【计算方法】#01 高斯消去法和列主元高斯消去法的原理简介及C++实现

满足以下条件任一即可

消元过程(第$k$次消元)：
{ a i j ( k ) = a i j ( k ? 1 ) ? a i k ( k ? 1 ) a k k ( k ? 1 ) a k j ( k ? 1 ) = a i j ( k ? 1 ) ? l i k a k j ( k ? 1 ) , i = k 1 , k 2 , . . . , n ; j = k 1 , k 2 , . . . , n b i ( k ) = b i ( k ? 1 ) ? l i k b k ( k ? 1 ) , i = k 1 , k 2 , . . . , n l i k ? a i k ( k ? 1 ) a k k ( k ? 1 ) , i = k 1 , k 2 , . . . , n \begin{cases} &a_{ij}^{(k)}=a_{ij}^{(k-1)}-\frac{a_{ik}^{(k-1)}}{a_{kk}^{(k-1)}}a_{kj}^{(k-1)}=a_{ij}^{(k-1)}-l_{ik}a_{kj}^{(k-1)}, \\ &i=k 1,k 2,...,n; j=k 1,k 2,...,n \\ \\ &b_i^{(k)} = b_i^{(k-1)}- l_{ik}b_{k}^{(k-1)}, i=k 1,k 2,...,n \\ \\ &l_{ik} \triangleq \frac{a_{ik}^{(k-1)}}{a_{kk}^{(k-1)}}, i=k 1, k 2, ..., n \end{cases} ????⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎧​​aij(k)​=aij(k−1)​−akk(k−1)​aik(k−1)​​akj(k−1)​=aij(k−1)​−lik​akj(k−1)​,i=k+1,k+2,...,n;j=k+1,k+2,...,nbi(k)​=bi(k−1)​−lik​bk(k−1)​,i=k+1,k+2,...,nlik​≜akk(k−1)​aik(k−1)​​,i=k+1,k+2,...,n​

回代过程：
$$\{\begin{array}{rl}{x}_n& =\frac{b_n^{(n-1)}}{a_{nn}^{(n-1)}},\\ x_k& =\frac{b_k^{(k-1)}-\sum _{j=k+1}^n{a}_{kj}^{(k-1)}{x}_j}{a_{kk}^{(k-1)}},k=n-1,n-2,...,2,1\end{array}$$

令运算量为乘除法个数，则

消元过程：$N_1=\sum _{k=1}^{n-1}(n-k{)}^2+2(n-k)=n^3/3+n^2/2-5n/6$

回代过程：$N_2=1+(n-1)+\sum _{k=1}^{n-1}(n-k)=n^2/2+n/2$

高斯消去法总运算量：$N=N_1+N_2=n^3/3+n^2-n/3$

时间复杂度：