矩阵行列式计算器
逐步计算矩阵的行列式
您的输入
计算 $$$\left|\begin{array}{ccc}1 & 2 & 2\\0 & 5 & 7\\1 & 1 & 1\end{array}\right|$$$。
解答
第$$$3$$$行减去第$$$1$$$行：$$$R_{3} = R_{3} - R_{1}$$$。
$$$\left|\begin{array}{ccc}1 & 2 & 2\\0 & 5 & 7\\1 & 1 & 1\end{array}\right| = \left|\begin{array}{ccc}1 & 2 & 2\\0 & 5 & 7\\0 & -1 & -1\end{array}\right|$$$
按第$$$1$$$列展开：
$$$\left|\begin{array}{ccc}1 & 2 & 2\\0 & 5 & 7\\0 & -1 & -1\end{array}\right| = \left(1\right) \left(-1\right)^{1 + 1} \left|\begin{array}{cc}5 & 7\\-1 & -1\end{array}\right| + \left(0\right) \left(-1\right)^{2 + 1} \left|\begin{array}{cc}2 & 2\\-1 & -1\end{array}\right| + \left(0\right) \left(-1\right)^{3 + 1} \left|\begin{array}{cc}2 & 2\\5 & 7\end{array}\right| = \left|\begin{array}{cc}5 & 7\\-1 & -1\end{array}\right|$$$
2x2 矩阵的行列式为 $$$\left|\begin{array}{cc}a & b\\c & d\end{array}\right| = a d - b c$$$。
$$$\left|\begin{array}{cc}5 & 7\\-1 & -1\end{array}\right| = \left(5\right)\cdot \left(-1\right) - \left(7\right)\cdot \left(-1\right) = 2$$$
答案
$$$\left|\begin{array}{ccc}1 & 2 & 2\\0 & 5 & 7\\1 & 1 & 1\end{array}\right| = 2$$$A