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Let the rank of n-dimensional vectors A1, A2 and as be r Then A. any R-1 vector in vector group is linearly independent B. any r vector in vector group is linearly independent C. If any R + 1 vector in a vector group is linearly independent of D, the number of vectors in the vector group must be greater than R
Choose B
B contains a,
C rank is the number of maximal linearly independent groups in a vector group
DR is OK
PX1 + x2 + X3 = 1, X1 + PX2 + X3 = P, X1 + x2 + PX3 = P & # 178; when p takes what value, the equation system has unique solution, no solution, infinite solution and general solution
Solution: coefficient determinant | a | = (λ + 2) (λ - 1) ^ 2. So when λ ≠ 1 and λ ≠ - 2, the equations have unique solution. When λ = 1, the augmented matrix = 1, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1
When we ask the value of λ, the equation system has a unique solution: λ X1 + x2 + x = 1 X1 + λ x2 + X3 = λ X1 + x2 + λ X3 = λ & #;
When λ = 1 and λ = 2, what are the solutions of the equations respectively It's a problem of linear algebra
When λ ≠ 1 or λ ≠ - 2, the equations have unique solutions
Because the coefficient determinant of the equation is as follows:
λ 1 1
1 λ 1
1 1 λ
=(λ-1)²(λ+2)≠0
λ ≠ 1 or λ ≠ - 2
The general solution of the equations X1 + x2-x3 = 0 is______ ．
Let X1 and X2 be free unknowns, then X3 = X1 + X2, let (x1, x2) = (1,0) t, or (0,1) t, then X3 = 1, then the basic solution system is: (1,0,1) t, (0,1,1) t, so the general solution of the original equation is: K1 (1,0,1) t + K2 (0,1,1) t, (where K1 and K2 are arbitrary constants), so the answer is: K1 (1,0,1) t + K (0,1,1) t
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