《高等代数学》（姚慕生），习题1.4：行列式的展开和转置

1. 通过计算证明下列行列式按第一行展开的值等于按第一列展开的值：
$$\left|\begin{array}{ccc}1& 0& 0\\ 2& 7& 3\\ 5& 2& 2\end{array}\right|，\left|\begin{array}{ccc}1& 0& -5\\ 6& 2& -1\\ 5& -11& 6\end{array}\right|$$

解
$$\{\begin{array}{rl}& \left|\begin{array}{ccc}1& 0& 0\\ 2& 7& 3\\ 5& 2& 2\end{array}\right|\underset{=}{\overset{按第1行展开}{\to }}1\times \left|\begin{array}{cc}7& 3\\ 2& 2\end{array}\right|-0\times \left|\begin{array}{cc}2& 3\\ 5& 2\end{array}\right|+0\times \left|\begin{array}{cc}2& 7\\ 5& 2\end{array}\right|=8.\\ & \left|\begin{array}{ccc}1& 0& 0\\ 2& 7& 3\\ 5& 2& 2\end{array}\right|\underset{=}{\overset{按第1列展开}{\to }}1\times \left|\begin{array}{cc}7& 3\\ 2& 2\end{array}\right|-2\times \left|\begin{array}{cc}0& 0\\ 2& 2\end{array}\right|+5\times \left|\begin{array}{cc}0& 0\\ 7& 3\end{array}\right|=8.\end{array}$$
$$\{\begin{array}{rl}& \left|\begin{array}{ccc}1& 0& -5\\ 6& 2& -1\\ 5& -11& 6\end{array}\right|\underset{=}{\overset{按第1行展开}{\to }}1\times \left|\begin{array}{cc}2& -1\\ -11& 6\end{array}\right|-0\times \left|\begin{array}{cc}6& -1\\ 5& 6\end{array}\right|+\left(-5\right)\times \left|\begin{array}{cc}6& 2\\ 5& -11\end{array}\right|=1+380=381.\\ & \left|\begin{array}{ccc}1& 0& -5\\ 6& 2& -1\\ 5& -11& 6\end{array}\right|\underset{=}{\overset{按第1列展开}{\to }}1\times \left|\begin{array}{cc}2& -1\\ -11& 6\end{array}\right|-6\times \left|\begin{array}{cc}0& -5\\ -11& 6\end{array}\right|+5\times \left|\begin{array}{cc}0& -5\\ 2& -1\end{array}\right|=1+330+50=381.\end{array}$$

2. 将下列行列式分别按第二行及第三列展开求值并比较其结果：
$$\left|\begin{array}{ccc}2& 3& 5\\ 1& 2& 0\\ 0& 3& 8\end{array}\right|，\left|\begin{array}{ccc}3& 2& -2\\ 2& -1& 3\\ 9& 6& -7\end{array}\right|$$

解
$$\{\begin{array}{rl}& \left|\begin{array}{ccc}2& 3& 5\\ 1& 2& 0\\ 0& 3& 8\end{array}\right|\stackrel{按r_2展开}{\to }1\times {\left(-1\right)}^{2+1}\times \left|\begin{array}{cc}3& 5\\ 3& 8\end{array}\right|+2\times {\left(-1\right)}^{2+2}\times \left|\begin{array}{cc}2& 5\\ 0& 8\end{array}\right|+0\times {\left(-1\right)}^{2+3}\times \left|\begin{array}{cc}2& 3\\ 0& 3\end{array}\right|=-9+32+0=23.\\ & \left|\begin{array}{ccc}2& 3& 5\\ 1& 2& 0\\ 0& 3& 8\end{array}\right|\stackrel{按c_3展开}{\to }5\times {\left(-1\right)}^{1+3}\times \left|\begin{array}{cc}1& 2\\ 0& 3\end{array}\right|+0\times {\left(-1\right)}^{2+3}\times \left|\begin{array}{cc}2& 3\\ 0& 3\end{array}\right|+8\times {\left(-1\right)}^{3+3}\times \left|\begin{array}{cc}2& 3\\ 1& 2\end{array}\right|=15+0+8=23.\end{array}$$
{ ∣ 3 2 − 2 2 − 1 3 9 6 − 7 ∣ → 按 r 2 展 开 2 × ( − 1 ) 2 + 1 × ∣ 2 − 2 6 − 7 ∣ + ( − 1 ) × ( − 1 ) 2 + 2 × ∣ 3 − 2 9 − 7 ∣ + 3 × ( − 1 ) 2 + 3 × ∣ 3 2 9 6 ∣ = 4 + 3 + 0 = 7. ∣ 3 2 − 2 2 − 1 3 9 6 − 7 ∣ → 按 c 3 展 开 ( − 2 ) × ( − 1 ) 1 + 3 × ∣ 2 − 1 9 6 ∣ + 3 × ( − 1 ) 2 + 3 × ∣ 3 2 9 6 ∣ + ( − 7 ) × ( − 1 ) 3 + 3 × ∣ 3 2 2 − 1 ∣ = − 42 + 0 + 49 = 7. \left\{ \begin{aligned} & \left| \begin{matrix} 3 & 2 & -2 \\ 2 & -1 & 3 \\ 9 & 6 & -7 \\ \end{matrix} \right|\xrightarrow[{}]{按{ {r}_{2}}展开}2\times { {\left( -1 \right)}^{2+1}}\times \left| \begin{matrix} 2 & -2 \\ 6 & -7 \\ \end{matrix} \right|+\left( -1 \right)\times { {\left( -1 \right)}^{2+2}}\times \left| \begin{matrix} 3 & -2 \\ 9 & -7 \\ \end{matrix} \right|+3\times { {\left( -1 \right)}^{2+3}}\times \left| \begin{matrix} 3 & 2 \\ 9 & 6 \\ \end{matrix} \right|=4+3+0=7. \\ & \left| \begin{matrix} 3 & 2 & -2 \\ 2 & -1 & 3 \\ 9 & 6 & -7 \\ \end{matrix} \right|\xrightarrow[{}]{按{ {c}_{3}}展开}\left( -2 \right)\times { {\left( -1 \right)}^{1+3}}\times \left| \begin{matrix} 2 & -1 \\ 9 & 6 \\ \end{matrix} \right|+3\times { {\left( -1 \right)}^{2+3}}\times \left| \begin{matrix} 3 & 2 \\ 9 & 6 \\ \end{matrix} \right|+\left( -7 \right)\times { {\left( -1 \right)}^{3+3}}\times \left| \begin{matrix} 3 & 2 \\ 2 & -1 \\ \end{matrix} \right|=-42+0+49=7. \\ \end{aligned} \right. ⎩⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎧​​∣∣∣∣∣∣​329​2−16​−23−7​∣∣∣∣∣∣​按r2​展开 ​2×(−1)2+1×∣∣∣∣​26​