用高斯消元法解方程组 {x1+3x2+5x3-4x4=1 {x1+3x2+2x3-2x4+x5=-1 {x1-2x2+x3-x4

用高斯消元法解方程组
{x₁+3x₂+5x₃-4x₄=1
{x₁+3x₂+2x₃-2x₄+x₅=-1
{x₁-2x₂+x₃-x₄-x₅=3
{x₁-4x₂+x₃+x₄-x₅=3
{x₁+2x₂+x₃-x₄+x₅=-1
【正确答案】：对它的增广矩阵作初等行变换，把它化成阶梯形． (A，β)= (1 3 5 -4 0 ┆ 1 1 3 2 -2 1 ┆ -1 1 -2 1 -1 -1 ┆ 3 1 -4 1 1 -1 ┆ 3 1 2 1 -1 1 ┆ -1) → (1 3 5 -4 0 ┆ 1 0 0 -3 2 1 ┆ -2 0 -5 -4 3 -1 ┆ 2 0 -7 -4 5 -1 ┆ 2 0 -1 -4 3 1 ┆ -2) → (1 3 5 -4 0 ┆ 1 0 -1 -4 3 1 ┆ -2 0 -5 -4 3 -1 ┆ 2 0 -7 -4 5 -1 ┆ 2） 0 0 -3 2 1 ┆ -2) → (1 3 5 -4 0 ┆ 1 0 -1 -4 3 1 ┆ -2 0 0 16 -12 -6 ┆ 12 0 0 24 -16 -8 ┆ 16 0 0 -3 2 1 ┆ -2) → (1 3 5 -4 0 ┆ 1 0 -1 -4 3 1 ┆ -2 0 0 8 -6 -3 ┆ 6 0 0 3 -2 -1 ┆ 2 0 0 -3 2 1 ┆ -2) → (1 3 5 -4 0 ┆ 1 0 -1 -4 3 1 ┆ -2 0 0 -1 0 0 ┆ 0 0 0 3 -2 -1 ┆ 2 0 0 0 0 0 ┆ 0) → (1 3 5 -4 0 ┆ 1 0 -1 -4 3 1 ┆ -2 0 0 1 0 0 ┆ 0 0 0 0 2 1 ┆ -2 0 0 0 0 0 ┆ 0) 即得同解方程组 {x₁+3x₂+5x₃-4x₄ =1 {-x₂-4x₃+3x₄+x₅=-2 { x₃ =0 { 2x₄+x₅=-2 以x₅为自由未知量，逐次解之，得 {x₁=-(1/2)x₅ {x₂=-1-(1/2)x₅ {x₃=0， {x₄=-1-(1/2)x₅ 故方程组的一般解为x₁=-(1/2)c，x₂=-1-(1/2)c，x₃=0，x₄=-1-(1/2)c，x₅=c(c为任意常数)．