用消元法求解下列线性方程组：
(1)
{2x1+3x2+x3=4，
{x1-2x2+4x3=-5，
{3x1+8x2-2x3=13，
{4x1-x2+9x3=-6；
(2)
{x1+2x2-x3=1，
{2x1+4x2+x3=5，
{x1+2x2+2x3=4；
(3)
{x1+2x2-x3=2，
{2x1-x2+2x3=10，
{x1+3x2=2；
(4)
{x1+x3=0，
{x1-x2+2x3=-1，
{2x1+x2+x3=2，
{5x1+x2+4x3=1．
(1)对方程组的增广矩阵进行初等行变换得 (2 3 1 4 1 -2 4 -5 3 8 -2 13 -4 -1 9 -6) → (1 -2 4 -5 0 7 -7 14 0 14 -14 28 0 7 -7 14) → (1 -2 4 -5 0 1 -1 2 0 0 0 0 0 0 0 0) → (1 0 2 -1 0 1 -1 2 0 0 0 0 0 0 0 0) 故原方程组化简为 {x1-2x2+4x3 =-5 {x2-x3 =2 通解为 {x1 =-2x3-1 {x2 =x3+2 (2)对方程组的增广矩阵进行初等行变换得 (1 2 -1 1 2 4 1 5 1 2 2 4) → (1 2 -1 1 0 0 3 3 0 0 3 3) → (-1 2 -1 1 0 0 1 1 0 0 0 0) → (-1 2 0 2 0 0 1 1 0 0 0 0 故原方程组化简为 {x1+2x2 =2 {x =1 通解为 {x1 =2-2x2 {x3 =1 (3)对方程组的增广矩阵进行初等行变换得 (-1 2 -1 2 2 -1 2 10 -1 3 0 2) → (1 2 -1 2 0 -5 4 6 0 1 1 0) → (1 2 -1 2 0 1 1 0 0 0 9 6) → (1 0 -3 2 0 1 1 0 0 0 1 2/3) 故原方程组化简为 {x1-3x3 =2 {x2+x3 =0 {x3=2/3 故方程组的解为 {x1 =4 {x2=-(2/3) {x3=2/3 (4)对方程组的增广矩阵进行初等行变换得 (1 0 1 0 1 -1 2 -1 2 1 1 2 5 1 4 3) → (1 0 1 0 0 -1 1 -1 0 1 -1 2 0 1 -1 3) → (1 0 1 0 0 1 -1 1 0 0 0 1 0 0 0 1) → (1 0 1 0 0 1 -1 1 0 0 0 1 0 0 0 0) 方程组变为 {x1+x3 =0 {x2-x3=1 {0=1 显然方程组无解
