x1-x2-x3+x4=0 x1-x2+2x4=1 x1-x2+2x3+3x4=-1
- 教育綜合
- 2023-06-23 12:59:41
X1-X2-X3+X4=0 X1-X2+X3-3X4=1 2X1-2X2-2X4=1 求通解
解: 增廣矩陣= 1 -1 -1 1 0 1 -1 1 -3 1 2 -2 0 -2 1 r2-r1,r3-2r1 1 -1 -1 1 0 0 0 2 -4 1 0 0 2 -4 1 r3-r2,r2*(1/2),r1+r2 1 -1 0 -1 1/2 0 0 1 -2 1/2 0 0 0 0 0 所以,方程組的通解為: (1/2,0,1/2,0)'+c1(1,1,0,0)'+c2(1,0,2,1)'. 滿意請采納^_^判斷非齊次線性方程組x1-x2-x3+x4=0,x1-2x2-x3+3x4=-1,x1+x2-x3-3x4=2,是否有解,如果有,求通解
2;2: x1- x2 - x3 + x4 = 0 ----(6) 2*x3 - 4*x4 = 1 ----(7) 以x4,和x2為自由變量, 2*x4+1/,,代入(2)(3)得: 2*x3 - 4*x4 = 1 ----(4) 2*x3 - 4*x4 = 1 ----(5) 由此可以看出: (x2+x4 +1/, x2解: 設(shè) x1- x2 = y,原方程組化為:y = x3-x4,4元方程組只有兩個約束條件,得到: x3 =2*x4 +1/2; x1 = x2+x4 +1/2; 因此,方程組通解為: y - x3 + x4 = 0 ----(1) y + x3 - 3x4 = 1 ---關(guān)于線性代數(shù)的題:求解方程組X1-X2-X3+X4=0,X1-X2+X3-3X4=1,X1-X2-
X1-X2-X3+X4=0(1) X1-X2+X3-3X4=1(2) X1-X2-2X3+3X4=-1/2(3)。 (2)-(1),得2x3-4x4=1, (2)-(3),得3x3-6x4=3/2. 上述兩個方程同解,所以x3=(4x4+1)/2,x4可以是任意數(shù); 代入(1),得x1=x2+(2x4+1)/2,x2可以是任意數(shù)。求線性方程組的解x1 x2-x3-x4=0 x1-2x2-x3 x4=1 x1 2x2-2x4=1 7x1-3x2 5x3-2x4=38
寫出方程組的增廣矩陣 4X5 1 1 -1 -1 0 1 -2 -1 1 1 1 2 0 -2 1 7 -3 5 -2 38 然后對矩陣進(jìn)行初等變換,使系數(shù)矩陣化為單位矩陣,則最后一列就是方程組的解! 精銳長寧天山數(shù)學(xué)組為您解答!線性代數(shù):求下列非齊次線性方程組的基礎(chǔ)解系:X1-X2-X3+X4=0 X1-X2+X3-3X4=0 X1-X2-2X3+3X4=0
這不是齊次嗎? 1 -1 -1 1 1 -1 1 -3 1 -1 -2 3 二三行減一行 1 -1 -1 1 0 0 2 -4 0 0 -1 2 哈哈 1 -1 0 -1 0 0 1 -2 解系基有4-2=2個 取x3=x4=0,得x1=x2=1 (1,1,0,0) 取x3=2,x4=1,得x1=1,x2=0 (1,0,2,1) 所以x=A(1,1,0,0)+B(1,0,2,1)展開全文閱讀