线性方程组的矩阵解法

设线性方程组为 { a 11 x 1 + a 11 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 \begin{cases} a_{11}x_1+a_{11}x_2+\cdots+a_{1n}x_n=b_1\\ a_{21}x_1+a_{22}x_2+\cdots+a_{2n}x_n=b_2\\ ......\\ a_{n1}x_1+a_{n2}x_2+\cdots+a_{nn}x_n=b_n\\ \end{cases} ⎩⎪⎪⎪⎨⎪⎪⎪⎧​a11​x1​+a11​x2​+⋯+a1n​xn​=b1​a21​x1​+a22​x2​+⋯+a2n​xn​=b2​......an1​x1​+an2​x2​+⋯+ann​xn​=bn​​ 其系数行列式为 D = ∣ a 11 a 12 ⋯ a 1 n a 21 a 22 ⋯ a 2 n ⋮ ⋮ ⋱ ⋮ a n 1 a n 2 ⋯ a n n ∣ ≠ 0 D=\begin{vmatrix} a_{11}&a_{12}&\cdots&a_{1n} \\ a_{21}&a_{22}&\cdots&a_{2n} \\ \vdots&\vdots&\ddots&\vdots\\ a_{n1}&a_{n2}&\cdots&a_{nn} \end{vmatrix} \not=0 D=∣∣∣∣∣∣∣∣∣​a11​a21​⋮an1​​a12​a22​⋮an2​​⋯⋯⋱⋯​a1n​a2n​⋮ann​​∣∣∣∣∣∣∣∣∣​​=0 则该线性方程组有且仅有唯一解： 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} x1​=DD1​​,x2​=DD2​​,⋯,xn​=DDn​​ 其中 D j ( j = 1 , 2 , … , n ) D_j(j=1,2,\ldots,n) Dj​(j=1,2,…,n)是把系数行列式 D D D中的第 j j j列的元素用常数项 b 1 , b 2 , … , b n b_1,b_2,\ldots,b_n b1​,b2​,…,bn​代替后得到的 n n n阶行列式，即 D j = ∣ a 11 ⋯ a 1 , j − 1 b 1 a 1 , j + 1 ⋯ a 1 n a 21 ⋯ a 2 , j − 1 b 1 a 1 , j + 1 ⋯ a 2 n ⋮ ⋮ ⋮ ⋮ a n 1 ⋯ a n , j − 1 b 1 a 1 , j + 1 ⋯ a n n ∣ D_j = \begin{vmatrix} a_{11}&\cdots&a_{1,j-1}&b_1&a_{1,j+1}&\cdots&a_{1n} \\ a_{21}&\cdots&a_{2,j-1}&b_1&a_{1,j+1}&\cdots&a_{2n} \\ \vdots&\vdots&\vdots&\vdots\\ a_{n1}&\cdots&a_{n,j-1}&b_1&a_{1,j+1}&\cdots&a_{nn} \end{vmatrix} Dj​=∣∣∣∣∣∣∣∣∣​a11​a21​⋮an1​​⋯⋯⋮⋯​a1,j−1​a2,j−1​⋮an,j−1​​b1​b1​⋮b1​​a1,j+1​a1,j+1​a1,j+1​​⋯⋯⋯​a1n​a2n​ann​​∣∣∣∣∣∣∣∣∣​

{ 2 x 1 + 3 x 2 − 5 x 3 = 3 x 1 − 2 x 2 + x 3 = 0 3 x 1 + x 2 + 3 x 3 = 7 \begin{cases} 2x_1+3x_2-5x_3=3\\ x_1-2x_2+x_3=0\\ 3x_1+x_2+3x_3=7\\ \end{cases} ⎩⎪⎨⎪⎧​2x1​+3x2​−5x3​=3x1​−2x2​+x3​=03x1​+x2​+3x3​=7​

解： D = ∣ 2 3 − 5 1 − 2 1 3 1 3 ∣ = ∣ 2 7 − 7 1 0 0 3 7 0 ∣ = − ∣ 7 − 7 7 0 ∣ = − 49 ≠ 0 D= \begin{vmatrix} 2&3&-5\\ 1 & -2 & 1\\ 3 & 1 &3\\ \end{vmatrix}= \begin{vmatrix} 2 & 7 & -7\\ 1 & 0 & 0\\ 3 & 7 & 0\\ \end{vmatrix}=- \begin{vmatrix} 7&-7\\ 7 &0\\ \end{vmatrix}=-49\not=0 D=∣∣∣∣∣∣​213​3−21​−513​∣∣∣∣∣∣​=∣∣∣∣∣∣​213​707​−700​∣∣∣∣∣∣​=−∣∣∣∣​77​−70​∣∣∣∣​=−49​=0

D 1 = ∣ 3 3 − 5 0 − 2 1 7 1 3 ∣ = ∣ 2 7 − 7 1 0 0 3 7 0 ∣ = − ∣ 3 − 7 7 7 ∣ = − 70 D_1= \begin{vmatrix} 3&3&-5\\ 0 & -2 & 1\\ 7 & 1 &3\\ \end{vmatrix}= \begin{vmatrix} 2 & 7 & -7\\ 1 & 0 & 0\\ 3 & 7 & 0\\ \end{vmatrix}=- \begin{vmatrix} 3 & -7\\ 7 & 7\\ \end{vmatrix}=-70 D1​=∣∣∣∣∣∣​307​3−21​−513​∣∣∣∣∣∣​=∣∣∣∣∣∣​213​707​−700​∣∣∣∣∣∣​=−∣∣∣∣​37​−77​∣∣∣∣​=−70

x 1 = D 1 D = − 70 − 49 = 10 7 x_1=\frac{D_1}{D}=\frac{-70}{-49}=\frac{10}{7} x1​=DD1​​=−49−70​=710​

其余两个解同理。