(x)v(x)}{d_k(x)}\in\F[x]\)，使得

\begin{equation*} \sum\limits_{j=1}^k(-1)^{j+1}\frac{v(x)f_j(x)}{d_k(x)}M_{1j}=v(x). \end{equation*}

注意到 \(k\)阶行列式

\begin{equation} D_k(x)=\begin{vmatrix} g_{21}(x)&g_{22}(x)&\cdots&g_{2k}(x)\\ \vdots&\vdots& &\vdots\\ g_{k1}(x)&g_{k2}(x)&\cdots&g_{kk}(x)\\ -\frac{v(x)f_1(x)}{d_k(x)}&-\frac{v(x)f_2(x)}{d_k(x)}&\cdots&-\frac{v(x)f_k(x)}{d_k(x)} \end{vmatrix}\tag{3.3.4} \end{equation}

\((k,j)\)元的余子式与(3.3.3) 左式 \((1,j)\)元的余子式 \(M_{1j}\)相同，所以将 (3.3.4)按第 \(k\)行展开，得

\begin{equation} \begin{array}{ccl} D_k(x)&=&\sum\limits_{j=1}^k(-1)^{k+j}\cdot\left(-\frac{v(x)f_j(x)}{d_k(x)}\right)\cdot M_{1j}\\ &= & (-1)^{k}\cdot\left(\sum\limits_{j=1}^k(-1)^{j+1}\frac{v(x)f_j(x)}{d_k(x)} M_{1j}\right)\\ &=& (-1)^{k}v(x). \end{array}\tag{3.3.5} \end{equation}

取

\begin{equation*} g_{2,k+1}(x)=g_{3,k+1}(x)=\cdots =g_{k,k+1}(x)=0, \end{equation*}

\begin{equation*} g_{k+1,j}(x)=-\frac{v(x)f_j(x)}{d_k(x)},\ j=1,\ldots ,k, \end{equation*}

\begin{equation*} g_{k+1,k+1}(x)=u(x), \end{equation*}

则

\begin{equation*} \begin{array}{cl} &\begin{vmatrix} f_1(x)&f_2(x)&\cdots&f_k(x)&f_{k+1}(x)\\ g_{21}(x)&g_{22}(x)&\cdots&g_{2k}(x)&g_{2,k+1}(x)\\ \vdots&\vdots& &\vdots&\vdots\\ g_{k1}(x)&g_{k2}(x)&\cdots&g_{kk}(x)&g_{k,k+1}(x)\\ g_{k+1,1}(x)&g_{k+1,2}(x)&\cdots&g_{k+1,k}(x)&g_{k+1,k+1}(x) \end{vmatrix}\\ =&\begin{vmatrix} f_1(x)&f_2(x)&\cdots&f_k(x)&f_{k+1}(x)\\ g_{21}(x)&g_{22}(x)&\cdots&g_{2k}(x)&0\\ \vdots&\vdots& &\vdots&\vdots\\ g_{k1}(x)&g_{k2}(x)&\cdots&g_{kk}(x)&0\\ -\frac{v(x)f_1(x)}{d_k(x)}&-\frac{v(x)f_2(x)}{d_k(x)}&\cdots&-\frac{v(x)f_k(x)}{d_k(x)}&u(x) \end{vmatrix}. \end{array} \end{equation*}

按第 \(k+1\) 列展开，

\begin{equation*} \begin{array}{cl} &\begin{vmatrix} f_1(x)&f_2(x)&\cdots&f_k(x)&f_{k+1}(x)\\ g_{21}(x)&g_{22}(x)&\cdots&g_{2k}(x)&g_{2,k+1}(x)\\ \vdots&\vdots& &\vdots&\vdots\\ g_{k1}(x)&g_{k2}(x)&\cdots&g_{kk}(x)&g_{k,k+1}(x)\\ g_{k+1,1}(x)&g_{k+1,2}(x)&\cdots&g_{k+1,k}(x)&g_{k+1,k+1}(x) \end{vmatrix}\\ =&(-1)^{k+2}f_{k+1}(x)D_k(x)+u(x)\begin{vmatrix} f_1(x)&f_2(x)&\cdots&f_k(x)\\ g_{21}(x)&g_{22}(x)&\cdots&g_{2k}(x)\\ \vdots&\vdots& &\vdots\\ g_{k1}(x)&g_{k2}(x)&\cdots&g_{kk}(x)\\ \end{vmatrix}, \end{array} \end{equation*}

