n n n n n n D n = ∣ a 11 ⋯ a 1 n ⋮ ⋮ a n 1 ⋯ a n n ∣ D_n=\left|\begin{array}{ccc}a_{11} & \cdots & a_{1 n} \\ \vdots & & \vdots \\ a_{n 1} & \cdots & a_{n n}\end{array}\right| D n = a 11 ⋮ a n 1 ⋯ ⋯ a 1 n ⋮ a nn n n n n n n a 1 = [ a 11 , a 12 , ⋯ , a 1 n ] , a 2 = [ a 21 , a 22 , ⋯ \boldsymbol{a}_1=\left[a_{11}, a_{12}, \cdots, a_{1 n}\right], \boldsymbol{a}_2=\left[a_{21}, a_{22},\cdots\right. a 1 = [ a 11 , a 12 , ⋯ , a 1 n ] , a 2 = [ a 21 , a 22 , ⋯ a 2 n ] , ⋯ , α n = [ a n 1 , a n 2 , ⋯ , a n n ] \left.a_{2 n}\right], \cdots, \alpha_n=\left[a_{n 1}, a_{n 2}, \cdots, a_{n n}\right] a 2 n ] , ⋯ , α n = [ a n 1 , a n 2 , ⋯ , a nn ] n n n n n n
行列式化简基本性质 性质1 行列互换, 其值不变, 即 ∣ A ∣ = ∣ A T ∣ |\boldsymbol{A}|=\left|\boldsymbol{A}^{\mathrm{T}}\right| ∣ A ∣ = A T 性质2 行列式中某行（列）元素全为零，则行列式为零。
性质3 行列式中的两行（列）元素相等或对应成比例，则行列式为零。
性质4 行列式中某行（列）元素均是两个元素之和，则可拆成两个行列式之和，即
∣ a 11 a 12 ⋯ a 1 n ⋮ ⋮ ⋮ a i 1 + b i 1 a i 2 + b i 2 ⋯ a i n + b i n ⋮ ⋮ ⋮ a n 1 a n 2 ⋯ a n n ∣ = ∣ a 11 a 12 ⋯ a 1 n ⋮ ⋮ ⋮ a i 1 a i 2 ⋯ a i n ⋮ ⋮ ⋮ a n 1 a n 2 ⋯ a n n ∣ + ∣ a 11 a 12 ⋯ a 1 n ⋮ ⋮ ⋮ b i 1 b i 2 ⋯ b i n ⋮ ⋮ ⋮ a n 1 a n 2 ⋯ a n n ∣ . \left|\begin{array}{cccc}
a_{11} & a_{12} & \cdots & a_{1 n} \\
\vdots & \vdots & & \vdots \\
a_{i 1}+b_{i 1} & a_{i 2}+b_{i 2} & \cdots & a_{i n}+b_{i n} \\
\vdots & \vdots & & \vdots \\
a_{n 1} & a_{n 2} & \cdots & a_{n n}
\end{array}\right|=\left|\begin{array}{cccc}
a_{11} & a_{12} & \cdots & a_{1 n} \\
\vdots & \vdots & & \vdots \\
a_{i 1} & a_{i 2} & \cdots & a_{i n} \\
\vdots & \vdots & & \vdots \\
a_{n 1} & a_{n 2} & \cdots & a_{n n}
\end{array}\right|+\left|\begin{array}{cccc}
a_{11} & a_{12} & \cdots & a_{1 n} \\
\vdots & \vdots & & \vdots \\
b_{i 1} & b_{i 2} & \cdots & b_{i n} \\
\vdots & \vdots & & \vdots \\
a_{n 1} & a_{n 2} & \cdots & a_{n n}
\end{array}\right| . a 11 ⋮ a i 1 + b i 1 ⋮ a n 1 a 12 ⋮ a i 2 + b i 2 ⋮ a n 2 ⋯ ⋯ ⋯ a 1 n ⋮ a in + b in ⋮ a nn = a 11 ⋮ a i 1 ⋮ a n 1 a 12 ⋮ a i 2 ⋮ a n 2 ⋯ ⋯ ⋯ a 1 n ⋮ a in ⋮ a nn + a 11 ⋮ b i 1 ⋮ a n 1 a 12 ⋮ b i 2 ⋮ a n 2 ⋯ ⋯ ⋯ a 1 n ⋮ b in ⋮ a nn . 性质5 行列式中两行 (列) 互换, 行列式的值反号.
