求线性代数方程X+2Y+3Z+4U+5V=1 X+3Y+4Z+5U+6V=1 2X+4Y+5Z+6U+7V=2...答:直接将增广矩阵做行变化即可,可解得 (X,Y,Z,U,V)=(0,1,-2,1,0)Y+(0,0,1,-2,1)V+(1,0,0,0,0)即, X = 1, Z = -2Y+V,U = Y-2V,
一道线性代数的题目,急答:3, 1, -1, +2| 2 2, 6, -3, -3|λ+1 -1,-11, 5, 4| -4→行初等变换→ 1, 0, -3/16, -9/16|9/16 0, 1, -7/15, -5/16|5/16 0, 0, 0, 0 |λ-2,0, 0, 0, 0 |0 λ=2时,方程有解,通解为:(a, b, c, d)...