因此将 (3.3.4)， (3.3.5) 代入，得

\begin{equation*} \begin{array}{cl} &\begin{vmatrix} f_1(x)&f_2(x)&\cdots&f_k(x)&f_{k+1}(x)\\ g_{21}(x)&g_{22}(x)&\cdots&g_{2k}(x)&g_{2,k+1}(x)\\ \vdots&\vdots& &\vdots&\vdots\\ g_{k1}(x)&g_{k2}(x)&\cdots&g_{kk}(x)&g_{k,k+1}(x)\\ g_{k+1,1}(x)&g_{k+1,2}(x)&\cdots&g_{k+1,k}(x)&g_{k+1,k+1}(x) \end{vmatrix}\\ =&f_{k+1}(x)v(x)+u(x)d_k(x)\\ =&\left(f_1(x),\ldots ,f_k(x),f_{k+1}(x)\right), \end{array} \end{equation*}

结论成立。

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19. 第四届全国大学生数学竞赛决赛.

设\(n\)阶实方阵\(A\)的每个元素的绝对值为\(2\)，证明：当\(n\geq 3\)时，\(|\det A|\leq \frac{1}{3}\cdot 2^{n+1}n!\)。

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解答.

令\(A_1=\frac{1}{2}A=\left(a_{ij}\right)_{n\times n}\)，则\(A_1\)是所有元素绝对值全为\(1\)的\(n\)阶实方阵，且

\begin{equation*} \det A=\det (2A_1)=2^n\det A_1. \end{equation*}

下面对\(n\)用数学归纳法证明 \(\left|\det A_1\right|\leq \frac{2}{3}n!\)即可。

当\(n=3\)时，

\begin{equation*} \det A= b_{123}+b_{231}+b_{312}+b_{321}+b_{132}+b_{213}, \end{equation*}

其中\(b_{j_1j_2j_3}=(-1)^{\tau(j_1,j_2,j_3)}a_{1j_1}a_{2j_2}a_{3j_3}\)绝对值都为\(1\)，且

\begin{equation*} \prod\limits_{j_1,j_2,j_3}b_{j_1j_2j_3}=-\prod\limits_{i,j=1}^3 a_{ij}^2=-1, \end{equation*}

这意味着\(b_{123},b_{231},b_{312},b_{321},b_{132},b_{213}\)中至少有一项为\(-1\)，也至少有一项为\(1\)。于是\(b_{123},b_{231},b_{312},b_{321},b_{132},b_{213}\)的和中至少有两项相消，因此

\begin{equation*} \left|\det A_1\right|\leq 4 =\frac{2}{3}\cdot 3!. \end{equation*}

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假设对于\(n-1\)阶矩阵结论成立，以下考虑\(n\)阶矩阵\(A_1\)的行列式。将\(\det A_1\)按第\(1\)行展开，得

\begin{equation*} \begin{array}{ll} \left|\det A_1\right| & =\left|\sum\limits_{j=1}^n a_{1j}A_{1j}\right|\\ & \leq\sum\limits_{j=1}^n \left|a_{1j}A_{1j}\right|\\ & =\sum\limits_{j=1}^n \left|A_{1j}\right|. \end{array} \end{equation*}

由于\(M_{1j}\)是所有元素绝对值均为\(1\)的\(n-1\)阶行列式，由归纳假设可知

\begin{equation*} \left|A_{1j}\right|=\left|M_{1j}\right|\leq \frac{2}{3}(n-1)!,\ j=1,\ldots ,n, \end{equation*}

因此\(\left|\det A_1\right|\leq n\cdot\left(\frac{2}{3}(n-1)!\right)=\frac{2}{3}n!\)，结论成立。

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20. Cauchy-Binet公式.

设\(A\)是\(m\times n\)，\(B\)是\(n\times m\)矩阵，且\(m\leq n\)，则

\begin{equation*} \det (AB)=\sum\limits_{1\leq j_1\leq\cdots\leq j_m}A\begin{bmatrix} 1 &\cdots & m\\ j_1 &\cdots & j_m \end{bmatrix}\cdot B\begin{bmatrix} j_1 &\cdots & j_m\\ 1 &\cdots & m \end{bmatrix}. \end{equation*}

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解答.