性质6 行列式中某行（列）的 k k k
性质7 行列式中某行（列）元素有公因子 k ( k ≠ 0 ) k(k \neq 0) k ( k = 0 ) k k k
∣ a 11 a 12 … a 1 n ⋮ ⋮ ⋮ k a i 1 k a i 2 … k a i n ⋮ ⋮ ⋮ a n 1 a n 2 … a n n ∣ = k ∣ a 11 a 12 … a 1 n ⋮ ⋮ ⋮ a i 1 a i 2 … a i n ⋮ ⋮ ⋮ a n 1 a n 2 … a n n ∣ . \left|\begin{array}{cccc}
a_{11} & a_{12} & \ldots & a_{1 n} \\
\vdots & \vdots & & \vdots \\
k a_{i 1} & k a_{i 2} & \ldots & k a_{i n} \\
\vdots & \vdots & & \vdots \\
a_{n 1} & a_{n 2} & \ldots & a_{n n}
\end{array}\right|=k\left|\begin{array}{cccc}
a_{11} & a_{12} & \ldots & a_{1 n} \\
\vdots & \vdots & & \vdots \\
a_{i 1} & a_{i 2} & \ldots & a_{i n} \\
\vdots & \vdots & & \vdots \\
a_{n 1} & a_{n 2} & \ldots & a_{n n}
\end{array}\right| . a 11 ⋮ k a i 1 ⋮ a n 1 a 12 ⋮ k a i 2 ⋮ a n 2 … … … a 1 n ⋮ k a in ⋮ a nn = k a 11 ⋮ a i 1 ⋮ a n 1 a 12 ⋮ a i 2 ⋮ a n 2 … … … a 1 n ⋮ a in ⋮ a nn . 注意行列式的性质和矩阵的区别，行列式是一行有公因数提取到外面，而矩阵是每个元素有公因数提取到外面。行列式的值是一个数，而矩阵的变换仍是一个矩阵。
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行列式考试容易失分的两点 易错1：性质4：下面行列式的拆法是错误的，很多考生会在这一步失分
∣ a 11 + b 11 a 12 + b 12 a 21 + b 21 a 22 + b 22 ∣ = ∣ a 11 a 12 a 21 a 22 ∣ + ∣ b 11 b 12 b 21 b 22 ∣ \left|\begin{array}{ll}
a_{11}+b_{11} & a_{12}+b_{12} \\
a_{21}+b_{21} & a_{22}+b_{22}
\end{array}\right|=\left|\begin{array}{ll}
a_{11} & a_{12} \\
a_{21} & a_{22}
\end{array}\right|+\left|\begin{array}{ll}
b_{11} & b_{12} \\
b_{21} & b_{22}
\end{array}\right| a 11 + b 11 a 21 + b 21 a 12 + b 12 a 22 + b 22 = a 11 a 21 a 12 a 22 + b 11 b 21 b 12 b 22 性质释义：性质4告诉我们再拆分行列式时一次只能拆分一个 。
易错2：
①一个常数k k k 行列式 ∣ A ∣ |A| ∣ A ∣ 一行 。
换言之，如果一个行列式有一个公约数k k k
∣ k A ∣ = k n ∣ A ∣ \boxed{
|kA|=k^n|A|
} ∣ k A ∣ = k n ∣ A ∣ 特别的，当k = − 1 k=-1 k = − 1