令\(M=\begin{pmatrix} A & {\bf 0}_{m\times m}\\ -E_n & B \end{pmatrix}\)，则 \(\begin{pmatrix} E_m & A\\ {\bf 0}_{n\times m} & E_n \end{pmatrix}M=\begin{pmatrix} {\bf 0}_{m\times n} & AB\\ -E_n & B \end{pmatrix}\)，两边同取行列式得

\begin{equation*} \det M=\det \begin{pmatrix} {\bf 0}_{m\times n} & AB\\ -E_n & B \end{pmatrix}. \end{equation*}

取定前\(m\)行，根据Laplace定理展开得

\begin{equation*} \begin{array}{ll} \det \begin{pmatrix} {\bf 0} & AB\\ -E_n & B \end{pmatrix} & =(-1)^{1+\cdots +m+(n+1)+\cdots (n+m)}\det (AB)\det (-E_n)\\ & =(-1)^{(m+1)n}\det (AB),\end{array} \end{equation*}

由此可知

\begin{equation} \det M=(-1)^{(m+1)n}\det (AB).\tag{3.3.6} \end{equation}

另一方面，由于\(m\leq n\)，所以直接将\(\det M\)按前\(m\)行展开，得

\begin{equation*} \det M=\sum\limits_{1 \leq j_1 < \cdots < j_m \leq n}A\begin{bmatrix} 1 &\cdots & m\\ j_1 &\cdots & j_m \end{bmatrix}\cdot \hat{A}\begin{bmatrix} 1 &\cdots & m\\ j_1 &\cdots & j_m \end{bmatrix}, \end{equation*}

其中\(\hat{A}\begin{bmatrix} 1 &\cdots & m\\ j_1 &\cdots & j_m \end{bmatrix}\)是\(A\begin{bmatrix} 1 &\cdots & m\\ j_1 &\cdots & j_m \end{bmatrix}\)在矩阵\(M\)中的代数余子式。

\begin{equation*} \hat{A}\begin{bmatrix} 1 &\cdots & m\\ j_1 &\cdots & j_m \end{bmatrix}=(-1)^{1+\cdots+m+j_1+\cdots+j_m}\det \left(-\varepsilon_{i_1} , \cdots , -\varepsilon_{i_{n-m}} , B\right), \end{equation*}

其中\(i_1,\dots,i_{n-m}\)是\(1,\dots ,n\)删去\(j_1,\dots,j_m\)后留下的数。记

\begin{equation*} N=\left(-\varepsilon_{i_1} ,\cdots , -\varepsilon_{i_{n-m}} , B\right), \end{equation*}

对于\(\det N\)，取定前\(n-m\)列，包含这些列的\(n-m\)阶子式只有一个非零，即\(N\begin{bmatrix}i_1 & \cdots & i_{n-m}\\1 & \cdots & n-m\end{bmatrix}\)，其值等于\(\det (-E_{n-m})=(-1)^{n-m}\)，\(N\begin{bmatrix}i_1 & \cdots & i_{n-m}\\1 & \cdots & n-m\end{bmatrix}\)在矩阵\(N\)中对应的余子式为\(B\begin{bmatrix}j_1 &\cdots & j_m\\ 1 &\cdots & m\end{bmatrix}\)，故

\begin{equation*} \det N=(-1)^{n-m}(-1)^{i_1+\cdots+i_{n-m}+1+\cdots+(n-m)}B\begin{bmatrix}j_1 &\cdots & j_m\\ 1 &\cdots & m\end{bmatrix}. \end{equation*}

注意到

\begin{equation*} i_1+\cdots+i_{n-m}+j_1+\cdots+j_m=1+\cdots +n, \end{equation*}

因此

\begin{equation*} \begin{array}{ll} & \hat{A}\begin{bmatrix} 1 &\cdots & m\\ j_1 &\cdots & j_m \end{bmatrix}\\ = & (-1)^{(1+\cdots+m)+(1+\cdots+n)+(1+\cdots+(n-m))+(n-m)}B\begin{bmatrix}j_1 &\cdots & j_m\\ 1 &\cdots & m\end{bmatrix}\\ = & (-1)^{(n-m)(m+1)}B\begin{bmatrix}j_1 &\cdots & j_m\\ 1 &\cdots & m\end{bmatrix}\\ = & (-1)^{n(m+1)}B\begin{bmatrix}j_1 &\cdots & j_m\\ 1 &\cdots & m\end{bmatrix}.\end{array} \end{equation*}

于是

\begin{equation} \det M=(-1)^{n(m+1)}\sum\limits_{1 \leq j_1 < \cdots < j_m \leq n}A\begin{bmatrix} 1 &\cdots & m\\ j_1 &\cdots & j_m \end{bmatrix}\cdot B\begin{bmatrix}j_1 &\cdots & j_m\\ 1 &\cdots & m\end{bmatrix}.\tag{3.3.7} \end{equation}

联立(3.3.6)与(3.3.7)得

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