∣ − A ∣ = ( − 1 ) n ∣ A ∣ |-A|=(-1)^n|A| ∣ − A ∣ = ( − 1 ) n ∣ A ∣ ②一个常数k k k 矩阵 A A A 每一行 。
换言之，如果一个矩阵有一个公约数k k k
[ k A ] = k [ A ] \boxed{
[kA]=k[A]
} [ k A ] = k [ A ] 行列式的基本性质 ∣ A T ∣ = ∣ A ∣ \left|\boldsymbol{A}^{\mathrm{T}}\right|=|\boldsymbol{A}| A T = ∣ A ∣
∣ λ A ∣ = λ n ∣ A ∣ |\lambda A|=\lambda^n|\boldsymbol{A}| ∣ λ A ∣ = λ n ∣ A ∣
∣ A B ∣ = ∣ B A ∣ = ∣ A ∣ ∣ B ∣ , ∣ A k ∣ = ∣ A ∣ k |\boldsymbol{A} \boldsymbol{B}|=|\boldsymbol{B} \boldsymbol{A}|=|\boldsymbol{A}||\boldsymbol{B}|, \quad\left|\boldsymbol{A}^k\right|=|\boldsymbol{A}|^k ∣ A B ∣ = ∣ B A ∣ = ∣ A ∣∣ B ∣ , A k = ∣ A ∣ k
∣ A − 1 ∣ = 1 ∣ A ∣ \left|\boldsymbol{A}^{-1}\right|=\frac{1}{|\boldsymbol{A}|} A − 1 = ∣ A ∣ 1 A \boldsymbol{A} A
∣ A ∗ ∣ = ∣ A ∣ n − 1 ( n ≥ 2 ) \left|\boldsymbol{A}^*\right|=|\boldsymbol{A}|^{n-1}(n \geq 2) ∣ A ∗ ∣ = ∣ A ∣ n − 1 ( n ≥ 2 )
∣ A ∣ = λ 1 λ 2 ⋯ λ n |\boldsymbol{A}|=\lambda_1 \lambda_2 \cdots \lambda_n ∣ A ∣ = λ 1 λ 2 ⋯ λ n λ 1 , λ 2 , ⋯ , λ n \lambda_1, \lambda_2, \cdots, \lambda_n λ 1 , λ 2 , ⋯ , λ n A \boldsymbol{A} A n n n
若 A \boldsymbol{A} A B \boldsymbol{B} B ∣ A ∣ = ∣ B ∣ |\boldsymbol{A}|=|\boldsymbol{B}| ∣ A ∣ = ∣ B ∣
伴随矩阵A ∗ A^* A ∗
上三角与下三角行列式 a_{11} & 0 & \cdots & 0 \\
0 & a_{22} & \cdots & 0 \\
\cdots & \cdots & \cdots & \cdots \\
0 & 0 & \cdots & a_{n n}
\end{array}\right|=\left|\begin{array}{cccc}
a_{11} & a_{12} & \cdots & a_{1 n} \\
0 & a_{22} & \cdots & a_{2 n} \\
\cdots & \cdots & \cdots & \cdots \\
0 & 0 & \cdots & a_{n n}
\end{array}\right|=\left|\begin{array}{cccc}
a_{11} & 0 & \cdots & 0 \\
a_{21} & a_{22} & \cdots & 0 \\
\cdots & \cdots & \cdots & \cdots \\
a_{n 1} & a_{n 2} & \cdots & a_{n n}
\end{array}\right|=a_{11} a_{22} \cdots a_{n n} .
副对角行列式 ∣ 0 λ 1 λ 2 ∴ λ n 0 ∣ = ∣ 0 λ 1 λ 2 ∗ ∴ ⋮ ⋮ λ n ⋯ ∗ ∗ ∣ = ∣ ∗ ⋯ ∗ λ 1 ∗ ⋯ λ 2 ⋮ . λ n 0 ∣ = ( − 1 ) n ( n − 1 ) 2 λ 1 λ 2 ⋯ λ n . \left|\begin{array}{cccc}
0 & & & \lambda_1 \\
& & \lambda_2 & \\
& \therefore & & \\
\lambda_n & & & 0
\end{array}\right|=\left|\begin{array}{cccc}
0 & & & \lambda_1 \\
& & \lambda_2 & * \\
& \therefore & \vdots & \vdots \\
\lambda_n & \cdots & * & *
\end{array}\right|=\left|\begin{array}{cccc}
* & \cdots & * & \lambda_1 \\
* & \cdots & \lambda_2 & \\
\vdots & . & & \\
\lambda_n & & & 0
\end{array}\right|=(-1)^{\frac{n(n-1)}{2}} \lambda_1 \lambda_2 \cdots \lambda_n . 0 λ n ∴ λ 2 λ 1 0 = 0 λ n ∴ ⋯ λ 2 ⋮ ∗ λ 1 ∗ ⋮ ∗ = ∗ ∗ ⋮ λ n ⋯ ⋯ . ∗ λ 2 λ 1 0 = ( − 1 ) 2 n ( n − 1 ) λ 1 λ 2 ⋯ λ n . 范德蒙行列式 D n = ∣ 1 1 1 ⋯ 1 x 1 x 2 x 3 ⋯ x n x 1 2 x 2 2 x 3 2 ⋯ x n 2 ⋮ ⋮ ⋮ ⋮ x 1 n − 1 x 2 n − 1 x 3 n − 1 ⋯ x n n − 1 ∣ = ∏ 1 ⩽ j < i ⩽ n ( x i − x j ) D_n=\left|\begin{array}{ccccc}
1 & 1 & 1 & \cdots & 1 \\
x_1 & x_2 & x_3 & \cdots & x_n \\
x_1^2 & x_2^2 & x_3^2 & \cdots & x_n^2 \\
\vdots & \vdots & \vdots & & \vdots \\
x_1^{n-1} & x_2^{n-1} & x_3^{n-1} & \cdots & x_n^{n-1}
\end{array}\right|=\prod_{1 \leqslant j < i \leqslant n}\left(x_i-x_j\right) D n = 1 x 1 x 1 2 ⋮ x 1 n − 1 1 x 2 x 2 2 ⋮ x 2 n − 1 1 x 3 x 3 2 ⋮ x 3 n − 1 ⋯ ⋯ ⋯ ⋯ 1 x n x n 2 ⋮ x n n − 1 = 1 ⩽ j < i ⩽ n ∏ ( x i − x j ) 请注意：考试时，有时候会把他转置，第一列全为1，这个时候也要能认出他是范德蒙行列式
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代数余子式的异乘变零定理 n n n 另一行 中对应元素的代数余子式乘积的和等于零，即
a i 1 A s 1 + a i 2 A s 2 + . . . + a i n A s n = 0 \boxed{
a_{i1}A_{s1}+a_{i2}A_{s2}+...+a_{in}A_{sn}=0
} a i 1 A s 1 + a i 2 A s 2 + ... + a in A s n = 0 拉普拉斯展开式 设 A \boldsymbol{A} A m m m B \boldsymbol{B} B n n n
① ∣ A O O B ∣ = ∣ A O C B ∣ = ∣ A C O B ∣ = ∣ A ∣ ∣ B ∣ \left|\begin{array}{ll}\boldsymbol{A} & \boldsymbol{O} \\ \boldsymbol{O} & \boldsymbol{B}\end{array}\right|=\left|\begin{array}{ll}\boldsymbol{A} & \boldsymbol{O} \\ \boldsymbol{C} & \boldsymbol{B}\end{array}\right|=\left|\begin{array}{ll}\boldsymbol{A} & \boldsymbol{C} \\ \boldsymbol{O} & \boldsymbol{B}\end{array}\right|=|\boldsymbol{A}||\boldsymbol{B}| A O O B = A C O B = A O C B = ∣ A ∣∣ B ∣
②∣ O A B O ∣ = ∣ O A B C ∣ = ∣ C A B O ∣ = ( − 1 ) m n ∣ A ∣ ∣ B ∣ \left|\begin{array}{ll}\boldsymbol{O} & \boldsymbol{A} \\ \boldsymbol{B} & \boldsymbol{O}\end{array}\right|=\left|\begin{array}{ll}\boldsymbol{O} & \boldsymbol{A} \\ \boldsymbol{B} & \boldsymbol{C}\end{array}\right|=\left|\begin{array}{ll}\boldsymbol{C} & \boldsymbol{A} \\ \boldsymbol{B} & \boldsymbol{O}\end{array}\right|=(-1)^{m n}|\boldsymbol{A}||\boldsymbol{B}| O B A O = O B A C = C B A O = ( − 1 ) mn ∣ A ∣∣ B ∣
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克拉默法则 如果线性方程组 { a 11 x 1 + a 12 x 2 + ⋯ + a 1 n x n = b 1 a 21 x 1 + a 22 x 2 + ⋯ + a 2 n x n = b 2 ⋮ a n 1 x 1 + a n 2 x 2 + ⋯ + a n n x n = b n \left\{\begin{array}{c}a_{11} x_1+a_{12} x_2+\cdots+a_{1 n} x_n=b_1 \\ a_{21} x_1+a_{22} x_2+\cdots+a_{2 n} x_n=b_2 \\ \vdots \\ a_{n 1} x_1+a_{n 2} x_2+\cdots+a_{n n} x_n=b_n\end{array}\right. ⎩ ⎨ ⎧ a 11 x 1 + a 12 x 2 + ⋯ + a 1 n x n = b 1 a 21 x 1 + a 22 x 2 + ⋯ + a 2 n x n = b 2 ⋮ a n 1 x 1 + a n 2 x 2 + ⋯ + a nn x n = b n D = ∣ a 11 ⋯ a 1 n ⋮ ⋮ a n 1 ⋯ a n n ∣ ≠ 0 D=\left|\begin{array}{ccc}a_{11} & \cdots & a_{1 n} \\ \vdots & & \vdots \\ a_{n 1} & \cdots & a_{n n}\end{array}\right| \neq 0 D = a 11 ⋮ a n 1 ⋯ ⋯ a 1 n ⋮ a nn = 0
x 1 = D 1 D , x 2 = D 2 D , ⋯ , x n = D n D , x_1=\frac{D_1}{D}, x_2=\frac{D_2}{D}, \cdots, x_n=\frac{D_n}{D}, x 1 = D D 1 , x 2 = D D 2 , ⋯ , x n = D D n , 其中 D j ( j = 1 , 2 , ⋯ , n ) D_j(j=1,2, \cdots, n) D j ( j = 1 , 2 , ⋯ , n ) D D D j j j n n n
行列式的判断方程的解 系数矩阵A 齐次方程 A X = 0 AX=0 A X = 0 非齐次方程 A X = b ( b ≠ 0 ) AX=b (b \ne 0) A X = b ( b = 0 ) 行列式 det A = 0 有非零解 无解 行列式 det A ≠ \ne = 只有零解 有唯一解
记忆技巧：类别 a x = 0 ax=0 a x = 0 a x = 1 ax=1 a x = 1 a = 0 a=0 a = 0 0 x = 0 0x=0 0 x = 0 0 x = 1 0x=1 0 x = 1 a ≠ 0 a \ne 0 a = 0 3 x = 0 3x=0 3 x = 0 3 x = 1 3x=1 3 x = 1
注意：上面A X = b AX=b A X = b b ≠ 0 b \ne 0 b = 0 A X = b AX=b A X = b b b b
注： 公式汇总主要参考 武忠祥 编制的 线性代数公式